Inverting OpAmp question

Discussion in 'General Electronics Chat' started by RFeng, Jun 26, 2012.

  1. RFeng

    Thread Starter New Member

    May 30, 2012
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    Hi,

    I simulated the below inverting OpAmp:
    [​IMG]

    and got the below results:
    [​IMG]

    I don't manage to understand why is it that for input vapc0 = 1.035V we get at the output vapc = 1.967V.

    It looks like that: vapc = 3V - vapc0; I don't understand why.

    Can you please help?

    Thank you.
     
    Last edited: Jun 26, 2012
  2. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
    12
    R7/R6 is a gain of 1, not 100dB, because they are both 1K ohms. To get "only" 80dB of gain, R7 would need to be a 10 Meg.

    What is the point of inducing +1.5V offset into the positive pin?
     
    Last edited: Jun 26, 2012
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  3. crutschow

    Expert

    Mar 14, 2008
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    The gain from the non-inverting input to output is (1 + R7/R6) = +2 (when R6 is grounded or connected to a low impedance source). The gain from the inverting input to output is -R7/R6 = -1. The two signals sum at the output accordingly.
     
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  4. RFeng

    Thread Starter New Member

    May 30, 2012
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    Hi crutschow, Thanks!

    I think I understand you.
    Is it correct to say that in general:

    [​IMG]

    Hi,
    Thank you too!
    I do not know the purpose of the 1.5V offset, as I'm currently learning the entire system (which the circuit is just part of it).
    The system is an Automatic Gain Control unit.
    How did you get the 80dB? (if R7 = 1MΩ).
     
  5. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
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    Sorry typo, 10 Meg -- the point being that generally it is stretching the usefulness of most op amps.
     
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  6. crutschow

    Expert

    Mar 14, 2008
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    You got it. ;)
     
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  7. RFeng

    Thread Starter New Member

    May 30, 2012
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    Thank you dear fellows :)
     
  8. upand_at_them

    Active Member

    May 15, 2010
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    What to observe about your circuit:

    1. R6 and R7 create a voltage divider.
    2. Because you have feedback the op-amp will try to make vapc such that the inverting input is the same as the non-inverting input.
    3. The non-inverting input is 1.5V.

    These three things will give you your answer.

    So you have a voltage divider with 1.035V, a 1k resistor, 1.5V, a 1k resistor, and then some unknown voltage vapc. Solving for vapc gives you the result you got in simulation.
     
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  9. WBahn

    Moderator

    Mar 31, 2012
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    Huh?

    If

    <br />
Vout = V_+ \left( 1\;+\; \frac{R_7}{R_6} \right) +\;V_-\left(-\frac{R_7}{R_6} \right)<br />

    Then that means that

    <br />
Vout = V_+ \;+\; (V_+\;-\;V_-) \frac{R_7}{R_6}<br />

    Since (V+) - (V-) is going to be a tiny voltage, this would say that Vout is always equal to V+.

    The simplest way (at least to me) to get the transfer characteristic is to analyze the circuit. The current flowing from the input is:

    I = (Vin-V+)/R6

    The output voltage is:

    Vout = (V-) - I*R7

    Assuming V+ ~= V-, we have

    Vout = (V+) - ((Vin-V+)/R6)*R7

    Vout = (V+)(1+ R7/R6) - (Vin)(R7/R6)

    NOTE: I now see how the error in the OP's equation was easily missed.

    For V+ = 1.5V and R7=R6, this reduces to

    Vout = 3V - Vin
     
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  10. crutschow

    Expert

    Mar 14, 2008
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    I don't follow the logic of your post. You start and end with the same equation while implying that there is some error.

    You say "Since (V+) - (V-) is going to be a tiny voltage, this would say that Vout is always equal to V+." I don't understand why you say that. (V+) - (V-) may or may not be a tiny voltage. It depends upon their relative value.

    What error in the OP's equation are you talking about? :confused:
     
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  11. WBahn

    Moderator

    Mar 31, 2012
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    I am assuming that V+ is the voltage at the noninverting input of the opamp and V- is the voltage at the inverting input. This is almost always what I see these terms to mean in an opamp circuit. Using one term to mean the voltage at one of the input pins and the other to mean the voltage someplace else seems a bit ackward to me, but I guess if the intent was to mean both to be the voltage someplace else and that it just turns out that V+ also happens to be the voltage at the noninverting input, then I see what you are saying.
     
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  12. JMac3108

    Active Member

    Aug 16, 2010
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    RFeng,

    The easiest way to analyze this kind of circuit is using superposition. The basic procedure is to consider one input at a time while grounding the other, then add the results.

    In your circuit, first ground the 1.5V source and conside the input. The circuit becomes a simple inverting amp with equation, Vout1 = -Vin(Rf/Rin).

    Then ground the input and consider the 1.5V source. In this case the circuit becomes a simple non-inverting amp with equation, Vout2 = Voffset (1+Rf/Rg).

    Combine the results and you'll get your equation. Vout = Vout1 + Vout2.

    If you do a few of these, you'll get where you can analyze them in your head. Its a really simple way to approach this kind of circuit.
     
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  13. RFeng

    Thread Starter New Member

    May 30, 2012
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    Hi,

    Thank you all for your inputs!
    You've given me a great help. :)
     
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