BatteryVampire

[rthorntn]

Hi,

https://www.kickstarter.com/projects/2ndlifetech/batteryvampire/description

Using the remaining power in a battery by boosting it to a useable voltage, wouldn't it chew through these half dead batteries at an ridiculously fast rate?

Cheers
Richard

[IanB]

Yes, it would tend to do so. The spent battery will have a higher internal resistance than when fresh and the voltage boosting circuit will try to draw more current from it than the powered device would normally require (doubly so since the boost circuit is boosting a single cell up to two cell levels). So the battery will not last much longer under such a load.

Furthermore, the "energy remaining" in a spent cell must be related to the measured voltage when tested under load. Testing a spent AA alkaline cell open circuit may give a higher voltage and a false impression of the real state of discharge. Battery testers always put a load on a cell when measuring to give a more accurate estimate of this.

Overall, this product is not really going to be value for money, especially since draining alkaline cells right down to the lowest voltage possible will increase the danger of leaking. Far better to use good low self discharge rechargeable cells instead. They won't leak and will last for hundreds of charges.

[a210210200]

Eneloops are a far better idea. With one battery being effectively a spacer there is very little capacity left from the get go.

Not to mention some devices will have boost converters already and adding another one is just going to increase wasted energy.

[rthorntn]

Thanks IanB and a210210200!

Is there any simple calculations we can make, say for a LED torch that normally runs on 2 AA cells?

Say for example one alkaline AA is 3000 mAh, so two AA give you 3000 mAh at 3v.

Cree XP-G2 at 350mA would last say 5 hours and leave 1250mAh in the cells, remove one cell and insert the batteryvampire, leaving you with 625mAh, so you would get about 90 mins out of it, use the second battery and get a second 90 mins, is this correct, what is realistic given the losses in boost circuitry and the actual voltage where the LED will stop working?

Thanks again!

Richard

[IanB]

You have to do the calculation with energy (mWh), not with capacity (mAh). For instance, here is a discharge curve taken from the Energizer E91 datasheet (AA alkaline cell):

http://data.energizer.com/PDFs/E91.pdf



With a single cell discharged at 100 mW down to 1.1 V you would get about 26 hours of service. At this point, discharging the cell further down to 0.9 V would only give about 3 hours more. But that's not considering how the battery vampire works.

If you had two cells providing 100 mW each, that would be a total load of 200 mW. Now remove one cell and replace it with the battery vampire. Our single cell now has to provide all of the 200 mW by itself. But it gets worse, because the battery vampire is not 100% efficient. Let's suppose it could be 80% efficient, then the required power is 200/0.8 = 250 mW. This puts us on the 250 mW discharge curve. On this curve from 1.1 V to 0.9 V gives us just 1 or 2 hours more service.

For the small amount of extra battery life you might get it hardly seems worth the bother.

[Rufus]

You have to do the calculation with energy (mWh), not with capacity (mAh). For instance, here is a discharge curve taken from the Energizer E91 datasheet (AA alkaline cell):

A crappy flash light which just sticks an LED across a pair of cells is an example of something that might benefit from this thing. The battery is 'flat' when it doesn't have enough voltage not when it doesn't have enough energy. You are kinda fitting a switch mode converter in a flash light which would already have one if it wasn't crappy.

The money would be better spent on a non-crappy flash light.

[rthorntn]

Thanks IanB and Rufus!

[PointyOintment]

Is this not just a joule thief in a AA-sized case?

[rthorntn]

Looks pretty similar, thanks PointyOintment!

[Corporate666]

Looks pretty similar, thanks PointyOintment!

Obviously you are Richard Thornton on the comments.  One thing that stood out to me is how did the "Kate" woman who was giving you crap know that you only backed for $1?  I looked through your past comments and you never said that.  Seems like she must be talking to the creator, or perhaps more likely, it's a shill account of the creators?  Seems like most of the accounts praising were created within a short period of each other and they all say the same thing... "wow, what an amazing project, I am so proud to back this!".

And I don't really see what is different in his device than in the Joule thief.  Even the name is virtually identical... power source + word that means 'to take'.  AmpMiser, VoltGrabber, ElectronStealer, PowerScrounger, WattPlucker.  Take your pick.  Thing is - the Joule Thief is a type of circuit that has been around for quite a while, as have DC-DC converters, so I don't see what the self proclaimed non-electronics guy thinks he is patenting?  No need for secrecy... patent applications are published by law, so if he has applied for one, there should be no issue telling everyone exactly how his device works. 

However ohms law is inescapable.  Power is volts and amps... and as one goes up, the other must go down.  With the losses of DC-DC conversion and using a single depleted cell to get 3V, that means quite a lot of current draw, which drops the voltage which increases current draw which drops the voltage - a snowball effect.  Not to mention the internal resistance of a battery goes up as it discharges, and with the nature of a DC-DC converter being that you're trading amps for volts... there will be an  decreasing current capability as the original cells disharge and you use this device to get your 3V.  I mean... what if you can only get 50mA out of it from a 1.1V battery - it won't work in most devices then.

He really needs to substantiate the claim that 50% of the juice is left.  I think that's "bullshit" as Dave would say.   It should be very simple to test and demonstrate. 

[IanB]

He really needs to substantiate the claim that 50% of the juice is left.  I think that's "bullshit" as Dave would say.   It should be very simple to test and demonstrate.

I posted a chart above. If a constant power drain of 100 mW gives 30 hours service down to 0.8 V, then the 50% point would be after 15 hours, or 1.25 V.

Now some devices are badly designed and have high low-voltage cutoff points, but few are so extraordinarily bad that they stop working at 1.25 V. Most would probably go down to about 1.1 V or 1.0 V, which is about 90% of the available energy.

[rthorntn]

Corporate666, correct, it's me in the project comments.

The project is daft, he may have the best of intentions, but it's pointless, it's not green at all.

He won't defend it technically, he just raves about changing the world, which says it all.

IanB thanks for all your help!

[0b01010011]

As others have said this seems very reminiscent of the Joule Thief (http://www.bigclive.com/joule.htm).

To be honest these can be easily built of, basically, rubbish - I made one using a ferrite out of an old tablet, an old transistor, and a LED and resistor from a scrap board. I now have a torch that runs on 'dead' batteries that lasts for about  12-15hrs.

[rthorntn]

Would a better idea be to make an AA case that houses an AAA battery and a boost circuit that took the AAA up to 1.5v?

[miguelvp]

Would a better idea be to make an AA case that houses an AAA battery and a boost circuit that took the AAA up to 1.5v?

AAA are 1.5v same as AA ones, but less capacity 1200 mAh on a AAA vs 2600 mAh in a AA one

[tom66]

Maybe if it held two AAA and tied them in series, then used the vampire circuit to get the last juice out.
Or just use rechargeables. I've not found a single device that doesn't work with them.

[a210210200]

Would a better idea be to make an AA case that houses an AAA battery and a boost circuit that took the AAA up to 1.5v?

More correct maybe would be to use a special sized rechargeable lithium cell and a tiny buck converter to get 1.5V. But from a cost/energy density/safety perspective low self discharge NiMH batteries seem to be good enough for now, you would probably sacrifice too much volume for the regulator to compensate for the lost capacity.

[rthorntn]

Thanks!

[rthorntn]

Jason Romney posted something very similar to this over on the project comments (I added a few bits for clarity):

The difference (between this and a joule thief) is what the battervampire is able to power with the circuit. It's giving the output of a boost converter while being able to operate at the low input voltages of a joule thief. So it appears that at the very least he's figured out a way to bridge those two circuit designs together. At best he's invented something entirely new. Either way he's contributing something new to the world of battery life extenders.

What is he talking about?

[IanB]

Jason Romney posted something very similar to this over on the project comments (I added a few bits for clarity):

The difference (between this and a joule thief) is what the battervampire is able to power with the circuit. It's giving the output of a boost converter while being able to operate at the low input voltages of a joule thief. So it appears that at the very least he's figured out a way to bridge those two circuit designs together. At best he's invented something entirely new. Either way he's contributing something new to the world of battery life extenders.

What is he talking about?

No idea what he's talking about, but it sounds like bullshit.

To paraphrase Montgomery Scott, you cannot change the laws of physics.

Almost all batteries, including alkaline batteries, have a discharge curve that looks like a sideways S. The voltage is flattish for most of the discharge, and it falls off a cliff when the battery is nearing end of life. So it really doesn't matter what the voltage is at that point, the remaining available energy is very little. It doesn't matter what clever tricks you play with a boost converter, you cannot change a little into a lot. If the battery is depleted, it is out of stored energy. You can design a boost converter to be 100% efficient and it will do you no good. Once the stored energy is gone,  it is gone. End of story.

[rthorntn]

Thanks again IanB!

[rthorntn]

In the BatteryVampire video we see him running a radio, could his circuit be storing power and releasing it, so he is basically getting a pulsing of power, say a tenth of a second off then a tenth of a second on (apologies for the poor terminology) that could power something analog without you really noticing?

[Jeroen3]

In the BatteryVampire video we see him running a radio, could his circuit be storing power and releasing it, so he is basically getting a pulsing of power, say a tenth of a second off then a tenth of a second on (apologies for the poor terminology) that could power something analog without you really noticing?

Thats how a switchmode converter works. Add an LC filter, and you have your regulator. But, the frequency should go in the upper KHz's to have some efficiency.

My logitech m705 mouse uses two paralleled AA batteries. When the mouse stops there is about .5 Volt left, about as depleted as possible. The bonus is, it also works on 1 cell so when you only find 1 new cell in that drawer of miscellaneous stuff, it still works.

[rthorntn]

Thanks Jeroen3!