EEVblog #751 - How To Debunk A Product (The Batteriser)

#150 Reply
Posted by Galenbo on 08 Jun, 2015 07:09
Quote from: Fungus on 05 Jun, 2015 18:09
You're not supposed to put this on a new battery and use it all the time, you're supposed to put it on a 'dead' battery so you can keep on using it a bit longer.

Quote from: Wytnucls on 05 Jun, 2015 20:17
The 800% increase in battery life seems far fetched though and would have to be confirmed independently.

They state otherwise, but if you put it on when it's nearly dead, you can also start counting battery life from that moment.
In that case, their 800% increase can be true, 90 seconds instead of 10 seconds use.

#151 Reply
Posted by Wytnucls on 08 Jun, 2015 07:16
The 800% extension is possible for a hypothetical device which cuts out near the 1.5V battery voltage.
I don't have such a device, but maybe there exists one in a dark corner of the world.

#152 Reply
Posted by Galenbo on 08 Jun, 2015 07:35
Quote from: Ian.M on 06 Jun, 2015 16:23

* Sell hundreds of thousands of units to non-techies looking for Christmas presents.

thanks for predicting this, it leaves me some time for thinking about:

-How to behave if I get this as a present from an illiterate.

#153 Reply
Posted by Galenbo on 08 Jun, 2015 07:51
As many devices already have a boost converter inside, how do multiple stages of boost converters behave combined?

I have very bad expieriences with that (another tech field), especially when at least one isn't designed for it, and the load goes up and down.

#154 Reply
Posted by Ian.M on 08 Jun, 2015 08:37
Quote from: Galenbo on 08 Jun, 2015 07:35
Quote from: Ian.M on 06 Jun, 2015 16:23

* Sell hundreds of thousands of units to non-techies looking for Christmas presents.

thanks for predicting this, it leaves me some time for thinking about:

-How to behave if I get this as a present from an illiterate.
Its still better than novelty socks - at least you can send it to Dave for a teardown.

#155 Reply
Posted by EEVblog on 08 Jun, 2015 09:21
Quote from: Galenbo on 08 Jun, 2015 07:51
As many devices already have a boost converter inside, how do multiple stages of boost converters behave combined?

You get twice the efficiency losses. And the DC-DC in the products usually present a constant power load, which the preceeding DC-DC (Batteriser) might not be as efficient at.

#156 Reply
Posted by MrAl on 08 Jun, 2015 12:12
Quote from: Mybetoostnedforths on 07 Jun, 2015 21:25
Hi from Sweden!
I just thought of something crazy aight.
Consider this
If applying the gizmo to your battery provides constant voltage at 1.5, by converting the stuff.
And if as Dave says, screws up battery-indicators,
wouldn't that mean that the indicator would just go to 100% (as in 1.5 volt), making it seem like it's full? As seen in their marketing video at 26 seconds.
That'd fool consumers into thinking it actually works, since it's full. Atleast for a while.

But hell, I'm just a very-low-time electronics hobbyist, and I sure as hell could just be wrong.
Regards
Mybetoostnedforths

Hi,

Yes, but to be fair, if you got 800 percent run time out of a battery you normally get only 100 percent run time (say 8 hours as compared to 1 hour) you would not care if the battery gauge was not correct anymore.

But to be more reasonable, it could be only 200 percent run time (2 hours instead of 1 hour for example) and that is if your home product cuts out normally at 1.3v, which is not typical at all. So in some strange device it might actually give you 2 hours run time instead of 1, and then you would not care as much about the battery gauge, even though you might miss it. That's if the battery gauge works right in the first place which some dont anyway.

So i think the battery gauge issue is just a small side issue, which would not matter if the device worked as advertised or even just 100 percent better than without it. But there dont seem to be many products that we will see a benefit for, unfortunately.

#157 Reply
Posted by EEVblog on 08 Jun, 2015 12:23
Quote from: MrAl on 08 Jun, 2015 12:12
Yes, but to be fair, if you got 800 percent run time out of a battery you normally get only 100 percent run time (say 8 hours as compared to 1 hour) you would not care if the battery gauge was not correct anymore.

Until it totally pisses you off when it dies on you unexpectedly. Personally I'd rather know were I stand with the battery level rather than have to guess/remember when I put the battery in last and how many hours it's been running for.
Murphy will ensure it dies at the most inconvenient time.

Quote
So i think the battery gauge issue is just a small side issue

I think it's a complete show-stopper for many people. Perhaps not if you only used it to get a bit more juice on an otherwise flat battery.

#158 Reply
Posted by RupertGo on 08 Jun, 2015 14:42
Battery gauges on many devices aren't that accurate anyway, and perhaps you could design the device to roll off its output with a bit of time to spare to give the user some warning, but I'm not sure if that's practicably possible given off-the-shelf converter chips and the limitations of the form factor. Especially given a dynamic load.

Something else that could be worrying: if you're using this thing on an NiMH cell and you put it and the cell into a charger, what happens? Do these circuits have protection against seeing a voltage like that on their output? How would the charger react? You probably don't want the sort of product liability issues that can come from mucking about with stuff like that unless you've designed for those cases.

#159 Reply
Posted by AG6QR on 08 Jun, 2015 16:06
If you really want to see some crazy claims, look at the founder's interview on CNN Money

http://money.cnn.com/2015/06/02/technology/make-battery-last-longer-batteriser/?sr=cnnmoneybin060615battery0900story

For example, a typical AA battery will stop working after 240 minutes of use powering a remote control, 95 minutes powering portable speakers, or just 38 minutes powering an RC toy. Roohparvar claims that the Batteriser can get 1,185 minutes out of a remote (5 times more energy), 570 minutes out of portable speakers (6x) or 355 minutes out of an RC toy (9x).

If you read the comments on that CNN Money story, you'll notice that the readers of CNN Money, on average, aren't as technically astute as those of EEVblog, but we knew that.

He doesn't have a product. He has an idea that he's pitching for crowdfunding investors. That's why he's promoting the story in CNN Money. On his website, he says his product will reduce the amount of battery waste going into landfills by a factor of eight. That's a very bold claim, and in order to meet it, his device would have to produce that claimed 8x improvement when averaged out over all devices that use alkaline batteries -- not just once in a while on the most special cases.

Not a chance.

#160 Reply
Posted by rr100 on 08 Jun, 2015 16:39
Dave, are feeling the load? You've got slashdotted!

I find it funny that for basic EE stuff like this the quality of discussion isn't that high on slashdot. I say that because I've been very often surprised by the collective knowledge of slashdot on anything slightly technical, from atomic power plants to 1800's technology, of course IT, etc. But they've got nothing, NOTHING on eevblog forum for EE stuff.

#161 Reply
Posted by MrAl on 08 Jun, 2015 17:19
Quote from: EEVblog on 08 Jun, 2015 12:23
Quote from: MrAl on 08 Jun, 2015 12:12
Yes, but to be fair, if you got 800 percent run time out of a battery you normally get only 100 percent run time (say 8 hours as compared to 1 hour) you would not care if the battery gauge was not correct anymore.

Until it totally pisses you off when it dies on you unexpectedly. Personally I'd rather know were I stand with the battery level rather than have to guess/remember when I put the battery in last and how many hours it's been running for.
Murphy will ensure it dies at the most inconvenient time.

Quote
So i think the battery gauge issue is just a small side issue

I think it's a complete show-stopper for many people. Perhaps not if you only used it to get a bit more juice on an otherwise flat battery.

Hello again,

Well, from personal experience with lots of battery operated devices, if i really did get 800 percent run time (to be clear that is 700 percent additional run time) i would easily accept giving up the battery gauge unless i could get a device that allowed it to still function properly and still get 800 percent. With only 200 percent (100 percent additional) i would have to think about it, but for me 100 percent additional time could be very useful, so i might go with it anyway.

As others have and will point out, battery gauges dont work that well anyway most of the time, or at least not with the devices i have. For example, my camera tells me 2 or 3 pictures before it dies, which i often miss anyway because i dont want to have to keep looking at the battery gauge. So my remedy is to carry a spare battery or set of batteries with me, especially if i intend to do a lot of shooting. That's the only way anyway. But yes, having the battery gauge could save that special shot that we miss otherwise, if we are diligent about watching the gauge, which we might be if every shot was very important. For me i guess my shots are not usually that important anyway (mostly still shots for important record keeping), but i could easily see that being a determining factor for sports fans.
What i always intended to do was to count the shots between recharges, and then check the picture number meter (numbers 0001 to 9999 i think) and that way i would know if i shot (say) 100 pictures it was time to change batteries, but i never got diligent enough to do this yet. Maybe some day :-)
Luckily this camera takes Li-ion so i just pop in one battery to get going again.

I am already a little pissed off that they had the nerve to state "800 percent" <chuckle>
But luckily they were not too smart about picking that number as i think that tips people off right away that something probably isnt right here.

#162 Reply
Posted by MyCo on 08 Jun, 2015 18:41
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. That way you could get some more lifetime in linear/unregulated devices by using the peak energy (when the battery is full, and the voltage is high) over a longer time instead of drawing more current than necessary.

#163 Reply
Posted by Herbert K. on 08 Jun, 2015 19:38
Hi,

i'm sorry if somebody else mentioned it already and i'm wasting your time, but i didn't read the whole thread, i'm just too tired right now.

If you leave a depleted alkaline cell alone, the voltage will soon recover to about 1.35V (measured only with a typical high impedance voltmeter) But if you put any load what so ever on that cell it will crap out instantly.

I think the "inventors" measured the cell voltage of a dead alkaline without any LOAD and therefore are saying most devices use the cells only down to 1.35V.

When i was a kid, i had the same idea and made the same mistake.

#164 Reply
Posted by EEVblog on 09 Jun, 2015 01:41
Here is some great data on used battery capacity:
http://www2.ife.ee.ethz.ch/~rolfz/batak/ICBR2003_Zinniker.pdf
About 33% wasted is their figure.
Perhaps Batteroo would be kind enough to publish their data?

#165 Reply
Posted by EEVblog on 09 Jun, 2015 01:42
Quote from: Herbert K. on 08 Jun, 2015 19:38
If you leave a depleted alkaline cell alone, the voltage will soon recover to about 1.35V (measured only with a typical high impedance voltmeter) But if you put any load what so ever on that cell it will crap out instantly.
I think the "inventors" measured the cell voltage of a dead alkaline without any LOAD and therefore are saying most devices use the cells only down to 1.35V.

That's the only conclusion I can come to as well.

#166 Reply
Posted by Poe on 09 Jun, 2015 02:52
Quote from: EEVblog on 09 Jun, 2015 01:41
Here is some great data on used battery capacity:
http://www2.ife.ee.ethz.ch/~rolfz/batak/ICBR2003_Zinniker.pdf
About 33% wasted is their figure.
Perhaps Batteroo would be kind enough to publish their data?

From the paper also, 10% were thrown away perfectly new.

Both figures can easily be attributed to just just human factors and multiple cell devices. The weak cell dies before the others get to zero and the user throws them all out.

I'm curious how much can be attributed to that, how much is due to an abnormally high device voltage cutoff, or simply a chemical battery recovery issue.

How much usable energy 'comes back' if a battery is discharged hard (like modern cameras) then allowed to 'recover'?

#167 Reply
Posted by EEVblog on 09 Jun, 2015 06:33
Quote from: Poe on 09 Jun, 2015 02:52
From the paper also, 10% were thrown away perfectly new.

Yes, the data will be skewed in terms of being used to backup the Batteriser claims. But it at least puts an upper number on it.

#168 Reply
Posted by EEVblog on 09 Jun, 2015 06:34
Quote from: Poe on 09 Jun, 2015 02:52
How much usable energy 'comes back' if a battery is discharged hard (like modern cameras) then allowed to 'recover'?

None. Energy does not and can not magically "come back". The cell voltage is the thing that "comes back".

#169 Reply
Posted by Wytnucls on 09 Jun, 2015 08:16
The cell I had discharged to 0.6V jumped back to 1.06V overnight.
A further discharge with the same load as before gave a start current of 115mA at 0.7V, ending at 90mA when it reached 0.6V again, after 4 minutes.
So, there is very little energy recovery in the chemical process (~7%), at that discharge level.

#170 Reply
Posted by Galenbo on 09 Jun, 2015 09:30
Quote from: EEVblog on 08 Jun, 2015 09:21
Quote from: Galenbo on 08 Jun, 2015 07:51
As many devices already have a boost converter inside, how do multiple stages of boost converters behave combined?

You get twice the efficiency losses. And the DC-DC in the products usually present a constant power load, which the preceeding DC-DC (Batteriser) might not be as efficient at.

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.

#171 Reply
Posted by MrAl on 09 Jun, 2015 14:12
Quote from: Poe on 09 Jun, 2015 02:52
Quote from: EEVblog on 09 Jun, 2015 01:41
Here is some great data on used battery capacity:
http://www2.ife.ee.ethz.ch/~rolfz/batak/ICBR2003_Zinniker.pdf
About 33% wasted is their figure.
Perhaps Batteroo would be kind enough to publish their data?

From the paper also, 10% were thrown away perfectly new.

Both figures can easily be attributed to just just human factors and multiple cell devices. The weak cell dies before the others get to zero and the user throws them all out.

I'm curious how much can be attributed to that, how much is due to an abnormally high device voltage cutoff, or simply a chemical battery recovery issue.

How much usable energy 'comes back' if a battery is discharged hard (like modern cameras) then allowed to 'recover'?

Hi there,

No energy really 'comes back', it is all there to begin with, just hiding a little.

In the more advanced electrical battery models there is a secondary storage component, a capacitor, that is smaller than the primary storage component which is a larger capacitor, and the resistance between the smaller cap and the larger cap is larger than the resistance from the battery terminal to the larger capacitor.
What this means is that when the battery is discharged, the larger capacitor voltage is what we measure when we measure the terminal voltage, while the smaller capacitor still has a slightly higher voltage across it because it does not discharge as much as the main capacitor due to the higher resistance.
When the load is finally disconnected, the smaller capacitor slowly recharges the larger main capacitor, and so we see a voltage rise across the battery terminals.
So the battery really looks sort of like two batteries, but they are not directly in parallel they are separated by a resistance that is larger than the so called series resistance of the battery, and the second smaller battery can charge the bigger main battery a little when the load is disconnected.
You might note however that the smaller battery gets it's energy from the regular charge process too just like the larger battery when the whole battery is charged normally, so there is no extra energy coming from anywhere.
This is still using only a simplified model with only two storage elements but that illustrates the main idea from a purely electrical perspective.
There are various battery models available on the web.

To get back to the main subject of the energy left in a battery when it is disconnected before all it's energy is depleted, here is a quick table of cutoff voltages and the total percent that could be attained if the rest of the energy was used too, compared to what is actually used up to that point...

1.4v 2.3hr 12.2x
1.3v 7.6hr 3.7x
1.2v 17.3hr 1.6x
1.1v 24.7hr 1.1x
1.0v 27.8hr 1.0x

For example in the above, if the cutoff voltage is 1.1v then we would only gain around 10 percent more if we depleted it all the way. If the cutoff voltage was 1.2v however then we would gain about 60 percent more run time.
These numbers come from a typical battery curve for a 100mw load but are not intended to be super accurate, just for comparison to get a feel for how important the voltage cutoff point really is.

#172 Reply
Posted by Brutte on 09 Jun, 2015 14:47
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*.

I have one technical remark regarding Dave's presentation.
When you compare integrals, the surfaces confined by some curves and horizontal axis, and the curves have an offset on the plot (as with the voltage=f(time) under load which was offset by 0.8V on that chart, 13:00), be aware not all of your viewers understand the marked region spans down to 0x axis. I feel that some people would think the energy|charge left in a cell is proportional to what you have covered with the marker, while it is not.

*The asterisk is the key.

#173 Reply
Posted by DanielS on 09 Jun, 2015 15:25
Quote from: MyCo on 08 Jun, 2015 18:41
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.

#174 Reply
Posted by MrAl on 09 Jun, 2015 21:09
Quote from: Brutte on 09 Jun, 2015 14:47
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*.

I have one technical remark regarding Dave's presentation.
When you compare integrals, the surfaces confined by some curves and horizontal axis, and the curves have an offset on the plot (as with the voltage=f(time) under load which was offset by 0.8V on that chart, 13:00), be aware not all of your viewers understand the marked region spans down to 0x axis. I feel that some people would think the energy|charge left in a cell is proportional to what you have covered with the marker, while it is not.

*The asterisk is the key.

Hi,

That's a good point. If the graph does not show anything below some voltage like 0.8 then we are not seeing the entire area, so we might envision the wrong area ratios.
In my post #174 i used a rough curve fit and then solved for the several cutoff voltages and then calculated the ratios, with all calculations using the full curve all th way down to the horizontal axis.