An equivalent circuit for a FET common-source amplifier or BJT common-emitter amplifier can be modeled by the circuit shown in Fig. 2.40a. We wish to determine an expression for the gain of the amplifier, which is the ratio of the output voltage to the input voltage.

Question

Accepted Answer

Note that although this circuit, which contains a voltage-controlled current source, appears to be somewhat complicated, we are actually in a position now to solve it with techniques we have studied up to this point. The loop on the left, or input to the amplifier, is essentially detached from the output portion of the amplifier on the right. The voltage across [latex]R_{2}[/latex] is [latex]v_{g}[/latex](t), which controls the dependent current source.

To simplify the analysis, let us replace the resistors [latex]R_{3},\,R_{4},[/latex] and [latex]R_{5}[/latex] with [latex]R_{L}[/latex] such that

[latex]\frac{1}{R_{L}}=\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}[/latex]

Then the circuit reduces to that shown in Fig. 2.40b. Applying Kirchhoff’s voltage law to the input portion of the amplifier yields

[latex]v_{i}[/latex](t) = [latex]i_{1}(t)(R_{1}\,+\,R_{2})[/latex]

and

[latex]v_{g}[/latex](t) = [latex]i_{1}(t)R_{2}[/latex]

Solving these equations for [latex]v_{g}[/latex](t) yields

[latex]v_{g}[/latex](t) = [latex]\frac{R_{2}}{R{1}\,+\,R{2}}v_{i}(t)[/latex]

From the output circuit, note that the voltage [latex]v_{o}[/latex](t) is give by the expression

[latex]v_{o}(t)\,=\,–g_{m}v_{g}(t)R_{L}[/latex]

Combining this equation with the preceding one yields

[latex]v_{o}(t)\,=\,\frac{-g_{m} R_{L} R_{2}}{R_{1}\,+\,R_{2}}v_{i}(t)[/latex]

Therefore, the amplifier gain, which is the ratio of the output voltage to the input voltage, is given by

[latex]\frac{v_{o}(t)}{v_{i}(t)}\,=\,–\frac{g_{m} R_{L} R_{2}}{R_{1}\,+\,R_{2}}[/latex]

Reasonable values for the circuit parameters in Fig. 2.40a are [latex]R_{1}[/latex] = 100 Ω, [latex]R_{2}[/latex] = 1 kΩ, [latex]g_{m}[/latex] = 0.04 S, [latex]R_{3}[/latex] = 50 kΩ, [latex]R_{4}\,=\,R_{5}[/latex] = 10 kΩ. Hence, the gain of the amplifier under these conditions is

[latex]\frac{v_{o}(t)}{v_{i}(t)}\,=\,\frac{–(0.04)(4.545)(10^{3})(1)(10^{3})}{(1.1)(10^{3})}[/latex]
= –165.29

Thus, the magnitude of the gain is 165.29.

Question 2.30: An equivalent circuit for a FET common-source amplifier or B...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions