Battery discharger - will this work?

The circut is here:

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The idea is to discharge a NiMH AA cell at 500ma down to the point where the cell voltage is 1V. I picked the specific parts because I already have them in my junque bin.

The idea is that R1 is the sense resistor which should have a voltage across it of 0.25V if 500ma is passing through it. Op amp B controls the pass transistor such that the current stays constant at that level.

Opamp A is supposed to keep the cell voltage from dropping below 1V.

I'm not very experienced at this, and would appreciate comments on whether this will work. I am most curious about:

  1. What will happen as the discharge voltage approaches 1V. As the output voltage of Opamp A drops, the current will be reduced. But when the battery load is reduced, the voltage will go back up again. So I wonder if it will start to oscillate, which I don't want.

  1. I have a blind spot with PNP transistors. I have R2 in the curcuit to limit the current that Opamp B tries to sink out of the transistor base. But I think maybe I don't need that resistor. And if so, it would be better without it since the opamp may have some trouble driving the transistor in the first place. So I guess the question is - if the voltage on the base is lower than the voltage on the collector, will current flow through the base?

And if it matters, there will be another duplicate circuit for the other cell, which will share the same quad opamp and 1V reference.

Thanks for any suggestions.

Reply to
George
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Well, the more I looked at it, the less I liked the PNP setup. So I found a couple NPN transistors on an old board, and have re-drawn it (v2) to use them. So that version is now the first one on the drop.io site - the brighter one.

Reply to
George

Why don't you just have one opamp turn off the transistor?

Tom

Reply to
Tom Biasi

Right. You might add hysteresis, or a pushbutton "start" thing, and have it shut off when it first hits 1 volt.

The NPN version is less likely to have loop stability problems. You might also add a resistor between R1 and the opamp B inv input (say,

5K) and a cap between opamp output and the inv input, .01 uF maybe. That will give the opamp local high-frequency feedback and prevent oscillations.

Make sure the opamp can supply enough base current.

John

Reply to
John Larkin

I don't understand what you mean.

Reply to
George

So, hysteresis would be a resistor between the + input of opamp B and the output? Positive feedback? Let me think about that. The way I have it now, I think it will discharge at 500 ma until it gets near 1V, and then it will discharge at ever decreasing currents until the unloaded voltage is down to 1V, which I think is where I want to be.

I know how to do a pushbutton start thing with a relay. Well, and with an SCR, but I don't have any of those. If you meant something different, could you link me to an example?

Ok.

Yes, I think the CMOS opamp may not do that. I'll look for another one. Maybe even an LM324 would do better at sourcing.

Thanks for your comments.

Reply to
George

Could you monitor the battery voltage with one op-amp and turn off the NPN when voltage is 1 volt?

Tom

Reply to
Tom Biasi

As I said originally, I don't have a lot of experience designing circuits. I guess I have to say I don't know how I would do that and still have a constant current discharge.

Reply to
George

Something like this maybe...

ftp://jjlarkin.lmi.net/Push.JPG

John

Reply to
John Larkin

Ok. Thanks very much.

Reply to
George

--
What I\'d do would be to detect the 1V and use that to set a latch which
would turn off the first opamp which would turn off the current sink.  
 Click to see the full signature
Reply to
John Fields

Thanks very much for the suggestion. I've drawn it out, and understand what you've done.

By the way, for future reference, one of the advantages of drop.io is that you can configure the drop so anyone can add content. So instead of spending time on ASCII drawings, you can just sketch it out, scan it or take a picture, and upload the jpeg to the same drop.

But back to the circuit:

Is the power-up state of the latch predictable? Well, if you power up without the battery in place, then I assume it would power up in the set state because that input would be low. But if you power up with the battery already in the ciruit, then it seems it could power up in either state because both the set and reset inputs would be high, which would permit either state, and the one you get would just be a race situation.

Also, with respect to the opamp, it may not matter if it will tolerate a shorted output, but it bothers me that if the battery isn't in place when the 5V power is on, for an instant the opamp will be trying to drive it's full high output through a .5 ohm resistor to ground. So I think I want to put a resistor into the base drive, even if it's only like 47 ohms or whatever still lets enough current through.

Reply to
Peabody

Thanks very much for the suggestion. I've drawn it out, and understand what you've done.

By the way, for future reference, one of the advantages of drop.io is that you can configure the drop so anyone can add content. So instead of spending time on ASCII drawings, you can just sketch it out, scan it or take a picture, and upload the jpeg to the same drop.

But back to the circuit:

Is the power-up state of the latch predictable? Well, if you power up without the battery in place, then I assume it would power up in the set state because that input would be low. But if you power up with the battery already in the ciruit, then it seems it could power up in either state because both the set and reset inputs would be high, which would permit either state, and the one you get would just be a race situation.

Also, with respect to the opamp, it may not matter if it will tolerate a shorted output, but it bothers me that if the battery isn't in place when the 5V power is on, for an instant the opamp will be trying to drive it's full high output through a .5 ohm resistor to ground. So I think I want to put some resistor into the base drive, even if it's only like 47 ohms or whatever.

Reply to
George

You'll need to use opamps that will swing all the way to ground, in both places.

John

Reply to
John Larkin

--
OK, assuming your TIP31\'s Vbe will be at about 0.7V with 7mA going into
it, then looking at "HIGH-LEVEL OUTPUT VOLTAGE vs HIGH-LEVEL OUTPUT
 Click to see the full signature
Reply to
John Fields

-snip-

Ok, I breadboarded the circuit depicted in Charger2.jpg on the drop site. But, it didn't work.

The problem was using the output of opamp A (the "1V test" opamp), which should be all the way high for a charged battery, to supply the top side of the pot used to provide the reference for the discharge opamp. I tried both the original TLC274 and an LM324, but neither worked - neither would drive the transistor sufficiently.

Well, it turns out that the so-called high opamp output level fluctuates all over the place, varying considerably depending on what else is going in the package. So an attempt to increase the discharge rate by adjusting the pot led to a voltage reduction at the top of the pot.

The first solution that came to me, while you don't see it much with an opamp, was a pullup resistor. But, you know, I don't really know how that would work when the output goes low.

So instead I changed that pot so it's supplied from +5V. Then I reversed the inputs of opamp A so the output is normally low for a charged battery. And then I connected that output to the inverting input of opamp B by way of a forward-biased diode.

All this is depicted in Charger3.jpg at the drop site. And this version actually works. It doesn't shut down abruptly when the battery falls below 1V. It gradually reduces the discharge current and presumably would continue to discharge at an ever lower current over an extended period, but the battery voltage would remain at about 1V.

And I also show in that drawing a way I think works for a latch function which adds just one more diode.

I really appreciate the comments, suggestions and drawings from John and John. And further comments woudl also be welcome.

Reply to
George

-snip-

Ok, I breadboarded the circuit depicted in Charger2.jpg on the drop site. But, it didn't work.

The problem was using the output of opamp A (the "1V test" opamp), which should be all the way high for a charged battery, to supply the top side of the pot used to provide the reference for the discharge opamp. I tried both the original TLC274 and an LM324, but neither worked - neither would drive the transistor sufficiently.

Well, it turns out that the so-called high opamp output level fluctuates all over the place, varying considerably depending on what else is going in the package. So an attempt to increase the discharge rate by adjusting the pot led to a voltage reduction at the top of the pot.

The first solution that came to me, while you don't see it much with an opamp, was a pullup resistor. But, you know, I don't really know how that would work when the output goes low.

So instead I changed that pot so it's supplied from +5V. Then I reversed the inputs of opamp A so the output is normally low for a charged battery. And then I connected that output to the inverting input of opamp B by way of a forward-biased diode.

All this is depicted in Charger3.jpg at the drop site. And this version actually works. It doesn't shut down abruptly when the battery falls below 1V. It gradually reduces the discharge current and presumably would continue to discharge at an ever lower current over an extended period, but the battery voltage would remain at about 1V.

And I also show in that drawing a way I think works for a latch function which adds just one more diode.

I really appreciate the comments, suggestions and drawings from John and John. And further comments woudl also be welcome.

Reply to
George

My first thought was a power transistor based constant current source with the battery as the load (between collector and V+). An op amp configured as a comparator sucks the base drive when the difference between V+ and the battery terminal drops below 1 V. It would need a diode to stop the comparator overdriving the transistor base drive when the battery voltage is over 1 volt.

Cheers

Reply to
Varactor

"George"

** You only need two components for that simple job.

A 1 amp silicon power diode and 1 amp Schottky diode wired in series.

Use diodes rated up to 3 amp if you like.

Such a combination will typically pass about 0.5 amps at 1.2 volts - dropping to 5 mA at 0.9 volts.

No oscillations possible.

Ever heard of the KISS principle ???

..... Phil

Reply to
Phil Allison

I appreciate your comment, but I just have no idea what this means.

Reply to
George

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