Battery discharging circuit not working

Discussion in 'The Projects Forum' started by Harnee15, Sep 14, 2015.

  1. Harnee15

    Thread Starter New Member

    Sep 14, 2015
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    I have a circuit that aims to discharge a battery and hence monitors it's voltage.
    The circuit either drains 4 uA current through a 680k resistor or 20mA current which is driven by the DAC. The switching between these two scenarios is done using a 74hc125 ( have tried N FET as well)

    The circuit diagram is attached.

    Problem: When the switch corresponding to 680k is enabled it gets grounded but when switch corresponding to 10 ohm resistor is enabled it doesn't gets grounded.
    I'm unable to spot the problem with the circuit .
    Please help!!
     
  2. ronv

    AAC Fanatic!

    Nov 12, 2008
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    The 74HC125 can't sink 20 ma.
    But since you replaced it with a FET and it still didn't work measure the voltages on the op amp driving the transistor. Try to set it for 3 ma.
     
  3. wayneh

    Expert

    Sep 9, 2010
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    Are the op-amps powered by the battery itself? It's not obvious (to me) in the drawing how the op-amps get biased relative to their common mode voltage range. For instance, is the DAC supplying a voltage that shares ground with the battery?
     
  4. Harnee15

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    Sep 14, 2015
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    [
    Thanks much for your reply. I have tried different current values but still both the ends of the 10 ohm resistor have the same value.
     
  5. ronv

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    OK, so what are the voltages on the op amp that drives the transistor?
     
  6. Harnee15

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    Sep 14, 2015
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    Thanks much for your reply.

    No, the opamps are not powered by the battery. They are powered by an external supply and yes you are right the voltage supplied by DAC shares ground with the battery. What implications can these things have on the circuit? I'm sorry I'm very naive in this field so don't have much understanding.
     
  7. Harnee15

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    Sep 14, 2015
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    Currently, the output from the potentiometer is 145mV. The opamp which drives the transistor has output as 30 mV, whereas both its inputs are around 149 mV.

    I have attached the schematic of the board as well.
     
  8. wayneh

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    Sep 9, 2010
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    My concern was partly the darlington. It's in emitter-follower configuration and so the base needs ~1.4V more than the emitter voltage before it will start to conduct. So the op-amp power supply, less the ~1-2V needed by the op-amp, must be high enough.

    And then the current gain needs to be high enough to not pulldown the op-amp output. The op-amp might not supply sufficient base current. That's a secondary concern, but a possibility.
     
    Last edited: Sep 14, 2015
  9. Harnee15

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    Sep 14, 2015
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    The power supply to the opamp is 3.3V .
    Sorry, to ask a silly question again, how do I ensure that the current gain is high enough to not to pulldown the op-amp output?
     
  10. ronv

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    The output of the op amp B should be higher so as to turn on the transistor as @wayneh pointed out.
    What is the part number of the op amp?
     
  11. Harnee15

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    Sep 14, 2015
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    Oh ok.. But the output I'm getting is low. What can be the issue? What can be done to rectify this problem?

    I'm using Microchip MCP 6002-E/SN. (Farnell part number : 1332118)
     
  12. ronv

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    Well, that's a good op amp for this so I guess more troubleshooting is in order. Here is what you might try:
    Since the 74HC125 won't work for 20 ma anyway lets rule it out by just grounding the bottom of the 10 ohm resistor. Remove it from the IC first of course.
    Then set the pot for a voltage a little higher say .4 volts so we can be sure what we are measuring.
    Then take the same measurements - op amp +/- output Vcc and ground to the op amp.
    Then measure emitter base and collector of the transistor (what is it's part # by the way?)
     
  13. Harnee15

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    Sep 14, 2015
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  14. Harnee15

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    Sep 14, 2015
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    Thanks much for your reply.
    I have removed 74hc125 and replaced with a FET. Should I remove this as well?
    Following are the reading as you suggested when the pot was set to 0.4 V:
    OPAMP:
    opamp + 400mV
    opamp- 400mV
    opamp Output- 30.34 mV
    Vcc opamp 3.3V
    Gnd opamp 29.04 mV

    Transistor:
    Collector: 1.61 V (battery connected)
    Emitter: 400mV
    Base: 31.25 mV

    Transistor is MMBTA14LT1G Part Number: 2317594

    Does this mean that this opamp is not working properly?
     
  15. ronv

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    Actually it looks like it is working. You set the input to .4 volts and .4 volts is on the 10 ohm resistor - so 40ma. The only thing that seems wrong is I would expect the output of the op amp and the base of the transistor to be about 1.7 volts.
     
  16. Harnee15

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    Sep 14, 2015
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    So, does that mean the opamp is not able to drive the darlington in a correct way. Should I consider replacing the opamp?
     
  17. wayneh

    Expert

    Sep 9, 2010
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    I'm getting a bit lost. Can you update your drawing with the voltage data you have and the changes you've made.

    The 29mV you see at the op-amp ground raises an eyebrow. I'm not saying this is the cause of your problem, but it indicates a voltage drop on the ground return line for the op-amp power. You might want a thicker and/or shorter wire. Your op-amp's output is very close to (only 1.3mV more than) the negative rail.

    How can there be 400mV on top of the shunt resistor to ground, when the darlington should not be conducting at all?
     
    Last edited: Sep 14, 2015
  18. InspectorGadget

    Active Member

    Nov 5, 2010
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    You have a lot of problems with this circuit.

    1. The 74HC125 will not sink the current you need, as described before. Why do you have the 74HC125 at all? Just run the resistors directly to ground. The Darlington will switch the 10 ohm one and the 680K one will be invisible at that point. I guess you can keep it on the 680K resistor if you want to switch on and off the low-current load.

    2. OpAmpA(6) in the upper left has the gain resistors configured incorrectly.

    3. You really should have a series resistor of about 1K or so in line to that Op Amp to current-limit any problem situations, like using too high a VBATT.

    4. Make sure OpAmpB(6) has the + input on the top and the - input on the bottom as drawn. It also might be good to have a series resistor of 1K or so to current limit on the + input should the battery exceed the Op Amp's input maximum or something.

    5. What is the free-floating "DAC" in the upper right? And what does that "Pot" do? The Pot won't adjust either voltage or current from the DAC because the OpAmpA(4) has such a high impedance input, it'll always see the voltage at the output of the DAC the way it's drawn no matter what the Pot is set to.

    6. Don't use a VBATT greater than your Op-Amp supply voltage (3.3V). You'll fry your Op Amps.

    7. TO answer one of your questions above, the Darlington base current is limited by the - feedback to OpAmpB(4) and the high gain of the Darlington itself. If you use a low gain transistor it'll pull more current from the Op Amp. The high gain of the Darlington pulls just enough current to conduct, which shunts higher current to the load resistor, which raises the load resistor top terminal voltage until it nearly equals the OpAmpB(4) + input. The + input will actually be slightly higher by the amount of the Darlington's voltage drop times the gain of the Op Amp (which is pretty freaking high) so the difference won't be much. Essentially, the Darlington will conduct more current to keep the OpAmpB(4) - terminal the same as the + terminal as you increase the + input (from the free-floating DAC in the upper right, presumably). This is the basic function of the voltage-to-current converter you have constructed.

    8. If it's not working, replace your Op Amps observing the comments I've made, and replace your Darlington being very mindful of the correct terminal identification. Then try it out with a 100 ohm or 1K resistor instead of the 10 ohm to make it more forgiving if there are any problems, and not immediately burn out other parts from too much current flowing.
     
  19. InspectorGadget

    Active Member

    Nov 5, 2010
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    One more note: Since the Darlington has a gain of around 10,000, if you run a 3V VBATT, it'll draw a max of about 300mA through the Darlington, which will draw only 30uA from the Op Amp.

    Also note that the inputs to OpAmpB(6) will track the load resistor; the output voltage of the OpAmp will be 1.4V higher than the top of the load resistor in order to drive the Darlington because of the gain of the Op Amp. But since this current controller is at the VBATT terminal, the Op Amp needs to generate up to 1.4V ABOVE the VBATT level in order to make the Darlington conduct. At maximum current, the Darlington will not drop any voltage from collector to emitter, so the emitter voltage will be at VBATT and the Op Amp will have to drive 1.4V above VBATT to maintain this condition.

    So the highest VBATT you can use with this circuit is actually about 1.9V, and the highest current you can get is about 200mA.
     
  20. ronv

    AAC Fanatic!

    Nov 12, 2008
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    But he only wants 20 ma.. So he has room.
     
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