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Low frequency FET response, determining f[sub]L[/sub] of Cs capacitor

Discussion in 'Electronics Homework Help' started by nvjnj, Aug 26, 2015.

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  1. nvjnj

    nvjnj

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    Sep 13, 2014
    I'm trying to get the low frequency cutoff value of the capacitor Cs of the following ac equivalent circuit using the equation:
    f(L)=1/(2*pi*Req*Cs)
    [​IMG]

    To do this, I'm supposed to reduce the circuit "seen" by capacitor Cs to an equivalent RC circuit where Req would be the equivalent resistance seen by Cs.

    Employing Thevenin's Theorem, open-circuiting all current sources and short circuiting all voltage sources gives me the equivalent resistance as:

    Req=RS||((RL||RD)+rd) which simpifies into the equation:
    Req=RS/(1+(RS/(RDRL+rd)))

    But the real equation stated in the internet and text books is:
    Req=RS/(1+(RS(1+gmrd)/(RDRL+rd)))

    So, I am missing the factor (1+gmrd) in my equation. Please help me figure out what I am doing wrong.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,363
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    Jan 21, 2010
    there is a current source there which is a dependent current source. You can't just open circuit that.
     
  3. nvjnj

    nvjnj

    11
    0
    Sep 13, 2014
    I see....I missed that.
    Thanks....I'll try to redo it
     
  4. nvjnj

    nvjnj

    11
    0
    Sep 13, 2014
    Okay, so I tried doing it by not removing the dependent current source. But I'm still not getting the answer. These are the steps I followed:

    Req=V(open-circuit)/I(short-circuit)

    V(open-circuit):
    Open circuiting Cs, Voltage across this open circuit is equal to voltage across Rs. Voltage across Rs = Is*Rs, where Is is the current flowing through Rs.
    Is=(rd/(rd+(Rs+RD||RL)))*gmVgs

    Therefore V(short-circuit)=(rd*Rs/(rd+(Rs+RD||RL)))*gmVgs

    I(short-circuit):
    Short circuiting Cs would make Rs redundant and is removed. Hence current accross this path is:
    Is=(rd/(rd+(RD||RL)))*gmVgs

    Hence Req=(rd*Rs/(rd+(Rs+RD||RL)))*gmVgs / (rd/(rd+(RD||RL)))*gmVgs

    This eliminates my gm terms, which is required in the final equation as stated in the original post. Please tell me what I am doing wrong?
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Why did you calculate V(short circuit)?
     
  6. nvjnj

    nvjnj

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    Sep 13, 2014
    Sorry, that was a typo, its actually V(open-circuit)=(rd*Rs/(rd+(Rs+RD||RL)))*gmVgs
     
  7. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
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