How do I intentionally drain my AA batteries?
December 19, 2011 2:46 AM Subscribe
What is the best way to regularly drain 6-8 AA batteries?
For a Christmas gift, I've been asked to make a battery drainer. The specifications I've been given is that it must drain either 6 or 8 AA batteries (As in, it needs to have the option to do either 6 or 8). I have some experience soldering things together, and some parts already.
So with that, what is the best way to quickly and safely drain 6-8 batteries? I suspect a bunch of LEDs or resistors would work, but for some reason the theory behind it to be certain eludes my mind at the moment (regrettably).
For a Christmas gift, I've been asked to make a battery drainer. The specifications I've been given is that it must drain either 6 or 8 AA batteries (As in, it needs to have the option to do either 6 or 8). I have some experience soldering things together, and some parts already.
So with that, what is the best way to quickly and safely drain 6-8 batteries? I suspect a bunch of LEDs or resistors would work, but for some reason the theory behind it to be certain eludes my mind at the moment (regrettably).
A dead short will drain them the fastest, but you're on the right track with the resistor idea. The role of the resistor would actually be to slow the discharge enough to keep things safely cool to avoid rupturing batteries or melting insulation. So, you're looking for just enough resistance to keep the temperature down. You could get there with a single low-ohm, high-wattage resistor, or a bunch of flashlight bulbs in parallel, or whatever.
I don't know enough about batteries to tell you how much current alkaline batteries can safely deliver. If you can find that information, designing the circuit should be easy.
posted by jon1270 at 3:14 AM on December 19, 2011
I don't know enough about batteries to tell you how much current alkaline batteries can safely deliver. If you can find that information, designing the circuit should be easy.
posted by jon1270 at 3:14 AM on December 19, 2011
Best answer: No battery will come to harm being drained at an 8 hour rate, and an AA cell is good for roughly 1 amp-hour. So you'd want to drain it at 125mA.
Since the only conceivable point of a battery drainer is to discharge NiCd cells fully to get rid of memory effects, and since a NiCd cell's nominal terminal voltage is about 1.2V, and since Ohm's Law says that the voltage across a resistor in volts is equal to its resistance in ohms times the current through it in amps, a resistor of 1.2V / 0.125A = roughly 10 ohms would do the job.
Power dissipated in a device in watts is current through it in amps times voltage across it in volts, which in this case is 1.2V × 0.12A = 0.14W. So in theory an ordinary quarter-watt carbon film resistor would be fine. But use half-watt resistors to give yourself some safety margin.
So if you get yourself a couple of four-cell AA battery holders and eight 10-ohm half-watt resistors, and solder one resistor across the terminals for each individual cell, then any cell placed in any holder will drain nicely flat overnight. The fact that the battery holders also connect the cells together will be of no consequence as long as you don't let resistor pigtails short-circuit to each other.
If you want to drain the cells a little more aggressively (say, at a 4 hour rate) then you'd go for 5.6 ohm resistors; these would dissipate 1.22 / 5.6 = 0.26W each, so half-watt resistors would still be fine. They'd run a little warm but even unventilated they should survive OK.
posted by flabdablet at 4:07 AM on December 19, 2011
Since the only conceivable point of a battery drainer is to discharge NiCd cells fully to get rid of memory effects, and since a NiCd cell's nominal terminal voltage is about 1.2V, and since Ohm's Law says that the voltage across a resistor in volts is equal to its resistance in ohms times the current through it in amps, a resistor of 1.2V / 0.125A = roughly 10 ohms would do the job.
Power dissipated in a device in watts is current through it in amps times voltage across it in volts, which in this case is 1.2V × 0.12A = 0.14W. So in theory an ordinary quarter-watt carbon film resistor would be fine. But use half-watt resistors to give yourself some safety margin.
So if you get yourself a couple of four-cell AA battery holders and eight 10-ohm half-watt resistors, and solder one resistor across the terminals for each individual cell, then any cell placed in any holder will drain nicely flat overnight. The fact that the battery holders also connect the cells together will be of no consequence as long as you don't let resistor pigtails short-circuit to each other.
If you want to drain the cells a little more aggressively (say, at a 4 hour rate) then you'd go for 5.6 ohm resistors; these would dissipate 1.22 / 5.6 = 0.26W each, so half-watt resistors would still be fine. They'd run a little warm but even unventilated they should survive OK.
posted by flabdablet at 4:07 AM on December 19, 2011
Just thinking about it a little more: if somebody were silly enough to put 1.5V alkaline cells in a drainer you made up with 5.6 ohm resistors, each resistor would initially dissipate roughly 1.5V2 / 5.6Ω = 0.4W, which is uncomfortably close to the line for an unventilated half-watt resistor; probably better to use 1W resistors if you're going to make a four-hour drainer.
posted by flabdablet at 4:21 AM on December 19, 2011
posted by flabdablet at 4:21 AM on December 19, 2011
I just have to ask why? I concur with the advice above. I would maybe use some old (ie incandescent) flashlight light(s) instead (or in series with) a boring 'ol resistor, then when the light is out, you know they're dead!
posted by defcom1 at 4:43 AM on December 19, 2011 [2 favorites]
posted by defcom1 at 4:43 AM on December 19, 2011 [2 favorites]
Best answer: I just have to ask why?
You could wire both 4-cell holders in series, then use a single resistor or a 12V automotive bulb to drain the string. But that's not as good as using one resistor per cell, for several reasons:
LEDs have a minimum voltage that needs to be reached before they will conduct at all (roughly 1.7V for a red one, 2.2V for green, can't remember but higher for blue) and you probably want at least 5mA through a LED to make it light up and no more than about 20mA to stop it blowing.
So if you've got a string of 6 cells in place, and each one is 1.2V, that's 7.2V in total; less 1.7V forward voltage for a red LED is 5.5V appearing across the LED's current-limiting resistor. If that's a 1k resistor, we get 5.5mA through it which is about right. Power dissipation will not be an issue for that resistor; any 1k resistor would do.
Six cells in place means there will be two empty cell bays, each with a 5.6Ω resistor across it. That will add a negligible 11Ω to the 1k current limiter.
Biggest possible LED current would happen if the drainer was fitted with 8 fresh alkaline cells by an idiot. That yields a total battery voltage of 1.5V × 8 = 12V; less the 1.7V LED forward voltage gives 10V across the current limit resistor; at 1kΩ that's 10mA LED current, which is still fine.
NiCd cells maintain their terminal voltage until very nearly fully discharged, so the LED should shine for most of the discharge cycle. However, discharge would continue even after the LED goes out completely.
The maximum possible reverse voltage across any empty cell bay would be the series-string LED current (under 10mA) multiplied by the resistance across the bay (5.6Ω) which comes out under 0.06V, not enough to cause reverse-charging damage to any cell (see list point 3 above).
posted by flabdablet at 5:11 AM on December 19, 2011
You could wire both 4-cell holders in series, then use a single resistor or a 12V automotive bulb to drain the string. But that's not as good as using one resistor per cell, for several reasons:
- One resistor per cell drains any cell you put in any holder, so the 6- or 8-cell requirement is fulfilled in a completely configuration-free manner.
- Heat causes less trouble if the source is not concentrated.
- Draining a series string of cells with unequal states of charge is bad for the lesser-charged cells, because they will spend some time completely flat but with current still flowing through them from the other cells; that current will be in the opposite direction to normal charging current, and this can do Bad Things to cell chemistry.
LEDs have a minimum voltage that needs to be reached before they will conduct at all (roughly 1.7V for a red one, 2.2V for green, can't remember but higher for blue) and you probably want at least 5mA through a LED to make it light up and no more than about 20mA to stop it blowing.
So if you've got a string of 6 cells in place, and each one is 1.2V, that's 7.2V in total; less 1.7V forward voltage for a red LED is 5.5V appearing across the LED's current-limiting resistor. If that's a 1k resistor, we get 5.5mA through it which is about right. Power dissipation will not be an issue for that resistor; any 1k resistor would do.
Six cells in place means there will be two empty cell bays, each with a 5.6Ω resistor across it. That will add a negligible 11Ω to the 1k current limiter.
Biggest possible LED current would happen if the drainer was fitted with 8 fresh alkaline cells by an idiot. That yields a total battery voltage of 1.5V × 8 = 12V; less the 1.7V LED forward voltage gives 10V across the current limit resistor; at 1kΩ that's 10mA LED current, which is still fine.
NiCd cells maintain their terminal voltage until very nearly fully discharged, so the LED should shine for most of the discharge cycle. However, discharge would continue even after the LED goes out completely.
The maximum possible reverse voltage across any empty cell bay would be the series-string LED current (under 10mA) multiplied by the resistance across the bay (5.6Ω) which comes out under 0.06V, not enough to cause reverse-charging damage to any cell (see list point 3 above).
posted by flabdablet at 5:11 AM on December 19, 2011
If I'm right about the motivation, it's not for shits and giggles; it's for extending the useful life of rechargeable cells. May or may not be effective but certainly not horrible.
posted by flabdablet at 5:36 AM on December 19, 2011
posted by flabdablet at 5:36 AM on December 19, 2011
Along with the LEDs, why not defer the rest of the current into a motor? You could have a blinky rotating xmas tree that lasts only 10 minutes. (thanks to this, I have a scene from Goonies playing in my head. "Oh Data....only problem is the batteries don't last so long")
posted by samsara at 5:37 AM on December 19, 2011
posted by samsara at 5:37 AM on December 19, 2011
why not defer the rest of the current into a motor?
Because the one thing that will absolutely destroy any conceivable value in a deliberate cell flattener is reverse charging current. You want most of your discharge current flowing in the individual discharge resistors, and as little flowing around the series string as possible. 5-10mA for a LED is probably fine, but 50-100mA through a motor could do damage.
posted by flabdablet at 5:51 AM on December 19, 2011
Because the one thing that will absolutely destroy any conceivable value in a deliberate cell flattener is reverse charging current. You want most of your discharge current flowing in the individual discharge resistors, and as little flowing around the series string as possible. 5-10mA for a LED is probably fine, but 50-100mA through a motor could do damage.
posted by flabdablet at 5:51 AM on December 19, 2011
OP never states what kind of batteries he's talking about. While we can probably assume he's talking about rechargables, without more information I think some confusion is understandable.
what is the best way to quickly and safely drain 6-8 batteries? [emphasis mine]
I know you want to build something, but IMO the answer to that is two of these since it only does 4 at a time. I checked and Maha doesn't seem to make an equivalent device with more slots. Another manufacturer might, but I own this particular model and love it and it is awesome, and it does way more than just drain them. You can probably find them for around $50 if you look.
posted by Edogy at 6:02 AM on December 19, 2011
what is the best way to quickly and safely drain 6-8 batteries? [emphasis mine]
I know you want to build something, but IMO the answer to that is two of these since it only does 4 at a time. I checked and Maha doesn't seem to make an equivalent device with more slots. Another manufacturer might, but I own this particular model and love it and it is awesome, and it does way more than just drain them. You can probably find them for around $50 if you look.
posted by Edogy at 6:02 AM on December 19, 2011
In terms of just draining them, I would just get a bunch of flashlights powered by one AA battery, put one battery in each, and turn them on until each battery dies. Cheap, safe, effective.
posted by thegears at 6:10 AM on December 19, 2011 [2 favorites]
posted by thegears at 6:10 AM on December 19, 2011 [2 favorites]
I assume you are discharging rechargeable batteries - if so, you don't want them to go down below a certain point.
Google up battery discharge trays - that's what the r/c hobbyists used to bring cells down without getting the voltage down to zero.
Example:
http://sarconastic.tripod.com/equaliser.html
posted by TrinsicWS at 6:55 AM on December 19, 2011
Google up battery discharge trays - that's what the r/c hobbyists used to bring cells down without getting the voltage down to zero.
Example:
http://sarconastic.tripod.com/equaliser.html
posted by TrinsicWS at 6:55 AM on December 19, 2011
Those LED flashlights are going to take a lot longer than 4 hours to run down an AA cell. Plus, eight of those at $3.79 each comes to thirty bucks. If you build it the way I suggested, it will cost you $1.72 for a battery holder and 85 cents for the resistors. Adding an indicator LED would cost another 19 cents plus 65 cents for a packet of 1k resistors.
posted by flabdablet at 7:14 AM on December 19, 2011
posted by flabdablet at 7:14 AM on December 19, 2011
you don't know what you're missing with the Maha smart chargers that auto discharge and recharge for you
Shortened battery life?
A full discharge is not something you should do every time.
posted by flabdablet at 8:08 AM on December 19, 2011
Shortened battery life?
A full discharge is not something you should do every time.
posted by flabdablet at 8:08 AM on December 19, 2011
Since we still don't know more about what the OP's actual goals are, or what kinds of battery he's using, I'm going to assume we're talking about today's most common format, NiMH, which have different characteristics from the NiCd type refered to in your link.
posted by Edogy at 8:39 AM on December 19, 2011
posted by Edogy at 8:39 AM on December 19, 2011
Same advice holds. Deep discharge should be occasional.
posted by flabdablet at 9:02 AM on December 19, 2011
posted by flabdablet at 9:02 AM on December 19, 2011
Ah, I originally read this as if it should drain alkaline batteries wastefully as a novelty...but for prolonging the life of rechargeables flab and trinsic are more spot on.
posted by samsara at 9:57 AM on December 19, 2011
posted by samsara at 9:57 AM on December 19, 2011
This thread is closed to new comments.
I'm not 100% sure if this checks out technically, but it's probably in the direction.
You can also consider adding a LED in series with the resistor, one that can accept 100mA over it ... If that even exists. It may be a little high for an LED.
posted by krilli at 3:14 AM on December 19, 2011