Question
A MOSFET in the common source configuration is modeled as a two-port network as illustrated in Figure 12.4. If $R_L=6 \mathrm{k} \Omega$, and $y_{11}=0 ; y_{12}=0 ; y_{21}=2 \times 10^{-3} \mathrm{~S}$; and $y_{22}=0$, determine the small-signal voltage gain of the amplifier.
Step 1
We have the following values: - Load resistance, \( R_L = 6 \, \text{k}\Omega = 6000 \, \Omega \) - Admittance parameters: - \( y_{11} = 0 \) - \( y_{12} = 0 \) - \( y_{21} = 2 \times 10^{-3} \, \text{S} \) - \( y_{22} = 0 \) Show more…
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Figure 3: Common-source amplifier with an LC resonator load (a) Draw a small-signal equivalent circuit of a common-source amplifier with a resonant load in Fig. 3. Hint: You will end up with a parallel RLC resonator formed by RL, Lp, and Cp at the drain node. (b) In (a), please prove that the small-signal voltage gain is given by Ay = -gmZp, where Zp is the total impedance of the parallel RLC resonator at the drain. Voltage gain in this case is defined as the ratio of input and output voltage phasors. (c) In (b), it follows that the magnitude of voltage gain is |Av| = gm|Z|. Z is the impedance of a parallel RLC resonator, so it is at its maximum at the resonant frequency f = 1/(2*pi*sqrt(LpCp)). What is the maximum value of |Zp|? Hint: At resonance, a parallel LC becomes an open circuit, leaving only the resistance.
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