Battery discharger - will this work?

[George]

The circut is here:

http://drop.io/kuv9yi5

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.

2. 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.

 

[George]

George says...

The circuit is here:

http://drop.io/kuv9yi5

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.

2. 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.

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.

 

[Tom Biasi]

"George" <gh424NO824SPAM@cox.net> wrote in message
news:u5fql.18457$Zp.11523@newsfe21.iad...

George says...

The circuit is here:

http://drop.io/kuv9yi5



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

 

[John Larkin]

On Sat, 28 Feb 2009 09:10:53 -0600, George <gh424NO824SPAM@cox.net>
wrote:

The circut is here:

http://drop.io/kuv9yi5

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.

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

2. 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?

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

 

[George]

Tom Biasi says...

The circuit is here:

http://drop.io/kuv9yi5

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?

I don't understand what you mean.

 

[George]

John Larkin says...

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.

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

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?

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.

Ok.

Make sure the opamp can supply enough base current.

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.

 

[Tom Biasi]

"George" <gh424NO824SPAM@cox.net> wrote in message
news:lcjql.43153$2O4.14615@newsfe03.iad...

Tom Biasi says...

The circuit is here:

http://drop.io/kuv9yi5

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?

I don't understand what you mean.

Could you monitor the battery voltage with one op-amp and turn off the NPN

when voltage is 1 volt?

Tom

 

[George]

Tom Biasi says...

The circuit is here:

http://drop.io/kuv9yi5

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?

I don't understand what you mean.

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

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.

 

[John Larkin]

On Sat, 28 Feb 2009 17:14:46 -0600, George <gh424NO824SPAM@cox.net>
wrote:

John Larkin says...

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.

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

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?

Something like this maybe...

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

John

 

[George]

John Larkin says...

Something like this maybe...

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

Ok. Thanks very much.

 

[John Fields]

On Sat, 28 Feb 2009 18:39:27 -0600, George <gh424NO824SPAM@cox.net>
wrote:

Tom Biasi says...

The circuit is here:

http://drop.io/kuv9yi5

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?

I don't understand what you mean.

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

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.

---
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] | | |<R |
R2| | R3| | | +--O | |
| | | | | | S1 |
GND>-+-----+-------+----+------+-----------------+----------+

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

 

[Peabody]

John Fields says...

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.

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.

 

[George]

Sorry - first post was at another sever wih a different name.

John Fields says...

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.

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.

 

[John Larkin]

On Sat, 28 Feb 2009 20:33:21 -0600, George <gh424NO824SPAM@cox.net>
wrote:

John Larkin says...

Something like this maybe...

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

Ok. Thanks very much.

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

John

 

[John Fields]

On Sun, 01 Mar 2009 10:44:51 -0600, George <ghNO343SPAM424@cox.net>
wrote:

Sorry - first post was at another sever wih a different name.

John Fields says...

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.

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.

---

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% | |
| | | [0.1ľF] [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] | | | |<R |
R2| | R3| | | | +-O | |
| | | | | | | S1 |
GND>-+-----+-------+----+----+---+--------------------+--------+

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

 

[George]

George says...

The circut is here:

http://drop.io/kuv9yi5

The idea is to discharge a NiMH AA cell at 500ma down to the point
where the cell voltage is 1V.

-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.

 

[Phil Allison]

"George"

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.

** 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

 

[Varactor]

On Mar 2, 1:16 pm, George <ghNO343SPAM...@cox.net> wrote:

George says...



The circut is here:

http://drop.io/kuv9yi5

The idea is to discharge a NiMH AA cell at 500ma down to the point
where the cell voltage is 1V.

-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]

Varactor says...

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.

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

 

[George]

George says...

The circut is here:

http://drop.io/kuv9yi5

The idea is to discharge a NiMH AA cell at 500ma down to the point
where the cell voltage is 1V.

-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.