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Homework Help: Finding Transfer Function from Circuit

  1. Nov 25, 2012 #1
    I need to find the transfer function from the attached circuit schematic.

    V_DD = 5V; R1 = 200kΩ; R2 = 300kΩ; and C1 = 900nF

    I used nodal analysis to attempt to get Vout over Vin but alas ended up with an equation like this:

    [itex]\frac{Vout}{Vin}[/itex] = [itex]\frac{-j}{120kΩ*ωC}[/itex] + [itex]\frac{5j}{200kΩ*ω*c*Vout}[/itex]

    I've run into a wall on how to derive the equation with V_DD. Any help would be great.

    Attached Files:

    Last edited: Nov 25, 2012
  2. jcsd
  3. Nov 25, 2012 #2
    It's not possible to find a transfer function that depends only on Vin :)

    Vout is a function of both Vin and Vout and its equation will therefore look like
    Vout = A*Vin + B*Vdd
    which can be seen immediately if you consider superposition to solve for Vout

    Are you doing a small signal analysis?


    By using superposition you can consider the portion of Vout caused by Vdd and Vin separately. Then:

    Vout = (Vout/Vin)*Vin + (Vout/Vdd)*Vdd


    Vout/Vin = transfer function due to Vin with Vdd=0
    Vout/Vdd = transfer function due to Vdd with Vin=0
    Last edited: Nov 25, 2012
  4. Nov 25, 2012 #3
    Yes, at least that's how the question is set up. So if I can't get rid of Vout on the right hand side of the equation how do I interpret the transfer function?

    i.e. plotting a bode plot
  5. Nov 25, 2012 #4
    Ok, I think I have it then. So the overall transfer function of the circuit is the addition of the DC and the AC portions of the circuit correct?

    (Wording might be a little off)
  6. Nov 25, 2012 #5
    Ah ok. In small signal analysis, the DC Vdd source is only setting up bias points. Because Vdd = constant, in this equation:

    Vout = A*Vin + B*Vdd

    The B*Vdd term settles to a constant value some short time after the circuit is switched on. We then only consider the remaining portion:

    Vout = A*Vin

    which you can write a transfer function for. The total result is understood to be the sum of the steady (dc) value supplied by Vdd and the variable part due to Vin.

    By superposition, we find Vout=A*Vin by setting the dc sources to zero (ie set Vdd=0).


    So for ac analysis:


    Find the DC portion of the response. This is called the bias point. We only consider steady inputs like your Vdd=5 volts in this part. If Vin had a dc component we would consider that dc part *only* as well. Superposition is going to allow us to find the total response by separating this DC part that doesn't change while the circuit is on from the more interesting part in 2 below.

    In your circuit, Vin is applied to a series capacitor. This is often done in audio circuits to block the dc component of Vin. In other words, if Vin has a dc component (and it will), it is blocked by that capacitor and will not affect the steady value of Vout determined by Vdd. This is a means to allow the dc bias point of the circuit generating Vin to be set without considering the following circuit.


    Find the part of the response that varies. By superpostion, we can set Vdd (and all other steady sources, including the dc portion of Vin) to zero in this part. This means in your circuit you would ground Vdd. Now you can find your transfer function Vout/Vin.

    It is understood that the total response will be a steady output voltage plus a varying part (Vout/Vin) added to it. In your bode plot, you are only interested in Vout/Vin. The steady part is what keeps the following transistors properly biased in their active regions.

    I explained it differently than is normally done in books so I hope I gave a different angle if you didn't pick up on it the other way :)
    Last edited: Nov 25, 2012
  7. Nov 25, 2012 #6
    that's what I was thinking when you mentioned it in your earlier post! (except you had better words. . . and descriptions. . .) that cleared the entire problem up for me!

    Thank you very much! :)
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