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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 C1 , RG , and RV.

V1 = VinRG / (RV + RG) + (1/jωC1)

V1 / Vin = RG / RV + RG(1 / (1 - j ( 1 / ω( RV + RG) C)))

V1 / Vin = RG / RV + RG(1 / (1 - j (ω1(input)) / ω))

(RV + RG) C = 1 / ω1(input) = 1 / 2 π f1(input)

Examining this expression reveals a pole in the low frequency response produced by the input elements, C1 , RG , and RV at a frequency of f1(input) = 1 / 2 π (RV + RG) C1

The output elements of the circuit can also be expected to produce a low frequency pole. Evaluating the affect separately by assuming RS in parallel with CS = ZS = 0. And rd in parallel with RD = 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:

Vout / Vin = Vout / gm Vgs R = RL / R + RL [ 1 - ( j / ω (R + RL)C2 ]

Vout / Vin = RL / R + RL [ 1 / 1 - ( j (ω1output / ω)) ]

So the transfer function becomes:

ω1output = 1 / (R + RL)C2

Notice that this transfer function reveals yet another Pole in the low frequency response, at a frequency determined by the output coupling capacitor C2 and the resistance the capacitor sees is (R + RL). Now, by disregarding the affects of RS and CS, the low frequency response looks like Figure 5.





CS in association with RS must be assumed to also affect the low frequency response. It is possible to select values of RS and CS that remove their effects sufficiently to allow C1 and C2 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 = -gmZL / 1 + ZS = -ZL / (1 / gm) + ZS

Sense only the effects of the source resistance RS, and the source capacitance CS are being considered, assume the load impedance ZL to be resistive.

Av = -R / (1 / gm) + ZS

And,

ZS = RS in Parallel with CS

ZS = [ RS ( 1 / j ω CS) ] / [ RS + ( 1 / j ω CS) ]

ZS = RS / (1 + j ω RSCS), substituting ZS into the derived gain equation produces:

Av = -R / [ 1 / gm + (RS / (1 + (1 / j ω RSCS))) ]

Rearranging the derivation with some basic algebraic relations produces the complete transfer function:

Av = - [ R / (1 / gm + RS) ] · [ (1 + j ω RSCS) / (1 + j ω (RS 1 / gmCS) / (RS + (1 / gm))) ]



Examination of the completed transfer function result shows a zero at a frequency denoted with the mathematical expression of:

ωSo = 1 / RSCS

And a pole at a frequency denoted as:

ωSp = 1 / (RS1 / GmCS) / (RS + 1 gm)

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, log-log, nonlinear effect, circuit analysis.


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