EEVblog #1284 - How Bad Product Design Kills The Environment

[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.