Effect of bypass capacitor on MOSFET amplifier circuit

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Discussion Overview

The discussion focuses on the effect of a bypass capacitor on a MOSFET amplifier circuit, particularly how it influences gain in the mid-band region. Participants explore the small signal model and the role of source resistance in determining gain, along with the implications of bypassing the source resistor.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that a bypass capacitor increases gain by allowing the source to be treated as grounded, affecting the equations used to calculate gain.
  • Another participant discusses the relationship between output voltage, current, and the voltage across the source resistor, suggesting that the gain can be approximated as Rd/Rs.
  • A different viewpoint highlights that without the bypass capacitor, the source resistor introduces negative feedback for AC signals, which can decrease gain but may improve linearity and input resistance.
  • One participant notes that a large bypass capacitor leads to gain instability and nonlinear behavior, providing a specific gain expression for mid-band and high-band signals.
  • There is a request for clarification regarding the origin of the factor 3/4 in the gain expression, with subsequent replies attempting to explain it through voltage divider principles, but corrections are made regarding the actual fraction involved.

Areas of Agreement / Disagreement

Participants express differing views on the implications of bypassing the source resistor and the resulting gain characteristics. There is no consensus on the exact origin of the factor 3/4, with corrections and alternative calculations presented.

Contextual Notes

Some participants mention specific values and configurations in their calculations, but there are unresolved aspects regarding the assumptions made in the gain expressions and the effects of the bypass capacitor on circuit behavior.

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Homework Statement


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


Homework Equations


IMG_20160802_113302_1.jpg

Rs = 1 k Ohm

The Attempt at a Solution


With the bypass capacitor, the source would be treated as grounded and the equations 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.
 
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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.
 
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)
 
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.
 
rude man said:
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.
Can you tell me please the origin of the factor 3/4? Many thanks.
 
tech99 said:
Can you tell me please the origin of the factor 3/4? Many thanks.
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:
rude man said:
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.
Thanks, I did not look closely at the circuit.
 

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