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Detecting down to 25mA-150mA on 0-20A line?

Discussion in 'General Electronics' started by John, Nov 15, 2005.

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  1. John

    John Guest

    Sounds like you're doing a manual adjust on a power supply with
    Yes. The end-of-charge voltage set to 4.20V (or a multiple of that)
    and the constant current limit is set to whatever the max. current
    value I want to use for charging. The current drops as the cell
    charges.


    Huh?
    In over my head because I wanted to find out if there was a better way
    to do this (than the ways I worked out on my own)?

    I already have a CC/CV supply with V and A readouts. I want to end
    the charge when the current has dropped down to a particular level
    (25mA-150mA). Sitting there monitoring the display is not what I had
    in mind.
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  2. Jasen Betts

    Jasen Betts Guest

    ["Followup-To:" header set to sci.electronics.]
    LM324 might be a better move then. they can handle input down to 0.6V _below_
    the negative supply. (so put the sense resistor in the negative circuit)

    For some reason this quad op-amp is cheaper than the 741 at jaycar...

    Bye.
    Jasen
     
  3. John

    John Guest

    This thing is probably going to be working as a comparator with some
    In my OP I mentioned that when the current drops down to a certain
    point (variable, from 25mA to 150mA), I want to flip a bit.
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  4. John

    John Guest

    Simulation now corrected...
    Ooohhh...I have the parts to play around with that circuit. Thanks!
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  5. John

    John Guest

    I was initially hesitant to use a sense resistor due to the playing
    around of the power supply voltage I'd have to do for every new cell
    (if the end-of-charge current was changing). But Jim's circuit has a
    nice linear drop across R1 for me to add to the power supply voltage
    so that the LiPo cell will get 4.20V at end of charge (i.e., a 100mA
    end-of-charge current needs 4.30V voltage at supply).

    Time to play!
    Thank you everyone for your input and recommendations. I really
    appreciate the time you've taken to get me going in a direction that
    looks like it will work well for my new toy. :)
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  6. Jim Thompson's circuit and simulation certainly has some merits, but there
    are some caveats to be aware of.

    In particular the circuit is most likely not stable, or at least very nearly
    unstable. There is a high probability that it may decide to burst into
    uncontrolled spontaneous and continuous oscillation under the expected
    circuit conditions.

    The first problem is op-amp outputs in general don't like driving capacitive
    loads directly. Unless they are specifically designed for direct capacitive
    loading, you can't place more than a few tens of picofarads (or maybe a few
    hundred at the most) directly on the output without risking instability and
    spontaneous oscillation. A MOSFET such as an IRF3704 has a larger gate
    capacitance than this, perhaps around 2.5nF. This isn't a very well
    advertised "feature" of op-amps since it isn't exactly a flattering
    characteristic, but it should be briefly mentioned in some of the op-amp ap
    notes out there.

    So how does one fix this? Well typically one would simply place a small
    resistor in series with the capacitor. 47 ohms or larger would typically
    work great for practically any size capacitance. So simply placing a 47 ohm
    resistor in series with the op-amp output of Jim Thompson's circuit should
    help this particular potential source of instability.

    Unfortunately, it also leads us to the next potential source of instability.

    The gain of the circuit is simply humongous (not that that is a real word).
    High gain isn't a friend of stability. As it is drawn the op-amp doesn't
    have any direct negative feedback, so it seems to me the op-amp is operating
    at it's maximum possible gain as if open loop. The op-amp is internally
    compensated and is supposed to be stable under these conditions, however,
    the IRF3704 MOSFET is also a part of the control loop, and it adds it's own
    gain to the control loop as well. The IRF3704 datasheet suggests it has a
    transconductance of 42 Siemens (minimum) at the specified test conditions of
    Vds=10V and Id=57A. I assume the Tran will be less at your expected
    operating conditions, but by exactly how much is uncertain since the
    datasheet graphs don't extend down into all of your expected regions of
    operation.

    So in other words if I'm not mistaken, it appears to me the entire control
    loop gain is higher than the open loop gain of the op-amp itself. As far as
    I'm aware, this by itself is a no-no and all bets are off as for
    expectations for stability. Additionally it seems to me the op-amp adds
    it's own inherent built in pole to the control loop, but the 47 ohm+
    resistor in series with the MOSFET gate capacitor makes another low pass
    filter and adds another pole to the control loop. Two poles=bad (unless
    counterbalanced by zeros or at frequencies that are not important) since
    this will push you up to your 180 degree phase shift limit. Note that the
    MOSFET gate+gate resistor pole probably occurs at quite high frequency, but
    given the uber huge gain of the circuit, it may not be sufficiently high
    frequency to be irrelevant.

    So in other words, use the circuit as drawn at your own risk. It may work,
    it may work some of the time, it may work some of the time with some types
    of parts at some temperatures, but then again, it may instead break into
    spontaneous oscillation at certain operating points.
     
  7. Jim Thompson

    Jim Thompson Guest

    And, testing your historical background, "And the horse may talk" ;-)

    You may be right, you may be wrong... stability of such large-gate
    circuits IS difficult to predict.

    However the 1 ohm sense resistor does help kill the loop gain.

    Capacitive loading depends on a lot of parameters... but 2.5nF is so
    huge that, using the model included with PSpice, the simulated Phase
    Margin = 87°

    However, since I trust no one but myself, when it comes to Spice
    modeling, I replaced the LM324 with my own modeling of an LM339, used
    a 9.1K pull-up resistor, and placed 0.33uF from gate-to-drain; then
    ran a pulse test.

    The result is _nearly_ perfect, but I see some slew-rate hunting going
    on (about 4mV P-P sawtoothing), so it probably needs a zero in there
    somewhere.

    But I think the basic concept is viable... maybe just back off and
    make a discrete low-gain diff-pair.

    ...Jim Thompson
     
  8. default

    default Guest

    Understood.

    I was thinking in terms of charging current in the real world. Under
    actual circuit conditions the temperature may affect the current even
    if it is (was) at its end-of-charge value.

    Does the bit flip back if the current rises above the set point? Does
    the bit control the charge voltage or just inform some other circuit
    that the battery is charged? Does it sit there and vacillate between
    on and off if it is at the threshold? Leave it to chance and chances
    are it will oscillate.
     
  9. Jim Thompson

    Jim Thompson Guest


  10. Well you certainly deserve an "A" for effort...

    But am I happy now? Well... Call me hard to please but I guess it would be
    a not too spectacularly enthusiastic happy. It appears to me the
    modifications should meet the OP's requirements without any major problems
    that I can see, but I guess it isn't quite the magic bullet solution I was
    hoping for.

    Losing the tight regulation of the prior circuit is a real bummer, although
    I suppose not really important in the OP's application. Gaining the extra
    three resistors and the capacitor are not unreasonably troublesome, although
    the new schematic obscures the real added complexity since the 200mV voltage
    reference can no longer be some high impedance resistive divider off some
    other regulated voltage.

    Hopefully you are more happy with your new circuit than I am, because
    someone should be happy with it. I'm reminded of the last fortune cookie
    fortune I got:

    "Ideas are like children, there are none so great as your own." (just don't
    add the obligatory "in bed" words at the end like one would normally do with
    fortune cookie fortunes)
     
  11. Jim Thompson

    Jim Thompson Guest

    You answered your own question... it's not germane... it's only
    protection. So I purposely killed the loop gain so low I didn't need
    to spend any great effort compensating it.
    Shirley you understand that you can make a divider from the OP's
    regulated supply that produces 0.2V with a 1K impednace ?:)
    It can be done with 2-3 small-signal transistors and 3-4 resistors but
    is even less tight.

    You're so picky-assed you must be an ASU professor ?:)

    ...Jim Thompson
     
  12. Zak

    Zak Guest

    You charge the battery from a voltage regulator; you probably do not
    want a diode or something like that in series with the battery.

    However, it is possible to bypass your voltage regulator with a current
    source equal to your setpoint.

    The moment that the output of the regulator rises to a level above
    normal, it is time to switch off.

    This is easily measured on whatever compares the output voltage with the
    reference, or on the output transistor drive for that matter.

    Do not forget to switch the current source off as well :)


    Thomas
     
  13. Fred Bloggs

    Fred Bloggs Guest

    I don't know what that 200mV reference is about when the standard charge
    voltage tolerance for a LiPo is +/-50mV per 3.7V standard cell. So using
    a 50mV reference means he can set the power supply to 4.20V and forget
    it. With that 200mV reference he still has at least 150mV to go through
    that 1 ohm resistor. You may not need any compensation with an emitter
    follower with diode pulldown driving the gate- the output circuit
    feeding back to IN(+) has nearly zero time constant:
    View in a fixed-width font such as Courier.
     
  14. John

    John Guest

    Does the bit flip back if the current rises above the set point? Does
    I agree.
    The end-of-charge current level is passed onto a PIC that latches the
    bit.
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  15. Fred Bloggs

    Fred Bloggs Guest

    I may have made a mistake, but I'm not getting any small signal gain
    through that MOSFET, the only instability should arise from the bulk
    Ciss loading of the OA:
    View in a fixed-width font such as Courier.

    . *
    . cutoff=150mA - Iadj x 168
     
  16. Jim Thompson

    Jim Thompson Guest

    Using National's LM324 model I was getting 67° of phase margin, but
    pulsing did show ringing, so I killed the loop gain and isolated Ciss.

    ...Jim Thompson
     
  17. Jim Thompson

    Jim Thompson Guest

    I was wrong. You're a young buck recent grad of the ASU playboy
    school of engineering, with virtually no real-world experience.

    But you seem to have good family background! That MAY make up for the
    ASU experience ;-)

    ...Jim Thompson
     

  18. Whoops. Silly me. Nix that little complaint.
     
  19. Zak

    Zak Guest

    Aagh... so many parts... just bridge the voltage regulator with a
    current source and detect when the voltage regulator drops out of
    regulation...


    Thomas
     

  20. For what it is worth, I happen to think your suggested solution is quite
    clever. Saving practically all of the power dissipation while eliminating
    the non-idealities of actual current measurements are real major advantages.
     
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