Build a cell discharger?

[tylernt]

How crazy would I be to put together a battery holder, zener diode, and an incandescent flashlight bulb (to act as both resistor and indicator) to discharge my NiMH cells to exactly 1.1v (or perhaps 1.0 or 0.9v) before putting them in my charger?

I have a couple of single-cell flashlights but my charger only does pairs (and has no discharge feature). My hope is by discharging both cells to the same level first, they'll both reach full charge at close to the same time, thus increasing longevity. Yes, a new charger would be better but this idea sounds cheaper and more fun.

So... is this a good idea, and would it work?

 

[Light Sabre]

I have thought of doing that too. Or maybe something a little bit more complicated with an op amp and a variable resistor to manually set the voltage cutoff point. I bleed down cells when I recycle or throw them away so that there's less of a fire danger if a few batteries happened to make contact. Very unlikely, but thought I would take the extra precaution.

I have a couple BC-900's which charge batteries individually, so no charging problem here.

 

[PeAK]

How crazy would I be to put together a battery holder, zener diode, and an incandescent flashlight bulb (to act as both resistor and indicator) to discharge my NiMH cells to exactly 1.1v (or perhaps 1.0 or 0.9v)...
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You can try this simpler circuit consisiting only of a regular diode and a resistor. The only downside is that the discharge current will is on the low side (70mA to 180mA), so it will basically be an overnight discharge circuit.

 

[tylernt]

Thanks guys. Since posting I went looking at Zeners and it appears that the lowest they go is about 2v, so they're no good for this application.

Using a regular diode hadn't occurred to me though. The images are broken in the linked-to thread, so I don't know if I should wire the diode for a 0.7V forward knee voltage, or a 0.7V reverse breakdown voltage? I ask because reverse-biasing a non-Zener diode tends to destroy it but perhaps that's why current was limited to such a small amount, requiring overnight discharge.

If, on the other hand, we're relying on a diode's forward knee voltage, can I use a much lower resistance to get a nice .5C discharge rate and discharge much more quickly?

If we have to rely on reverse breakdown, could a transistor be added so that a lower-rated diode can be used merely as a "sense" to switch the transistor, which would be conducting a full .5C discharge?

I don't want to get too complex but as long as it's simple enough to hot glue the components to the side of a battery holder, we're cool.

 

[Sub_Umbra]

If you have a dead AAA-AA-C-D charger like an 8 channel Maha to build it into the whole project is a snap...

 

[TakeTheActive]

...Using a regular diode hadn't occurred to me though. The images are broken in the linked-to thread, so I don't know if I should wire the diode for a 0.7V forward knee voltage, or a 0.7V reverse breakdown voltage? I ask because reverse-biasing a non-Zener diode tends to destroy it

...that's a 2.7 Ohm resistor and a 1N4001 diode in series, connected to both poles of the cell...

Reference: Cell Recovery: Homemade Low-Cost Automatic 0.7VDC Discharge Circuit.

How do the diodes in half-wave and full-wave AC-to-DC circuits survive then? :thinking:

EXPERIMENT! Do you own a DMM/VOM? Build one circuit and measure what's happening. :tinfoil:

CLICK on my GREEN Sig Line LINK and read some of the threads with the Keywords: Definition, Theory

For example: Definition: Please explain V)olts, A)mps, R)esistance, W)atts, and C)apacity

...can I use a much lower resistance to get a nice .5C discharge rate and discharge much more quickly?

You can adjust whatever parameters you like AS LONG AS you also account for the MAXIMUM rated capacities of the components you choose.

You could also try: Cell Recovery: DIY Constant Current Load finished (pics)

Personally, if I didn't already own a BC-900 or C9000, I'd build a few:

• 0.2C (for various Capacities)
• 100mA
• ~25-50mA for DEEP DISCHARGE / breaking up large crystal formations

Have fun - be safe! :wave: :welcome:

 

[tylernt]

How do the diodes in half-wave and full-wave AC-to-DC circuits survive then? :thinking:

By never exceeding the reverse breakdown voltage, of course.

EXPERIMENT! Do you own a DMM/VOM? Build one circuit and measure what's happening. :tinfoil:

I'd love to but if I'm going to spend some money just to blow stuff up, I might as well buy a real charger. I'd kind of like to know ahead of time if what I'm building actually has some chance of working.

So with that said, how does this look?

(+) --- R1 --- D1> --- R2 --- (-)

R1 would be sized to drop about .3 volts
D1 would drop out about .7 volts (1.0v battery voltage) forward knee voltage
R2 would be sized to allow about a .5C rate for whatever cell we're discharging (2000mAh AA or 800mAh AAA)

Something like a 1N5401 looks like it would work for D1. Diodes from different manufacturers seem to have different voltage drops so R1 would probably need to be sized after measuring D1. I played around with some resistor calculators but didn't really know what I was doing so I could use some guidance on how to size R1 and R2.

R1 could also be replaced by a germanium diode which drops .3v... might be easier to calculate R2 that way.

 

[PeAK]

...
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So with that said, how does this look?

(+) --- R1 --- D1> --- R2 --- (-)
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It would looks like this

              |\ |
 plus    o____| \|___/\/\/\/\_____o  minus
              | /|             
              |/ |

The resistors can be combined into one without changing the circuit by just summing the values together. The diode can also be shifted to the right of the resistor without chaning the circuit. To analyze, you'll have to make some assumptions about the diode. This should be done so that the worst case power dissipation across the resistor is accounted for: Assume 0.7V across the diode and a battery putting out about 1.4 V nominally. The drop across the resistor would be the difference or 0.7V and for a current of 1A (0.5C), the resistor value should be about 0.7 ohms. Power dissipation should be about 1 watt.

A value of 2 ohms would allow 1/4 watt resistors to be used and allow for faster discharge at about 1/6 C.

 

[tylernt]

EDIT: Nevermind, I think I figured it out.

My only concern then is over-discharging the cell. NiMHs don't like to go below 0.9v, correct? Won't cell damage result from taking it all the way down to the diode's 0.7v?

 

[PeAK]

...My only concern then is over-discharging the cell. NiMHs don't like to go below 0.9v, correct? Won't cell damage result from taking it all the way down to the diode's 0.7v?

Good question that is not straightforward to answer. After further reading, I think the circuit should be modified to a 1n4148 diode instead of the In4001 diode to raise the diode voltage to 0.75V.

I generated a graph for the IN4148 showing the total voltage across the discharge circuit Vtotal plotted against the current I for different values of series resistors Rs equal to 1, 2 and 2.7 ohms.

If you extrapolate all the lines back towards the origin, they intercept roughly at 0.75V and 10mA. In this region the diode behaviour dominates the resistor makes little difference. As the voltage drops to the 0.75V level, all of the discharge circuits drop to about 10mA of current. An approximation to the diode

​

When you draw a horizontal line at 0.9V, the 2 and 2.7 ohm curves intercept it around 50mA. This corresponds to a discharge rate of about 1/40 C for 2000mAh LSD cell. This represents a 40 hour charge rate. If the battery had about 10% of the charge remaining, the discharge circuit would require another 4 hours to do the jobs if the current were constant. However, as the battery runs down further, the discharge rate slows even more to a lower current.

At the higher voltages, near 1.2V where a fully charged battery begins to discharge, the slope is equal to the resistance value and the resistor dominates the behaviour of the discharge characteristic. The 1 ohm curve has about 300mA of current while the 2/2.7 ohm gives a current of about 150mA.

I'll look into generating a simple discharge circuit that will discharge at a constant rate down to 0.9V and post it back under this link. Should be fairly straightforward.

P.S. If you want the EXCEL spreadsheet from which this data was generated, PM me. You can modify it to see what happens when a 1N4001 diode is used.

 

[tylernt]

It sounds like the practical upshot of all this is, the diode will prevent the total destruction of the cell, but if you accidentally left it going too long you're still going to end up at a rather low voltage. So it's best to only use it when you'll be around to keep an eye on the voltage and stop it when it gets to a desired point.

For a more foolproof design, how about one resistor connected to the base of a transistor darlington pair, which output is then connected to a power resistor. The base resistor would be small, just enough to drop 0.9v battery voltage to the transistor's natural 0.7v cutoff, and the darlington should give you a nice hard OFF at that point with only micro-current leakage after that.

EDIT: If not a darlington, then a http://en.wikipedia.org/wiki/Sziklai_pair

I think I can still find the room to hot glue four components to the side of a 1xAA or 1xAAA battery holder.

 

[PeAK]

It sounds like the practical upshot of all this is, the diode will prevent the total destruction of the cell, but if you accidentally left it going too long you're still going to end up at a rather low voltage...

I did some digging and the the 0.9V target is suggested for a 1C (i.e. 2A). At lighter loads, the end point voltage should be higher as in the case of this DIY diode discharge circuit. Take a look at an excerpt from the post below:

...The IEC standard uses the following cut off voltages...

Discharging at 0.2C, discharge down to 1.0 volts
Discharging at 1C, discharge down to 0.9 volts
Discharging at 5C, discharge down to 0.8 volts, and
Discharging at 10C, discharge down to 0.7 volts...
Tom

​

It would be better and less time to use a 2.5 ohm resistor by itself that would draw roughly a 0.2C constant current of about 400 mA (1V/2.5ohm=.4A) and to then terminate the charge when the voltage reaches 1.0 Volts under load but only with the use of a timer to monitor the progress. A comparator could fire trigger a flop at 0.9V to disconnect the charge to eliminate the need for monitoring.

The main advantage of the diode/resistor combination (over just a resistor) is, as Tylernt mentioned, to prevent the complete destruction of the NiMH cell but that the extent of how much damage is caused by a diode/resistor combination has yet to be ascertained.

 

[tylernt]

I've been playing with a circuit simulator and came up with this:

The cell's voltage is shown in blue and the discharge current in green:

Looks good on "paper", think it'll work in real life?