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Battery discharger - will this work?

G

George

Jan 1, 1970
0
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.
 
G

George

Jan 1, 1970
0
George says...
The circuit is here:

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.
 
T

Tom Biasi

Jan 1, 1970
0
George said:
George says...



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
 
G

George

Jan 1, 1970
0
John Larkin says...
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.
 
T

Tom Biasi

Jan 1, 1970
0
George said:
Tom Biasi says...



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
 
G

George

Jan 1, 1970
0
Tom Biasi says...
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.
 
P

Peabody

Jan 1, 1970
0
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.
 
G

George

Jan 1, 1970
0
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.
 
G

George

Jan 1, 1970
0
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.
 
G

George

Jan 1, 1970
0
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.
 
V

Varactor

Jan 1, 1970
0
George says...




-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
 
P

Phil Allison

Jan 1, 1970
0
"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
 
G

George

Jan 1, 1970
0
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.
 
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