Analysis of a FET Amplifier

by Tory W. Vaughan III, BEE Georgia Institute of Technology

NOTE: This worksheet computes the high and low 3dB frequencies of a single-stage field-effect transistor amplifier.

Introduction

In this worksheet we derive the high and low 3dB frequencies of a single stage F ield E ffect T ransistor (FET) amplifier in a Common Source configuration. The stage consists of the input voltage (Vvs) in series with a resistance (R1- the voltage source's resistance) feeding the FET through a coupling capacitor (C1). The resistor R2 is across the gate to source terminals of the FET and determines the input resistance. Capacitor C2 is the gate to drain capacitance of the FET. The amplification factor for the FET is mu. R3 is the drain to source resistance of the FET. C3 is a coupling capacitor connecting the output of the FET to the load resistance (R4). The complex frequncy variable is s. V1, V2, V3, V4, and Vo are the node voltages for the circuit. The Miller approximation will not be used showing the power of Maple to solve a system of equations and allowing us to obtain the exact frequency response of the FET.

> restart:

The Frequency Response of a Single Stage FET Amplifier

We write 4 node equations for the circuit.

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We now solve these for the node voltages (not displaying the long & complicated result)

> solve({eq1,eq2,eq3,eq4},{V1,V2,V3,Vo}):

We assign the expressions given in this solution to the variables Vo, V1, V2 and V3.

> assign(%);

We now solve for the system transfer function H.

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To express H as a function of the frequency, we substitute the s (the complex frequency variable) for an expression in f (frequency):

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> H;

The transfer function H can now be separated into a magnitude (AVM) and a phase component (PhaseAV).

> AVM:=evalc(abs(H)):

> PhaseAV:=evalc(argument(H)):

We now convert the magnitude of H into decibels or dB,

> AVdB:=20*log10(AVM):

and the phase component of H from radians to degrees:

> R2D:=evalf(360/(2*Pi));

> AVdeg:=R2D*PhaseAV:

We now define the values of circuit components:

> R1:=1000: R2:=10000: R3:=1000: R4:=10000:
C1:=1*10^(-6): C2:=10*10^(-12): C3:=1*10^(-6):
mu:=100:

Now we can calculate the magnitude and phase at various frequencies. We then load Maple's plotting package "plots", and plot the frequency dependence of the magnitude and phase components.

> gaindata:=NULL: phasedata:=NULL:
for a from 0 to 8 do:
for i from 2*10^a to 10^(a+1) by 10^a do
gaindata:=gaindata, [i, evalf(subs(f=i,AVdB))];
phasedata:=phasedata, [i, evalf(subs(f=i,AVdeg))];
od:
od:

> with(plots):

> loglogplot([gaindata], thickness=2, color=red, style=line, axes=boxed, title="Gain vs. Frequency", labels=["Frequency (Hz)","Gain(dB)"]);

> loglogplot([phasedata], thickness=2, color=blue, style=line, axes=boxed, title="Phase (in degrees) vs. Frequency", labels=["Frequency (Hz)","Phase"]);

The High and Low 3dB Frequencies

To find the low and high 3dB frequencies, we first determine the mid-frequency gain and subtract 3dB from it. We then use Maple's numerical solver, fsolve, to determine the frequency corresponding to this value of the gain. From the magnitude plot we determine a frequency in the mid-band region, say 4000 Hz.

> AVmid:=evalf(subs(f=4000, AVdB));

> eq5:=AVmid-3=20*log10(AVM);

Solve for the value of frequency which gives the value of gain to be 3dB below the mid-frequency value. From the graph of the magnitude plot we determine the ranges of frequency which give the respective 3dB points to narrow the search for fsolve.

> `f3dB low`:=fsolve(eq5,f, f=10..2000);

> `f3dB high`:=fsolve(eq5,f, f=2000..100*10^6);

Let us plot these, having first loaded Maple's "plottools" package.

> with(plottools):
aplot:= loglogplot([gaindata], thickness=2, color=red, style=line, axes=boxed,
title=" Low and High 3dB Points", labels=["Frequency (Hz)","Gain(dB)"]):
bplot:=line([0.1,log10(35.33556394)], [7.1,log10(35.33556394)], color=red, linestyle=3):
cplot:=line([log10(22.50999550),0.58], [log10(22.50999550),1.6], color=blue, linestyle=3):
dplot:=line([log10(188219.0685),0.58], [log10(188219.0685),1.6], color=green, linestyle=3):
display([aplot,bplot,cplot,dplot]);

Conclusions

In this worksheet we have shown how Maple can be used in the symbolic analysis of an electronic circuit which involves a single-stage field-effect transistor amplifier. Maple's "solve" command is used to solve the set of node equations and hence expressions are found for the gain and phase of the circuit as a function of frequency. When values are assigned to the components of the circuit, we can then plot the responses and find the high and low 3dB frequencies using Maple's numerical solver, "fsolve".

Disclaimer: While every effort has been made to validate the solutions in this worksheet, Waterloo Maple Inc. and the contributors are not responsible for any errors contained and are not liable for any damages resulting from the use of this material.