Common source amplifier low frequency resonse

TheRedDevil18 · Jul 31, 2016

Here's the amplifier in question:

Circuit schematic (b) of a common-source MOSFET amplifier with source bypass capacitor CS, drain resistor RD, coupling capacitors CC1 and CC2, load RL, gate resistor RG, and signal source Vsig with Rsig, showing drain current Id and output current Io.

I'm not understanding this part:

$Printed text followed by an equation: I_o = -I_d times R_D over R_D + 1/(sC_C2) + R_L, with the current divider rule used to find the fraction of I_d flowing through R_L.$

Why is Cs not included in the current divider ?
To my understanding of current divider, It = Id+Io, where It is the current flowing through Rd
So, Io = It*(1/Cs)/(1/Cc2)+RL

Jony130 · Jul 31, 2016

But Cs is in series with Id current. Please explain to me how Cs affects Io.

TheRedDevil18 · Jul 31, 2016

Jony130 said:

But Cs is in series with Id current. Please explain to me how Cs affects Io.

Yes, it's correct. I must have gotten confused somewhere but it's fine now

Jony130 · Jul 31, 2016

Do you have any more qestion about this amp?

Small-signal circuit schematic of a common-source MOSFET amplifier with drain resistor R_D, load R_L, coupling capacitor C_C2, source capacitor C_S, and labeled currents I_D, I_o, I_r and output V_o.

TheRedDevil18 · Aug 1, 2016

Jony130 said:

Do you have any more qestion about this amp?
View attachment 104139

No, I'm good, thanks

Common source amplifier low frequency resonse

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