Balancing supercapacitors-Error in Circuit

Discussion in 'The Projects Forum' started by Ezio, Aug 2, 2012.

  1. Ezio

    Thread Starter New Member

    Aug 2, 2012
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    Hello,

    I am doing a supercapacitors circuit and I need to balance them so I use the specific topology : http://www.energyharvestingjournal....to-manage-your-power-00001921.asp?sessionid=1 (Figure 5)

    However due to the fact that the opamp drives a high load I suppose that a parallel feedback High pass filter needs to be added.

    So I made a modification resulting into this circuit with compensation : my_circuit.png

    But it does not work :/ Can anyone spot any mistake I have done and help me?

    Thank you in advance
     
  2. #12

    Expert

    Nov 30, 2010
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    You have failed to allow ANY DC current at the inverting input. Parallel the feedback cap with a resistance so the amp can equalize its inputs.
     
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  3. Ezio

    Thread Starter New Member

    Aug 2, 2012
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    Thank you #12

    However after adding the resistor I still have the same problems..I did a pole zero analysis and I see that each zero tries to cancel the respective pole.Any recommendations on choosing the right values?
     
  4. #12

    Expert

    Nov 30, 2010
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    The problem here is that I don't simulate. I build real circuits with real parts and I'd expect that circuit of yours to work if you added a 2.5 meg resistor in parallel with C5.

    I can only guess at why a simulator says it won't work. It might be that the output has a time constant of more than 1000 seconds and you didn't wait a couple of hours for the simulation to finish, or maybe it is because you have given the circuit (a DC balancer) an AC input.

    I can only advise you now to build the circuit for real or wait for the

    NEXT HELPER, PLEASE.

    ps, here's the datasheet
     
  5. Ezio

    Thread Starter New Member

    Aug 2, 2012
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    The thing is that I build this circuit for real and I used 1-2 simulators in order to find out why it's not working and do the compensation needed .

    When I say "it's not working" I mean that as soon as I connected the opamp output to the supercapacitors junction (through the 330 Ω resistor) VR1≠VR2 and when I say not equal I mean big difference like 200 mV .This is supposed to be stable at Vdd/2 and if not the circuit does not work. And this is because the big load of the supercapacitors.

    I will try with the suggestion you said and I will report you back with my findings.

    Thanks again!
     
  6. #12

    Expert

    Nov 30, 2010
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    That's good information. 200mv is not what is expected.

    I should have said 1.25 meg for the negative feedback resistor because that balances the input bias currents as well as you can without an adjustable balancing circuit. Still, that error is in the range of 20mv, not 200 mv. Without the negative feedback resistor, the magnitude could be as high as 93 millivolts.

    Vr1 not equal to Vr2 this suggests the input bias currents are not within spec or you have a leakage on the board. After all, you are working below the microamp range. Try reducing the voltage divider and feedback path by a power of 10 and see what happens. 250k, 250k, 120k

    Edit: you have trim pins at 1 and 8. You can dial out the offset!
    Edit: You could move the negative feedback resistor so it connects from the center of the caps to the inverting input. That will get the chip to sense the final results instead of just sensing its own output. That will make it stabilize faster.
    Edit. You better be using precision resistors. 200mv/2.5V = 8% error. That is entirely reasonable if using 10% resistors.
     
    Last edited: Aug 3, 2012
  7. Ezio

    Thread Starter New Member

    Aug 2, 2012
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    I was referring to the performance of the circuit before applying your suggestion.It seems it work now :)Now as you predicted the error is in the range of 20mV.
    Behind the practical part I want to know how you calculated/decided this value,because I want to understand the way it works.

    Well I do not use this amplifier in my circuit I just use it in simulations.the one I use is MAX4470 as proposed in the pdf I gave the link to.

    So I understand that you suggest integrating the 330 Ω and 1.25 MΩ resistor into 1 resistor,right?

    All, in all thank you for the insight , you really helped me :)
     
  8. #12

    Expert

    Nov 30, 2010
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    No. I didn't say anything about changing or moving the 330 ohm resistor. Just connect the 1.2 meg feedback resistor from the center of the big capacitors to the inverting input instead of from the chip output to the inverting input.

    first chart, input offset current = 18 nanoamps.
    That's the "unbalance" of the input currents.
    18 E-9 x 1.25 megohms = 22.5 millivolts (maximum)

    Theoretically, you can change the value of the 1.2M feedback resistor and use that as a centering adjustment.
    Got math?
     
    Last edited: Aug 3, 2012
  9. Ezio

    Thread Starter New Member

    Aug 2, 2012
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    My misunderstanding ,I have already done that in the first place ,without taking out the resistor and it works.

    The weird thing is that .As I told you I use the MAX4470 chip .Here is the datasheet which says that the input offset current is 12.5 pA http://datasheets.maxim-ic.com/en/ds/MAX4464-MAX4474.pdf .However it works(?!) with 1.25 M resistor.

    I guess you suggest decreasing the value of the resistor to a point where R*input current offset ≈ 0 mV . I am not that greedy 20mV is fine with me but I will try :)Are there any drwabacks with this?Will it become less stable?

    I have to say that ,before talking with you ,I approached it differently.With the big load at the output of the resistors a low pass filter is created and my intention was to create a high pass parallel feedback feedback (the result we have now).However I tried to set the values of the HPF with pole-zero analysis and AC response,to check whether is it stable, (that's why I used simulators) which did not work for me.
    I am happy your approach does :)
     
    Last edited: Aug 4, 2012
  10. #12

    Expert

    Nov 30, 2010
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    The mistake you made was in DC. Every op-amp has some current flow into or out of the input pins. No matter how microscopic, it must be allowed to flow somewhere. Each input pin will develop a voltage across its input resistance because of that current flow. Each input pin has a basic bias current and a "difference from the other pin" current.

    First run design assumes the difference current is a minor problem and calculates the larger Ibias current through the input resistors such that equal resistance on each input pin will cause equal input offset voltage of each input pin. Equal offset voltage = no output error.

    If that isn't perfect enough, the difference currents are compensated with things like centering adjustments provided by the chip manufacturer, a centering circuit built of resistors and a potentiometer that adds or subtracts an idle current to one of the input pins, or manual adjustment of a feedback path.

    In this case, I was thinking the feedback resistor must be carrying Ibias +/- the unbalanced Ibias. The centering position might be either more or less resistance, depending on which way the current is unbalanced, and it changes from chip to chip in the same part number.

    In practical terms, when you get to chasing nanoamps or pico amps, you get into things like guard rings, air mounting the parts, teflon standoffs, baking the humidity out of the circuit board and adding conformal coatings. This is the level where part of your troubleshooting equipment is a hair dryer!

    Input bias current doesn't give a rat sass about poles and frequencies. It's a DC phenomenon.
     
  11. Ezio

    Thread Starter New Member

    Aug 2, 2012
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    Thank you for your detailed explanation :)
     
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