

Low Frequency Response of a FET Circuit


The Field Effect Transistor circuit low frequency response can be evaluated by analyzing the transfer functions
of the elements which affect the response at frequencies below midband. If there is more then one transfer function,
the resultant overall response can be determined from the product of the individual response; usually done
graphically with a Bode plot.
Figure 2 shows the low frequency equivalent of the circuit of the FET configuration in Figure 1. When examining
the input circuit, the input transfer function is developed from C_{1} , R_{G} , and R_{V}.
V1 = V_{in}R_{G} / (R_{V} + R_{G}) + (1/jωC_{1})
V1 / V_{in} = R_{G} / R_{V} + R_{G}(1 / (1  j ( 1 / ω( R_{V} + R_{G}) C)))
V1 / V_{in} = R_{G} / R_{V} + R_{G}(1 / (1  j (ω_{1}(input)) / ω))
(R_{V} + R_{G}) C = 1 / ω_{1}(input) = 1 / 2 π f_{1}(input)
Examining this expression reveals a pole in the low frequency response produced by the input elements, C_{1} ,
R_{G} , and R_{V} at a frequency of f_{1(input)} = 1 / 2 π (R_{V} + R_{G})
C_{1}
The output elements of the circuit can also be expected to produce a low frequency pole. Evaluating the affect
separately by assuming R_{S} in parallel with C_{S} = Z_{S} = 0. And r_{d} in
parallel with R_{D} = R the following equivalent circuit in Figure 3 can be produced.
Applying a Theven solution to the left of X  X we can redraw the equivalent circuit in Figure 4.
Therefore, the transfer function of Figure 4 becomes the following:
V_{out} / V_{in} = V_{out} / g_{m} V_{gs}
R = R_{L} / R + R_{L} [ 1  ( j / ω (R + R_{L})C_{2} ]
V_{out} / V_{in} = R_{L} / R + R_{L}
[ 1 / 1  ( j (ω_{1output} / ω)) ]
So the transfer function becomes:
ω_{1output} = 1 / (R + R_{L})C_{2}
Notice that this transfer function reveals yet another Pole in the low frequency response, at a frequency
determined by the output coupling capacitor C_{2} and the resistance the capacitor sees is
(R + R_{L}). Now, by disregarding the affects of R_{S} and C_{S}, the low
frequency response looks like Figure 5.


C_{S} in association with R_{S} must be assumed to also affect the low frequency
response. It is possible to select values of R_{S} and C_{S} that remove their
effects sufficiently to allow C_{1} and C_{2} to establish dominant poles. Note,
this is not always the case.
Recall the gain equation (i.e. transfer function) of the drain loaded amplifier with source impedance:
Av = g_{m}Z_{L} / 1 + Z_{S} = Z_{L} / (1 / g_{m}) + Z_{S}
Sense only the effects of the source resistance R_{S}, and the source capacitance C_{S}
are being considered, assume the load impedance Z_{L} to be resistive.
Av = R / (1 / g_{m}) + Z_{S}
And,
Z_{S} = R_{S} in Parallel with C_{S}
Z_{S} = [ R_{S} ( 1 / j ω C_{S}) ] / [ R_{S} + ( 1 / j ω C_{S}) ]
Z_{S} = R_{S} / (1 + j ω R_{S}C_{S}), substituting Z_{S}
into the derived gain equation produces:
Av = R / [ 1 / g_{m} + (R_{S} / (1 + (1 / j ω R_{S}C_{S}))) ]
Rearranging the derivation with some basic algebraic relations produces the complete transfer function:
Av =  [ R / (1 / g_{m} + R_{S}) ] · [ (1 + j ω R_{S}C_{S}) /
(1 + j ω (R_{S} 1 / g_{m}C_{S}) / (R_{S} + (1 / g_{m}))) ]
Examination of the completed transfer function result shows a zero at a
frequency denoted with the mathematical expression of:
ω_{So} = 1 / R_{S}C_{S}
And a pole at a frequency denoted as:
ω_{Sp} = 1 / (R_{S}1 / G_{m}C_{S}) / (R_{S} + 1 g_{m})
The frequency response plot of the derived transfer function can be represented graphically as shown in Figure 6.
Engineering key words: frequency response, fet, low frequency, amplifier, amp, freq, bode plot, poles and zeros,
bandwidth, bandpass, rolloff, transfer function, derivation, mathematical expression, gate, drain, source, gain,
Av, specification, manufacture, input function, output, equivalent circuit, elements, Theven, substitution, load,
dB, decade, semilog, loglog, nonlinear effect, circuit analysis.
