Biasing for single supply to op amp

Discussion in 'The Projects Forum' started by EpicFails, May 7, 2013.

  1. EpicFails

    Thread Starter New Member

    Sep 7, 2012
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    0
    Hi,

    I'm struggling with a biasing arrangement for an audio preamp I'm building.

    The circuit has to be run off a 9v battery and also requires a high input impedance.

    I'm using an op amp for the impedance buffer, so have created Vcc/2 for virtual ground.

    The problem is that the DC operating point is not right at the +input of the op amp - it's at c.0.95V from a supply of 8.72V and VRef of 4.15V. I have attached the circuit. All other points are working at the DC op point I expected.

    I'm almost 100% sure that this problem is something to do with the high value of R3. I have altered the values for R1 and R2 to compensate for the value of R3 and can get a bit closer to Vcc/2 but in order to get there properly I would need to increase R1 / R2 by a lot. This is a problem, especially as this is an audio circuit, in terms of Johnson noise contribution it would make.

    So, I was wondering if anyone had any ideas on how to get around this?

    I was thinking about using a rail splitter op amp like the TLE2426, but I don't know if this would make any difference when faced by such a high R3 value. I have already tried buffering the voltage divider and that didn't help...

    Any thoughts much appreciated.

    Thanks
     
  2. wayneh

    Expert

    Sep 9, 2010
    12,101
    3,033
    So you're saying the voltage on the +Vin is correct at ~4V but the output of the op-amp is at <1V?

    I'm not sure but I'll speculate that you need an op-amp that can sense much closer to the negative rail than this one.

    And what do you mean by the "high" value of R3? It's very low compared to R1 and R2 and to the input impedance of the TL081.
     
    Last edited: May 7, 2013
  3. EpicFails

    Thread Starter New Member

    Sep 7, 2012
    18
    0
    It's the other way around. The output from the op amp is at ~4.35V as is the inverting input of the op amp. However, the non-inverting input is at ~0.95V.

    Vref (Vcc/2) is 4.15V. Vcc+ is 8.72V.

    Sorry - R3 should read 10Mohms, I was playing around with the simulator before exporting it and forgot..!

    And, the op amp being used is actually a TL072, not a TL081.

    Sorry for the confusion there...
     
  4. wayneh

    Expert

    Sep 9, 2010
    12,101
    3,033
    Ah, so there a voltage drop across the 10MΩ R3, from Vref = 4.12 down to ~0.95V at +Vin.

    I assume that you measured 0.95V with a multimeter? The meter's input impedance is probably low enough to pull down that measurement. I think you're looking at an artifact.

    If the output is 4.35V, what's the problem? The output voltage likely drops when you probe +Vin with your meter. Check it if you have a second meter, or hook up an LED.
     
  5. EpicFails

    Thread Starter New Member

    Sep 7, 2012
    18
    0
    It was a measurement with a multimeter. I'll try connecting an LED and see what happens.

    I was only worried because my understanding is that both inputs should be at the same operating point.

    Incidentally, I have also tried the same circuit out with a 1M ohm R3 as well. This also gave a lower figure than the Vref - I think it was about 1.5V. Would a 1M ohm R3 be big enough to cause the multimeter to react in the way you described?

    Cheers for your help!
     
  6. EpicFails

    Thread Starter New Member

    Sep 7, 2012
    18
    0
    Just tried the LED and you were totally right - lit up when no probe connected, or when probing other parts of circuit and then goes out totally when probe +Vin of op amp and that's with a 1MΩ R3.

    Thanks so much for your help with that wayneh. So am I correct in assuming that the voltage at +Vin is probably roughly the same voltage as at Vout and at -Vin?
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Vout and +Vin will be identical, within the offset spec of the op amp.
    Your input capacitor C3 can be much smaller. The cutoff frequency is
    Fc=1/(2*pi*R3*C3).
    Solve for C3.
     
  8. EpicFails

    Thread Starter New Member

    Sep 7, 2012
    18
    0
    Thanks for clarifying the +Vin and Vout relationship Ron H and thanks for the reminder re. input cap. I rushed the circuit drawing, but will resize at a later date.

    Thank you both for all your help. Very much appreciated.
     
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