Help analysing opamp input offset adjustment ciruit

Discussion in 'The Projects Forum' started by Stin, Mar 11, 2014.

  1. Stin

    Thread Starter New Member

    Jul 26, 2012
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    Hi guys. Im including a circuit to adjust the offset voltage of an opamp in one of my projects, and as the title says im having a bit of trouble analysing it. It's on page 6 of this article, top left.

    http://www.ti.com/ww/en/bobpease/assets/AN-31.pdf

    Can anyone help me get the equations?

    From my workings i got this,

    Vout = Vin(R4/R2 + R4/R1 + 1) +- V(R4/R1)

    Any help would be appreciated.
     
  2. #12

    Expert

    Nov 30, 2010
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    Vo = (Vos - Vin) (R3/R4) + Vos

    I can't think in your terms, so I used the fact that there is no current from the output going through R1 or R2. R1 and R2 are not in the gain equation. They only present an offset voltage (Vos).

    Pin -in is going to be at the voltage of the offset circuit (R5, R1, and R2). That voltage compared to Vin is the current through R4. That current must arrive through R3, PLUS the amount of voltage on pin +in. There is an active current through the resistors R3, R4 plus a voltage offset equal to the Vos you dialed in at the +in pin.

    Now, I might have fumbled the math, but the concept is where you fell off the horse. It looks like you're trying to work this as a differential amp, but I can't see it that way.

    Somebody double check me?
     
    Last edited: Mar 11, 2014
  3. ronv

    AAC Fanatic!

    Nov 12, 2008
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    You don't need to worry about the gain of the amplifier. The circuit allows you to cancel any offset voltage at the inputs of the op amp. These are usually in the order of a few millivolts or less.
    So R5 allows for the wiper to be set anywhere between the + & - supply voltages. This voltage is then applies to the input - but only a very small amount 100 ohm/200k. With zero volts applied to the input you can then adjust until you get 0 volts at the output.
     
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  4. #12

    Expert

    Nov 30, 2010
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    That's the intended purpose of this circuit, to make the +in exactly zero, but Stin asked for the equation, so I went crazy on it.

    Too literal?
     
  5. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Yea, I understand. I can't do them.
     
  6. #12

    Expert

    Nov 30, 2010
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    I can't do them either. :D
    Literally, I only know 2 opamp equations, but I know how they work and I'm good with algebra, so I can figure out most any feedback loop.
     
  7. LvW

    Active Member

    Jun 13, 2013
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    Hi Stin, just a question:

    I`ve got the impression that you want to see
    * how the input offset voltage Voff influences the output, and
    * how this influence can be compensated by the extra circuitry as shown in the referenced figure.

    Is this correct?

    In this case, the expression for the output voltage must include a term Voff instead of Vin, don`t you think so?
    So what do you want us to do?
     
  8. Stin

    Thread Starter New Member

    Jul 26, 2012
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    Thanks for the replies guys. Without the addition of R1 and R3, the circuit is just a non-inverting amplifier with
    Vout = Vin(1+ R5/(R4+R2))

    What im wondering is how this equation changes with the addition of R1 and R3. What is the equation for Vout, which would depend both on Vin and on the voltage from the additional parts. Ill just go through my workings again there and post them.
     
  9. #12

    Expert

    Nov 30, 2010
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    I thought I covered that in post #2. Please point out how I failed you.
     
  10. Stin

    Thread Starter New Member

    Jul 26, 2012
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    Ah im sorry! I made a mistake. Its the circuit on the top right of page 6 that im trying to understand. Sorry for wasting peoples time.
     
  11. Stin

    Thread Starter New Member

    Jul 26, 2012
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    So there is a current across R5 equal to (Vout-Vin)/R5. So is this the current that flows through R4 and R2 as a result of Vout?
    Im thinking that there's a current across R3 equal to 2V/R3. Will this current effect what happens across R1?

    The app note says that the Gain is 1 + R5/(R4+R2) and the adjustment range
    is +- V(R2/R1) so i think the final equation would be
    Vout = Vin (1 + R5/(R4+R2) +- V(R2/R1).

    It is deriving this equation that im having trouble with.
     
  12. #12

    Expert

    Nov 30, 2010
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    R1 in series with R3 must be seen as a resistance in parallel with R2 so the gain equation changes to Vo = Vin x 1+R5 /(R4 + 99.95 ohms)

    Assume the 50k R3 is at mid point so each end is 25k
    Two 25k's in parallel is 12.5k
    200k is in series with the 12,5k so that makes 201250 ohms

    201250//R2 is your new term for R2
    201250//100 ohms is 99.95 ohms

    Of course, if the pot is not centered, this resistance decreases, but the difference is so small that it won't show up on my calculator, even if I assume R3 is zero ohms.
     
  13. Stin

    Thread Starter New Member

    Jul 26, 2012
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    Ok, so for a large R1 and small R2, R2 is pretty much unchanged, so the gain equation stays roughly the same. That makes sense, thanks. How then does the +-V(R2/R1) come into it?

    Would it be that the current flow from +-V through R1 and R2 is equal to +-V/(R1+R2), which for large R1 is pretty much +-V/R1. So the voltage across R2 as a result of +-V is equal to +-V(R2/R1), and it is the addition/subtraction of this voltage a the junction between R1, R2 and R4 which allows you to trim the offset voltage?
     
  14. #12

    Expert

    Nov 30, 2010
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    Tiny correction: The DC offset at the top of R2 is
    +-V x R2/(R1+R2) where R1 = 201250 ohms in this case

    and...yes.

    That tiny voltage on the top of R2 is usually in the range of less than +/- 5 millivolts.
    That 5 millivolts through R4 and R5 changes the output by a tiny bit.
    .005V x -R5/R4
     
  15. Stin

    Thread Starter New Member

    Jul 26, 2012
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    Ah excellent. Thanks you!
     
  16. LvW

    Active Member

    Jun 13, 2013
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    Again a tiny correction:
    I think, the quoted DC offset is the maximum possible voltage at the top of R2 (because +-V is the voltage at both ends of the pot).
    The equation is correct if V is replaced by the voltage Vm at the midpont of the pot. Only in this case the offset correction can be zero (ideal case).

    Because the input offset voltage Voff appears at the output with an amplification factor identical to the noise gain An=1+R5/(R4+R2) we can calculate the midpoint voltage Vm at the pot, which is necessary to compensate Voff:

    Vm=Voff*[R4/R5 + R4/(R2+R4)]*(1+R1/Rp) with Rp=R2||R4.

    (This formula assumes that R1>>R3/2 and R1>>Rp)
     
    Last edited: Mar 13, 2014
  17. LvW

    Active Member

    Jun 13, 2013
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    Perhaps it is helpful to add that the considerations as given in my last post are - more or less - theoretical, because only the offset voltage is treated.
    But there is another loss of symmetry which is to be added to Voff: Offset current and quiescent currents.
    These currents produce another offset effect which depends on the elected resistor values.
     
  18. #12

    Expert

    Nov 30, 2010
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    Ah yes. Ibias and delta I bias. Well, that's why people build offset adjusters.:p
     
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