Simple 1A single-cell discharger?

[meteor]

Looking for ideas on how to build a super simple unattended discharger for my loose AA and AAA NiMh cells.

If you have kids, you will have far too may rechargeable cells lying about in all sorts of levels of discharge!

Far too often, I mix up which cells are charged or not, and end up baking them in the OEM charger, as these cheaper wall-wart style chargers (Rayovac, Energizer) seem to have a hard time with mixed level cells and can badly overheat them if you're not careful! (when I say "wall-wart", I mean the ones that have an AC plug on the back, and room for four AA or AAA cells on the front...)


I do use my Orbit 6.2 and CE Turbo 35 to discharge cells individually to 0.9V, but it's a very tedious operation! :(


Basically, I'd like to hack a wall-wart charger to have an easy cell-by-cell discharger you can just pop cells into and then have a light go off when they reach 0.9V.

Most likely, the device would have to be internally-powered, as I would not want to rely on a specific number of cells to be present to start a discharge cycle. (maybe a couple of button lithium cells?)

Also, the load need not be constant-current. A simple resistor with a FET or transistor is fine, as I don't expect to measure capacity.


Any ideas on the lowest component-count solution to detect 0.9V, and trip off the discharge circuit?

I'm after specific applications here, hopefully with a basic circuit diagram and component suggestions, not generalized suggestions! :D

PICs would also be fine, as I have a programmer, etc...

[Rodney]

For single cells, either NiCad or NiMh, it will not damage them to be discharge all the way to zero. If you are not in a hurry, just clip a 5 ohm 5 watt resistor accross the terminals and leave it sit overnight. If you have two or more cells in series, this will NOT work as you will wind up reverse charging the weaker of the cells which will most likely do them harm. Of course, if you are measureing the capacity, this will not work either as they are usually speced out by being discharged to 0.9 volts at a rate not exceeding 0.3C. However, to just discharge a single cell, it will work without any harm to the cell.

[PaulVi]

I just use a auto tail lamp with clip leads or a cell holder attached when the lamp goes out it is done.

The lamp can get hot so keep it away from things that can ignite...

[quax]

In the magazine "elektor" in Europe there was a little more complex solution as a single resistor. As I remember it was a simple multvibrator made with 2 transitors and a coil to increase the voltage to drive a LED. If the LED is off, the accu is empty.

I cannot find the schematic, but I have build it and can reengineer it, if someone is interested.

cul
quax

[meteor]

For single cells, either NiCad or NiMh, it will not damage them to be discharge all the way to zero...


From what I've read over the years, you actually should not discharge individual NiMh cells to zero volts, they can be damaged by doing so, shortening their service life.

See: http://www.hardingenergy.com/pdfs/NiMH.pdf

Extreme over discharge of a NiMH cell results in excessive gassing of the electrodes resulting in permanent
damage in two forms. First, the negative electrode is reduced in storage capacity when oxygen permanently
occupies a hydrogen storage site, and second, excess hydrogen is released through the safety vent reducing
the amount of hydrogen inside the cell. To protect against the damaging effects of over discharging, proper end
of discharge terminations must be used.


NiCd cells can usually be discharged to zero volts without any issues.




Just a FYI.

[JohnMuchow]

Also, the load need to be constant-current. A simple resistor with a FET or transistor is fine, as I don't expect to measure capacity.This requirement complicates things significantly. :)

You'll need a power supply, op-amp, reference voltage, MOSFET to act as a variable-resistor, and a current-sense resistor to create a constant-current load for the cell and a cutoff voltage detector to turn off the load when the cutoff voltage is reached. Or the on-board ADC of a PIC, a DAC, and a MOSFET or transistor to act as the variable resistor to control the discharge current. Not hard to do but IMHO overkill for discharging single cells if you're not going to be comparing them to other cells.

There are several circuits for constant-current loads on the web or you can PM me and I can send you one.

A 1N400X diode in series with a resistor that can handle the power dissipation required and limit the current thru the diode, and both across the cell, can give you anywhere from a 0.7V to 1.1V cutoff (stick under 1A). The higher the discharge current level, the higher the voltage drop across the diode and the higher the cutoff. Or use two germanium diodes in series (and in series with the resistor) to give you about a 0.8V cutoff.

[MarkusN]

This requirement complicates things significantly. :)
I think meteor forgot a "not" there. If you read the second part of the statement it becomes clear.

[meteor]

I think meteor forgot a "not" there. If you read the second part of the statement it becomes clear.

Wups! :D

Yes, I missed the "not". A resistor is perfectly fine...


To get back on track, I know the solar bug robot people use discrete 3-pin Panasonic voltage detectors in their designs, is there anything similar that could sense around 0.7-0.9V to trip off a FET/transistor?


Other ideas?



Thanks!

[meteor]

A 1N400X diode in series with a resistor that can handle the power dissipation required and limit the current thru the diode, and both across the cell, can give you anywhere from a 0.7V to 1.1V cutoff (stick under 1A). The higher the discharge current level, the higher the voltage drop across the diode and the higher the cutoff. Or use two germanium diodes in series (and in series with the resistor) to give you about a 0.8V cutoff.

Does this sugestion rely on the diode's forward voltage drop, or reverse breakdown volatge?

Also, in all cases, is there not leakage loss in the "off" state of both of the above cases? (I guess it depends on magnitude, to determine if it's relevant to this application...)

If you were to use a diode/resistor, could you have some sort of visual indicator as well? (not enough voltage for LED's, right?)


Thanks.

[JohnMuchow]

I think meteor forgot a "not" there. If you read the second part of the statement it becomes clear.Ahh, yup. But at 6:00am it wasn't so clear. :)

[JohnMuchow]

Does this sugestion rely on the diode's forward voltage drop, or reverse breakdown volatge?

Also, in all cases, is there not leakage loss in the "off" state of both of the above cases? (I guess it depends on magnitude, to determine if it's relevant to this application...)

If you were to use a diode/resistor, could you have some sort of visual indicator as well? (not enough voltage for LED's, right?)

Thanks.It depends on the forward voltage drop. When the battery voltage drops enough, the diode turns off (since the voltage across it drops to less than the diode's internal drop) and you only have the diode leakage current, which is very low.

The 11DQ05's I often use have approx a 0.35V-0.5V forward drop and essentially no forward current when the diode is off. But, the reverse leakage current ranges from 1mA to 11mA, depending on the reverse voltage. But, you wouldn't be using the diode biased this way.

Actually, you could use LEDs in series instead of a diode. That way the LEDs are a great indicator of when the discharge is still going on. But, you'd probably have to experiment a bit since LED forward v-drops can vary (IIRC approx. 1.7V each, roughly).

[orraman]

I'm after specific applications here, hopefully with a basic circuit diagram and component suggestions, not generalized suggestions! :D

PICs would also be fine, as I have a programmer, etc...

Having an 8 pin Picaxe and a latching relay already on a solderless breadboard I added a single cell alongside a dummy, a 2.4V 0.5A bulb, and an LED.

Not the lowest parts count, but one Picaxe should? handle 2 relays at the same time.

Picaxe 08M $3.30
http://www.phanderson.com/picaxe/

Relay $1
http://item.express.ebay.com/LATCHING-RELAY-NAIS-5VDC-DUAL-COIL-DPDT-10-PIN-DIP-PKG_W0QQitemZ170127877011QQihZ007QQfromsoiZ1QQcmdZ ExpressItem

2N7000 driver $0.10

I have not taken the cell down to 0.9V but on a number of runs with progressively lower ADC numbers the cut-off has worked reliably with a simplistic programme. B2>66 approximates to 1.2V on this lash-up which includes download resistors. The relay reset 'switch' is the red wire near the wire loop beside the relay lower contact. Many improvements suggest themselves, switch off battery instead of LED, averaging V, regulated PSU, bigger chips for banked cells?

start:
readadc 4,b2
debug 'to show ADC value
pause 5555
if b2>66 then start
high 1
pause 40
low 1
goto start

Dave

[JohnMuchow]

To get back on track, I know the solar bug robot people use discrete 3-pin Panasonic voltage detectors in their designs, is there anything similar that could sense around 0.7-0.9V to trip off a FET/transistor?

Other ideas?
Thanks!We've used TI's TL7700 Voltage Supervisor very successfully for detecting a voltage level and gating a transistor to turn a load on/off. Any input below 0.5V will trip the output so we just use a voltage divider on the input to dial-in the right cutoff voltage.

Using a NAND latch on the output will keep the load from turning back on as the battery voltage rises again after you remove the load. Or, have the transistor control the power to the TL700 too so it gets powered down (and the load stays off) when the cutoff is reached.