Op-amp DC offset question...

Discussion in 'General Electronics Chat' started by foxfire3, May 26, 2013.

  1. foxfire3

    Thread Starter Member

    Jan 29, 2013
    34
    0
    I'm trying to use the AD AD828 op-amp as simple noninverting audio linestage amplifier with a gain of 5.

    AD828 data sheet:
    http://www.analog.com/static/imported-files/data_sheets/AD828.pdf

    When I place a volume control in front of the 828 the DC offset measures a whopping 1.3V.:eek:

    When I place the 828 in a noninverting circuit without a volume pot., such as a CD player or tuner, the offset only measures 200mV.

    Why is the volume pot. causing the DC offset to be so much greater and how can I reduce it?:confused:

    Thank you...
     
  2. LDC3

    Active Member

    Apr 27, 2013
    920
    160
    Please post you circuit so we can answer your questions.
     
  3. foxfire3

    Thread Starter Member

    Jan 29, 2013
    34
    0
    I have a sneaky suspicion that my problem may be related to unequal impedances on the positive and negative inputs of a bipolar op-amp, but I'm not sure.

    Anyway here's the schematic:
     
  4. LDC3

    Active Member

    Apr 27, 2013
    920
    160
    The DC offset is caused by the input and feedback to the amplifier. The inputs to the amplifier have a very high impedance and act as a virtual ground. Therefore, when no signal is applied to the inputs, the output moves towards the midpoint of Vcc and ground. This causes the feedback to have a positive potential on the -ve input, which causes a potential on the +ve input. The DC offset is what you measure from this feedback loop.
    There is nothing wrong with having a DC offset.
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    What???



    Foxfire3, what are your supply voltages?
    I think you may be correct. The input bias current is 6.6uA max. With a 10k pot, that will generate up to about 66mV of input offset. With a gain of 5.e, the output could be as much as about 350mV. 200mV is about right with typical input bias current of 3.3uA. Using lower resistance values will help. Why do you need low offset? Does your signal have a DC component that you need to preserve, or could you use capacitor coupling?
     
  6. crutschow

    Expert

    Mar 14, 2008
    13,048
    3,244
    Looking into the wiper of a 10k pot, the maximum equivalent impedance is 1/4 of the pot value or 2.5KΩ, assuming both sides of the pot are at DC ground potential (which may not be true here). This makes the voltage offset from the pot for 6.6μA maximum bias current to be 16.5mV at the input or 82.5mV at the output.
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    True. I was assuming he had nothing connected to the input when he made the offset measurement.
     
  8. foxfire3

    Thread Starter Member

    Jan 29, 2013
    34
    0
    Supply voltage is 15 volts on both the positive and negative rails.

    With no signal going through it and the volume pot at near its minimum position, I measure 1.3VDC at the outputs.
    This is without shorting the inputs. I'm not sure if they're supposed to be shorted or not when measuring offset voltages.

    Anyway, over 1 volt seems like way too much at the outputs.

    I need low offset because I'm going to be feeding the signal into a power amp.
     
  9. foxfire3

    Thread Starter Member

    Jan 29, 2013
    34
    0
    I didn't have anything connected to the inputs when I measured it.

    I still measure 1.3 volts.
     
  10. crutschow

    Expert

    Mar 14, 2008
    13,048
    3,244
    Then it would appear you have a miss-wired circuit or a faulty op amp.
     
  11. foxfire3

    Thread Starter Member

    Jan 29, 2013
    34
    0
    I have two and they both measure the same.

    The circuit is fine as I can try other op-amps with much lower offset values.

    This is the most DC I've ever measured out of an op-amp before. I've measure quite a few both FET and bipolar.
     
  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If the output changes when the pot is adjusted one would agree this is related to bias current. Does the change in pot position produce a change in output?
     
  13. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    A wrong resistor value could also do it.
    Why are you using a video amp? What are you trying to amplify?
     
  14. foxfire3

    Thread Starter Member

    Jan 29, 2013
    34
    0
    I just checked the output voltage while varying the volume pot's setting and it has no effect on the voltage. The output voltage stays the same.

    I thought it was 1.3VDC, but it's actually 1.7VDC...:eek:

    I also tried the same IC's in a totally different factory-made preamp. board(with different but similar resistor values) and sure enough I measure 1.7 volts at the outputs.

    Could the resistor values I'm using be too large?
     
  15. foxfire3

    Thread Starter Member

    Jan 29, 2013
    34
    0
    I knew the video amp question was coming...I'm using it because it sounds quite good.

    I'm amplifying audio.
     
  16. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Have checked the output on a CRO? Maybe it's oscillating.
     
    Ron H likes this.
  17. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Good point!

    If you don't have a 0.1uF ceramic cap from each power pin to ground, close to the IC, with short leads, I'll practically guarantee that it will oscillate.
     
  18. foxfire3

    Thread Starter Member

    Jan 29, 2013
    34
    0
    No, I don't have a CRO.

    I have .1uF decoupling caps. installed.
     
  19. foxfire3

    Thread Starter Member

    Jan 29, 2013
    34
    0
    Alright, I think I've figured out what's going on here.

    In my circuit, I don't have a resistor going from the noninverting input to ground.
    I see some people using resistors around 47K ohm and some don't.
    With this particular op-amp, if you don't have a resistor from the noninverting input to ground, the DC offset is very high.
    The lower the resistor's value, the lower the offset seems to be.
    With a 1K ohm resistor the offset measures 30mV.

    But with a 1K ohm resistor(or lower) in parallel with the volume pot, it seems to me the input impedance would be too low.
    Perhaps that's why some people use very large value resistors in this position so the input impedance is closer to the value of the pot's position?
    I don't know...perhaps someone can explain the true purpose of the resistor in parallel with the volume pot.
     
  20. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    The schematic you posted shows 1k in series with the 10k pot to ground. Now you say you don't have a resistor from the +input to ground.
    If you are changing the circuit, you need to post a schematic of what you are describing. We can't read your mind.

    We can tell you the purpose of each resistor in an op amp circuit, but we have to know what the circuit looks like.
     
Loading...