EEVblog #1284 - How Bad Product Design Kills The Environment

[EEVblog]

Let's save 30 cents/unit and stuff the environment!
How can bad and cheap product design kill the environment by requiring vastly more energy generation, delivery and PF correction infrastructure?
Dave looks at the shockingly energy wasteful design of mains powered smoke alarms.

[ogden]

Mains supply have to be designed for max power. You forgot to measure consumption while alarm is on.

[langwadt]

Let's save 30 cents/unit and stuff the environment!
How can bad and cheap product design kill the environment by requiring vastly more energy generation, delivery and PF correction infrastructure?
Dave looks at the shockingly energy wasteful design of mains powered smoke alarms.

isn't 90% of the led lights BigClive take a part the same capacitive dropper and lousy power factor?

[tszaboo]

I think you should review how reactive power works. The 100MW is not generated by the wind turbine. You generate 5MW by the wind, and place a compensation on the power grid, that will generate the reactive power. It is just a big capacitor. The grid already has a bunch of inductive load, because of motors.

Here is an example of 120MVAR capacitor bank.
https://trench-group.com/wp-content/uploads/2019/07/Air-Core-Shunt-Reactors.pdf

[berniwa]

You can acutally use a depletion mode MOSFET to modify any "normal" regulator to support mains voltage.



In the given configuration, the MOSFET will reduce the drain, source current to a point, where the regulator only sees VOUT - VGSoff.
Simple, easy and cost effective solution and hv depletion mode MOSFETs are readily available.

[EEVblog]

So this is an interesting discussion, and I expected it.
I think I'm right in that ultimately at some point there has to be no free lunch here, but happy to be proven wrong.
I did simplify it in the video to highlight to potential magnitude of the problem.
#NotAPowerGenerationEngineer

[EEVblog]

I think you should review how reactive power works. The 100MW is not generated by the wind turbine. You generate 5MW by the wind, and place a compensation on the power grid, that will generate the reactive power. It is just a big capacitor. The grid already has a bunch of inductive load, because of motors.

See my comment above.

[fki82]

I think NANDBlog is right. The generator doesn't need to generate all the reactive power but only the active power + losses.
The efficiency of this power supplies is still horrible.

Is it required to connect the smoke alarm to the grid in Australia?
Here in Germany we just put a smoke alarm with a 10 Year battery on the ceiling and that's it.

[EEVblog]

Is it required to connect the smoke alarm to the grid in Australia?

For all new home and renovations, yes, mandatory. (might vary by state, don't know)

[EEVblog]

I think NANDBlog is right. The generator doesn't need to generate all the reactive power but only the active power + losses.

There is reason why commercial companies are charged in kVA. They aren't charged kW+"copper losses"

[langwadt]

https://www.cashmanequipment.com/cashmancat/media/Cashman-Equipment/generator-power-factor.pdf

[Maloxyl]

Is it required to connect the smoke alarm to the grid in Australia?

For all new home and renovations, yes, mandatory. (might vary by state, don't know)

I don't understand that requirement at all. In addition to being hardwired you have a backup-battery that has to be changed every year or two (see manual / regs) that could power the whole thing alone anyways. In addition after 10 years you have to replace the whole alarm.

Here in Austria / Germany the standard is alarms with hardwired battery - lifetime 10 years. Less hassle and probably cheaper.


Regulations: http://mfb.vic.gov.au/Community/Home-Safety/SmokeAlarms.html
Manufacturer manual: https://www.quell.com.au/download/quell-q1300-240v-photoelectric-alarm-installer-manual/

[fki82]

With a bad power factor, the current in the wires is going to be higher, which requires thicker wires, bigger transformers, bigger generators and produces more loss.
That's why power factor correction is necessary, and companies get charged extra if they don’t do it properly. It’s about compensating for the additional cost on the grid for reactive power.
In this specific case, the smoke alarm is a capacitive load.
Most machines an appliances will be inductive loads.
They are usually not compensated completely, so in reality, the capacitive load of the smoke alarm won't cause additional current on the wires to the power plant.
Because of this, only the active power should be used for calculating the energy consumption (which is still horribly high).

[Dundarave]

Given that domestic power consumption is overwhelmingly inductive (fridge motors, fans, wall-warts of all descriptions), the capacitive load of these smoke detectors might actually be the only thing in the house serving to “even up” the reactive load in the household...  I.e. I’m thinking that they may be doing the power company a favour...

[EEVblog]

They are usually not compensated completely, so in reality, the capacitive load of the smoke alarm won't cause additional current on the wires to the power plant.

Not if you suddenly added a million of them to the grid.

Because of this, only the active power should be used for calculating the energy consumption (which is still horribly high).

Nope, you can't assume the PF is corrected close to the loads(s).

[Psi]

Here's an interesting idea to keep the capacitor dropped circuit but remove all the inefficiency, and make the product last longer too.

1) Swap the primary battery for a super capacitor.
2) Only enable the capacitor dropper circuit when the super capacitor is low and needs a recharge. (Maybe using an SSR/relay)

If you only enable the capacitive dropper when you can accept all the energy into the super cap then the wasted energy is practically zero.

A capacitive dropper circuit is only inefficiency because it's always burning off a constant amount of energy based on the cap size.
Since the cap size has to be selected for the products peak current requirements you're always wasting that current even if you're not using it.

[ogden]

Here's an interesting idea

Your idea is already implemented in the chip Dave mentions in the video.

[Psi]

Here's an interesting idea

Your idea is already implemented in the chip Dave mentions in the video.

Sorry, that will teach me for skipping through the video   

[floobydust]

I changed over to mains-powered smoke alarms strictly due to the cost of ownership.
9V alkaline batteries are ridiculously expensive at $6-$7 each, per year.
A new (AC) smoke alarm costs $14, can tie into a home alarm system, and the wasted watt of heat is recovered heating the house over winter for the $0.35/year electricity.
I break even in two years and earn $6 a year afterward (each) so I can buy a new multimeter 

[ogden]

9V alkaline batteries are ridiculously expensive at $6-$7 each, per year.

BS. For 22$ you get 12pieces of 9V batteries that will last at least 2 years in properly designed smoke detector.

https://www.amazon.com/Energizer-Industrial-Batteries-Alkaline-Battery/dp/B000099SKB

[langwadt]

Given that domestic power consumption is overwhelmingly inductive (fridge motors, fans, wall-warts of all descriptions), the capacitive load of these smoke detectors might actually be the only thing in the house serving to “even up” the reactive load in the household...  I.e. I’m thinking that they may be doing the power company a favour...

most wall warts are switching supplies now, is a bridge rectifier on mains an inductive load?

[langwadt]

I changed over to mains-powered smoke alarms strictly due to the cost of ownership.
9V alkaline batteries are ridiculously expensive at $6-$7 each, per year.

you use gold plated batteries? 

[Dundarave]

Given that domestic power consumption is overwhelmingly inductive (fridge motors, fans, wall-warts of all descriptions), the capacitive load of these smoke detectors might actually be the only thing in the house serving to “even up” the reactive load in the household...  I.e. I’m thinking that they may be doing the power company a favour...

most wall warts are switching supplies now, is a bridge rectifier on mains an inductive load?

Good question, which brings to mind the impact of bazillions of idle wall-warts, compared to merely millions of AC-powered smoke detectors.

[langwadt]

Given that domestic power consumption is overwhelmingly inductive (fridge motors, fans, wall-warts of all descriptions), the capacitive load of these smoke detectors might actually be the only thing in the house serving to “even up” the reactive load in the household...  I.e. I’m thinking that they may be doing the power company a favour...

most wall warts are switching supplies now, is a bridge rectifier on mains an inductive load?

Good question, which brings to mind the impact of bazillions of idle wall-warts, compared to merely millions of AC-powered smoke detectors.

wallwarts now have requirements on no-load power consumption and PFC above a certain size

[temperance]

@EEVblog

-Can you find the surge rating of the series resistor in the cap dropper?
-Can you find any information on the capacitor being used in the cap dropper and it's expected life time or it's construction? I can see it's an X2 capacitor which a bad sign.
-Can you measure how much droop in capacitance this thing can withstand?

Perhaps the thing can withstand a very large droop in capacitance before the thing fails. So it's power consumption goes down after a while...

An app note on the lifetime of those capacitors can be found here:
https://ec.kemet.com/wp-content/uploads/sites/4/2019/10/RFI-X2-Capacitors-for-High-Humidity-Enviornments-WP1022.pdf

Vishay also has an app note detailing the construction differences between long life and low cost X capacitors. But Somehow I can't find it.

I see MOV's but no fuses.

How much energy has been wasted in producing this garbage, putting it in stores...? But who cares about that if the thing survives the warranty period?

[floobydust]

9V alkaline batteries are ridiculously expensive at $6-$7 each, per year.

BS. For 22$ you get 12pieces of 9V batteries that will last at least 2 years in properly designed smoke detector.
https://www.amazon.com/Energizer-Industrial-Batteries-Alkaline-Battery/dp/B000099SKB

I quoted $CDN from Home Depot. Amazon.ca is $2.67 each if I buy that box of 12 for $32 and they aren't knockoffs. But I don't use that many 9V batteries! WTF I have to waste $32 and the last straw was the chirping low battery alert at 3AM, so I'm choosing to destroy the power factor of the grid. Zero purchases or hassles for 4 years now over smoke alarms in the house. NFPA advises to replace smoke alarm batteries every year at Christmas.

I have filed complaints to an "importer" and UL over a mains-powered smoke alarm due to the capacitive-dropper being a cheap hot running POS and smoking and false-triggering the alarm. They ruled it's not a danger recall-worthy because the units are still detecting smoke ok. I have pics. The units have an aux relay contact for alarm bells, so extra power wasted as heat.

I saw a cheap chinese night-light with capacitive dropper, that shorts out the LED during daytime. They could have just left the LED on 

[Ed.Kloonk]

My two cents.

I've put in a few of these smoke detectors and replaced too many. I wish to Christ they would have a standard foot print access all smoke detectors so you don't have to wreck the ceiling when you replace it with a different brand.

Why do they fail early? 1, they're cheap and 2, rental occupants hit the thing with a broom stick to shut it up becuase they don't know how to cook without setting fire to food.

I've had renters hack the battery switch so the routine inspector might not notice it needs to be fixed for above reason.

I cringe a little bit when Dave waves the non-insulated screwdriver at the terminals. If it's 240 or ever had 240 in it, use sleeved tools. Please.

I've had renters buy/acquire the paint/reno covers for them too. Sometimes the 9V only ones are the solution.

[Ed.Kloonk]

The other reason I think these horrid things are so prevalent in Oz is because of Bunnings. That company is price driven on the cheapest shit and the real deal smoke detectors don't get a look in.

[gslick]

I just happened to have replaced the 9V batteries in the 6 smoke detectors in my house before watching this video. They had all been giving off low battery chirps for quite a while before I got around to dealing with the nuisance of replacing the batteries.

These smoke detectors are old enough (at least 15 years, maybe 20) that they are due to be replaced. Any thoughts on the models that are AC powered with a non-replaceable "10 Year" lithium backup battery? I like the idea of not having to deal with replacing batteries every year, as long as they actually last close to the rated 10 year lifetime. But the real question is, how bad is it that when the "10 Year" lithium backup battery expires you throw away and replace the whole smoke detector instead of just the battery? How wasteful is that? Or if the backup battery does last close to 10 years, is that somewhat near the lifetime of a smoke detector before it should be replaced anyway for reliability reasons?

[dzseki]

Given that domestic power consumption is overwhelmingly inductive (fridge motors, fans, wall-warts of all descriptions), the capacitive load of these smoke detectors might actually be the only thing in the house serving to “even up” the reactive load in the household...  I.e. I’m thinking that they may be doing the power company a favour...

most wall warts are switching supplies now, is a bridge rectifier on mains an inductive load?

Good question, which brings to mind the impact of bazillions of idle wall-warts, compared to merely millions of AC-powered smoke detectors.

wallwarts now have requirements on no-load power consumption and PFC above a certain size

Uncorrected SMPS mostly acts as capacitive load, the rectifier charges a capacitor afterall In some mediocre PC power supplies you can see an iron core yoke around the input, that is for "static" PFC.
But switcmode power suppies do worse: because they rectify the mains and the capacitor is charged to the peak voltage of the mains, therefore the diodes only conduct current when the capacitor's voltage is less than the actual mains voltage  v(t). Therefore most current is flowing when the sine is near at its peak (otherwise no current is flowing), this can lead to harmonic distortion of the mains, ie in flattening out the peak of the sine.

[SparkyFX]

The alternative of solely battery powered ones probably has the bigger impact on the environment, while the grid powered ones have the backup battery and depend on the grid. The value seems to be too low for battery salvaging and refurbishing.
OTOH last thing you want is the smoke detector catching fire because of bad design/part failure

Discussions about the environmental impact of grid power always suffer when it comes to the question if people speak about the current or future energy mix of renewable or fossil fueled ones.

A low PF just means more phase shift of current vs. voltage, which conventional meters do not measure, but still needs to be provided by the grid. It really is just about the metered value and actual current, not about the environment. Any inductive load on the same grid will do the phase shift to the other direction, so such devices need to be seen in context of the whole household.

[filssavi]

I think most of the confusion in this thread comes from improper terminology in the power systems field

While it is said that reactive power is generated/consumed, this is not really physically happening, as reactive power is just an effect that stems from the current exchanged between inductances and capacitances on the grid, these being purely reactive don’t consume any power per se (the resistance on the lines does, but the losses are much smaller then the whole reactive power)

The control of reactive power on the grid is of the utmost importance as it determines the voltage. For this reason several methods for “generating” and “consuming” reactive power need to be installed by utilities and grid controllers (e.g. field excitation control on wound rotor generators, synchronous condensers, statcoms, SVCs etc.)

As for the matter of the video, you must keep in mind that a single 150W fridge compressor (I’m assuming a power factor of 0.7 inductive) exchanges 45 var with the grid each cycle, this will offset probably dozens if not hundreds of smoke detectors

Last but not least businesses are charged for reactive power for 2 reasons:
1) it is a deterrent from the utilities, so that they have to install the minimum amount of compensation possible
2) it is a measurable quantity as opposed to the losses caused by reactive power (which are spread all over the grid)

[EEVblog]

The alternative of solely battery powered ones probably has the bigger impact on the environment, while the grid powered ones have the backup battery and depend on the grid.

The mains powered ones use exactly the same alkaline battery that last essentially the same amount of time. So the mains powered units saves zero batteries.

[EEVblog]

I cringe a little bit when Dave waves the non-insulated screwdriver at the terminals. If it's 240 or ever had 240 in it, use sleeved tools. Please.

I'm a professional 

[Muttley Snickers]

These smoke detectors are old enough (at least 15 years, maybe 20) that they are due to be replaced. Any thoughts on the models that are AC powered with a non-replaceable "10 Year" lithium backup battery? I like the idea of not having to deal with replacing batteries every year, as long as they actually last close to the rated 10 year lifetime.

An option many seem to overlook are smoke detectors powered by and connected to the security system if one is present. These are ideal for both residential and smaller commercial premises as the security system incorporates a backup battery anyway. Regulations changed down here a couple of years ago which meant that even these 12 volt alarm powered detectors required an onboard battery.

Some of the other benefits with alarm interfaced units were:
1. Option for both internal and external sirens to alert neighbours.
2. Notification of events to security monitoring or direct via SMS.
3. Ability to isolate or bypass problematic zones if necessary.
4. Ability to turn off the output powering the smoke detectors.

We generally used the Brooks or System Sensor brands but I've been out of the game for a few years now so there might be some other options as well.

https://www.brooks.com.au
https://www.systemsensor.com
 
 

[daqq]

Another solution: Use one of those crazy small tranasformers:

http://www.farnell.com/datasheets/65263.pdf

such as:
https://uk.farnell.com/block/vb-0-35-1-15/transformer-0-35-va-15v/dp/1711394

Size: 22x23x15 mm, 20g weight

[SparkyFX]

The mains powered ones use exactly the same alkaline battery that last essentially the same amount of time. So the mains powered units saves zero batteries.

Sorry, did not point this out correctly, the Lithium 10 year type vs. the 9V alkaline vs. the mains powered w/ 9V alkaline backup. The 9V type required replacement every 2 years or so, the Lithium 10year so far is just 3 years old.

Mine usually go off when i cook or forgot the pizza in the oven. My cooking skills improved a bit... and the habit to use a timer helped a bit too.

[Ice-Tea]

Going to agree with some others here: I object to the random mixing of VA and W. I understand your point but watching the movie you can leave with the idea that it's all the same. It's not.

Yes, the capacity has to be available on the net and yes, without compensation you're using up that capacity *but* you're not burning a Joule of fuel to generate a VA.*

* Yeah, yeah, the ohmic losses in the wiring and all that...

[mikeselectricstuff]

Another solution: Use one of those crazy small tranasformers:

http://www.farnell.com/datasheets/65263.pdf

such as:
https://uk.farnell.com/block/vb-0-35-1-15/transformer-0-35-va-15v/dp/1711394

Size: 22x23x15 mm, 20g weight

IME these things are terrible - very poor efficiency and they run warm enough to distort their  plastic case over time even with minimal load.

[filssavi]

Another solution: Use one of those crazy small tranasformers:

http://www.farnell.com/datasheets/65263.pdf

such as:
https://uk.farnell.com/block/vb-0-35-1-15/transformer-0-35-va-15v/dp/1711394

Size: 22x23x15 mm, 20g weight

IME these things are terrible - very poor efficiency and they run warm enough to distort their  plastic case over time even with minimal load.

Not to mention that the power factor of the transformer with such small loads might be just as terrible

[mikeselectricstuff]

Another solution: Use one of those crazy small tranasformers:

http://www.farnell.com/datasheets/65263.pdf

such as:
https://uk.farnell.com/block/vb-0-35-1-15/transformer-0-35-va-15v/dp/1711394
The datasheet mentions no-load losses around 1W


Size: 22x23x15 mm, 20g weight

IME these things are terrible - very poor efficiency and they run warm enough to distort their  plastic case over time even with minimal load.

Not to mention that the power factor of the transformer with such small loads might be just as terrible

[GeorgeOfTheJungle]

It takes 5W to charge a 1µF capacitor 100 times a second.

100*1e-6*((240*sqrt(2)-15)^2)/2 ~= 5.2 joules/s

Where are the other 15W up to 20 VA dissipated then, in 100R and the zeners? 15W are many watts, they should be quite toasty. No?

[Poe]

But Dave, you waste more energy than even a million smoke detectors.

n*m/60=Tmh
Tmh*p=Twh

Where
n = number of people viewing this video
m = number of minutes wasted by not editing
p = power consumption of YouTube viewing device
Tmh=Total man-hours wasted
Twh = Total watt hours wasted vs a polished/edited video

Assuming 100k views where 20 out of a 30min video is waffling..
100,000 * 20 / 60 = 33,333 man-hours wasted
33,333 * 10 = 333,333wh wasted

That's actual wasted power, not reactive, for ONE video.

[EEVblog]

It takes 5W to charge a 1µF capacitor 100 times a second.
100*1e-6*((240*sqrt(2)-15)^2)/2 ~= 5.2 joules/s
Where are the other 15W up to 20 VA dissipated then, in 100R and the zeners? 15W are many watts, they should be quite toasty. No?

@79.3mA and 1.36W total measured:
100R = 0.63W
So that leaves 0.73W for the zeners.
Ignore the load which is naff all.
Maybe a smidge in the cap ESR.
The rest will be copper losses down the system unless compensated for.

[Brutte]

I think most of the confusion in this thread comes from improper terminology in the power systems field

+1

Dave, please do not confuse Watts and Volt*Amperes.
These are two different animals, like apples and oranges.
You cannot add them, subtract or compare.

Please split the video into two parts:
-Part One: You are ranting only about low efficiency of this power supply (do not even mention "volt*amperes" word in this part).
-Part Two: You are ranting only about low power factor of this power supply (do not even mention "watts" word in this part).

By far the least profesional eevblog so far.
Very close to some YT blogs where people start to compare power(kW) to energy(kWh).
 

[Jr460]

Wow people have some crappy smoke alarms.

Mine is an old one that has the radioactive stuff in it.   Still works fine, i know the wife set it off a month or so ago cooking and not turning on the vent fan.   You couldn't ever "see" anything in the air, but the alarm went off.   No need to replace it if it keeps being that sensitive.   

One 9V battery and that lasts, and lasts, and lasts.   2 years plus out of one battery, yes I do test the alarm.

[coppice]

I think NANDBlog is right. The generator doesn't need to generate all the reactive power but only the active power + losses.

There is reason why commercial companies are charged in kVA. They aren't charged kW+"copper losses"

Where have you seen companies charged for kVA? Around the world. commercial users are normally charged for actual consumed (i.e. active) energy. In some places a commercial user's reactive power or power factor is measured continuously, and if its really bad (or sometimes if its really bad specifically during peak hours) the customer is charged a punishment factor on top of their active energy, to encourage them to improve their power factor.

[Domagoj T]

Where have you seen companies charged for kVA? Around the world. commercial users are normally charged for actual consumed (i.e. active) energy. In some places a commercial user's reactive power or power factor is measured continuously, and if its really bad (or sometimes if its really bad specifically during peak hours) the customer is charged a punishment factor on top of their active energy, to encourage them to improve their power factor.

This is Croatian power company price list for industrial consumers:
https://www.hep.hr/ods/korisnici/poduzetnistvo/tarifne-stavke-cijene-161/161
The table is an image so it won't translate, but the column "Prekomjerna jalova energija" (excessive apparent power) is what we're talking about.

Also this:
https://translate.google.com/translate?sl=auto&tl=en&u=https%3A%2F%2Fwww.hep.hr%2Fods%2Fkorisnici%2Fsavjeti-kupcima%2Fnaknada-za-prekomjernu-jalovu-energiju%2F579%23
Basically, if the customer keeps their power factor above 0,95, there is no penalty. If it drops below it, then there is a penalty.

[Jeff1946]

As noted in Dave's video, photoelectric detectors detect visible smoke.  Those based on radioactive sources produce ions in the detection chamber which produce a current between two charged plates.  Particles, including those to small to see, in the air decrease this current.  That's why I replaced the ionization detector nearest my kitchen which would often alarm when we were cooking with a photoelectric one.  No more false alarms.

[nctnico]

I think you should review how reactive power works. The 100MW is not generated by the wind turbine. You generate 5MW by the wind, and place a compensation on the power grid, that will generate the reactive power. It is just a big capacitor. The grid already has a bunch of inductive load, because of motors.

See my comment above.

Nope. Nandblog is right. Your reasoning doesn't make sense. If you input 100MW into 5 million devices then each device would dissipate 20W and -given the parts inside- something would go up in flames. But that isn't happening. This is the basic law of physics which says 'the amount of power going in is equal to the amount of power going out' in action. And no, the losses don't occur somewhere in between because your meter says there is about 20VA going into the smoke detector.

The windturbines only deliver the actual power but the current and voltage may be out of phase and/or have different waveform shapes. The only problem with too much out-of-phase / non-sinusoidal current is that the losses in the wires are higher and cores of distribution transformers may get saturated at lower power levels.

[Bud]

As noted in Dave's video, photoelectric detectors detect visible smoke.  Those based on radioactive sources produce ions in the detection chamber which produce a current between two charged plates.  Particles, including those to small to see, in the air decrease this current.  That's why I replaced the ionization detector nearest my kitchen which would often alarm when we were cooking with a photoelectric one.  No more false alarms.

Ionization vs photoelectric alarms serve specific use cases. Ionization ones better detect flames and photoelectric ones are for smoke. They should be used based on risk of type of fire.

[DBecker]

Capacitive reactance is a benefit to the grid, not a negative.

The grid is generally inductive, not capacitive.

Back in the "good old days" (which were never as good as memory suggests) the grid was dominated by inductively reactive devices, mostly lightly-loaded motors.  Large capacitor banks were added in industrial areas when the power factor was absurdly bad, but that wasn't cost effective when the power factor was only modestly bad.

Even in a present-day residential area it's quite unlikely that capacitive supplies come close to locally compensating for refrigerator, fan and compressor motors.

Many areas have a requirement that smoke alarms et al (fire, monoxide, fume) both be wired to the grid and have battery backup.  This is because of the perceived risk that a battery-only alarm won't have its battery replaced.  The requirement has a high lifetime cost with marginal benefit, since hard-wiring costs $50-$250 per location.  The same people that wouldn't replace a battery in a battery-only alarm will just unplug the dual-power alarm when its battery dies.

[rrinker]

 I'm pretty sure it's a requirement for new construction, and has been for a number of years now, but I don't think there is any requirement for it in existing structures. In my area I don't even think they REQUIRE you to have smoke detectors, at least in single family residences, but you're pretty stupid if you don't.

 One thing the hard wired ones buy you, they usually have an extra wire which gets run between all units, so that when one trips, they all sound off.

[coppice]

Where have you seen companies charged for kVA? Around the world. commercial users are normally charged for actual consumed (i.e. active) energy. In some places a commercial user's reactive power or power factor is measured continuously, and if its really bad (or sometimes if its really bad specifically during peak hours) the customer is charged a punishment factor on top of their active energy, to encourage them to improve their power factor.

This is Croatian power company price list for industrial consumers:
https://www.hep.hr/ods/korisnici/poduzetnistvo/tarifne-stavke-cijene-161/161
The table is an image so it won't translate, but the column "Prekomjerna jalova energija" (excessive apparent power) is what we're talking about.

Also this:
https://translate.google.com/translate?sl=auto&tl=en&u=https%3A%2F%2Fwww.hep.hr%2Fods%2Fkorisnici%2Fsavjeti-kupcima%2Fnaknada-za-prekomjernu-jalovu-energiju%2F579%23
Basically, if the customer keeps their power factor above 0,95, there is no penalty. If it drops below it, then there is a penalty.

Demanding the power factor always stay above 0.95 is stricter than the schemes I've seen, but this is basically what I said. Customers are not specifically charged for VA. They are charged for actual watts, and incur an additional penalty if they don't keep their power quality within defined limits. Another interesting power quality related thing you'll find in some commercial tariffs is payment for only the fundamental power consumed. This is related to neighbouring customers having poor power factor loads, since they cause excessive resistive voltage drops, which results in substantial THD In the supplied voltage waveform, and hence harmonic energy.

[DBecker]

I'm pretty sure it's a requirement for new construction, and has been for a number of years now, but I don't think there is any requirement for it in existing structures. In my area I don't even think they REQUIRE you to have smoke detectors, at least in single family residences, but you're pretty stupid if you don't.

 One thing the hard wired ones buy you, they usually have an extra wire which gets run between all units, so that when one trips, they all sound off.

The requirements for alarms vary considerably by region, and even between towns.  In California they are stricter for rental units than owner-occupied homes.
Here interconnection of hard-wired units is required, but my impression is that there is lots of cheating on this point.  The extra conductor doubles the cost of the power cable (NM-B 14-2 to -3) so it's only installed if they know the AHJ inspector will be checking.

[hendorog]

Since there are already smoke detectors which last 10 years, I think there is an opportunity to combine that with energy harvesting.

RF or solar, it wouldn't need to snag very much power to remove the battery problem forever.

[HackedFridgeMagnet]

A few things.

Those chips are interesting I didn't know they existed. Looking forward to using the TI chip the next chance I can.

Reading the video title Dave is completely right. The zener based capacitive dropper is extremely inefficient, time to use the latest technology. Even if it is a very small percentage of the grid power, it is an easy fix for the next generation of smoke alarms.

As to the power factor, it may actually improve the transmission losses being a capacitive load while most of the grid is slightly inductive. (edit: Just had a look at some large office power meters, some leading some lagging, residential seems to be a bit random too?)

re: smoke alarm rules. In Australia as usual it varies state to state but these type of rules tend to be fairly consistent across the country.

from: https://www.legislation.tas.gov.au/view/html/inforce/current/sr-2012-140#GS8@EN

8.   Smoke alarm to be mains-powered after 3 years after commencement of regulations

    On and from the standards upgrade day, a smoke alarm that is required to be in place for the purposes of section 36C of the Act must be –

            (a) a mains-powered smoke alarm; or

            (b) powered by a 10-year non-removable battery.

[thm_w]

Big Clive is overly paranoid about poor power factor products, in that he thinks the power company will start charging you the 15VA a 1W LED takes for example. I have yet to see any evidence of this being the case for residential, and as two people have pointed out its probably balancing out the motor loads. Even for industrial, they don't blanket charge you the VA they generally charge a fee.

Anyway, the environmental aspect is still valid. Here is an article that should back up Daves point:

Ergon Energy is undertaking a power factor pilot project with nearly 30 major business and industrial customers in and around Toowoomba.
The aim of the Queensland Government funded project is to use incentive payments to reduce peak demand by a total 4.7 MVA, with subsequent customer savings and carbon emissions reduction.
Improving the power factor of appliances such as motors and other equipment at individual premises will help improve energy efficiency both for the customers and for the Ergon Energy network1.
An appliance with a low power factor draws more current from the available power source than an appliance with a high power factor. The aim is to increase the power factor to 0.9 or more, that is, so electricity use is 90-plus per cent efficient.
For Ergon Energy the success of this project can eliminate or defer the need for new infrastructure such as transmission lines, sub-stations and transformers

https://web.archive.org/web/20140214055620/https://www.ergon.com.au/energy-conservation/demand-management/electricity-demand-trials/power-factor-correction-pilot-project
https://web.archive.org/web/20130903162112/http://www.ergon.com.au/__data/assets/pdf_file/0014/80123/Power-factor-fact-sheet-12mar2012.pdf
https://electronics.stackexchange.com/questions/85291/why-do-power-companies-never-bother-residential-customers-about-power-factors
https://www.powerelectronics.com/technologies/power-electronics-systems/article/21860956/backtobasics-on-power-factor-and-why-we-correct-it

Although they did seem to conflate efficiency and power factor, which is not correct.

[coppice]

Big Clive is overly paranoid about poor power factor products, in that he thinks the power company will start charging you the 15VA a 1W LED takes for example. I have yet to see any evidence of this being the case for residential, and as two people have pointed out its probably balancing out the motor loads. Even for industrial, they don't blanket charge you the VA they generally charge a fee.

In most countries it is illegal not to base charges on actual consumed energy. These things are specified by WELMEC in Europe, the related OIML in much of the rest of the world, and ANSI standards in the US. Most metrology related to the sale of stuff, from the weight of cheese, to the dispensing of gasoline, to the consumption of electrical energy, follows a tight legal framework in most countries. The requirement to base electrical energy charges on actual consumed energy is a key reason why when there are charges related to power quality issues, like poor power factor, harmonics, etc., these are usually structured as some kind of supplementary fees on top of the tightly controlled charges for actual energy.

[scopeman]

This has been a really eye opening and interesting discussion especially considering the certifications that these devices have to go through. Obviously energy savings was not in the specification.

There was a comment on "rolling your own" better power supply which I agree with from a designers perspective, but lets say you do that and something goes wrong and it fails and burns your house down.

Then I wonder if that would void your home insurance policy? Just a thought.

I do agree that we should demand the the designers do better than what I have seen here. Seems like that bad design should create a market for good design.

Sam
W3OHM

[EEVblog]

I think you should review how reactive power works. The 100MW is not generated by the wind turbine. You generate 5MW by the wind, and place a compensation on the power grid, that will generate the reactive power. It is just a big capacitor. The grid already has a bunch of inductive load, because of motors.

See my comment above.

Nope. Nandblog is right. Your reasoning doesn't make sense. If you input 100MW into 5 million devices then each device would dissipate 20W and -given the parts inside- something would go up in flames. But that isn't happening.

Of course it's not happening. Any potential extra power above what I calculated above (and as measured in W) is dissipated outside the smoke alarm in the form of I²R losses in the system.
It doesn't have to be 100MW, that's just the theoretical worst system case. If the current is compensated for and filtered (for harmonic based PF) then it will be lower. But the fact is that 80mA is real current from the grid, it doesn't magically vanish.

[DBecker]

I do agree that we should demand the the designers do better than what I have seen here. Seems like that bad design should create a market for good design.

It's a nice concept, but there is no market feedback about crappy power supplies.  With complex products it's one spec among thousands, most of them not revealed.  You can't make an informed decision, you can't even get most info before buying and tearing it down.

While government regulations such as Energy Star are intrusive interference in commerce, and I feel I should oppose them on pure philosophical grounds, they are the only way to get the market to switch to efficient power supplies.  Just like safety standards, if there are not enforced regulations some companies will build the cheapest possible product even if it's dangerous for consumers.

[EEVblog]

For those who think this bad PF can be corrected by the motors in your home etc, think again, this is the current waveform.



The poor PF is caused mostly by the terrible harmonic content, that cannot be just lag/lead corrected out, it needs to be filtered out. I believe this is done in stages in the grid, but someone mentioned in the Youtube comments that it's not done until the 110kV stage of the transmission system. This means huge I²R losses in the grid up to that point.

[EEVblog]

Bonus screenshot:

[floobydust]

I worked in the utility industry and big scale power factor correction was rare.

One substation 144kV to 25kV (going from transmission to distribution system) with a couple 40MVA transformers. The system is overstressed mainly due to the largest loads being inductive with a lagging PF, loads that have ever increased as businesses and the city expand. Big electric motors in plants. Upsizing to 240kV lines is ultra expensive due to changing all connected substations equipment to support the higher voltage class, land owner disputes etc. due to the bigger towers.

The usual patch is to crank up the system voltage, for the resulting lower current on the power lines and transformers, so they are not as stressed. But this has limits.

It was decided to install 20MVAR capacitor banks on the 25kV bus to help the power factor. The price for capacitor banks, 34.5kV vaccuum circuit breakers to switch-in on zero-cross, and all the SCADA equipment was over $2M. Just too expensive to be commonplace and they weren't so reliable, the capacitors failed. They experience very high ripple current and lots of transients. Was it a decent solution? It bought the utility a few more years.
Power quality is degrading, demand is increasing and the utility infrastructure is boxed in.

[Ed.Kloonk]

Bonus screenshot:
(Attachment Link)

The old Audio engineer in me goes... farken Hay Zues.

Introducing this into to power is like putting fluoride in the water. (it turns the frogs gay!)

[floobydust]

Is there a DC offset? The scope is AC coupled yet the trace is up. Audiophools are worried about it and AC-couple mains into power amplifiers, seems a bit overboard. https://sound-au.com/articles/xfmr-dc.htm

At the coal-fired power plant 430MW the generator sound is a nice pure sine hum. Further away you go on the grid, those harmonics make transformers growl quite loud.
Most of my equipment buzzes during the day when power quality is worst and go quiet at night. Even LED light fixtures run noisy with dirty power.

I notice mains transformers really don't work well with flattened sine waves, their regulation seems to degrade?

[EEVblog]

Is there a DC offset? The scope is AC coupled yet the trace is up.

The scope is DC coupled.
Although I am a bit perplexed why the negative half current is the same as the positive half current. But kinda makes sense in that the zener is just a diode in the half cycle, and sensor circuit is doing whatever
all the time to generate that crap.
BTW, the other Quell brand is near identical.

[jackbob]

Hello Dave! It's Jacob from the comments section of the video. Thanks for posing this, it looks exactly as I thought it would. This is perhaps worthy of another video. The harmonic content here is horrendous, this is what is causing the poor PF. This is also the worst type of poor power factor not only because it's difficult to correct, but because it causes interference and a host of other noise-related issues. As I stated in the comments section it can be corrected with filters in the power system but these are usually not found until a transmission substation after all the losses have occurred.

Here is a quick wikipedia page describing the components which will ultimately end up removing the harmonic content. The line reactor section describes the use as harmonic filters.

https://en.wikipedia.org/wiki/Current_limiting_reactor

Why does harmonic content create low PF??
Most people are familiar with P = Vrms*Irms*cos(theta) where theta is the angle between the voltage and current waveform. This equation holds true HOWEVER it can only be applied (without thought) to pure sinusoids of the same frequency. That means the voltage and current waveforms are of the same frequency and are both sinusoidal. For any non-sinusoidal waveforms such as this current waveform, they must be decomposed into sinusoids using the Fourier series and analyzed using the same equation for every harmonic. I will post an image of this formula from one of my power electronics textbooks.

One important realization is if the voltage source is a sinusoid but the current is not, ONLY the fundamental frequency of the current waveform can draw real power. The harmonics are orthogonal to the voltage fundamental frequency and will never draw real power unless they have a voltage harmonic to pair with. These harmonics do still draw current and contribute to the overall RMS current draw. These harmonics of different frequencies are responsible for the apparent power draw and the power factor is calculated using its fundamental definition PF=P/S. This is how PF must always be calculated for non-sinusoidal waveforms. P = Vrms*Irms*cos(theta) WILL NOT work in this ball field. Therefore adding inductance or capacitance to compensate for the PF will not work, the harmonic content must be filtered out.

This one of the reasons is why there is generally so much regulation on harmonic content. If these power supplies were a little larger (say for a TV or PC) they likely would not be permissible by many regulating entities with harmonic content of this magnitude.

[nctnico]

I think you should review how reactive power works. The 100MW is not generated by the wind turbine. You generate 5MW by the wind, and place a compensation on the power grid, that will generate the reactive power. It is just a big capacitor. The grid already has a bunch of inductive load, because of motors.

See my comment above.

Nope. Nandblog is right. Your reasoning doesn't make sense. If you input 100MW into 5 million devices then each device would dissipate 20W and -given the parts inside- something would go up in flames. But that isn't happening.

Of course it's not happening. Any potential extra power above what I calculated above (and as measured in W) is dissipated outside the smoke alarm in the form of I²R losses in the system.
It doesn't have to be 100MW, that's just the theoretical worst system case. If the current is compensated for and filtered (for harmonic based PF) then it will be lower. But the fact is that 80mA is real current from the grid, it doesn't magically vanish.

But it isn't power. Please think back about how a generator works, the equations involved (induced voltage due to moving a wire through a magnetic field) and go from there. Your numbers don't add up and it is never ever going to be anywhere near 100MW (not even including I2R losses). Just do the math. For starters let your scope multiply the V and I trace to calculate the power and then calculate the RMS power from that. That is the real power and that is what the mains needs to supply.

[jackbob]

But it isn't power. Please think back about how a generator works, the equations involved (induced voltage due to moving a wire through a magnetic field) and go from there. Your numbers don't add up and it is never ever going to be anywhere near 100MW (not even including I2R losses). Just do the math. For starters let your scope multiply the V and I trace to calculate the power and then calculate the RMS power from that. That is the real power and that is what the mains needs to supply.

Not necessarily true. That is the only power the mains needs to supply at your house maybe. But the I^2R losses in the grid are a result of the current not real power. If the PF is really low, you are drawing lots of useless current, in this case due to harmonics. My previous post describes that harmonics will work their way rather high in the electrical system before being suppressed by filters. All this useless current supplying the reactive power is being moved through the transmission and distribution system while creating I^2R losses in the system. These I^2R losses do create real power losses. If the PF is low enough (in this case it was quite low), then you could literally be using more real power (watts) in the transmission and distribution losses than in the product itself. The generator must supply all the real power associated with the device being connected to the grid, not just the power draw measured at the product.

[jackbob]

Also, I have seen some ask in the youtube comments about harmonics created by switchmode power supplies. Many switch mode power supplies today have active power factor correction. Active PFC does a fairly decent job of getting the current waveform to track the voltage waveform, making the power supply appear almost as a resistive load. It is however not perfect, active PFC relies on a boost converter operating at 100's of kHz which creates some harmonics. It should be noted that harmonics this high in frequency can be easily suppressed with a small LC filter at the mains input of the power supply. This is a common component of every switching power supply and can clean up most of the harmonics because of the high frequency. Inductors have a high impedance to high frequency currents and capacitors shunt harmonics well at high frequencies. The harmonic content pictured in the scope image can easily be seen on the same time scale as the fundamental frequency. In fact, the harmonics are multiples of the line frequency. These low frequency harmonics 100's of Hz are much more difficult to filter and cannot be filtered by a small LC filter on the power supply. They are so close to the fundamental frequency that they make their way through the grid with ease.

[EEVblog]

I think you should review how reactive power works. The 100MW is not generated by the wind turbine. You generate 5MW by the wind, and place a compensation on the power grid, that will generate the reactive power. It is just a big capacitor. The grid already has a bunch of inductive load, because of motors.

See my comment above.

Nope. Nandblog is right. Your reasoning doesn't make sense. If you input 100MW into 5 million devices then each device would dissipate 20W and -given the parts inside- something would go up in flames. But that isn't happening.

Of course it's not happening. Any potential extra power above what I calculated above (and as measured in W) is dissipated outside the smoke alarm in the form of I²R losses in the system.
It doesn't have to be 100MW, that's just the theoretical worst system case. If the current is compensated for and filtered (for harmonic based PF) then it will be lower. But the fact is that 80mA is real current from the grid, it doesn't magically vanish.

But it isn't power. Please think back about how a generator works, the equations involved (induced voltage due to moving a wire through a magnetic field) and go from there. Your numbers don't add up and it is never ever going to be anywhere near 100MW (not even including I2R losses). Just do the math. For starters let your scope multiply the V and I trace to calculate the power and then calculate the RMS power from that. That is the real power and that is what the mains needs to supply.

Nope, I repeat my position.
I don't care about the generator, all I care about is the 80mA real rms current that comes from the 240V supply. The 80mA creates in theory real I²R system losses all along the chain, you can't just ignore that, it's not imaginary.
If this is not compensated for then it must ultimately come from a source in order to maintain that voltage.
I said in the video that 100MW is not a real figure, it's hyperbole based on that 80mA current and no correction. I admit 100MW this is not a practical figure.
And we are not talking sinusoidals here.

[EEVblog]

Bonus screenshot:
(Attachment Link)

Turns out all that funny business is not coming from the product consumption, and with a moments thought that is obvious, more investigation required, likely a proby thing.
A pure R-C-Zener circuit should not create that mid frequency stuff, just the front and back porches.



Well this is strange. Using my AIM current probe I get the exact same waveform. And if I bypass the power monitor for mains input I also get the same waveform.
More investigation required...

[HackedFridgeMagnet]

Nice for a video down the track if you could build up those other two solutions (TI and OnSemi iirc?)  and then measure the three different options. Four if you count the full bridge.

[EEVblog]

Nice for a video down the track if you could build up those other two solutions (TI and OnSemi iirc?)  and then measure the three different options. Four if you count the full bridge.

Don't think I'll go that far, it's obvious what results a HV regulator will give in terms of current vs the zener solution, no need to build and measure to prove that.

[jackbob]

Turns out all that funny business is not coming from the product consumption, and with a moments thought that is obvious, more investigation required, likely a proby thing.
A pure R-C-Zener circuit should not create that mid frequency stuff, just the front and back porches.

The zero voltage crossings look alright. The mid frequency on the top is a bit odd. Nevertheless, the waveform will be non-sinusoidal. Grab the mid frequency with the cursors. Is it an exact multiple of the line frequency? Could very well be a quirk with the probes and this particular measurement.

[EEVblog]

The zero voltage crossings look alright. The mid frequency on the top is a bit odd. Nevertheless, the waveform will be non-sinusoidal. Grab the mid frequency with the cursors. Is it an exact multiple of the line frequency? Could very well be a quirk with the probes and this particular measurement.

Yeah, I think she's'a'ringing.
Will try my current probe, but not sure how it goes on currents that low.

[nctnico]

But it isn't power. Please think back about how a generator works, the equations involved (induced voltage due to moving a wire through a magnetic field) and go from there. Your numbers don't add up and it is never ever going to be anywhere near 100MW (not even including I2R losses). Just do the math. For starters let your scope multiply the V and I trace to calculate the power and then calculate the RMS power from that. That is the real power and that is what the mains needs to supply.

Not necessarily true. That is the only power the mains needs to supply at your house maybe. But the I^2R losses in the grid are a result of the current not real power.

But still the I2R losses from the smoke alarm will be extremely low. Mains impedance is very low anyway in order to keep I2R losses low by design. Ofcourse you can think of theoretical devices which draw 100A near the zero crossing but that is not the case with normal devices. Mains impedance for a house connection seems to be typically below .5 Ohm so the I2R losses of the smoke alarm are around 3mW when using a mains impedance of .5 Ohm. For 5 million devices it would add up to about 17kW. That is a far cry from needing 95MW extra.

[EEVblog]

Not necessarily true. That is the only power the mains needs to supply at your house maybe. But the I^2R losses in the grid are a result of the current not real power.

But still the I2R losses from the smoke alarm will be extremely low. Mains impedance is very low anyway in order to keep I2R losses low by design. Ofcourse you can think of theoretical devices which draw 100A near the zero crossing but that is not the case with normal devices. Mains impedance for a house connection seems to be typically below .5 Ohm so the I2R losses of the smoke alarm are around 3mW when using a mains impedance of .5 Ohm. For 5 million devices it would add up to about 17kW. That is a far cry from needing 95MW extra.
[/quote]

I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.

I'll repeat for the last time, I said in the video that 100MW is not a practical figure, it's hyperbole based on that 80mA current and no correction. I admit 100MW this is not a practical figure.
I will not debate this further.
If you want to talk real interesting and pertinent issues like harmonic power factor, or low power direct mains supply design then I'm happy to hear it.

[Ice-Tea]

I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.

0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong. Grid losses amount to 5-10% or so, I believe. So, 10MW is probably closer to a usable number.

Truth be told, this video is about as close to clickbait as I've ever seen you go. Sure, you've said it's not a practical number but it's all a bit questionable...

[EEVblog]

I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.

0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.

Think about it again, and how I did the calcs in the video for a home, and then get back to me.

[nctnico]

I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.

0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.

Think about it again, and how I did the calcs in the video for a home, and then get back to me.

Sorry for my critical thingking but the calculations shown in the text of the video make no sense to me because you keep mixing Watts and VA as if they are equal. So please enlighten us with a real calculation including a good estimate of the actual grid losses. Hint: if small devices with low PF are deemed a problem then they would be subject to regulations regarding power factor. My simple, off the cuff, calculation already shows that the I2R losses for the smoke alarm are neglectible even if my resulting number is 10 times too low.

[Ice-Tea]

I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.

0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.

Think about it again, and how I did the calcs in the video for a home, and then get back to me.

Sorry for my critical thingking but the calculations shown in the text of the video make no sense to me because you keep mixing Watts and VA as if they are equal. So please enlighten us with a real calculation including a good estimate of the actual grid losses. Hint: if small devices with low PF are deemed a problem then they would be subject to regulations regarding power factor. My simple, off the cuff, calculation already shows that the I2R losses for the smoke alarm are neglectible even if my resulting number is 10 times too low.

I think the discussion is much better served with a general, bulk number for grid losses. I agree with the VA and W mixing, that said, he's probably hinting at the fact that this load comes on top of a certain "base load"? Imagine in your scenario, there's already 10 amps running around then the additional I2R losses would be 10.08*10.08*0.5 - 10*10*0.5 = 800mW.  But then you'd have to take into account the multitude of circuits etc. Easier is the lump transmission losses...

[EEVblog]

I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.

0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.

Think about it again, and how I did the calcs in the video for a home, and then get back to me.

Sorry for my critical thingking but the calculations shown in the text of the video make no sense to me because you keep mixing Watts and VA as if they are equal.

They are not equal. BUT if a product takes 80mA in measured RMS current, and that is NOT compensated for in the grid by filtering and/or phase correction, then to maintain the same mains voltage that power MUST ultimately be generated at some point, there is no free lunch there.

Thought experiment for you: Assume your grid is at almost maximum capacity (minus the 1.2MW real device power (1.2W * 1,000,000 devices)), and there is no more harmonic filtering or phase correction left to give, and you suddenly connect a million of these devices to the grid, where do you think the almost 100MVA magically comes from to maintain that 240V at the home? The current fairy?

So please enlighten us with a real calculation including a good estimate of the actual grid losses. Hint: if small devices with low PF are deemed a problem then they would be subject to regulations regarding power factor.

Just because they didn't bother to do <75W devices in EN61000-3-2 and EnergyStar doesn't mean it doesn't matter. The entire point of the video is to consider this in product design.
If it's a problem at 75W for a single device, why shouldn't it be a problem with 75 devices of 1W each (localised EMC excluded).

My simple, off the cuff, calculation already shows that the I2R losses for the smoke alarm are neglectible even if my resulting number is 10 times too low.

Seeing as that you couldn't figure out what was wrong with your own calculation, let me oblige. 5 units per house (assume 0ohms in-house wiring) = 0.4A² * 0.5ohms grid impedance = 80mW, not 3mW. Be careful of your assumptions before you square stuff.

Again, I'm done on this 100MW thing, I will ignore all further questions on it. If you want to talk harmonic PF, filtering, and low power design, please do so.

[Ice-Tea]

Thought experiment for you: Assume your grid is at almost maximum capacity (minus the 1.2MW real device power (1.2W * 1,000,000 devices)), and there is no more harmonic filtering or phase correction left to give, and you suddenly connect a million of these devices to the grid, where do you think the almost 100MVA magically comes from to maintain that 240V at the home? The current fairy?

Obviously not. But, once more ignoring I2R, you won't be burning a Joule of fuel more anywhere. Which means that the claim that it "kills the environment" is a hyperbole at best, pure clickbait at worst. The content between, say, 8:00 and 11:00 or so is confusing, misleading and sometimes just wrong.

Nobody is debating that it's a bad design, not even that it wastes too much power (as the power in the zener *is* lost for real). But some of the content is no good.

Thought experiment for you: ping Thunderfoot (whom you seem to respect and admire), ask him how he feels about that part of the video.

Again, I'm done on this 100MW thing, I will ignore all further questions on it. If you want to talk harmonic PF, filtering, and low power design, please do so.

Your prerogative.

[EEVblog]

Thought experiment for you: Assume your grid is at almost maximum capacity (minus the 1.2MW real device power (1.2W * 1,000,000 devices)), and there is no more harmonic filtering or phase correction left to give, and you suddenly connect a million of these devices to the grid, where do you think the almost 100MVA magically comes from to maintain that 240V at the home? The current fairy?

Obviously not. But, once more ignoring I2R, you won't be burning a Joule of fuel more anywhere.

You can't ignore I2R at that point, that's the whole point 

Which means that the claim that it "kills the environment" is a hyperbole at best, pure clickbait at worst.

I admitted it's hyperbole 

The content between, say, 8:00 and 11:00 or so is confusing, misleading and sometimes just wrong.

I admit it was poorly explained.

[nctnico]

I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.

0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.

Think about it again, and how I did the calcs in the video for a home, and then get back to me.

Sorry for my critical thingking but the calculations shown in the text of the video make no sense to me because you keep mixing Watts and VA as if they are equal.

They are not equal. BUT if a product takes 80mA in measured RMS current, and that is NOT compensated for in the grid by filtering and/or phase correction, then to maintain the same mains voltage that power MUST ultimately be generated at some point, there is no free lunch there.

Thought experiment for you: Assume your grid is at almost maximum capacity (minus the 1.2MW real device power (1.2W * 1,000,000 devices)), and there is no more harmonic filtering or phase correction left to give, and you suddenly connect a million of these devices to the grid, where do you think the almost 100MVA magically comes from to maintain that 240V at the home? The current fairy?

The 80mA from each smoke detector is just out of phase so the only thing you'll see in the generator is that the current through the windings is not in phase with the voltage. For a minute assume zero losses in the windings of the generator so it has zero impedance. The voltage induced depends on the speed at which the wire is travelling through the (-let's assume- homogenous) magnetic field. That gets you a sinusoidal voltage waveform. If a load decides to take it's maximum current at a different point of the sine wave the total power will still be the real part of the voltage waveform multiplied by the current waveform over one cycle. You can't just multiply RMS current by RMS voltage without taking the phase relationship into account. An extreme situation would be where a load pulls 100A around the zero crossing; the amount of power is near zero so the extra power needed to turn the generator is near zero as well. So, yes, the extra current comes from the current fairy; it is called imaginary for a reason 

In a real generator you'll have copper losses (resistive and skin effect) and things like core saturation which eventually will limit the amount of current the generator can physically handle. Suffice to say the generator delivers best bang per buck if the load is resistive; it can use the thinnest windings and smallest core.

All in all the I2R losses aren't very extreme compared to the real power consumption of the device so this won't be a problem at all. I think this can be demontrated using a relatively simple test setup.

[EEVblog]

The content between, say, 8:00 and 11:00 or so is confusing, misleading and sometimes just wrong.

I admit it was poorly explained.

I watched it again, and yeah, it's just too confusing I think. I have removed that part with the edit tool, I hope it works. Will take many hours to process.

[Poe]

But Dave, you waste more energy than even a million smoke detectors.
...

...

I watched it again, and yeah, it's just too confusing I think. I have removed that part with the edit tool, I hope it works. Will take many hours to process.

Assuming 3min was saved from a video that's going to get 100k views....
100,000views * 3min / 60 =  5,000 man-hours saved
5,000 * 10W per device = 50kwh saved

That's about the same annual energy consumption as all the smoke detectors in Australlia, for one 3min edit.

So we can conclude that influencers putting out unpolished videos do more harm to the environment than poorly designed electronic devices.

[ckambiselis]

What is the failure rate of the crappy solution "they" have used compared to the ones you showed, could that be a reason they went with it?

Oscar

[DBecker]

I think a point that is missed is that the I2R losses aren't extending to the grid.  They probably aren't even getting to the pole pig.  They are getting to the nearest traditional electric motor (non-inverter-driven), where the capacitive reactance is a benefit and any harmonics are being absorbed.

Look at it from the other side: try to deliberately inject a signal into the power line.  The powerline Ethernet guys will tell you how difficult that is, even when you get to dynamically pick the frequency and modulation.

Edit: fixed typeo

[floobydust]

Is there a DC offset? The scope is AC coupled yet the trace is up.

The scope is DC coupled.
Although I am a bit perplexed why the negative half current is the same as the positive half current. But kinda makes sense in that the zener is just a diode in the half cycle, and sensor circuit is doing whatever
all the time to generate that crap.
BTW, the other Quell brand is near identical.

Try an FFT to see what those higher harmonics are about. I had power quality problems at one site and ended up making a VLF receiver and driving around to narrow down the source, following the distribution power lines. Higher harmonics get transmitted airborne from power lines as antennas.

[tszaboo]

You can't ignore I2R at that point, that's the whole point 

Ok, so lets just do a quick, back of the envelope calculation, based on real data.
Australia has 5% grid losses (2014 data), source: https://data.worldbank.org/indicator/EG.ELC.LOSS.ZS?name_desc=false
Lets take this as a worst case value, the grid has 0 VAR on it. So the delivery of each watt dissipates 0.05W on the grid. The same way, each VAR will dissipate 0.05W on the grid. OK, now the device needs 1W real power, and it uses 20VAR to have it delivered. So there is 20x0.05W= 1W dissipated on the grid, for the delivery of 1W.
And all the smoke alarms need 10MW real power delivered to the grid, instead of 5MW with perfect power factor, and instead of 100MW stated on the video. Our problem is, that the video is off by an order of magnitude.

This will be different for different countries, and different customers, depending how far away they are from the power plant. And that waveform is indeed horrendous.

[gf]

Exactly, reactive power does not cost you money, but it fully contributes to grid losses (unless it happens to be compensated by other inductive loads nearby). It is still not clear, though, how much real power the device consumes. Given that other devices can last 10 years with a battery I find 1W still pretty high - that would be 2.60 Euro/year, assuming the electricity price in my country.

[ogden]

Guys, cool down. Everything is fine. Dave is just doing business, following YT trends.

[nctnico]

Exactly, reactive power does not cost you money, but it fully contributes to grid losses (unless it happens to be compensated by other inductive loads nearby). It is still not clear, though, how much real power the device consumes. Given that other devices can last 10 years with a battery I find 1W still pretty high - that would be 2.60 Euro/year, assuming the electricity price in my country.

I agree. There has to be a more clever way to power such a gadget from mains without dissipating hundred's of times more energy from mains compared to what it draws from a battery. I guess that a lot has to do with the ability to deliver enough power to the beeper when necessary.

[GeorgeOfTheJungle]

It takes 5W to charge a 1µF capacitor 100 times a second.
100*1e-6*((240*sqrt(2)-15)^2)/2 ~= 5.2 joules/s
Where are the other 15W up to 20 VA dissipated then, in 100R and the zeners? 15W are many watts, they should be quite toasty. No?

@79.3mA and 1.36W total measured:
100R = 0.63W
So that leaves 0.73W for the zeners.
Ignore the load which is naff all.
Maybe a smidge in the cap ESR.
The rest will be copper losses down the system unless compensated for.

I imagine the rest is power that simply flows in and out of the capacitor, the joules that enter end flowing back into the grid and dilute into the other zillion grid loads that are in parallel. Similar to what a grid tied solar inverter does, only that this one pulls half a cycle and injects it back during the other half. If so, the net load to the grid would be ~= zero, except for the losses in 100R and the zeners. No?

[Kleinstein]

The capacitive loading is not a real problem. Usually there are extra capacitors near the distribution transformers to compensate for the inductive loads and the magnetizing current of the transformer. So in most cases they can just make those caps a little smaller. Smoke detectors tend to be on all the time, so no need to adjust that very often.

However the real power taken from the grid is still rather high. So he design is still bad. It is odd to see only half wave rectification - with full wave rectifier they could have reduced the expensive capacitor to half. - though with a little more loss.

The capacitor drop supply is just not good for something that usually needs low power, but sometimes needs much more.

[HackedFridgeMagnet]

To get the current waveform, probe the voltage across the droppers resistor, using one of your differential probes.
Or to Ch1 - Ch2.

Don't think I'll go that far, it's obvious what results a HV regulator will give in terms of current vs the zener solution, no need to build and measure to prove that.

Not that obvious to me? I will try it myself if/when I can.

[TerminalJack505]

So far as I understand it, all of the current that flows through the capacitor has to be burned off (by the load, Zener, in-line resistor, rectifier, etc.) so that is energy lost.

This is a particularly bad circuit so far as efficiency is concerned.  The smoke detector will need--in the worst case--the current to drive the Piezo element in the event of a fire.  It will likely need around 10mA to sample for smoke every 10 seconds, or so.  All other times it is likely burning no more than 1mA.  They likely draw 50mA of current from the battery during a low battery test but--unless their circuit is incorrectly designed--that should come exclusively from the battery and never from mains.

[gf]

The capacitor drop supply is just not good for something that usually needs low power, but sometimes needs much more.

Unfortunately a zener is a shunt regulator. On the other hand this design can provide about 70mA when needed. Would be interesting how much current it needs when crying, an how long it is supposed to cry at minimum. Would a low-current series regulator + supercap suffice?

[EEVblog]

What is the failure rate of the crappy solution "they" have used compared to the ones you showed, could that be a reason they went with it?

Bordering on nothing in it really with a zener vs a HV regulator. I guarantee it's because of cost.

[EEVblog]

I think a point that is missed is that the I2R losses aren't extending to the grid.  They probably aren't even getting to the pole pig.  They are getting to the nearest traditional electric motor (non-inverter-driven), where the capacitive reactance is a benefit and any harmonics are being absorbed.
modulation.

How are the harmonics being absorbed by the motor?
The harmonics need matching harmonics in the voltage waveform in order to be compensated, if that's not the case then the harmonics will contribute I²R losses right up the delivery system until they are filtered out.
This is why non-linear loads are a huge problem and standards like 61000-3-2 and EnergyStar exist.
https://en.wikipedia.org/wiki/IEC_61000-3-2
and why harmonic filters like this exist:
https://www.nhp.com.au/files/editor_upload/File/Brochures/Power-Quality/NSCHAFFC.pdf

[EEVblog]

The capacitor drop supply is just not good for something that usually needs low power, but sometimes needs much more.

Unfortunately a zener is a shunt regulator. On the other hand this design can provide about 70mA when needed. Would be interesting how much current it needs when crying, an how long it is supposed to cry at minimum. Would a low-current series regulator + supercap suffice?

That's in the video, I measured 15mA peak current from the 9V battery.

[EEVblog]

So far as I understand it, all of the current that flows through the capacitor has to be burned off (by the load, Zener, in-line resistor, rectifier, etc.) so that is energy lost.

Yes, zener regulators are the traditional "worst case design", which is why they have gone the way of the dodo, unless you want to save a couple of cents.

This is a particularly bad circuit so far as efficiency is concerned.  The smoke detector will need--in the worst case--the current to drive the Piezo element in the event of a fire.  It will likely need around 10mA to sample for smoke every 10 seconds, or so.  All other times it is likely burning no more than 1mA.  They likely draw 50mA of current from the battery during a low battery test but--unless their circuit is incorrectly designed--that should come exclusively from the battery and never from mains.

Again, this is in the video.
The measured current draw is 50uA maximum during standby and 15mA max during sounding.

[TerminalJack505]

The capacitor drop supply is just not good for something that usually needs low power, but sometimes needs much more.

Unfortunately a zener is a shunt regulator. On the other hand this design can provide about 70mA when needed. Would be interesting how much current it needs when crying, an how long it is supposed to cry at minimum. Would a low-current series regulator + supercap suffice?

I worked on a wireless smoke detector for the US/Canada market and I seem to remember that the minimum time the battery needed to support the sounder was 15 minutes.  Which is to say that the low-battery indication needed to factor this in.  Once smoke is detected and the sounder goes off, it won't stop sounding until either 1) the user intervenes, 2) smoke is no longer detected, or 3) the battery gives out.

[coppice]

So far as I understand it, all of the current that flows through the capacitor has to be burned off (by the load, Zener, in-line resistor, rectifier, etc.) so that is energy lost.

Yes, zener regulators are the traditional "worst case design", which is why they have gone the way of the dodo, unless you want to save a couple of cents.

Zener regulation is also valuable when you want a low cost regulator that clamps - e.g. when your design needs to deal with high EMI conditions, and energy can get in post regulator. An LDO won't stop the Vcc rising dangerously, but a zener will.

[gf]

At
This is why non-linear loads are a huge problem and standards like 61000-3-2 exist.

And not to forget millions of LED lamps w/o PFC...

[EEVblog]

So please enlighten us with a real calculation including a good estimate of the actual grid losses. Hint: if small devices with low PF are deemed a problem then they would be subject to regulations regarding power factor.

Just because they didn't bother to do <75W devices in EN61000-3-2 and EnergyStar doesn't mean it doesn't matter. The entire point of the video is to consider this in product design.
If it's a problem at 75W for a single device, why shouldn't it be a problem with 75 devices of 1W each (localised EMC excluded).

Small correction: It's 75W (or 100W) input power, so that includes device efficiency. So if your widget takes 37.5W real power, but it's only 50% efficient, then it falls under the regulation for power factor and harmonic content.

[thm_w]

Unfortunately a zener is a shunt regulator. On the other hand this design can provide about 70mA when needed. Would be interesting how much current it needs when crying, an how long it is supposed to cry at minimum. Would a low-current series regulator + supercap suffice?

70mA capable makes sense:
- 15mA peak as Dave stated, add 2x for safety factor, then add another 2x for degradation of the capacitor over time (10yr).
Big clive has seen a number of these capacitors where they have degraded to the point of not being able to supply enough current.

[EEVblog]

Zener regulation is also valuable when you want a low cost regulator that clamps - e.g. when your design needs to deal with high EMI conditions, and energy can get in post regulator. An LDO won't stop the Vcc rising dangerously, but a zener will.

Good point, but still it's ultimately a cost driven choice at the expense of efficiency. Has been since the 1970's.

[EEVblog]

Unfortunately a zener is a shunt regulator. On the other hand this design can provide about 70mA when needed. Would be interesting how much current it needs when crying, an how long it is supposed to cry at minimum. Would a low-current series regulator + supercap suffice?

70mA capable makes sense:
- 15mA peak as Dave stated, add 2x for safety factor, then add another 2x for degradation of the capacitor over time (10yr).
Big clive has seen a number of these capacitors where they have degraded to the point of not being able to supply enough current.

The better quality more expensive design is much lower than the 79mA in the cheapo though.
For capacitor degradation, if you halve the capacitance you halve the current, so that could be a thing.

[dcac]

You may find this “cheap” capacitor dropper circuit in almost any product where there’s a MCU controller board and just some LED indicators or i.e. a LCD display that only require a low current DC supply.

While you (perhaps) can debate how much power losses it really is resulting in, in my experience it is also a quite high failure rate design. And when they fail the product has still often been working but just intermittently, not exactly the behavior you’d want in a smoke alarm.

I've only worked with battery powered smoke alarms and they will warn also for the battery running low and needs changing, but in those mains powered smoke alarms - will they also warn if the dropper capacitor is about to fail, or any other failure from the mains circuit?

[langwadt]

So please enlighten us with a real calculation including a good estimate of the actual grid losses. Hint: if small devices with low PF are deemed a problem then they would be subject to regulations regarding power factor.

Just because they didn't bother to do <75W devices in EN61000-3-2 and EnergyStar doesn't mean it doesn't matter. The entire point of the video is to consider this in product design.
If it's a problem at 75W for a single device, why shouldn't it be a problem with 75 devices of 1W each (localised EMC excluded).

Small correction: It's 75W (or 100W) input power, so that includes device efficiency. So if your widget takes 37.5W real power, but it's only 50% efficient, then it falls under the regulation for power factor and harmonic content.

afaiu there is a minium required efficiency for wall warts and such in that power range of ~85%

[TerminalJack505]

Now that I think about it--if Aussie detectors are the same as US/Canada detectors--the detector will check for smoke while the Piezo sounder is going, so that is likely the worst-case so far as current draw goes.  I'm guessing that would be about 25mA.

[EEVblog]

Now that I think about it--if Aussie detectors are the same as US/Canada detectors--the detector will check for smoke while the Piezo sounder is going, so that is likely the worst-case so far as current draw goes.  I'm guessing that would be about 25mA.

No, the detector only draws 50uA maximum when checking, it's practically nothing. This is why a 9V battery lasts for a couple of years, and they can get 10 years lithium ones.

[EEVblog]

but in those mains powered smoke alarms - will they also warn if the dropper capacitor is about to fail, or any other failure from the mains circuit?

Just the usual low battery indicator, which could be low battery, or mains cap failure reducing the zener current and ultimately voltage.
https://datasheetspdf.com/pdf/1082256/Allegro/5366/1
But in theory I could imagine a scenario where the cap has failed but the zener just has enough voltage to not trip the low battery warning, but then doesn't have enough current in order to maintain the voltage during the alert, so the alarm never goes off.

[EEVblog]

I imagine the rest is power that simply flows in and out of the capacitor, the joules that enter end flowing back into the grid and dilute into the other zillion grid loads that are in parallel. Similar to what a grid tied solar inverter does, only that this one pulls half a cycle and injects it back during the other half. If so, the net load to the grid would be ~= zero, except for the losses in 100R and the zeners. No?

No, the 80mA is real current that will have I²R losses in the transmission system until it is compensated for (lead/lag and harmonic). If you have say 5 in your home at 80mA a pop that's 0.4A that must be delivered from the grid infrastructure.
The lead/lag compensated part can be done locally at your home with other leading phase device, but the harmonic content usually cannot, so it will go back down the grid resulting in losses until it's filtered out.

[richnormand]

Enjoyed the video Dave, thanks.
No reason to put up with inordinate power usage when solutions exist, even if they initially cost a few cents more.
The idea of "built to a cost" and "cheap as possible" from mains power for safety equipment that should be reliable made me think of this pic.

[EEVblog]

Enjoyed the video Dave, thanks.
No reason to put up with inordinate power usage when solutions exist, even if they initially cost a few cents more.
The idea of "built to a cost" and "cheap as possible" from mains power for safety equipment that should be reliable made me think of this pic.

Can't get that sort of energy readily from a 9V battery!

[TerminalJack505]

Now that I think about it--if Aussie detectors are the same as US/Canada detectors--the detector will check for smoke while the Piezo sounder is going, so that is likely the worst-case so far as current draw goes.  I'm guessing that would be about 25mA.

No, the detector only draws 50uA maximum when checking, it's practically nothing. This is why a 9V battery lasts for a couple of years, and they can get 10 years lithium ones.

I'll have to re-watch the video to see what technology your detector uses.  The one I worked on used photoelectric technology for smoke detection which consisted of an (ultraviolet?) LED and phototransistor.  So both the detection chamber LED as well as the "sampling indicator" LED (as seen by the user) would have to be powered when the sample is done.

Disclaimer: I only worked on the embedded code for the wireless transceiver so I don't claim to be an expert in smoke detector technology.

[EEVblog]

Bonus screenshot:
(Attachment Link)

Turns out all that funny business is not coming from the product consumption, and with a moments thought that is obvious, more investigation required, likely a proby thing.
A pure R-C-Zener circuit should not create that mid frequency stuff, just the front and back porches.

Well this is strange. Using my AIM current probe I get the exact same waveform. And if I bypass the power monitor for mains input I also get the same waveform.
More investigation required...

Double strange - Exactly the same result using my Micsig portable scope with either my HV probe and sense resistor or my AIM current probe.
So everything is completely isolated.

I can't see how a shunt zener regulator can produce this? I can only imagine it's some sort of harmonic resonance thing happening because of the sudden high slew waveform rise.

[EEVblog]

It has to be some sort of mains system resonance caused by the harmonic distortion. If so, brilliant example of the problems harmonic power factor can cause.
Need to find another way that it doesn't resonate somehow though

[EEVblog]

My next hunch is resonance with a PFC capacitor bank inside the building. I might go to the old lab in another building and try again. Worth a shot.
And BTW, those wiggles are not random, that waveform above is an average, they are completely consistent across scopes and probing.

[EEVblog]

Well, there you go, that's the difference between the two labs. Just lots of high frequency crap, otherwise identical.

[ali_asadzadeh]

you can use an ultra low cost Flyback chip in the non-isolated buck mode, ST have app notes on that, for example you can use Viper12 part, and it can supply in the order of 100mA or more,total cost would be under 1$

[GeorgeOfTheJungle]

Well, there you go, that's the difference between the two labs. Just lots of high frequency crap, otherwise identical.

Looks much nicer in LTSpice:

[GeorgeOfTheJungle]

Isn't this the power that's being fed back into the grid every 1/4 cycle?

[ckambiselis]

What is the failure rate of the crappy solution "they" have used compared to the ones you showed, could that be a reason they went with it?

Bordering on nothing in it really with a zener vs a HV regulator. I guarantee it's because of cost.

Was thinking more of a government regulation on safety systems, for them to be able to sell it as a smoke alarm, it must have a failure rate smaller than x%.

Oscar

[gf]

Looks much nicer in LTSpice:

Your current has double frequency of voltage. This was not the case in the posted oscillogram. Something must be different in your circuit, IMO.

[GeorgeOfTheJungle]

Your current has double frequency of voltage. This was not the case in the posted oscillogram. Something must be different in your circuit, IMO.

It's power V*I not current. Think about it: every cycle, the capacitor charges to +V, discharges to 0, charges to -V, discharges to 0, ergo 4 power peaks, two positive, two negative, and double the frequency.

[gf]

It's power V*I not current.

Sorry, misssed that. Reactive power alternates of course between positive and negative, with zero average (the average is not supposed to be zero here, as there is also a real power component in the play).

[tszaboo]

What is the failure rate of the crappy solution "they" have used compared to the ones you showed, could that be a reason they went with it?

Bordering on nothing in it really with a zener vs a HV regulator. I guarantee it's because of cost.

Was thinking more of a government regulation on safety systems, for them to be able to sell it as a smoke alarm, it must have a failure rate smaller than x%.

Oscar

They should require SIL for smoke or CO alarms. Honestly, they should also require gas detectors, for houses with gas heating, and the gas detector should have SIL, ATEX/IECEX. And all these alarms would need a remote alarm for the nearest fire brigade.

[TerminalJack505]

The capacitive loading is not a real problem. Usually there are extra capacitors near the distribution transformers to compensate for the inductive loads and the magnetizing current of the transformer. So in most cases they can just make those caps a little smaller. Smoke detectors tend to be on all the time, so no need to adjust that very often.

However the real power taken from the grid is still rather high. So he design is still bad. It is odd to see only half wave rectification - with full wave rectifier they could have reduced the expensive capacitor to half. - though with a little more loss.

The capacitor drop supply is just not good for something that usually needs low power, but sometimes needs much more.

BBM.

Maybe they add two more diodes and do full-wave rectification for the 120VAC version but keep all of the other components the same?

I don't see how that would be more economic than changing the cap but that's one reason I can think of for using half-wave rectification.

[SeanB]

Current waveform distortion is from all your local grid tie solar installations, feeding power back into the grid. If you want a higher capacitor current to use in the measurement simply use a 20-50uf motor run capacitor across the mains, it will have a very high current flow through it, and will survive direct connection for a long time. Higher current, better resolution on the higher current ranges on the current probe, and you can use a standard current transformer as well to measure, as the current will be in the order of 10A or so for the 50 uF unit.

The solar inverters probably have had some cost cutting done on the output filtering, removing the high cost LC filtering  by removing half of the LC sections, meaning they cause more AC current distortion.

[floobydust]

This is the schematic of capacitive-divider for the smoke alarm I felt recall-worthy. It's a special smoke alarm having an aux. relay for alarm-bell tie in, so it wastes over 1W. This cooked the pcb phenolic and plastics over a couple years. It is imported from china by "American Sensors"/First Alert and quite a POS with bodges etc. on the board. Note sim has the relay on, 120VAC mains.

You design a capacitive-divider for highest load. All of us use mains frequency for the math- which is not correct when you have HF harmonics present. So the zener and resistor cook even more because they waste the excess energy purely as heat.

[coppice]

This is the schematic of capacitive-divider for the smoke alarm I felt recall-worthy. It's a special smoke alarm having an aux. relay for alarm-bell tie in, so it wastes over 1W. This cooked the pcb phenolic and plastics over a couple years.

Did it cook because it was dissipating 1W, or because the voltage waveform in that place was quite distorted and the dissipation was way up? Well designed things with cap drop supplies use large resistors well spaced from the board and case, even when those resistors normally run very cool. That way when there is a lot of harmonic distortion on the supply waveform they don't overdissipate and cause damage or fires.

[floobydust]

Here's evidence smoke alarms contributing to global warming. The resistor is a steady 1W. Something started giving off a tiny bit of smoke and false triggering itself, seems to be high line, as harmonics only show up when a huge industrial phase-control heater is running. That may have aggravated it.

[David Hess]

It is fascinating to think of the various ways to make a suitable high efficiency switching regulator.  The high switching voltage increases charge pumping losses according to 1/2cv^2 which becomes overwhelming at low power.

A series resonate switching regulator might be suitable but one thing I have seen in very old blocking oscillator based inverter designs, which are commonly used for electronic ballast now, is a charge recovery circuit which extracts residual charge after the switching cycle and dumps it back into the supply.  I do not think it would help the power factor except through being orders of magnitude more efficient.  But just the output transformer would probably cost more than all of the parts in the existing poor circuit.

[kcbrown]

That TI part sounds interesting and all, but at $1 each in quantity, I can't help but wonder how a solution built around it can outcompete a low-power SMPS, which I'd expect to have roughly the same BOM cost in quantity (since SMPS controllers are cheap as chips these days, and the rest is jellybean stuff).

Of course, the SMPS might not have the reliability you want for a mission critical device like this.  Then again, isn't that what the battery is for?   

[GeorgeOfTheJungle]

No, the 80mA is real current that will have I²R losses in the transmission system until it is compensated for (lead/lag and harmonic). If you have say 5 in your home at 80mA a pop that's 0.4A that must be delivered from the grid infrastructure.
The lead/lag compensated part can be done locally at your home with other leading phase device, but the harmonic content usually cannot, so it will go back down the grid resulting in losses until it's filtered out.

I wonder, what would that power analyzer say if you connected it to the output of a solar inverter? Have you ever checked that?

[EEVblog]

Nice for a video down the track if you could build up those other two solutions (TI and OnSemi iirc?)  and then measure the three different options. Four if you count the full bridge.

Just for you!

[coppice]

That TI part sounds interesting and all, but at $1 each in quantity, I can't help but wonder how a solution built around it can outcompete a low-power SMPS, which I'd expect to have roughly the same BOM cost in quantity (since SMPS controllers are cheap as chips these days, and the rest is jellybean stuff).

Of course, the SMPS might not have the reliability you want for a mission critical device like this.  Then again, isn't that what the battery is for?   

Have you talked to TI about buying a million parts? If not, you have no idea whatsoever what the volume price really is. However, you know by inspection that its not as high as $1, as there would be no market for it at that price.

[Per Hansson]

70mA capable makes sense:
- 15mA peak as Dave stated, add 2x for safety factor, then add another 2x for degradation of the capacitor over time (10yr).
Big clive has seen a number of these capacitors where they have degraded to the point of not being able to supply enough current.

Enjoyed the video Dave, thanks.
No reason to put up with inordinate power usage when solutions exist, even if they initially cost a few cents more.
The idea of "built to a cost" and "cheap as possible" from mains power for safety equipment that should be reliable made me think of this pic.

This whole mains powered smoke alarms just sounds like a solution to a problem that does not exist to me.
There are so many ways it can fail: lightning strike, X2 capacitor degrading, or just overheating and catching fire as shown earlier in the thread because zeners are quite unreliable components!
But really I think the X2 capacitor degrading is the worst part, so you have this mains powered device with a weak battery.
And the X2 capacitor degrades so that it works fine in ordinary use but when the buzzer kicks in it just browns-out due to insufficient current.
Yup, great design for a device supposed to save lives!

[ogden]

Nice for a video down the track if you could build up those other two solutions (TI and OnSemi iirc?)  and then measure the three different options. Four if you count the full bridge.

Just for you!

You just demonstrated typical trap for young players of "Design By Inspection". Engineers who made 80mA dropper instead of 1mA, made it for for some quite obvious reason - ALARM. You definitely can't produce audible alarm using 15v @50uA supply, unless you are building "smoke alarm earbud". Only actually making whole device and measuring SPL, you will see how much power is actually needed. Could be so that 10.5mA @ 15V of NCP785A is not even enough.

[gf]

You just demonstrated typical trap for young players of "Design By Inspection". Engineers who made 80mA dropper instead of 1mA, made it for for some quite obvious reason - ALARM. You definitely can't produce audible alarm using 15v @50uA supply, unless you are building "smoke alarm earbud". Only actually making whole device and measuring SPL, you will see how much power is actually needed. Could be so that 10.5mA @ 15V of NCP785A is not even enough.

What about a low-current dropper (say 1mA or less) + a couple of Farad supercap which can store enough energy for 15min sounding?

[coppice]

You just demonstrated typical trap for young players of "Design By Inspection". Engineers who made 80mA dropper instead of 1mA, made it for for some quite obvious reason - ALARM. You definitely can't produce audible alarm using 15v @50uA supply, unless you are building "smoke alarm earbud". Only actually making whole device and measuring SPL, you will see how much power is actually needed. Could be so that 10.5mA @ 15V of NCP785A is not even enough.

It seems like he forgot that he has already measured the peak consumption during beeping as 15mA.

[coppice]

You just demonstrated typical trap for young players of "Design By Inspection". Engineers who made 80mA dropper instead of 1mA, made it for for some quite obvious reason - ALARM. You definitely can't produce audible alarm using 15v @50uA supply, unless you are building "smoke alarm earbud". Only actually making whole device and measuring SPL, you will see how much power is actually needed. Could be so that 10.5mA @ 15V of NCP785A is not even enough.

What about a low-current dropper (say 1mA or less) + a couple of Farad supercap which can store enough energy for 15min sounding?

You would need a 15V supercap. That gets pricy.

[ogden]

What about a low-current dropper (say 1mA or less) + a couple of Farad supercap which can store enough energy for 15min sounding?

You would need a 15V supercap. That gets pricy.

Right. Supercap is "no go", dropper as well. NCP785A  is excellent, only it has max 10.5mA specs As a solution I see either external power transistor (?) or electrolytic cap + sound "chirps" with let's say 20% duty cycle resulting peak ~50mA sound driver current. - Obviously if such sound chirps allowed by regulations.

[thm_w]

"UL 217 – Single and Multiple Station Smoke Alarms, requires an A-weighted sound pressure level of at least 85 decibels (dBA) when measured at a distance of 10 feet from the horn" 10ft = 3m

This one is claiming 85dB 2mA at 5Vpp:
https://www.alibaba.com/product-detail/SS3038T3P-Pin-Type-Piezo-Piezoelectric-Buzzer_62050104559.html

But that would likely be at 1m, so you'd need something louder.. 90dB or so?

[nctnico]

What about a low-current dropper (say 1mA or less) + a couple of Farad supercap which can store enough energy for 15min sounding?

You would need a 15V supercap. That gets pricy.

Right. Supercap is "no go", dropper as well. NCP785A  is excellent, only it has max 10.5mA specs As a solution I see either external power transistor (?) or electrolytic cap + sound "chirps" with let's say 20% duty cycle resulting peak ~50mA sound driver current. - Obviously if such sound chirps allowed by regulations.

Better yet... run the beeper from rectified mains directly through a dropper capacitor. The circuit itself only needs 50uA.

[langwadt]

You just demonstrated typical trap for young players of "Design By Inspection". Engineers who made 80mA dropper instead of 1mA, made it for for some quite obvious reason - ALARM. You definitely can't produce audible alarm using 15v @50uA supply, unless you are building "smoke alarm earbud". Only actually making whole device and measuring SPL, you will see how much power is actually needed. Could be so that 10.5mA @ 15V of NCP785A is not even enough.

What about a low-current dropper (say 1mA or less) + a couple of Farad supercap which can store enough energy for 15min sounding?

use two capacitive dropper supplies. a low power one for normal operation, a high power one driving the sounder
only enabled hen needed

[thm_w]

use two capacitive dropper supplies. a low power one for normal operation, a high power one driving the sounder
only enabled hen needed

Wonder if a straight up dropper resistor would be any cheaper or not, probably not much difference. 50uA at 240V = ~12mW.
They have 220V rated piezos, but the one I found is not that cheap (~$5), active circuitry inside: https://www.cuidevices.com/product/resource/cpe-422ac.pdf

[nctnico]

use two capacitive dropper supplies. a low power one for normal operation, a high power one driving the sounder
only enabled hen needed

Wonder if a straight up dropper resistor would be any cheaper or not, probably not much difference. 50uA at 240V = ~12mW.
They have 220V rated piezos, but the one I found is not that cheap (~$5), active circuitry inside: https://www.cuidevices.com/product/resource/cpe-422ac.pdf

That one has internal electronics. The beeper in the smoke alarm is probably just a piezo disk. Connected in series with a capacitor so it doesn't get full mains and all you need to drive it is a mains rated MOSFET (and some other components).

[gf]

"UL 217 – Single and Multiple Station Smoke Alarms, requires an A-weighted sound pressure level of at least 85 decibels (dBA) when measured at a distance of 10 feet from the horn" 10ft = 3m

This one is claiming 85dB 2mA at 5Vpp:
https://www.alibaba.com/product-detail/SS3038T3P-Pin-Type-Piezo-Piezoelectric-Buzzer_62050104559.html

But that would likely be at 1m, so you'd need something louder.. 90dB or so?

The referenced page sais 85dB at 10cm, so it would need to be even more louder...
[ In Europe, EN 14603 also requires 85dB @3m. ]

Better yet... run the beeper from rectified mains directly through a dropper capacitor. The circuit itself only needs 50uA.

In case of mains power failure it still must run from the battery, though.

[kcbrown]

Have you talked to TI about buying a million parts? If not, you have no idea whatsoever what the volume price really is. However, you know by inspection that its not as high as $1, as there would be no market for it at that price.

No.  But neither have I talked to anyone about buying the parts for a million low-power SMPS sections, either.  The comparison is about as apples-to-apples as can be made.  Cheap SMPS controllers seem to be in the mid teens to low tens of cents each in quantities of 1000, from sites like Mouser.  The rest is jellybean parts.

Perhaps TI's thousands-quantity price is a substantially higher multiple of their millions-quantity price than the multiplier that applies to low-power SMPS components.  On that, I just can't say.  The only thing I have to go by is the thousands-quantity prices.  Why do you believe that those prices aren't comparable, or aren't indicative of the millions-quantity relative price?

As for the market for the TI part, well, if board space is at a premium and low standby dissipation is also a requirement, the TI part looks like just the thing, so it does fill a niche.

[nctnico]

Better yet... run the beeper from rectified mains directly through a dropper capacitor. The circuit itself only needs 50uA.

In case of mains power failure it still must run from the battery, though.

No problem. Feed the beeper circuit from the battery through a diode or so. With a couple of days to work on this I'm sure someone can come up with a clever circuit which is both cheap and efficient.

[coppice]

Perhaps TI's thousands-quantity price is a substantially higher multiple of their millions-quantity price than the multiplier that applies to low-power SMPS components.  On that, I just can't say.  The only thing I have to go by is the thousands-quantity prices.  Why do you believe that those prices aren't comparable, or aren't indicative of the millions-quantity relative price?

This is how the semiconductor industry works. If you have a part with some novelty there are small volume users who will pay a good price for it, so you set the small volume price high. However, high volume users are ALWAYS driven by the BOM above all other factors. You need to offer attractive volume pricing to win any high volume design. The markets for this device are high volume things like smoke detectors, where you can lose a design over a fraction of a cent on the BOM, and utility meters, which will be attracted to the lack of any magnetic components (they are sensitive to tampering issues), but still need to meet aggressive BOM goals.

[splin]

use two capacitive dropper supplies. a low power one for normal operation, a high power one driving the sounder
only enabled hen needed

This is a cost effective solution requiring the addition of a triac or back to back mosfet switch (approx $.05), along with a dirt cheap and (likely much more reliable) 30uA dropper to power the sensor.  All more than paid for by reducing the main dropper capacitance by 50% or more as it no longer has to have degradation factored in.  The terrible power factor will be the least of your concerns when it triggers for real...

[Alti]

I'd say that to find "the best" solution for powering smoke detector we need to rigorously specify requirements first and "the best" is going to be the one with the lowest total cost of ownership (design+investment+runningcost+recycling) that meets all requirements.

Otherwise we are going to have next discussion about preferences and not about solution of real life problem.

-I bet for some "the best" is a networked solution tied to fire department, that costs 1k$ in investments but hey, this smoke detector is for fire brigade to get here asap.
-Others prefer resistive dropper (ok, it is 1W but do not want to have a safety critical device that heavily reles on capacitors or SMPS. Besides, 1W heater just adds up to heating cost, if you happen to live in Norway).
-Then others will "the best" with Viper (bridge + buck) because it saves electricity at the cost of BOM.
Etc.

Can anyone specify a starting point of requirements that a smoke detector has to meet to switch discussion about personal preferences to the one about engineering?

[nctnico]

The primary function of a smoke detector is to wake people up so they don't suffocate from smoke while their home is burning. See it as an early warning system so you can get out of your home safely after a fire has started.

[TerminalJack505]

I'd say that to find "the best" solution for powering smoke detector we need to rigorously specify requirements first and "the best" is going to be the one with the lowest total cost of ownership (design+investment+runningcost+recycling) that meets all requirements.

Otherwise we are going to have next discussion about preferences and not about solution of real life problem.

-I bet for some "the best" is a networked solution tied to fire department, that costs 1k$ in investments but hey, this smoke detector is for fire brigade to get here asap.
-Others prefer resistive dropper (ok, it is 1W but do not want to have a safety critical device that heavily reles on capacitors or SMPS. Besides, 1W heater just adds up to heating cost, if you happen to live in Norway).
-Then others will "the best" with Viper (bridge + buck) because it saves electricity at the cost of BOM.
Etc.

Can anyone specify a starting point of requirements that a smoke detector has to meet to switch discussion about personal preferences to the one about engineering?

BBM.

Well, to start with, the idea that only 50uA is required at all times other than when powering the sounder, is incorrect.  There are other high-current components in the circuit.

There is the IR LED used in the detection chamber and (I'm guessing) two indicator LEDs--likely one green LED and one red LED.

If the unit is mains powered then the green LED will be always on.  (This is the case for the US/Canada market.)  This will be about 5mA.  I don't think Dave has factored this in since I think he did his measurements while the unit was under battery power--in which case the green LED will not be illuminated.

The red LED will come on and stay on when the alarm sounds.  This might account for about 5mA of the 15mA drawn as reported by Dave.

The IR LED (and red LED) will come on when sampling for smoke every 10 seconds.  The IR LED likely draws about 5mA.  The IR LED and red LED don't necessarily have to be on simultaneously but the red LED needs to flash to indicate that a sample was done.

Sampling will be done even while the alarm is sounding so that it can turn the alarm off automatically once the smoke clears.  The sample can be done during a quiet period of the sounder.  If, for example, the temporal-3 tone pattern is being used then it can be done during one of the 1/2 second quiet periods or the 1.5 second quiet period.

Finally, there is the MCU.  Most of the time it will be in an energy efficient sleep mode.  It will obviously have to come out of sleep mode for sampling, low battery check, etc.  If the alarm goes off then the MCU will have to drive the Piezo sounder.  The 50uA is likely the current drawn when the MCU is sleeping.  The current drawn by the MCU when it is awake has likely already been accounted for in the 15mA drawn when in alarm.

So, the circuit could be designed so that it uses no more than ~20mA at once--powering the sounder, the green LED, and the red LED.  Or 15mA when under battery power.  (No green LED.)

My guess is that this detector wasn't designed so that it could guarantee that all of the high current components aren't all powered at once.  They basically have to assume that all of them are going to be active at once.  They then added a safety margin of what seems to be about 100%.

There is one other high-current event that I can think of.  When a low battery check is done the MCU will engage a circuit that will draw about 50mA from the battery.  This current will not be supplied by the mains, however.

All of this is based on my experience with a smoke detector project.

The project I worked on was a battery-only (not mains powered) wireless 'smart' detector.  In addition to the smoke detection circuit, it had a wireless transceiver that would talk to a security/fire panel.  It required much more power than the one Dave has.  It was powered by a single 3V CR123 lithium battery (not the best choice in my opinion) and the battery would need to be replaced yearly.

The high power requirements were mostly due to the fact that the device would have to wake up frequently and check for commands coming over the air.  (If any of the units went into alarm then ALL of them would start sounding--in sync. with each other.)  It would also have to transmit a check-in message to the security/fire panel every three minutes.  If the unit ever stopped talking to the panel then it would constantly try to reconnect with no regard battery usage so this could kill a battery fast if the panel is down for a long period of time.

[dcac]

Found this article from 2011: "99% Waste: The Unexpected Energy Consumption of Smoke Alarms:"
https://reductionrevolution.com.au/blogs/news-reviews/5842566-99-waste-the-unexpected-energy-consumption-of-smoke-alarms

Here's a working link to the " comprehensive report":
https://prod-energyrating.energy.slicedtech.com.au/sites/new.energyrating/files/documents/sb200405-smokealarms_0.pdf


I think someone will recognize this:

[ogden]

Can anyone specify a starting point of requirements that a smoke detector has to meet to switch discussion about personal preferences to the one about engineering?

At least 10000x more energy-efficient smoke detector having same functions as one shown in the original video. Detector that actually works - not only during standby, but producing nominal alarm sound as well.

[TerminalJack505]

Allegro has what looks like a very efficient smoke detector IC.  (Digikey link provided since I couldn't find a link on Allegro's site.)  The datasheet outlines some of the requirements for such a product.

[DBecker]

Allegro has what looks like a very efficient smoke detector IC.  (Digikey link provided since I couldn't find a link on Allegro's site.)  The datasheet outlines some of the requirements for such a product.

That's the detector only.  It has logic level outputs to the alarm, which requires an external driver.  It is intended to be run from a 3V battery and thus doesn't address the issue here, which is a requirement for a line-powered alarm with battery backup and external interconnect wire.

You might argue that a line powered alarm is obsolete when 10 year battery life is achievable, and the sensor should be replaced after 10 years anyway.  But a technical argument doesn't change that there are widespread requirements for a hardwired alarm, and that's not likely to change for decades.

As a side note, the interconnect signal wire works with activated alarm applying a 5V-12V DC level (typically 9V) with the input sensing threshold is 4-5V for other alarm to echo the alert.  An alarm needs to generate that voltage at the same time it is operating the alert.

[TerminalJack505]

Allegro has what looks like a very efficient smoke detector IC.  (Digikey link provided since I couldn't find a link on Allegro's site.)  The datasheet outlines some of the requirements for such a product.

That's the detector only.  It has logic level outputs to the alarm, which requires an external driver.  It is intended to be run from a 3V battery and thus doesn't address the issue here, which is a requirement for a line-powered alarm with battery backup and external interconnect wire.

You might argue that a line powered alarm is obsolete when 10 year battery life is achievable, and the sensor should be replaced after 10 years anyway.  But a technical argument doesn't change that there are widespread requirements for a hardwired alarm, and that's not likely to change for decades.

As a side note, the interconnect signal wire works with activated alarm applying a 5V-12V DC level (typically 9V) with the input sensing threshold is 4-5V for other alarm to echo the alert.  An alarm needs to generate that voltage at the same time it is operating the alert.

I'm not sure what you think my intentions were of making that post.  Others were asking what the requirements are for a smoke detector product.  That datasheet gives a general idea about the requirements.  Also, see my earlier post.

[EEVblog]

Nice for a video down the track if you could build up those other two solutions (TI and OnSemi iirc?)  and then measure the three different options. Four if you count the full bridge.

Just for you!

You just demonstrated typical trap for young players of "Design By Inspection". Engineers who made 80mA dropper instead of 1mA, made it for for some quite obvious reason - ALARM. You definitely can't produce audible alarm using 15v @50uA supply, unless you are building "smoke alarm earbud". Only actually making whole device and measuring SPL, you will see how much power is actually needed. Could be so that 10.5mA @ 15V of NCP785A is not even enough.

It was in my first video, I measured 15mA peak buzzer current.
The entire point of the video (both of them) was not selecting a suitable replacement regulator for this exact design, it was about showing the design topology differences. And the 2nd video was about showing why I didn't have to build up and test the regulator circuit in order to know it was 1/1000th the consumption of the zener circuit.

[ogden]

It was in my first video, I measured 15mA peak buzzer current.

It means that NCP785A with it's max 10.5mA is not suitable for the product.

The entire point of the video (both of them) was not selecting a suitable replacement regulator for this exact design, it was about showing the design topology differences.

Right. It's like facepalming about high idle consumption of the car, then proving that you can "solve problem" using motor from the moped. For me such "engineering" do not make any sense.

[Kleinstein]

Nice for a video down the track if you could build up those other two solutions (TI and OnSemi iirc?)  and then measure the three different options. Four if you count the full bridge.

Just for you!

You just demonstrated typical trap for young players of "Design By Inspection". Engineers who made 80mA dropper instead of 1mA, made it for for some quite obvious reason - ALARM. You definitely can't produce audible alarm using 15v @50uA supply, unless you are building "smoke alarm earbud". Only actually making whole device and measuring SPL, you will see how much power is actually needed. Could be so that 10.5mA @ 15V of NCP785A is not even enough.

It was in my first video, I measured 15mA peak buzzer current.
The entire point of the video (both of them) was not selecting a suitable replacement regulator for this exact design, it was about showing the design topology differences. And the 2nd video was about showing why I didn't have to build up and test the regulator circuit in order to know it was 1/1000th the consumption of the zener circuit.

The linear regulator circuit is not 1/1000 the power: the linear regulator version takes some 70 µA from 300 V and thus some 21 mW. The capacitive dropper was ca. 1 W real power and maybe 15 VA.
The the current spikes from the rectifier gives an AC current that is considerably higher than the 70 µA DC, more like 200 µA RMS and also a relatively poor power factor, though for a different reason.

[EEVblog]

Right. It's like facepalming about high idle consumption of the car, then proving that you can "solve problem" using motor from the moped. For me such "engineering" do not make any sense.

Then go make your own videos.
I could make half a dozen videos on all the engineering factors that go into selecting, testing, and proving a simple regulator solution like this if you want to get into real practical engineering of it. I chose not to  in this case and had a different intention. You don't have to like that, and that's ok, I can't please everyone.
But if you can't at least understand my intention, even if you don't like it, then 

[EEVblog]

It was in my first video, I measured 15mA peak buzzer current.

It means that NCP785A with it's max 10.5mA is not suitable for the product.

If you cared to actually watch the video:

[ogden]

Then go make your own videos.

What?! Did not expect such "argument" from you.

But if you can't at least understand my intention, even if you don't like it, then 

I said it already - you are doing business. Sometimes engineering excellence have to be traded for clicks. That's OK especially for professional youtuber. BTW existence of argument do not prove that I don't like video. Your videos are fine, they get better over time as well.

If you cared to actually watch the video:

Text overlay that looks like afterthought? - Nah. Sad that you consider critique as an attack rather than idea for your future video(s).

[Alti]

Gentlemen, either we get back on the topic or the topic risks being locked, I am afraid.

Thank you for valuable inputs, regarding requirements a "EEVblog #1284 considered" Smoke Detector (SD) has to meet.

Now we can see Australian requirements for SD are quite stringent, w.r.t upper half of the planet, my interpretation of the presented data indicates that:

SD has to be locally (battery) powered and remotely (mains) powered.

This "and" means the choice is not optional, if either one of power sources fails, the sensor stays functional using second source.
Understandably "locally" and "remotely" means there is no allowance to run the SD from just any two power sources. The case where you power smoke detector from lets say (mains 230VAC wiring) and (24VDC wired to central backup battery) interferes with whole concept. This is because during fire the whole wiring dies first, the SD has to beep till it melts on a ceiling in such scenario.

Also, powering this thing with two batteries (main + auxilary) is not legal because - well, if you forget about SD hanging there, after several years you are left without any SD protection.

Of course the regulation does not prohibit SD having an on-board battery and 24VDC wiring or 4-20mA current loop. Or on-board battery and two independent wirings, etc.


So,  the primary requirement is:

Single point of (power) failure does not compromize core SD functionality.

That is what I think is a good starting point in a discussion about fullfiling SD requirements.

[ogden]

Gentlemen, either we get back on the topic or the topic risks being locked, I am afraid.

Which exactly topic? "Design topology differences of standby supplies" is well-covered in the video already. Other topics are not supported by Dave as you can see.

[EEVblog]

Then go make your own videos.

What?! Did not expect such "argument" from you.

But if you can't at least understand my intention, even if you don't like it, then 

I said it already - you are doing business. Sometimes engineering excellence have to be traded for clicks. That's OK especially for professional youtuber. BTW existence of argument do not prove that I don't like video. Your videos are fine, they get better over time as well.

I had a point I wanted to make in that video and I think I made it, it's got nothing to do with "trading engineering excellence for clicks", if you think that then you are demonstrably wrong.
Again, you don't have to like my intention of that video

If you cared to actually watch the video:

Text overlay that looks like afterthought? - Nah. Sad that you consider critique as an attack rather than idea for your future video(s).

The problem is you that you are arguing based on something I had no intention of doing in that video.
If you don't know how I do videos, let me explain. I don't have a script, I don't plan them, I have an idea for an intention for a video and I hit record and start talking. I then join the clips together and upload and hopefully something useful comes out of it.

In this case my intention with the first video was to show that zeners based solutions waste a lot of power and here's a couple of alternative that could have possibly been used instead. It was not about redesigning that actual product and doing the engineering to select the best part to do the job. When editing that video I thought some people like yourself might pick up on the 10mA part, so I measured actual buzzer current and added that overlay saying that this would require a proper engineering solution, and that was the end of that. I wasn't going to go and shoot another hour worth of material exhaustively going through the design process of selecting the exact correct part to meet the requirements, that would have been stupid as it was not the intent of the video.

For the 2nd video my intention was the explain why I didn't have to test such a solution, and explain what "design by inspection" is. Again it was not about redesigning that actual product and doing the engineering to select the best part to do the job.

Again, you don't have to like my approach, and that's fine, but at least try and understand that I have a different intention to what you wanted, and made the content accordingly.

[kcbrown]

This is how the semiconductor industry works. If you have a part with some novelty there are small volume users who will pay a good price for it, so you set the small volume price high. However, high volume users are ALWAYS driven by the BOM above all other factors. You need to offer attractive volume pricing to win any high volume design. The markets for this device are high volume things like smoke detectors, where you can lose a design over a fraction of a cent on the BOM, and utility meters, which will be attracted to the lack of any magnetic components (they are sensitive to tampering issues), but still need to meet aggressive BOM goals.

Sure, I get that, and it makes sense.  But the manufacturer can't sell at a loss, either, especially at the highest quantities.  Their price has to be enough to (in aggregate) recover their R&D plus the manufacturing costs (and sales/advertising costs, etc.).

In this case, their competition is the wide array of SMPS controllers that are on the market, with offerings ranging from those produced by first-tier manufacturers of the same caliber as TI to the cheapest offerings from Chinese vendors/manufacturers.  Against that kind of competition, it seems to me that TI would have to sell a device like this on its novelty, not just to small volume users but even to high-volume purchasers.

Which is to say, if a high-volume purchaser can get away with using an SMPS solution then it seems to me they probably will use that in favor of using a solution built around this TI part, except perhaps under one condition: when the cost of the controllers is of the same order of magnitude as the jellybean parts, such that the difference in the cost of the jellybean parts is at least enough to compensate for the difference in the cost of the controllers, thus turning the space advantage of the TI part (due to fewer jellybean parts) into a cost advantage.  Maybe TI can sell these things cheaply enough to meet that condition and still make a profit.  No idea.

[Nominal Animal]

I am really glad Dave made this video, because it does matter.

How do I know?  Because it exactly mirrors the standby power issues a decade or two ago.  It was common for devices to waste up to tens of watts in standby power, because designers and engineers didn't think it mattered.

But it did.  I recall a time here in Finland, where those of us with devices like TVs and computers on extension cords with power switches could save on their electricity bills simply by completely unplugging the devices when not in use.  For me, dwelling in a city apartment but with lots of electronics in the late nineties, it was about a third of my electricity bill.

Some of you might have seen my Beginner thread about USB-to-3.3V, too.  That is this exact same situation in another form.  Most of the USB gadgets not running on battery power use an LDO to drop the USB 5V to 3.3V.  Sure, that keeps the BOM cost low.  But it also turns a third of the consumed power into waste heat.  In some cases, like LTE modem dongles (the USB sticks in particular), that extra heat is a real issue, and shortens the devices' lifespan and usability.

I am more than a little annoyed at so many people trying to convince others that this is a non-issue, because the BOM cost is too high to make this kind of a change.  Yet, we basically eliminated the standby power issues within a decade, as knowledge about it grew; and I'm absolutely sure the EEs at that time tried to make the same argument, that significantly lowering the standby power would never be cost effective or competitive, because of the increased cost of the designs.  Fuck that kind of statist thinking: wasting power is silly, and sooner or later people will understand its effects on their finances, and will switch.

Before any of you respond with "but Nominal, making a better but slightly more expensive product makes no business sense, so you're wrong", I'll just say that unless you have earned millions from your EE designs, your opinion has zero weight in real life.  The trajectory I am seeing has happened several times in the past (different technologies, from steam forwards), and most recently, just over a decade ago with standby power.  So I am not proposing this is anything out of the ordinary; I am saying this has happened before, will happen again, and Dave pointing out one possible (albeit small) way we could take right now along this trajectory, is a very good point.



I only wish Dave would attack USB and LiPo power supplies to 3.3V projects, because there is very little talk of this right now on the internet, we're talking about 34% of losses (5V to 3.3V using LDO is only 66% efficient; step-down converters with low enough ripple and noise are black magic to us hobbyists), and even a bumblefuck like I can find chips and datasheet designs that can reach 90%-95% for ≲ 5€ total cost per unit for a set of ten, using resources available for us hobbyists.  Yes, small Chinese step-down modules are ubiquitous, but because their ripple and noise are unknown factors (and one really needs an oscilloscope to determine them in practice), so many avoid them, and opt to use wasteful and hot LDOs instead, because at least their characteristics are easy to understand and determine, needing just a multimeter, really.

Yet, thousands of us hobbyists are doing these projects, and would love to know how to do it better/properly; and realize that just like standby power or bad product design that wastes power, this too is one way we can reduce the overall energy costs of our projects: many do not even realize how many of the practical problems of their projects are due to waste heat or too high current draw.  (The latter is more of an issue with single-board computers; desktop machines and laptops tend to have better filtering and power budgets for USB power lines.  There is a reason why Olimex has an USB module with just bulk capacitors between VUSB and ground; that kind of thing was necessary with the power-hungry 3G USB modem dongles.)  For these USB gadgets, the vastly reduced waste heat means one can use cheap 3D printed closed enclosures with much fewer issues -- which was kinda the way I stumbled on this myself.

[Ed.Kloonk]

I am really glad Dave made this video, because it does matter.

How do I know?  Because it exactly mirrors the standby power issues a decade or two ago.  It was common for devices to waste up to tens of watts in standby power, because designers and engineers didn't think it mattered.

But it did.  I recall a time here in Finland, where those of us with devices like TVs and computers on extension cords with power switches could save on their electricity bills simply by completely unplugging the devices when not in use.  For me, dwelling in a city apartment but with lots of electronics in the late nineties, it was about a third of my electricity bill.

Some of you might have seen my Beginner thread about USB-to-3.3V, too.  That is this exact same situation in another form.  Most of the USB gadgets not running on battery power use an LDO to drop the USB 5V to 3.3V.  Sure, that keeps the BOM cost low.  But it also turns a third of the consumed power into waste heat.  In some cases, like LTE modem dongles (the USB sticks in particular), that extra heat is a real issue, and shortens the devices' lifespan and usability.

I am more than a little annoyed at so many people trying to convince others that this is a non-issue, because the BOM cost is too high to make this kind of a change.  Yet, we basically eliminated the standby power issues within a decade, as knowledge about it grew; and I'm absolutely sure the EEs at that time tried to make the same argument, that significantly lowering the standby power would never be cost effective or competitive, because of the increased cost of the designs.  Fuck that kind of statist thinking: wasting power is silly, and sooner or later people will understand its effects on their finances, and will switch.

Before any of you respond with "but Nominal, making a better but slightly more expensive product makes no business sense, so you're wrong", I'll just say that unless you have earned millions from your EE designs, your opinion has zero weight in real life.  The trajectory I am seeing has happened several times in the past (different technologies, from steam forwards), and most recently, just over a decade ago with standby power.  So I am not proposing this is anything out of the ordinary; I am saying this has happened before, will happen again, and Dave pointing out one possible (albeit small) way we could take right now along this trajectory, is a very good point.



I only wish Dave would attack USB and LiPo power supplies to 3.3V projects, because there is very little talk of this right now on the internet, we're talking about 34% of losses (5V to 3.3V using LDO is only 66% efficient; step-down converters with low enough ripple and noise are black magic to us hobbyists), and even a bumblefuck like I can find chips and datasheet designs that can reach 90%-95% for ≲ 5€ total cost per unit for a set of ten, using resources available for us hobbyists.  Yes, small Chinese step-down modules are ubiquitous, but because their ripple and noise are unknown factors (and one really needs an oscilloscope to determine them in practice), so many avoid them, and opt to use wasteful and hot LDOs instead, because at least their characteristics are easy to understand and determine, needing just a multimeter, really.

Yet, thousands of us hobbyists are doing these projects, and would love to know how to do it better/properly; and realize that just like standby power or bad product design that wastes power, this too is one way we can reduce the overall energy costs of our projects: many do not even realize how many of the practical problems of their projects are due to waste heat or too high current draw.  (The latter is more of an issue with single-board computers; desktop machines and laptops tend to have better filtering and power budgets for USB power lines.)  For these USB gadgets, the vastly reduced waste heat means one can use cheap 3D printed closed enclosures -- which was kinda the way I stumbled on this myself.

My phone has a charger that probably offends even the most ambivalent EE designer. And how has my phone manufacturer dealt with it? A big fat message that takes up two thirds of the screen "Unplug Charger!". When I followed this up, cos it's really annoying when I just want to glance notifications while it's in it's little cradle, they don't mean the plug going into the phone.

There is actually an animated graphic depicting the user pulling the plug out of the wall socket. I suspect that this 'solution' has been shoved into the phone OS to obviously appease somebody, wouldn't it have been easier to include another 20c worth of parts to make the charger go to sleep when nothing is connected at the USB end?

[TerminalJack505]

Like it, or not, it will likely take legislation to change this.  That's why you no longer see big, inefficient wall warts included with new devices.  The US has minimum efficiency requirements of external power supplies.

Because they are internal, these capacitor dropper power supplies don't fall under any of the requirements.  The only incentive manufacturers have to use anything more efficient is cost.

Sadly, I don't imagine it would increase the cost by more than 5% to 10% to include a switching regulator.  You can get something like a UCC28881 for $0.54 in volume, for example.  This is a high voltage AC-DC non-isolated buck converter with fairly low external part count.  It could easily power a smoke detector.

[Nominal Animal]

Like it, or not.  It will likely take legislation to change this.

Sure, that's how the standby power problems were solved as well.

But first, the enthusiasts, hobbyists, and ordinary users must realize and understand the situation.  Then it may become a political issue, and as it is such a simple thing, it may go through quite rapidly, just as the standby power stuff did.  The only reason legislation was necessary, was because the companies and their EEs didn't want to spend the extra to produce devices users wanted.  Think about that for a second, when rereading the comments in this thread.

Like I said, this kind of step forward is not new; it has happened dozens of times in the last two centuries already, and will happen again.

What is important, is someone like Dave making these easily understood videos so more people realize this is an issue.
(And my USB-to-3.3V thing is related, so I'm hoping Dave or one of the other youtubers with EE design experience could show how the same applies to current 3.3V USB gadgets, and how hobbyists can avoid that pitfall.)

Let me elaborate a little.

I have an Odroid-HC1 that I'd like to use as an LTE firewall, with a Huawei ME909s-120 LTE modem (a real modem, not an embedded Linux system!) as the uplink.  The modem is one of many modules that use a miniPCIe interface -- or, actually just the USB and SIM card pins on the interface.  These modems run on 3.3V, and tend to be a bit power-hungry: the average current consumption of this one is only 200mA, but it can be a bit spikey.  (Older 3G/4G/LTE modems and dongles, like the ZTE 823 embedded USB model I also have, is very power-hungry, but I'm not exactly sure of its consumption as it has been in continuous use for over a year already.)

MiniPCIe-WWAN adapters (an USB connector and a SIM card slot, connecting to such a module via MiniPCIe; most modules have very good support in Linux with ModemManager -- in my case, plug-and-play) are easy to obtain at e.g. Ebay, and there are even companies making their own for use in ad displays, indoor advertisements, et cetera, but they all have the same issue: they use an LDO to drop the 5V to the required 3.3V, and the waste heat is wreaking havoc with the enclosures.  (I even know of a Finnish company that is having this issue with this in their own designs, having observed the change in LDO to one with a serious heatsink on it!)

I now have four of these adapters (plus one either in, or lost in the mail).  Only one has a reasonable step-down DC-DC converter, a Silergy SY8009B (in SOT23-6).  It's not bad for this particular application, as it is 90% efficient at the nominal 200mA (5V to 3.3V), and can do 2A max, but the datasheet doesn't even list the recommended inductor values, so I am quite suspicious of it.

In fact, I am considering making my own.  Which is, given my level of competence in these matters, rather ridiculous.  And that just shows how too stuck to their ways most EEs designing these jellybean modules are, not considering the issues Dave highlighted here -- and the issues like nasty waste heat due to silly component selection causing real use risks.  And that is why this kind of videos -- showing the issue in plain terms, then a few suggestions on how to do better -- are so good: it is food for thought.  That thought leads to better designs and wider understanding of the issue, and may lead to the legislation that eventually enforces the change, as it did with wasteful standby power a decade or two ago in most Western countries.

(And lots of traps for us new players, too.  Which is why I'm going slow and careful, trying to work out the reasonable design, before I commit to testing it in real life, risking my 60€ modem.)

The point BigClive makes about apparent power -- that eventually, we will be charged by apparent power use, instead of the actual power use --, is simply a continuation of the same trajectory.  It is just much further in the future; I'd say a decade at least, perhaps two, but it will come.
It would be interesting to see what kind of a circuit you could use right now to create a LED light with a reasonable (say, 0.95 or higher) power factor, and how much it would cost.

Or, better yet, how much a 2.4A (12 watt) USB charger would cost, if it had efficient components and actual power factor correction.  You know: useful information, food for thought.

[nctnico]

I only wish Dave would attack USB and LiPo power supplies to 3.3V projects, because there is very little talk of this right now on the internet, we're talking about 34% of losses (5V to 3.3V using LDO is only 66% efficient; step-down converters with low enough ripple and noise are black magic to us hobbyists), and even a bumblefuck like I can find chips and datasheet designs that can reach 90%-95% for ≲ 5€ total cost per unit for a set of ten, using resources available for us hobbyists.  Yes, small Chinese step-down modules are ubiquitous, but because their ripple and noise are unknown factors (and one really needs an oscilloscope to determine them in practice), so many avoid them, and opt to use wasteful and hot LDOs instead, because at least their characteristics are easy to understand and determine, needing just a multimeter, really.

It is not that simple. A switching DC-DC converter uses a few to several tens of mA for itself. Often this means it is actually less efficient compared to an LDO in that situation. If you want to power a microcontroller from USB an LDO is the most efficient solution in many cases.

[Nominal Animal]

It is not that simple. A switching DC-DC converter uses a few to several tens of mA for itself.

What you are claiming is that TI Webench is lying, and that you know better.  For some reason, I'm not convinced.

The particular circuit I am talking about is the one straight off the TPS82084 datasheet, except with a 510k:162k voltage divider to get the 3.3V instead of the 200k:162k to get 1.8V shown in the datasheet.  The figures I quoted are from the TI webench design report that I first found; I only later noticed it is really a direct copy of the datasheet circuit.

So, which one is it, then?  TI straight out lying and you right, or you too stuck in your familiar patterns to admit there might be something here?

[nctnico]

It is not that simple. A switching DC-DC converter uses a few to several tens of mA for itself.

What you are claiming is that TI Webench is lying, and that you know better.  For some reason, I'm not convinced.

The particular circuit I am talking about is the one straight off the TPS82084 datasheet, except with a 510k:162k voltage divider to get the 3.3V instead of the 200k:162k to get 1.8V shown in the datasheet.  The figures I quoted are from the TI webench design report that I first found; I only later noticed it is really a direct copy of the datasheet circuit.

So, which one is it, then?  TI straight out lying and you right, or you too stuck in your familiar patterns to admit there might be something here?

Well, this didn't stand out in your wall of text. But this converter looks like it (finally) addresses the downside of switching converters at low load. And it seems cheap too (which is also new to me when it comes to integrated switching modules). Nice catch!

[Nominal Animal]

Well, this didn't stand out in your wall of text.

Yeah right.  You're very talented in avoiding admitting any error or culpability, instead hiding it inside a snipe at the accuser.  The trick with the compliment at the end was a masterful stroke; it makes you look like a reasonable person, if one isn't very observant.

Have I mentioned I absolutely detest people who try to manipulate others like that?

[ogden]

Well, this didn't stand out in your wall of text. But this converter looks like it (finally) addresses the downside of switching converters at low load.

Power Save Mode (PSM) of TPS82084/5 is similar to what pulse-skipping/burst_mode converters do for long time already. Check TPS62743 for example (360 nA Operational Quiescent Current, Up to 90% Efficiency at 10-µA Output Current).

[nctnico]

Well, this didn't stand out in your wall of text.

Yeah right.  You're very talented in avoiding admitting any error or culpability, instead hiding it inside a snipe at the accuser.  The trick with the compliment at the end was a masterful stroke; it makes you look like a reasonable person, if one isn't very observant.

Ofcourse it is a masterfull stroke. Carefully designed to piss you off so the next time you write a shorter text and put a link in there with an actual part we can check out (instead of a writing a long rant nobody is interested in). Take an example from how I can do so much with so little text. 

[Nominal Animal]

Ofcourse it is a masterfull stroke. Carefully designed to piss you off so the next time you write a shorter text and put a link in there with an actual part we can check out (instead of a writing a long rant nobody is interested in). Take an example from how I can do so much with so little text. 

I am a self-confessed unexperienced hobbyist with a cheap multimeter, an Analog Discovery 2, and a few odd projects.

My point in the post was not to advertise existing products, but to explain why this video by Dave was very on point.
You disagreed, and were proven wrong.  You didn't even admit to that, just stated you were only wrong because I write too much. 

When you need people like me to point out what the reality is and what the existing products can do, you proved you are stuck in your incorrect "undestanding" (beliefs), and need these videos to get you to do your job properly.  I showed why it is so important Dave makes these videos for us ordinary people, you showed why it is so important Dave makes these videos for you who work in this field.

Stuck like fly in amber, you are, and still smiling like an idiot.

Perhaps you should learn to read more.

[Nominal Animal]

Reported to moderators as rude hijacking idiot ^

Try convincing Simon to ban me.  He already hates me for pointing out how he blatantly lied in one thread.  That might work.

[David Hess]

It is not that simple. A switching DC-DC converter uses a few to several tens of mA for itself. Often this means it is actually less efficient compared to an LDO in that situation. If you want to power a microcontroller from USB an LDO is the most efficient solution in many cases.

There are many ways to lower the quiescent and operating current of a switching regulator.  Quiescent and operating currents of 10s of microamps or lower are possible:

1. Hysteretic regulators have no oscillator and because they require no frequency compensation, they also have no error amplifier.  Note that because hysteretic and constant off-time regulators require no frequency compensation, they do not care about operating in discontinuous conduction mode which is common at low output current.

2. Shut down the oscillator and error amplifier between "burps".  This is a combination of a conventional switching regulator and hysteretic regulator.  Linear Technology calls this "burst mode".

3. Bootstrap the regulator's supply from its low voltage output after startup and then disconnect the startup supply.  Bootstrapping from the output or a dedicated winding is common in off-line regulators which start from a small capacitor charged from the high input voltage.  This technique is as old as transistors.

Switching regulators can be competitive in micropower applications but often a linear regulator is used for lower cost and ease of use.

[EEVblog]

Like it, or not, it will likely take legislation to change this.

Almost certainly.

[floobydust]

Engineers gone down the rabbit hole. The power supply that is say 25% more efficient is better for the environment?
You have to compare a high and low efficiency PSU from end to end. The fact that the capacitors fail after 2 years while the low efficiency PSU is still running for 5+ years, consider the manufacturing, recycling, disposal costs and the coal burned to do that.
You save 25% on energy and global warming during the PSU's running life - but had to use two power supplies after the high efficiency one failed due to cheap parts.

The local Eco Electronics Recycling Center is full of cheap consumer electronics that had a short life. I dumpster dive, put in new electrolytics and it's good for years.
I think longer lifetime electronics benefits the planet more than a bit higher efficiency.

[EEVblog]

The local Eco Electronics Recycling Center is full of cheap consumer electronics that had a short life. I dumpster dive, put in new electrolytics and it's good for years.
I think longer lifetime electronics benefits the planet more than a bit higher efficiency.

Probably, but full proper life cycle analysis is hard to calculate.
Any good engineer will shoot for both though.

[Nominal Animal]

Small USB gadgets using an LDO to drop 5V to 3.3V and consuming 200-400mA (pretty common for anything with a radio) in small enclosures definitely have shorter lifespans due to heat issues.  (I don't have any details, but I seem to recall many of the early USB WiFi dongles had issues with that too, and quickly switched to buck converters.  Not so much for energy savings, but because in such small plastic enclosures (no airflow inside, with the plastic acting essentially as an insulator), the heat was causing them to fail.  I could recall wrong, though.)

[thm_w]

Engineers gone down the rabbit hole. The power supply that is say 25% more efficient is better for the environment?
You have to compare a high and low efficiency PSU from end to end. The fact that the capacitors fail after 2 years while the low efficiency PSU is still running for 5+ years, consider the manufacturing, recycling, disposal costs and the coal burned to do that.
You save 25% on energy and global warming during the PSU's running life - but had to use two power supplies after the high efficiency one failed due to cheap parts.

The local Eco Electronics Recycling Center is full of cheap consumer electronics that had a short life. I dumpster dive, put in new electrolytics and it's good for years.
I think longer lifetime electronics benefits the planet more than a bit higher efficiency.

I'm not sure why you instantly imply the high efficiency PSU is less reliable. The low efficiency has two clear wear items, main capacitor and electrolytic. A high efficiency version may be more or less reliable, there is no way to say for sure without going into more details.

BTW I tried to get electronics from my city recycling center, but they won't let you remove anything from there, sadly.

[amyk]

More components, more complexity, more points of failure.

My experience with LED lamps shows the ones with capacitive droppers and otherwise minimal circuitry have outlasted the more expensive ones with a full SMPS, and the failure mode has always been some active component in the latter.

[Alti]

I'm not sure why you instantly imply the high efficiency PSU is less reliable.

In the context of a smoke detector: power, power factor, reliability, installation cost - these are all factors that allow customers to decide whether they want Kwality or Miele, based on their subjective oppinions. Whether they are aware of power factor or reliability or installation cost at purchase time - that is another topic for another discussion. The undisputamle fact is that every smoke detector draws some power, has some reliability, etc.

From systems engineering we know that (with fixed reliability of the designs considered) an increase of design complexity imposes using more reliable components, w.r.t. a base design. So the net effect is that with smoke detector the BOM cost includes a factor that is a square function of complexity. First you pay for higher count of components and then you pay for higher grade of those components (usually achieved by underrating).

So the outcome is that (for unaware customers that are purely interested in purchase price) this pushes the smoke detector market in the direction of simple, less efficient designs that barely meet warranty period.
Surprised?