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Effect of bypass capacitor on MOSFET amplifier circuit

  1. Aug 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A bypass capacitor increases gain in the mid-band region. Explain how with the figure and small signal model. Assume gm=0.5mS.

    IMG_20160802_113253.jpg IMG_20160802_113253.jpg

    2. Relevant equations
    Rs = 1 k Ohm

    3. The attempt at a solution
    With the bypass capacitor, the source would be treated as grounded and the eqautions to find gain would be:
    Vo= -gm x Vgs x Rd
    Vgs= Vsig x R1||R2 / (R1||R2 + Rsig)
    How does the source resistor effect the circuit? Vgs will be different because Vs will no longer be 0. Vs will be gm x Vgs x Rs.

    Thank you.
    Last edited by a moderator: Apr 17, 2017
  2. jcsd
  3. Aug 2, 2016 #2
    If the output is 1 volt, then the current Id in Rd is 1/Rd.
    The same current flows in Rs. So the voltage Vrs across Rs is Id x Rs.
    The voltage across source to gate (not ground referenced, actually at the transistor) Vsg must be Id / Gm.
    So the total voltage between gate and ground Vg is the sum of Vrs and Vsg.
    And the gain is output voltage/ input voltage = 1/Vg.

    Roughly speaking, the gain is Rd / Rs because Vrs is very small.
  4. Aug 3, 2016 #3


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    Without the capacitor C3 the resistor RS causes a negative feedback effect not only for DC (that is his purpose!) but also for ac signals. Sometimes this signal feedback is a desired effect (better linearity, increase of signal input resistance), but it decreases the signal gain. For this reason, the resistor RS sometimes is bypassed by a capacitor which cancels signal feedback for frequencies above the corresponding highpass cut-off frequency (roughly wc~gm/C3)
  5. Aug 4, 2016 #4

    rude man

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    The very large value of C3 means this is a very gain-unstable circuit with grossly nonlinear behavior. But the gain is now (3/4) gmRD which is >> (3/4) RD/RS which it would be in the absence of C3.
    The above applies for mid-band and high-band signals only, thanks to C1's blocking of low-frequency and dc signals.
  6. Aug 4, 2016 #5
    Can you tell me please the origin of the factor 3/4? Many thanks.
  7. Aug 4, 2016 #6

    rude man

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    30K/(30K + 10K) going from the input to the source.
    EDIT: sorry, that's not quite right. The correct fraction is 0.69, not 0.75. I missed the 90K going from the gte to Vcc.
    You have a voltage divider made by 22.5K shunt and 10K series.
    Last edited: Aug 5, 2016
  8. Aug 5, 2016 #7
    Thanks, I did not look closely at the circuit.
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