Author Topic: EEVblog #751 - How To Debunk A Product (The Batteriser)  (Read 2299895 times)

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MrAl

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #175 on: June 09, 2015, 09:14:09 pm »
I'm not so much familiar with battery powered device design. But looking at this product and daves video, I thought: wouldn't do a buck converter a better job in this? So basically it converts the battery voltage always down to let's say 1.1V and when the input is too low it gets bypassed.
Since the batteriser's claim is that many devices stop working at 1.35V, bucking the voltage down to 1.1V would mean the intended devices would never work in the first place now that the fresh battery voltage is lower than the dead battery threshold.

Hi,

The buck converter idea is a very interesting idea, for sure.  If a product could work with less voltage and NOT simultaneously draw more current, then the power consumption would be reduced and so the battery (now properly bucked) would last longer.  For devices with built in boost circuits it would be a waste however because they would just draw more current to make up for the voltage loss :-)

mrkev

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #176 on: June 09, 2015, 09:50:59 pm »
I'm not so much familiar with battery powered device design. But looking at this product and daves video, I thought: wouldn't do a buck converter a better job in this? So basically it converts the battery voltage always down to let's say 1.1V and when the input is too low it gets bypassed.
Since the batteriser's claim is that many devices stop working at 1.35V, bucking the voltage down to 1.1V would mean the intended devices would never work in the first place now that the fresh battery voltage is lower than the dead battery threshold.

Hi,

The buck converter idea is a very interesting idea, for sure.  If a product could work with less voltage and NOT simultaneously draw more current, then the power consumption would be reduced and so the battery (now properly bucked) would last longer.  For devices with built in boost circuits it would be a waste however because they would just draw more current to make up for the voltage loss :-)

It looks like anything you do, it's generally a bad idea to do it on the battery side. Since you have no way to know what kind of input and even V-A characteristic the device has (f.e. they often consume slightly more current on 1.5V than they would at almost discharged batt).

EEVblog

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #177 on: June 09, 2015, 10:41:53 pm »
The buck converter idea is a very interesting idea, for sure.  If a product could work with less voltage and NOT simultaneously draw more current, then the power consumption would be reduced and so the battery (now properly bucked) would last longer.  For devices with built in boost circuits it would be a waste however because they would just draw more current to make up for the voltage loss :-)

Buck converters aren't magic, they have efficiency characteristic curves too.
If you want something seamless like this, you need a SEPIC.

MyCo

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #178 on: June 09, 2015, 11:46:19 pm »
For devices with built in boost circuits it would be a waste however because they would just draw more current to make up for the voltage loss :-)

I wrote "I'm not so much familiar with battery powered device design", that doesn't mean I don't know anything about electronics. Therefor I explicitely mentioned "in linear/unregulated devices"!

Buck converters aren't magic, they have efficiency characteristic curves too.
If you want something seamless like this, you need a SEPIC.

A SEPIC would be too big in size... there are simpler buck-boost solutions. But that wasn't the point I wanted to make at all. Because the efficiency characteristic wouldn't matter that much in the buck converter idea, because the energy wasted by the buck converter would be lost anyway in linear/unregulated devices as I mentioned before.

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #179 on: June 10, 2015, 12:02:16 am »
But that wasn't the point I wanted to make at all. Because the efficiency characteristic wouldn't matter that much in the buck converter idea, because the energy wasted by the buck converter would be lost anyway in linear/unregulated devices as I mentioned before.

So what's the point of the buck converter idea then? Waste the energy in the linear reg in the product, or waste it in the efficiency of the buck converter, same thing either way.
Sure, just like the Batteriser you could find a sweet spot where it's advantageous, but overall it's going to be quite pointless.
And you still have the downside of the battery gauge always showing empty.

MyCo

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #180 on: June 10, 2015, 12:12:30 am »
Waste the energy in the linear reg in the product, or waste it in the efficiency of the buck converter, same thing either way.

Oh, c'mon Dave. You know that's not true.

And you still have the downside of the battery gauge always showing empty.

There are a lot of products that don't have a gauge, and in many others you actually don't care. And in most linear/unregulated products that's the case.

Someone

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #181 on: June 10, 2015, 01:08:51 am »
I really like the idea. It is not a scam, it is a smart BS.
Mind this is one of these devices whose guaranteed performance cannot be verified. Mostly many in average by 800% often frequently and rarely. OTOH, I could swear I have seen a reputable alkaline battery manufacturer claiming their new product means 7x improvement*.
*The asterisk is the key.
Energizer certainly promote their Lithium line as having up to 6x and 9x the life span of quality alkaline batteries, based only on the industry standard "camera" discharge test. Notable is their lack of application data for the less reputable A91 eveready gold alkaline, which could have arbitrarily high internal impedance to keep it as a low performance tier while still being an "alkaline" cell.

westfw

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #182 on: June 10, 2015, 08:49:50 am »
Quote
what's the point of the buck converter idea then?
A lot of unregulated linear devices will have ohms-law-like current consumption, drawing more current at 1.5V than at 1.2V.   If they actually operate acceptably at 1.2V, you could extend battery life.  A flashlight bulb will run longer with 1.2V cells than 1.5V cells, assuming the same WH in the actual battery and no conversion losses.   It will also run dimmer, which is where "operate acceptably" comes in.
Similarly, a motor would run longer and slower (but at more constant speed as the battery runs down), which might make sense in some toys.

Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #183 on: June 10, 2015, 09:10:46 am »
Quote
what's the point of the buck converter idea then?
A lot of unregulated linear devices will have ohms-law-like current consumption, drawing more current at 1.5V than at 1.2V.   If they actually operate acceptably at 1.2V, you could extend battery life.  A flashlight bulb will run longer with 1.2V cells than 1.5V cells, assuming the same WH in the actual battery and no conversion losses.   It will also run dimmer, which is where "operate acceptably" comes in.
Similarly, a motor would run longer and slower (but at more constant speed as the battery runs down), which might make sense in some toys.
It would probably work better as a battery extender than this 'boost' nonsense.

EEVblog

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #184 on: June 10, 2015, 09:29:01 am »
Waste the energy in the linear reg in the product, or waste it in the efficiency of the buck converter, same thing either way.

Oh, c'mon Dave. You know that's not true.

Assume 1.3V input to a 1V linear reg, with 1A output.
That's 0.3W drop in the voltage reg.
Use a buck converter to covert same 1.3V to 1V output at 1A. Assume 70% efficient converter and you still drop 0.3W in the switching reg.
Looks like I'm right in the ballpark.

PeterL

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #185 on: June 10, 2015, 10:59:23 am »
A lot of unregulated linear devices will have ohms-law-like current consumption, drawing more current at 1.5V than at 1.2V.   If they actually operate acceptably at 1.2V, you could extend battery life.  A flashlight bulb will run longer with 1.2V cells than 1.5V cells, assuming the same WH in the actual battery and no conversion losses.

His entire marketing strategy run's on the assumption that devices don't work on a 1.2V cell voltage.
But there is more, just look at a graph from his own patent.

Here he is actually stepping up a bit to 1.8V. This might not be the case in the final product, but I could imagine that a small step is actually required for the circuit to properly work.
So if you have a linear device, it could use more power right from the start.

And with this graph his also showing himself how much of the advantage of this batteriser disappears if the cut-off voltage is only dropping from 1.4 to 1.35V.

Edit:
Little (ok big) misinterpretation from my side:
He is actually comparing the time he can generate a fixed voltage against the time that a battery would otherwise be above 1.39 or 1.35 Volt. That gives him approx 7 times more battery life. I was confused by the fact that he showed a voltage of 1.8V in this graph. The text clearly states that he generates 1.5V no matter what the battery does:
Quote
A fresh AA battery provides a voltage to regulator 105 in the range of 1.5V to 1.6V. Output 102 of regulator 105 is then regulated to 1.5V,
So the difference between the text and the graph is a bit strange.

But you can also see that if your device otherwise goes down to 1V, then the batteriser will kindly half your battery life.
« Last Edit: June 10, 2015, 02:08:58 pm by PeterL »

MrAl

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #186 on: June 10, 2015, 01:40:15 pm »
Hello again,

Back to the buck idea again for a moment...

If the input is 1.5v and the output is 1.1v and the device can tolerate that, the efficiency using a linear regulator to go from 1.5 down to 1.1 would be about 36 percent better than without it assuming the load is resistive.  This is mostly because the resistive load draws less current at the lower voltage.

For the same operating conditions and load, the buck converter operating at 90 percent efficiency would raise the efficiency by about 69 percent, which is about twice that of the linear.  This is because of the two factors: the load current falls because of the lower voltage and the buck circuit provides a true power conversion.

They both fall short though as the battery voltage falls down to 1.4v, 1.3v, etc., as the efficiency would not be as good anymore.

Brutte

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #187 on: June 10, 2015, 02:34:17 pm »
I have one technical remark regarding Dave's presentation.
That's a good point.(..) In my post #174 i used a rough curve fit and then solved for the several cutoff voltages and then calculated the ratios
Another thing worth noting is that the integral of u(t) on a power=const curve does not represent any meaningful quantity (that I know of). The area is proportional to the integral of reciprocal of i(t).
The energy left that you/we were interested in is simply proportional to the distance left on the time axis on that chart - no need to integrate anything (that is the core idea of that chart anyway). I have just noticed that you follow Dave's mistake (13:00). Oops.

If you would like the area to be proportional to something meaningful, I'd suggest integrating either voltage of constant current discharge curve (then the area is proportional to the energy) or simply integrating current on a current=f(something) chart (then the area is proportional to the charge).
Still, my previous offset remark applies.

MrAl

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #188 on: June 11, 2015, 02:17:36 pm »
I have one technical remark regarding Dave's presentation.
That's a good point.(..) In my post #174 i used a rough curve fit and then solved for the several cutoff voltages and then calculated the ratios
Another thing worth noting is that the integral of u(t) on a power=const curve does not represent any meaningful quantity (that I know of). The area is proportional to the integral of reciprocal of i(t).
The energy left that you/we were interested in is simply proportional to the distance left on the time axis on that chart - no need to integrate anything (that is the core idea of that chart anyway). I have just noticed that you follow Dave's mistake (13:00). Oops.

If you would like the area to be proportional to something meaningful, I'd suggest integrating either voltage of constant current discharge curve (then the area is proportional to the energy) or simply integrating current on a current=f(something) chart (then the area is proportional to the charge).
Still, my previous offset remark applies.

Hi,

I am not sure why but you must have misinterpreted my post, or else i did not explain what i did clearly enough.

First, i never integrated anything.  All i did was do a quick curve fit of the constant power curve, then solved for specific voltage set points that are easy to relate to such as 1.4, 1.3, 1.2, and 1.1 volts.  So there was no area calculation of any kind, just one calculation for the exact voltage set point, then another calculation for the ratio of two time values in hours.
The first calculation is just like using the inverse function of the constant power discharge curve, which given a voltage, yields a time in hours.  The ratio is then the time in hours to discharge to 0.8v divided by the time to discharge to the given voltage set point (such as 1.3v).  This ratio is then a number like 1.6, and because that is a ratio that expresses the entire gain the actual gain over a normal run without any circuit would be 1, so the gain with a circuit (that works that is) would be 1.6-1 and of course this equals 0.6 and that is the percent gain in time expressed as a fraction so this would mean 60 percent more run time.
Also just to be clear, this would be with a boost circuit that is 100 percent efficient with any input voltage.  We'd have to cut this back of course for a boost circuit that ranges from 50 to 90 percent efficient, and if they got 95 percent efficiency from their circuit i'd be very surprised seeing as they dont know how to calculate anything else <little chuckle> :-)

Short form of the math i used:
For v(t) = approximate constant power discharge curve,
t = T(v), where T is the inverse function of v(t), and t in hours, v in volts,
t1=T(0.8 ), the time to discharge to 0.8v, in hours,
then the total gain ratio is:
r(t)=t1/t for any given time t, or:
r(v)=t1/T(v) for any given cutoff voltage.

I'll also point out that the curve from 0.8v to 0.6v is almost a straight line down so there is very little gain in time to add there if we went all the way down to 0.6v.

So we see ratios of about 1.0 to 12 or so, but many of these cutoff points are not realistic that's what makes this whole product seem like a bunch of bool crap :-)
« Last Edit: June 11, 2015, 02:26:50 pm by MrAl »

HighVoltage

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #189 on: June 13, 2015, 02:18:44 pm »
Today the Batteriser hit the German mainstream news:
http://www.welt.de/wissenschaft/article142416810/So-sollen-Batterien-acht-Mal-laenger-halten.html

At least they are asking some very good and critical questions.
There are 3 kinds of people in this world, those who can count and those who can not.

dizzwold

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #190 on: June 17, 2015, 08:45:43 pm »
I also wonder about the figures quoted. My trackpad '2x 1.5v AA's', drops off at about 1.1v. So realistically, we're all get more value for our money than the Batteriser states.
The one thing that i do find odd in the demo video 'used with a wireless Mac keyboard', is they use 2 batteries in series, then 2 batteries again with the devices fitted in series? If these are simple 1.5v step up converters on each battery, how do they get the required 3v. The batteries on there own, i can appreciate that, but then having a circuit on each battery, then put them in series, your surly adding a reverse polarity to 1 of those circuits.
Does that make sense, or am i having a senior moment?
How is this done?
« Last Edit: June 18, 2015, 03:14:13 pm by dizzwold »

DanielS

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #191 on: June 18, 2015, 10:15:21 pm »
The batteries on there own, i can appreciate that, but then having a circuit on each battery, then put them in series, your surly adding a reverse polarity to 1 of those circuits.
Each battery gets "boosted" to 1.5V by its own boost converter, no reverse polarity there unless one of the cells is too weak to operate its booster.

This gets really silly since:
1- boosting individual cells is horribly inefficient compared to boosting (or bucking) the total voltage only once
2- I bet the majority of semi-recent Apple and other decent quality products have their own internal buck/boost converters

dizzwold

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #192 on: June 19, 2015, 01:23:52 pm »
@DanileS,

Thanks for that, regarding 'boost the total voltage', not each cell, but;
If each cell has it's own booster, powered from the + & - of the cell, when you put 2 in series your supplying the 2nd cell with 1.5v's from the 1st cell, your also supplying 1.5v's to the ground of the second cells booster circuit?

ruffy91

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #193 on: June 19, 2015, 02:20:47 pm »
Yes. So you add 1.5V of the second booster to the 1.5V of the first booster and get 1.5V+1.5V=3V

Smokey

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #194 on: June 22, 2015, 02:10:43 am »
Yes. So you add 1.5V of the second booster to the 1.5V of the first booster and get 1.5V+1.5V=3V

Wait... wait... two batteries?!?!?!  all you have to do is add a wheel and some magnets and you get......

https://www.eevblog.com/forum/blog/eevblog-708-free-energy-bullshit!/

Where is DaveWing when you need him!

apis

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #195 on: June 25, 2015, 05:18:02 pm »
Since many cheap devices use more energy at higher voltage (eg if it has a linear regulator inside or is completely unregulated), boosting the voltage to a constant 1.5 might actually reduce battery life, add to that the losses in the converter itself.
Well designed devices will already have a custom tailored boost converter inside if there's any benefit to it. I would be surprised (and even impressed) if this thing doesn't reduce battery life in most devices. So yeah:

Also there could be performance problem because the boost regulator will introduce a lot of ripple noise, some battery operated devices rely on the low noise characteristic of a battery voltage source.

The only benefit one should hope to get from this is to squeeze a few extra drops out of an already "dead" battery: the question is if this gadget will be able to generate enough extra operating time in that case that it would be worth the trouble. Will be interesting to see some actual tests of this thing.

There really should be laws against marketing scams like this.

The problem we had is the phenomen known as dynamic energy exchange.
At a certain load, the system swapped between 2 ocillations at 2 eigenfrequencies.

As a separate system, the stability was good, but if we put them together, seems like the stabilisations were playing with eachother.
Another field, but I ask myself if it's maybe possible when putting several DCDC converters after eachother.
Interesting so they were acting sort of like coupled oscillators or some such?

UPDATE:
They now also have a video "proving" that it works by showing how the battery gauge goes up from 13% to 100% like magic when you add the batterizer! This is definitely 100% bs.

« Last Edit: June 30, 2015, 10:17:55 pm by apis »

Fungus

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #196 on: June 26, 2015, 09:37:29 pm »
UPDATE:
They now also have a video "proving" that it works by showing how the battery gauge goes up from 13% to 100% like magic when you add the batterizer! This is definitely 100% bs.

vimeo.com/130292451

"Comments have been disabled for this video..."

EEVblog

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #197 on: June 26, 2015, 10:54:30 pm »
UPDATE:
They now also have a video "proving" that it works by showing how the battery gauge goes up from 13% to 100% like magic when you add the batterizer! This is definitely 100% bs.

vimeo.com/130292451

"Comments have been disabled for this video..."

Gee, I wonder why...

EEVblog

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #198 on: June 26, 2015, 11:10:14 pm »
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?

apis

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Re: EEVblog #751 - How To Debunk A Product (The Batteriser)
« Reply #199 on: June 28, 2015, 11:43:53 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? Not sure if that's what the batteriser does but it could explain why there is no large inductor visible. Still couldn't work as advertised though since Daves debunking arguments are valid no matter how it operates.

Smf