Transfer function

Discussion in 'Homework Help' started by tim510, Aug 13, 2012.

  1. tim510

    Thread Starter New Member

    Jan 11, 2012
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    Hello all,

    I'm trying to get the transfer function of this om amp example:
    [​IMG]

    Combining the capacitor and resistor in series gives \frac {1+sRC}{sC}=Z1

    Usually I would use KCL but for this particular example it would need a load on the output to work out an equation. For example, if I put a load Z2 connected between Vout and ground the transfer function can than be solved and would come out like this \frac {Vout} {Vin} = \frac {Z2} {Z2-Z1}

    This om amp [​IMG]
    has the transfer function \frac {Vout} {Vin} = \frac {1} {sRC}

    My question is why is it necessary to have a load on the output of the first om amp example? Is it because there is no reference to ground?

    Please excuse me if I'm not seeing the obvious. :confused:
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    In the first circuit there is no means by which output current may flow back towards the op-amp negative input. Providing a load to signal common / ground at the op-amp output won't help either. It wont work unless there is some additional impedance from the -ve input to signal common / ground. Plus, even if you provide that path to ground [say via a resistor] you would still have a pole at DC which would mean the circuit behaves somewhat like an integrator with lead compensation.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    How did you come up with this?

    You are saying that if Z1=Z2 that Vout is infinite (or at least the max output of the opamp). Does that make sense?

    If you are assuming that Vin is appearing at the -ve input of the opamp, then you would get

    \frac {Vout} {Vin} = \frac {Z2} {Z2+Z1}

    But this is a case of throwing an equation at a circuit where it doesn't apply. The voltage divider equation is for a circuit in which Vin is applied to the top of Z1 and then Vout is a result of the current from Vin flowing through the series connected Z1 and Z2. Note that to be 'series connected', the assumption must be that the current flow in or out of the opamp output. So, where is the current that is flowing in Z1 and Z2 coming from?

    In general, don't try to apply KCL at the output of an opamp unless you specifically trying to determine how much current the amp is supplying. And, most especially, do not assume that it is not supplying current; you can usually assume that the inputs don't have any current in or out of them, you cannot make that assumption about the output.

    What you should find is that this configuration will act mostly like a voltage follower but probably with a drifting DC value as the leakage current of the -ve input slowly charges the capacitor in one direction or the other.
     
  4. tim510

    Thread Starter New Member

    Jan 11, 2012
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    Thanks for your responses. I just want a clear understanding of what I'm doing wrong here.

    Going back to the first example, in my analysis I violated the ideal opamp equations. Zero current into the inputs. Right?
     
    Last edited: Aug 15, 2012
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    WBahn has already pointed out the probable practical circuit behavior. Assuming the op-amp is ideal the input resistance at either +/- terminal is infinite. So no current can flow in the negative feedback path. In that case the voltage at the -ve input is simply the same as the output voltage. This means the circuit behaves in the ideal case as a voltage follower.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    Yes, I would say that was the case. But also, I think you violated KCL at the output by assuming that there WAS zero current into/out of the op amp output, which is not a valid assumption, except by extreme coincidence. It required BOTH bad assumptions in order to get the result you did.
     
  7. ramancini8

    Member

    Jul 18, 2012
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    Adding an impedance from Vout to ground does nothing because Vout is a zero impedance source. Add Rin from the inverting input to ground and the gain equation becomes ((Rin+R)Cs + 1)/RinCs. Let Rin go to infinity as is the case and the gain is 1.The ac response is much like an integrator. The circuit would never work in the real world because the inverting input has no bias current except for C1 leakage current.
     
  8. tim510

    Thread Starter New Member

    Jan 11, 2012
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    I have a circuit here that I took from an application note that I'm trying to understand. It was what my original post was leading to. The output of the opamp is driving a bjt.

    [​IMG]

    It's like the first opamp in my original post but with an impedance connected to the input -ve and signal ground/common.
    t_n_k's original post suggest that this is would act like an integrator with lead compensation.

    I want to find the transfer function to understand what it's doing. I'm not sure how to go about analyzing the circuit (KCL at node Vf?). My first step would be to add the impedances of the series resistor and capacitor and call it Z1.

    Any suggestions before I violate rules :D
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The transfer function is given by

    G(s)=\frac{\( R_1+R_2 \parallel R_3 \)C_1s +1}{R_2 \parallel R_3 C_1 s}

    Another way to look at the response is to consider the output to be the summation of a term proportional to the input and a term related to the integral of the input as would be clear with re-casting the transfer function as

    G(s)=\frac{\( R_1+R_2 \parallel R_3 \)}{R_2 \parallel R_3 }+\frac{1}{R_2 \parallel R_3 C_1 s}
     
    Last edited: Aug 24, 2012
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