Battery discharger - will this work?

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.

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

Tom

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

I don't understand what you mean.

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.

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

Tom

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.

Something like this maybe...

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

John

Ok. Thanks very much.

--- 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.

Then if the battery voltage started to rise again after it was disconnected from the sink it wouldn't matter.

View in Courier:

+---------------------------------------------------+ +5>--+--|--+-------------------+-----------------+ | | | | |1% | | | | | [4020] [10K] | |1%| | R4| |R6 | [4750]| | Q1 +---|-\\U1B | | R1 | | | TIP31 | | >---A U2A | R7 C Q2 +--+-|+\\U1A C----+------|---|+/ HC00 Y-+-|--[1K]--B 2N3904 | | >---B | |1% +-B | | E | +-|-/TLC E | [1K] | | | | | | | 274 | |+ |R5 | A-+ | | | +--|-------+ [CELL] | +-Y HC00 | | |1% | | |BT1 | U2B B---+--O | | [249] | [0.5R] | | |-+-----+-------+----+------+-----------------+----------+

Your TIP31 has a current gain of about 75 minimum at the current you want it to limit from the cell, so the opamp needs to supply at least:

Ic 0.5A Ib = ----- = ------ = 0.0066A ~ 7mA Hfe 75

The opamp is rated to source 35mA max, so you should be OK in that regard.

I've shown 1% resistors for the reference dividers, but if you don't need the precision, 5% carbon films will work OK.

The RS latch is made from 2 NANDs out of 4 in a 74HC00, so one chip will be enough for both of your circuits.

The way it works is that R1 and R2 comprise a voltage divider which sets the voltage on U1A+ at 0.25V, then U1A drives Q1 as hard as it needs to in order to make the voltage dropped across R3 the same as the voltage on U1A+.

That will happen when the cell is pushing 500 mA through R3, and U1A will adjust the drive into Q1 to keep that happening no matter what BT1's output voltage happens to be.

That is, until the voltage falls to ~1V.

When that happens, the output of U1B (which is being used as a comparator with a 1V reference on the inverting input) will go low and set the latch U2A-U2B.

That will turn on Q2, driving it into saturation, and since the collector of Q2 is connected to the reference input of U1A, will pull it close to 0V.

That will force the outout of U1A low, removing the base drive from Q1 and effectively disconnecteng the + end of BT1 from the circuit.

Now, even if the output voltage of BT1 rises because it's resting, and causes the putput of U1B to go high, the latch will stay set and BT1 will stay disconnected until S1 (a normally-open momentary action swich) is pressed.

When that happens, the latch will be reset and the discharge cycle will begin anew.

JF

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.

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.

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

John

---

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 CURRENT" on page 22 of the TLC724 data sheet shows that with a Vdd of 5V and an output current of 7mA, the opamp can output about 2.8V.

since you only need 0.7V to drive the TIP31, then you can drop the remainder in a base resistor, if you want, and that resistance would be:

Vout - Vbe 2.8V - 0.7V R = ------------ = ------------- = 300 ohms Ib 0.007A

If you want a power-on reset you could add C1 and change R4 and R5 to

40.2K and 10K, respectively, like this:

+--------------------------------------------------------+

+5>--+--|--+------------+--------+--------------------+ | | | | |C1 |1% | | | | | [0.1µF] [40k2] [10K] | |1%| | | R4| |R6 | [4750]| | Q1 +--------+-------|-\\U1B | | R1 | | | TIP31 | R9 | >--A U2A | R7 C Q2 +--+-|+\\U1A C----+----+---|-[10k]-|+/ HC00 Y-+-|-[1K]-B 2N3904 | | >---B | R8| | +-B | | E | +-|-/TLC E | [1M] | | | | | | | | 274 | |+ | |1% | A-+ | | | +--|-------+ [CELL] | [10k] +-Y HC00 | | |1% | | |BT1 | |R5 U2B B---+-O | | [249] | [0.5R] | | | |-+-----+-------+----+----+---+--------------------+--------+

Also, you should add R8 to keep U1B+ from floating if there's no cell in there and it's powered up, and R9, just in case...

JF

-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.

-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.

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

"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

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