(d) Figure $20

Question

(d) Figure $20 < 1$ -Refer to Figure 20-1 (a). If $\mathrm { R } _ { 1 } = \mathrm { R } _ { 2 } = 28 \mathrm { k } \Omega$ and $R _ { G } = 100 \Omega$ , the A_cl would be: A) 551 . B) 560 . C) 56.1 . D) 5.61 .

Quizplus · Accepted Answer

The closed-loop gain is given by:$$A_{cl} = \frac{A_v}{1 + \beta A_v}$$where:$$A_v = -g_mR_L$$and$$\beta = \frac{R_2}{R_1 + R_2}$$Substituting the given values, we get:$$A_v = -100 	imes 28 \mathrm{k} \Omega = -2.8 \mathrm{M}\Omega$$$$\beta = \frac{28 \mathrm{k} \Omega}{28 \mathrm{k} \Omega + 28 \mathrm{k} \Omega} = 0.5$$Therefore,$$A_{cl} = \frac{-2.8 \mathrm{M}\Omega}{1 + 0.5 	imes (-2.8 \mathrm{M}\Omega)} = 560$$So, the correct answer is B) 560.

(D) Figure 20<120 < 120<1 -Refer to Figure 20-1 (A) R1=R2=28kΩ\mathrm { R } _ { 1 } = \mathrm { R } _ { 2 } = 28 \mathrm { k } \OmegaR1​=R2​=28kΩ

(D)

Figure $20 < 1$ -Refer to Figure 20-1 (A) $\mathrm { R } _ { 1 } = \mathrm { R } _ { 2 } = 28 \mathrm { k } \Omega$