Simple Op Amp problem

Discussion in 'Homework Help' started by rishid, Sep 12, 2006.

  1. rishid

    Thread Starter New Member

    Oct 24, 2005
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    Hi,

    Seem to have forgotten how to do this stuff. Trying to find Vo / Vi

    Circuit attached as image.

    Thanks for any help.

    RishiD
     
  2. pebe

    AAC Fanatic!

    Oct 11, 2004
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    Depending on Vi, current will flow through the feedback resistor from Vo to always try to maintain the voltage at the inverting input at its status quo.

    So with the source and feedback resistors both of 10K, Vo will equal -Vi.
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
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    From looking at your diagram, you will need to take account of the 10K resistor to ground from the output. Since the inverting opamp configuration has a virtual earth at the inverting input (i.e. in this case the inverting input is at ground potential) then the 10k resitsor can be effective viewed as being in parallel with the 100k resistor.

    For the inverting opamp:

    Vout/Vin = -Rf/R1

    Where Rf = 100k//10k
    And R1 = 10k

    From there you should be able to calculate the gain.

    Dave
     
  4. pebe

    AAC Fanatic!

    Oct 11, 2004
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    Oops! Sorry! I viewed the circuit then took it off the screen to write a reply. I seemed to remember the feedback was 10K - not 100K. :mad:
     
  5. hgmjr

    Moderator

    Jan 28, 2005
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    After studying the circuit, I guess I don't see the 10K between the opamp's output and ground playing any significant role in computing the gain of this particular "ideal" opamp configuration.

    I always revert back to the basic assumptions for "ideal" opamps when I tackle these circuits.

    1. The open-loop gain approaches infinity.
    2. The impedance of the output of the opamp approaches zero.
    3. The positive and negative input impedance approaches infinity.

    In this opamp, the power source is not stated. In that case, I usually assume the opamp is being powered by a +V/-V power source.

    I view an opamp as having one mission in life. That mission is to continuously adjust its output so that the voltage at the negative and positive terminals are always equal. Stated another way, the opamp (in the presense of negative feedback) attempts to adjust its output so that the difference between the voltages at its positive and negative input terminals is as close to zero as possible.

    With all this in mind, I look at the subject circuit and see that for a positive input signal, the output is going to need to swing negative in order to keep the voltage at the negative input at the same voltage as the positive input which is connected to ground. From this observation it is possible to say that the output of this opamp is going to be inverted from the input signal.

    I next note that with the negative input adjusted to ground by the action of the opamp's output, it is possible to use Ohm's Law to compute the current flowing in Ri for a given input voltage Vi. Since the input impedance looking into the negative terminal is infinite then the current flowing in Ri must of necessity being flowing in Rf.

    This leads me to conclude that, just as pebe has stated, the gain of this opamp is

    Vo/Vi = -Rf/Ri.

    Did I overlooked something in my analysis, Dave?

    hgmjr
     
  6. Dave

    Retired Moderator

    Nov 17, 2003
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    Hi Harry,

    Your analysis is correct for the inverting configuration, however I beg to differ that the 10k resistor at the output plays no part. Consider the following mathematical analysis of the circuit.

    We agree that the gain for the inverting opamp is given as thus:

    Vo/Vi = -Rf/R1

    We agree that R1 = 10k

    However I say that Rf = 100k//10k because of the 10k resistor to ground at the output.

    For the subsequent analysis, lets call the 100k resistor R2, and the 10k resistor to ground R3. Therefore, Rf = R2//R3.

    Now consider that R3 can assume any resistance value between 0 and infinity. Considering the limits:

    If R3 → 0, i.e. there is a short circuit between Vo and ground.

    Then R2//R3 = 0 and the gain Vo/Vi = -0/R1 = 0

    As expected since the output is grounded there would be no gain.

    If R3 → ∞, i.e. we now have a situation which is the same as the typical inverting opamp configuration with just R1 and R2, which is now Rf. (Ref: Typical inverting opamp).

    Then R2//R3 → R2 and the gain Vo/Vi → -R2/R1

    As expected we would ascertain the gain for the typical inverting opamp configuration.

    This leads to the conclusion, that the 10k resistor (R3) affects the gain of the inverting configuration as such:

    1) Increasing R3 causes the gain of the inverting configuration to tend towards -R2/R1, with a maximum where R3 = ∞.
    2) Decreasing R3 causes the gain of the inverting configuration to tend towards 0, with the gain equal to 0 where R3 = 0 (i.e. open circuit).

    That is how I look at the circuit. Seem a reasonable analysis?

    Dave
     
  7. pebe

    AAC Fanatic!

    Oct 11, 2004
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    Hi Dave,
    I agree with hmjnr on this. The attachment shows the equivalent circuit.

    IC o/p is the output of the voltage source of the output stage, and Rs is its source resistance. Rl is the 10K load, R1 is the 10K input resistor, and Rf is the 100K feedback resistor.

    If Vi increases, current will flow through R1 and Rf to Vo. Whether it then goes through Rl or through Rs to the output stage is not relevant and doesn’t affect gain calculations. So for the inverting i/p to stay almost constant, Vo/Vi must equal Rf/R1.

    That will apply as long as Rl is fairly large compared with Rs and IC o/p can still operate in its linear range (Rs is about 50Ω in a 741).
     
  8. n9352527

    AAC Fanatic!

    Oct 14, 2005
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    I tend to agree with hgmjr, that the 10K resistor between the output and ground does not affect the gain in normal operation and the op-amp gain is given by -Rf/Ri.

    Of course, Dave's analysis above on the effect when R3 = 0 is quite right, that the output voltage would be 0V. But, this is not due to the interaction between paralled R3 and Rf, more on the driving capability of the op-amp output. This will not be linear with decreasing R3 value. In my opinion, the gain will be -Rf/Ri when R3 is decreased until the op-amp runs out of output drive and can not provide more current, then the output voltage would depend on R3 and Imax (~ Imax*R3) unless the op-amp didn't like it and decided to quit instead.
     
  9. Dave

    Retired Moderator

    Nov 17, 2003
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    Hi all,

    Having sat and mused this over on the train, I am going to jump ship and agree with hgmjr, pebe and n9352527. My above analysis was based on an inconsistency I saw with Kirchoff's Current Law, however (in my stupidity I oversaw one critical element) I made one grave mistake.

    With reference to the schematic in the opening post and pebe's equivalent circuit, if we apply +10V at the input then we can see that the current through R1 is 1mA. If we disregard the 10k resistor at the output, then it is easy to see the current through Rf is 1mA, therefore Vo = -100.

    This should come as no surprise since the gain, Vo/Vi = -Rf/R1 = -100k/10k = -10.

    If we now consider the 10k resistor at the output, we can see that the voltage across it is 0 --100V = 100V. Therefore a current is drawn from the ground equal to 100/10k = 10mA. This is where I became confused, since I knew that the currents flowing into node Vo must equal 0 (from KCL). By considering KCL, some simple Maths could be employed to show that Vo = 0.909V would satisfy the requirements of KCL and hence be proved correct by my earlier assertion that Vo/Vi = -(Rf//R2)/R1. The big mistake I made was not taking into consideration that the current at node Vo can flow back into the opamp. Thus the following conclusions can be made for Vi = +10V:

    1) The voltage gain Vo/Vi = -Rf/R1
    2) Current through R1 and Rf = 1mA
    3) Current through R2 is 10mA
    4) Because there is nothing applied at the output of the opamp, the current flowing back through the opamp output = 11mA (from KCL).
    5) The 10k resistor at the output has the effect of increasing the current gain to 10.

    Sorry for the inaccuracies before, hope this has been a valauable lesson!

    Dave
     
  10. hgmjr

    Moderator

    Jan 28, 2005
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    Dave,

    No harm, no foul.

    I think your analysis would hold up nicely in the case where the opamp in question was a transconductance opamp. In that case the output of the opamp would have been a current source and thus the output impedance would have been set to 10K by the 10K resistor to ground.

    Is that true?

    hgmjr
     
  11. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Actually, Harry and Dave are both correct. It depends on whether you take the Vi as a voltage input or if you decide to use a current input. KCL amd KVL still rules. I assumed Vi to be a Voltage and immediately agreed with Harry.

    Sorry about using a pdf file.
     
  12. Dave

    Retired Moderator

    Nov 17, 2003
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    In essence, yes. Effectively the opamp in the opening post will always have the same voltage gain, Vo/Vi. Further to my previous post, the important point to note is that the potential at the inverting input is a virtual earth, not a physical earth. If it were a physical earth, then we would have another violation of KCL because the 10k resistor at the output would be drawing 10mA from earth, and the sum of currents at the earth potentials in the circuit would not be zero.

    The resistor at the output serves to control the current gain. Say for example the resistor were a 5k resistor, then it it would draw 20mA from earth and the current gain would be -20mA/1mA = -20. Therefore:
    1) Increasing the value of the output resistor reduces the current gain of the configuration.
    2) Decreasing the value of the output resistor increases the current gain of the configuration.

    Good work JoeJester, that pretty much shows what I was gesturing at. By varying R3, the value of AM4 and hence AM2 changes accordingly. And the current gain is equal to AM4/AM5.

    Dave
     
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