Author Topic: EEVblog #751 - How To Debunk A Product (The Batteriser)  (Read 2300121 times)

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Offline Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #200 on: June 29, 2015, 10:12:04 am »
On some other forum someone mentioned a "joule thief 2.0" which allegedly works sort of like a voltage doubler/charge pump. An IC would switch capacitors so that they charge in parallel and then unload in series so that you get higher voltage out. Would that work in this case?
It might, but ... in this case you definitely don't want to double the voltage. It would wreck stuff.
 

Offline Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #201 on: June 29, 2015, 10:19:13 am »
From their FAQ:
Quote
Are there limitations on devices that can use the Batteriser based on the current consumptions? No, the Batteriser sleeve is designed to deliver as much current as a battery is able to supply to the device.

Really, no limitations on current? At all?
No change in the expected life gain based on current? At all?
I think this statement is the least of their sins. Most devices won't be discharging at a current that would cause problems.

Maybe that's why their inductor is so small - they assume very small currents (and, hey, a battery is quite a big heatsink if you make it work hard).

Their whole strategy isn't based on this thing working well. It's about selling as many as they can to the sort of people who'll just shrug when it doesn't work out as promised.
 

Offline apis

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #202 on: June 29, 2015, 04:55:15 pm »
On some other forum someone mentioned a "joule thief 2.0" which allegedly works sort of like a voltage doubler/charge pump. An IC would switch capacitors so that they charge in parallel and then unload in series so that you get higher voltage out. Would that work in this case?
It might, but ... in this case you definitely don't want to double the voltage. It would wreck stuff.
Yes, but with a little bit extra you should be able to regulate the output to between 1x and 2x the input voltage, that should work down to approx 0.75V which is what the batteriser claims to do. I googled a little and found this IC from maxim: MAX682. You should be able to do the same for approximately 0.75-1.5V in and get regulated 1.5V out. I have never used this type of converter so not sure if there's some downside to it that prevents it from working in this case but it could explain how they got it so small and why there is no inductor visible, only requires three relatively small capacitors. Of course there will be current limitations and losses in this type of regulator as well and the 1.4V cutoff assumption is false no matter how it works.
 

Offline Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #203 on: June 29, 2015, 05:21:22 pm »
On some other forum someone mentioned a "joule thief 2.0" which allegedly works sort of like a voltage doubler/charge pump. An IC would switch capacitors so that they charge in parallel and then unload in series so that you get higher voltage out. Would that work in this case?
It might, but ... in this case you definitely don't want to double the voltage. It would wreck stuff.
Yes, but with a little bit extra you should be able to regulate the output to between 1x and 2x the input voltage, that should work down to approx 0.75V which is what the batteriser claims to do.
How is that better than just regulating it directly to 1.5V?
 

Offline apis

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #204 on: June 29, 2015, 06:09:16 pm »
I think the benefit compared to typical boost circuit is that it is easier to miniaturize since you don't need the inductor.
« Last Edit: June 29, 2015, 06:38:53 pm by apis »
 

Offline Galenbo

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #205 on: June 29, 2015, 06:22:04 pm »
They guy popped up in the youtube comments, in full-retard repetitive mode.
Saying something about ESR and internal resistance, patents and PhD's.....
If you try and take a cat apart to see how it works, the first thing you have on your hands is a nonworking cat.
 

Offline apis

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #206 on: June 29, 2015, 06:41:44 pm »
It's a safe bet someone is wrong when their only argument is they have a PhD.  ::)

If it really does work they could always send Dave a prototype for testing, would be interesting. :popcorn:
« Last Edit: June 29, 2015, 06:58:47 pm by apis »
 

Offline Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #207 on: June 29, 2015, 07:03:32 pm »
I think the benefit compared to typical boost circuit is that it is easier to miniaturize since you don't need the inductor.
But if you double the voltage you have to regulate it down again so that needs an inductor.

(Unless you're thinking of using a linear regulator and throwing away half the battery power...)

 

Offline bktemp

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #208 on: June 29, 2015, 07:04:42 pm »
On some other forum someone mentioned a "joule thief 2.0" which allegedly works sort of like a voltage doubler/charge pump. An IC would switch capacitors so that they charge in parallel and then unload in series so that you get higher voltage out. Would that work in this case?
It might, but ... in this case you definitely don't want to double the voltage. It would wreck stuff.
Yes, but with a little bit extra you should be able to regulate the output to between 1x and 2x the input voltage, that should work down to approx 0.75V which is what the batteriser claims to do. I googled a little and found this IC from maxim: MAX682. You should be able to do the same for approximately 0.75-1.5V in and get regulated 1.5V out. I have never used this type of converter so not sure if there's some downside to it that prevents it from working in this case but it could explain how they got it so small and why there is no inductor visible, only requires three relatively small capacitors. Of course there will be current limitations and losses in this type of regulator as well and the 1.4V cutoff assumption is false no matter how it works.
The main problem with all those switched capacitor converters is the efficiency: It can only adjust the voltage in multiples of the input voltage. If you generate 1.5V from a 1.4V battery this voltage gets doubled to 2.8V and the difference to 1.5V wasted as heat. The MAX682 datasheet has no efficiency vs input voltage diagram, but it shows on page 4 the efficiency for a number of different input voltages. Unlike inductor bases converters, the efficiency drops when the input voltage increases.
 

Offline Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #209 on: June 29, 2015, 07:06:13 pm »
They guy popped up in the youtube comments, in full-retard repetitive mode.
Saying something about ESR and internal resistance, patents and PhD's.....
I'm just hoping it blows up in his face before he has time to rip anybody off.

Seriously, why isn't the news all over this? I thought new outlets loved uncovering fraudsters.

There should be reporters hounding him for answers and making him look stupid on national TV. I'd pay to watch that.  :popcorn:
 

Offline apis

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #210 on: June 29, 2015, 07:46:58 pm »
But if you double the voltage you have to regulate it down again so that needs an inductor.

(Unless you're thinking of using a linear regulator and throwing away half the battery power...)
Based on the datasheet it looks like they charge the output capacitor only when it's below a threshold voltage (i.e. the desired output voltage), so doesn't use a linear regulator. But as bktemp says efficiency is low anyway?

---
So they have also put up a "batteriser fan page ::)" on google plus, claiming that engineers don't know about battery ESR
Quote
Most electronic engineers are not very familiar with intricate details of the inter-workings of batteries and its equivalent circuit models. Batteries have an internal impedance or resistance (ESR) that plays a major role in operation of a battery in a system.
« Last Edit: June 29, 2015, 08:08:49 pm by apis »
 

Offline apis

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #211 on: June 29, 2015, 08:27:30 pm »
I'm not sure what datasheet you're referring to
I was looking at the MAX682 on page 6 there's a block diagram (Figure 3. Skip-Mode Regulation). Yes there would be a lot of ripple.
 

Offline Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #212 on: June 30, 2015, 06:10:05 am »
I'm not sure what datasheet you're referring to
I was looking at the MAX682 on page 6 there's a block diagram (Figure 3. Skip-Mode Regulation).
OK.

Quick glance says it needs 1µF caps per 100mA of output current. This thing needs to put out more than 100mA so would it really be smaller than an inductor?

Yes there would be a lot of ripple.
I imagine so...  :o
 

Offline Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #213 on: June 30, 2015, 09:23:23 am »
To keep the ripple in check you you might need a fourth capacitor across the battery so that the two parallel-series-switched capacitors in the middle will charge up faster (it will have a lower ESR than the battery).
I almost got it...  :)

The way they do it is to permanently connect one of the two 'charge transfer' capacitors across the source (ie. it doesn't switch). The chip only switches the other charge transfer capacitor between series/parallel to charge up the main output capacitor.

Pity it needs a 2.7V input. That means you can't run it from 2 batteries, it really limits its usefulness.  :(

Wait! I heard about this thing called 'The Batteriser' that can fix that problem!!


Update: I checked out Maxims other chips and there's one that can use an extra capacitor to triple the input voltage and get 5V from two batteries - the MAX619. Not very efficient though, only 60-70% with two batteries.

 

Offline bktemp

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #214 on: June 30, 2015, 09:34:18 am »
Update: I checked out Maxims other chips and there's one that can use an extra capacitor to triple the input voltage and get 5V from two batteries - the MAX619. Not very efficient though, only 60-70% with two batteries.
On page 3 there is a nice EFFICIENCY vs. INPUT VOLTAGE diagram showing the efficiency jump hat 2.5-3.0V when it switches from 3x to 2x mode.
If you want higher efficiency use a boost converter like MCP1640 or TPS6109x and many others. They all have a higher efficiency and lower ripple current. The inductors can be really small at >1MHz switching frequency. That is why charge pumps are rarely used in such applications.
 

Offline Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #215 on: June 30, 2015, 09:48:58 am »
Update: I checked out Maxims other chips and there's one that can use an extra capacitor to triple the input voltage and get 5V from two batteries - the MAX619. Not very efficient though, only 60-70% with two batteries.
On page 3 there is a nice EFFICIENCY vs. INPUT VOLTAGE diagram showing the efficiency jump hat 2.5-3.0V when it switches from 3x to 2x mode.
Yep, I saw that. It's where I got my 60-70% figure from.

I'm sure these chips have their place. I was interested because they come in DIP packages so it would be easy for a hacker like me to add one to some perf-board in a battery-powered Arduino project.

OTOH I can get little inductor-based booster boards on eBay for a lot less money (MAXIM pricing!). And they work down to 0.7V. And they need less soldering.

If you want higher efficiency use a boost converter. The inductors can be really small at >1MHz switching frequency.
I wonder if that's what The Batteriser uses...?
 

Offline EEVblog

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #216 on: June 30, 2015, 01:12:07 pm »
Someone pointed me to this response from Batteriser, apparently in response to my video and criticism from others:
https://plus.google.com/+BatteriserBatteroo/posts/8emCWUBJbve
Apparently it's a non-affiliated "fan page"  ::)
https://www.youtube.com/channel/UCFr9QYi8Yq2ckWYPx_Vui8A
Yeah right, someone went to all the trouble to make a fan page for a startup tech, and they go around defending the product in comments, like on this video here, yeah that sounds totally legit  ::)
  :-DD

It's late and I couldn't be bothered reading it all, so go for it...

Quoted here in full in case it vanishes:
Quote

After seeing all the skepticism online, the Batteriser inventors answered a Q and A! We got a hold of the that Q and A with Batteriser! Here it is and I hope it helps you all get a better understanding of the product:
“Question 1: In the articles I have seen about Batteriser, your CEO claims that most battery operated devices stop working at around 1.3 or even higher. I have seen videos online that someone connects a power supply and shows that some typical devices work down to 1.1V or even lower, showing LED’s on a device still blink at that level. What is the explanation?
Answer:  There are multiple aspects to this issue. Let’s start with the less complicated aspect:
1.   Just because LED in a device is blinking, it doesn’t mean the device is fully functional. In one of our experiments with an RC remote, at 1.3 volt the car would only go forward, but not backwards. We use the voltage that the device is no longer fully functional as the cut off voltage rather than the voltage that the device’s LED is still blinking. In yet another example that most people can relate to, we all have had experience with our TV remotes. Once a battery gets to a low level, even though the LED blinks, you literally have to walk right up to the TV for it to work. Again, the batteries are considered dead by most, long before LED’s stop blinking. So the levels demonstrated in experiments like that are artificially too low.
2.   Using a Power Supply to detect, where a battery operated device stops working is wrong and misleading at best. A power supply has 0 ohm impedance and can supply high current at a constant voltage. A typical AA battery has internal resistance called ESR (Equivalent Series Resistance) which is vastly different than an ideal power supply.  Most electronic engineers are not very familiar with intricate details of the inter-workings of batteries and its equivalent circuit models. Batteries have an internal impedance or resistance (ESR) that plays a major role in operation of a battery in a system. To Quote from Wikipedia: In use, the voltage across the terminals of a disposable battery driving a load decreases until it drops too low to be useful; this is largely due to an increase in internal resistance rather than a drop in the voltage of the equivalent source (source: Wikipedia). In other words, when a battery is perceived to be dead in a device, it is most likely NOT because the battery is fully depleted of energy, rather it is largely due to voltage drop across this internal resistance of the battery. I will get into more details below.
3.   The third aspect may be the difference in definition of the voltages being quoted. There are two distinct ways of looking at voltages that people discuss but sometimes mistakenly interchange.   One is the Open Circuit Voltage (referred to voltage at no load condition) and the other one is the Closed Voltage Circuit (referred to voltage under load condition). The two numbers that are quoted from the CEO and your question is an example of this incorrect interchange.
To fully appreciate the totality of the picture, Let us talk about ESR (Equivalent Series Resistor). To understand how ESR interacts at a circuit level, let’s go to Wikipedia:
A practical electrical power source which is a linear electric circuit may, according to Thévenin's theorem, be represented as an ideal voltage source in series with an impedance. This impedance is termed the internal resistance of the source. When the power source delivers current, the measured voltage output is lower than the no-load voltage; the difference is the voltage drop (the product of current and resistance) caused by the internal resistance.
A battery may be modeled as a voltage source in series with a resistance. In practice, the internal resistance of a battery is dependent on its size, chemical properties, age, temperature, and the discharge current. It has an electronic component due to the resistivity of the component materials and an ionic component due to electrochemical factors such as electrolyte conductivity, ion mobility, and electrode surface area. Measurement of the internal resistance of a battery is a guide to its condition, but may not apply at other than the test conditions.
In use, the voltage across the terminals of a disposable battery driving a load decreases until it drops too low to be useful; this is largely due to an increase in internal resistance rather than a drop in the voltage of the equivalent source.
To learn more, please, visit Energizer’s technical Bulletin at http://data.energizer.com/PDFs/BatteryIR.pdf. The data from this bulletin shows that the typical effective resistance of fresh Energizer alkaline cylindrical batteries will be approximately 150 to 300 m-ohms at room temperature. It further shows that at very cold temperatures, this initial internal resistance value could be as high as 900 m-ohm, or roughly 1 ohm. This resistance further increases as the battery is used. Therefore the ESR value of a fresh new AA battery has an approximate range of 150 m-ohm to 900 m-ohm depending on temperature, chemistry, types and brands of batteries. While a resistance of 1 ohm may be considered insignificant and ignored by some people, we will see that it plays a major role in the final analysis as highlighted above in the Wikipedia description in the paragraph above.
For the sake of this analysis, assume 0.5 ohm as the internal resistance, which is a typical value for a battery at 1.3v (Open Circuit Voltage). If a device draws 400mA of current, the drop across this internal resistance is around 0.2v. This means that the device would see only 1.1V at the terminal of the battery.
The third aspect of the question is that when Batteroo’s CEO talks about batteries stop functioning at 1.3V, if you place that battery in a device described above, under load, the terminals of the battery would provide only 1.1V to the device and that is where the questioner agrees the device stops functioning. This is most likely the source of the discrepancy between what you stated as the low range of operation of devices vs. what is quoted as the open voltage of the batteries by the company.
Going back to your question that there are devices that operate at 1.1V but based on what we just showed this device will not function because it needs to have current supplied  from a battery with terminal voltage of 1.1 V. This means that the device would not be functioning properly around 1.3 V open voltage circuit (i.e. open circuit voltage of 1.3 v minus about 0.2v drop across ESR would produce a battery terminal voltage of 1.1 v seen by the device) …. Therefore the battery open voltage circuit must be 1.3 v or higher.
This is an important point which plays a significant role and has been missed and ignored. In other words, once a device with the operating point of 1.1v stops working, and the battery is pulled out and measured without load, the meter would show 1.3V. This has been a source of confusion for some people that hopefully is cleared.
To emphasize the point, ignoring the internal resistance of the battery leads to wrong conclusions. For example, some who ignored the ESR, would wrongly assume that a device that has an operating cut off voltage of 1.1V, can be serviced by a battery at 1.1V Open Circuit Voltage. It is noteworthy that the internal resistance of the battery increases to over 1.5  ohm in a non-linear fashion at room temperature (depending on many factors, and could be significantly higher at colder temperature. 
Question 2: I measured a used battery voltage using volt-meter and showed 1.2v. This toy can work all the way down to 1.1v but it is not working. Why?
Answer:  The fact that you battery is not working indicates that the closed circuit voltage is seen by the toy when is turned on is less than 1 volt. This suggests that your voltage drop across the internal resistance (ESR) is greater than 0.2 volts.
This 0.2 voltage drop is multiplication of load current and the ESR value of battery which indicated that the battery has high ESR and/or your toy is drawing a few hundred milliamps of current…
Let us say your toy is drawing 300 ma and the battery ESR is about 0.7 ohm, then your toy sees the terminal battery voltage to be under 1 volt; (i.e. open circuit of 1.2v minus the voltage drop across ESR  0.7 ohm times 300 ma is equal to closed circuit voltage of about  0.99 volt which is under 1 volt)   
Question 3: The relationship between current and voltage makes it impossible for a boosted battery to deliver such great gains, up to 8x. Besides the example below, there’s this one from Reddit: “In order to increase the voltage supplied to the target electronics, you would have to draw more current. Therefore as the battery voltage droops, the current draw will increase. Alkaline battery capacity is greatly affected by the current the cell is subjected to, with effective capacity dropping off a cliff as the current consumption increases.”
Answer:  Every system, boosted or not, has voltage and currents relationship and to say: "The relationship between current and voltage makes it impossible for a boosted battery to deliver such great gains, from 5x to 8x" is meaningless. However the most important aspect of extending the life of the battery has to do with how much of the energy left in the battery and how much of that energy can be harnessed after the battery is considered dead, depending on the specifics of devices and load conditions. Obviously to get these types of gains implies that there is significant amount of energy left is the battery as it is discarded. That could be improved by better device design that allow the circuitry to work at lower voltages, and there are devices in the market that are better than others and therefore the mileage would vary based on such  factors.
It is true that as the voltage drops, there is some increase in the current. However that current is being consumed from the energy that was in the battery and was going to be discarded at the point it was perceived to be dead.
Question 4: How does flash light benefit from Batteriser?
Answer: There are two types of devices, one with Passive load and the other with Active load:
 1.   Passive loads such as Flashlights that draw the current out of the battery until it is depleted, but the intensity of the light becomes too low for it to be useful. We measured the intensity of the light from a flashlight on a side by side comparison of, with and without Batteriser. They both started at 60 lux and after two hours, the flashlight utilizing the Batteriser maintained its 60 lux light intensity while the flashlight without Batteriser decreased its intensity to 25 lux.
2.  Active Loads such as electronic devices that usually have a cut off voltage.
Question 5: There are some well-built devices have dc-dc conversion (or similar power management) built in, so Batteriser wouldn’t help. For example, this comment was made on Reddit: “For something like a GameBoy, it actually DOES include a good switching power supply, which is why it got great runtime out of those batteries. The DMG01 was quite happy down to nearly 3V (less than 0.8V per cell, anyway).”
Answer: The question above is an actual proof of why the Batteriser is going to extend the life of the battery. The assertion that a GameBoy device is benefiting from a DC-DC and getting “great runtime” is a validation of the concept of the Batteriser. There are systems that have or may have designed these types of converters inside their electronics. Such devices are usually expensive and putting the added cost of the electronics is offset by the competitive advantage gained relative to their competition. As well, this example implies that the system must have 4 AA batteries. There are regulators in the market that would boost voltages at these levels. Batteriser uses a boost circuitry that can work down to 0.5v. You will not find any solutions in the market that allows systems with single battery boosting capabilities. Batteroo had to design a custom IC that allows boosting of the voltage from 0.5V and currents of over one Amp steady state at very high efficiency. In many systems that use one AA battery, there is no solutions in the market to provide the same benefits of extended life.
Question 6: In low-drain devices like (I assume) a TV remote, the actual shelf life of the battery will be over before the Batteriser delivers noticeable gains. For example, this comment on Macworld: “Since most batteries (excluding some lithium types) have shelf lives of say 5 years or less, then taking a low drain application scenario where the batteries will naturally last two years or more (i.e. a remote…), then boosting the battery life by the claimed 8X would mean your 5 years shelf-life battery would be “good-to-go” for up to 40 years!  Doesn’t take an engineer to tell you “it ain’t gonna happen!!”
Answer: Batteriser does work with voltage and current, regardless of the chemistry. We have seen improvement even with lithium battery. For those folks that change a battery every 5 years, we recommend not to utilize Batteriser technology.  However if you are like most of us living in an average household having 25 battery operated devices and having to continuously change batteries, Batteriser would certainly be a good choice to extend the time between battery changes by 8x depending on the end device.”?
« Last Edit: June 30, 2015, 01:15:15 pm by EEVblog »
 

Offline EEVblog

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #217 on: June 30, 2015, 01:19:19 pm »
 :-DD  :-DD
Anyone who can identify these models/actors, err, sorry, "fans", wins a lifetime supply of Batterisers!

« Last Edit: June 30, 2015, 01:21:42 pm by EEVblog »
 

Offline EEVblog

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #218 on: June 30, 2015, 01:29:56 pm »
It gets even funnier!
 

Offline Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #219 on: June 30, 2015, 04:06:13 pm »
It gets even funnier!
Australian accents add comedy to anything they touch.

 

Offline apis

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #220 on: June 30, 2015, 10:17:11 pm »
Yeah right, someone went to all the trouble to make a fan page for a startup tech, and they go around defending the product in comments, like on this video here, yeah that sounds totally legit  ::)
Yeah, those videos look totally fan made. :palm: Maybe the "fans" just accidentally hired the same pr-firm that the batterizer uses! :-DD

It's late and I couldn't be bothered reading it all, so go for it...
That is probably the point, it's important to put up a rebuttal, but they don't wan't anyone to read it too carefully or else they might spot the BS. I especially like how they claim most electrical engineers don't know about ESR.  :-DD
 

Offline EEVblog

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #221 on: June 30, 2015, 10:24:37 pm »
I especially like how they claim most electrical engineers don't know about ESR.  :-DD

The funniest thing about them claiming I didn't take into account ESR in my PSU test, is that if I did take ESR into account (i.e. the drop in the test leads and connection in the PSU case), then the results would have been better for my case and worse for their case  :palm:
By not taking the test leads and connection drop into account I was cutting them some slack. Of course, the battery discharge curves do take into account ESR, so any argument there is silly and it's clear they have no idea what they are talking about. And that's obvious to any competent engineer without having even reading their (I'm sure) waffle.
 

Offline apis

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #222 on: June 30, 2015, 10:46:21 pm »
Yeah, and if it was legit it would be so easy for them to prove it, just send out a device for testing. I would happily think it's a success if it gives only 25% extra battery-life (in most devices)! (But the typical scammer response to this is that they can't because then you will steal their trade secrets...)

EDIT:
Yes, they seem to be arguing the 1.4V cutoff voltage refers to unloaded battery voltage, but then the claim of 80% wasted energy makes no sense, just proves you were right. :-//  :palm:

And that's obvious to any competent engineer without having even reading their (I'm sure) waffle.
it's probably not intended to fool engineers, it's some complicated sounding waffle to comfort those in doubt who don't know any electronics.
« Last Edit: June 30, 2015, 11:06:41 pm by apis »
 

Offline EEVblog

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #223 on: June 30, 2015, 11:33:24 pm »
Yeah, and if it was legit it would be so easy for them to prove it, just send out a device for testing. I would happily think it's a success if it gives only 25% extra battery-life (in most devices)! (But the typical scammer response to this is that they can't because then you will steal their trade secrets...)

They claim on their Facebook page they will release the data and prove their claims.
They claim they are still compiling the data. So they've spent several years developing this, patents applications, are rpactically on the verge of selling production ready parts, have had countless time and energy to produce slick marketign campaign, but they haven't been able to compile data that proves beyond doubt that their product actually works as well as claimed?  :-//
Skeptical engineers like us are very easy to shut up, just show us the data that proves it and we'll happily admit we were wrong and that this is the best thing since sliced bread.



Quote
Yes, they seem to be arguing the 1.4V cutoff voltage refers to unloaded battery voltage, but then the claim of 80% wasted energy makes no sense, just proves you were right. :-//  :palm:

That's the only conclusion any engineer can too, they are measuring the unloaded battery voltage. In whcih case they aren't taking into account the ESR during operation, and it's a fundamentally wrong way to measure battery voltage for remaining capacity calculations.
They are running out of legs to stand on.
« Last Edit: June 30, 2015, 11:36:30 pm by EEVblog »
 

Offline EEVblog

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #224 on: July 01, 2015, 12:12:23 am »
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The third aspect of the question is that when Batteroo’s CEO talks about batteries stop functioning at 1.3V, if you place that battery in a device described above, under load, the terminals of the battery would provide only 1.1V to the device and that is where the questioner agrees the device stops functioning. This is most likely the source of the discrepancy between what you stated as the low range of operation of devices vs. what is quoted as the open voltage of the batteries by the company.
Going back to your question that there are devices that operate at 1.1V but based on what we just showed this device will not function because it needs to have current supplied  from a battery with terminal voltage of 1.1 V. This means that the device would not be functioning properly around 1.3 V open voltage circuit (i.e. open circuit voltage of 1.3 v minus about 0.2v drop across ESR would produce a battery terminal voltage of 1.1 v seen by the device) …. Therefore the battery open voltage circuit must be 1.3 v or higher.
This is an important point which plays a significant role and has been missed and ignored. In other words, once a device with the operating point of 1.1v stops working, and the battery is pulled out and measured without load, the meter would show 1.3V. This has been a source of confusion for some people that hopefully is cleared.

Holy crap!
They actually did admit that they are measuring the open circuit voltage of the battery!  :-DD  :palm:
Wow, just wow.
They are done. Their product is bunk right there, they have admitted it, no doubt about it.
If I made that newbie error I'd have to quit the industry in shame, yet they have based an entire company and product around a fundamentally wrong measurement.

All anyone has to do it look at the curves


For there example of 1.1V product voltage dropout under load, there is only about 12% of energy wasted in the battery for a 100mA constant current load.
 


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