# Effect of bypass capacitor on MOSFET amplifier circuit

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1. Aug 2, 2016

### bl965

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.

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. Aug 2, 2016

### tech99

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.

3. Aug 3, 2016

### LvW

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)

4. Aug 4, 2016

### rude man

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.

5. Aug 4, 2016

### tech99

Can you tell me please the origin of the factor 3/4? Many thanks.

6. Aug 4, 2016

### rude man

Sure.
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
7. Aug 5, 2016

### tech99

Thanks, I did not look closely at the circuit.