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

nvjnj

Sep 13, 2014
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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)
EET211-Ch6a-Figure2.jpg


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.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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there is a current source there which is a dependent current source. You can't just open circuit that.
 

nvjnj

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

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Why did you calculate V(short circuit)?
 
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