Questions: The classic feedback amplifier in the following figure has beta=0.125. Find the loop gain T, ideal closed-loop gain Av^Ideal, actual closed-loop gain Av, and the fractional gain error (FGE) for A=72 dB. The loop gain T=498 ± 2%. (Round the final answer to the next whole number.) The ideal closed-loop gain Av^Ideal=8 ± 2%. (Round the final answer to the nearest whole number.) The actual closed-loop gain AV=8 ± 2%. (Round the final answer to the nearest whole number.) The fractional gain error (FGE) =0.2010 ± 0.0001%. (Round the final answer to four decimal places.) Explanation Av^Ideal=1/beta=1/0.125=8 T=A beta=10^(72 / 20) x 0.125=498 Av=Av^Ideal T/(T+1)=8 498/(498+1)=8 FGE=100%/(1+A beta)=100%/(1+498)=0.2010 %

$The classic feedback amplifier in the following figure has beta=0.125. Find the loop gain T, ideal closed-loop gain Av^Ideal, actual closed-loop gain Av, and the fractional gain error (FGE) for A=72 dB. The loop gain T=498 ± 2%. (Round the final answer to the next whole number.) The ideal closed-loop gain Av^Ideal=8 ± 2%. (Round the final answer to the nearest whole number.) The actual closed-loop gain AV=8 ± 2%. (Round the final answer to the nearest whole number.) The fractional gain error (FGE) =0.2010 ± 0.0001%. (Round the final answer to four decimal places.) Explanation Av^Ideal=1/beta=1/0.125=8 T=A beta=10^(72 / 20) x 0.125=498 Av=Av^Ideal T/(T+1)=8 498/(498+1)=8 FGE=100%/(1+A beta)=100%/(1+498)=0.2010 %$

Transcript text: The classic feedback amplifier in the following figure has $\beta=0.125$. Find the loop gain $T$, ideal closed-loop gain $A_{v}^{\text {Ideal }}$, actual closed-loop gain $A_{v}$, and the fractional gain error (FGE) for $A=72 \mathrm{~dB}$. The loop gain $T=498 \pm 2 \%$. (Round the final answer to the next whole number.) The ideal closed-loop gain $A_{v}^{\text {Ideal }}=8 \pm 2 \%$. (Round the final answer to the nearest whole number.) The actual closed-loop gain $A_{V}=8 \pm 2 \%$. (Round the final answer to the nearest whole number.) The fractional gain error (FGE) $=0.2010 \pm 0.0001$ \%. (Round the final answer to four decimal places.) Explanation \[ \begin{array}{l} A_{v}^{I d e a l}=\frac{1}{\beta}=\frac{1}{0.125}=8 \\ T=A \beta=10^{72 / 20} \times 0.125=498 \\ A_{v}=A_{v}^{I d e a l} \frac{T}{T+1}=8 \frac{498}{498+1}=8 \\ F G E=\frac{100 \%}{1+A \beta}=\frac{100 \%}{1+498}=0.2010 \% \end{array} \]

Solution

Solution Steps

Step 1: Calculate the Ideal Closed-Loop Gain $ A_v^{\text{Ideal}} $

The ideal closed-loop gain $ A_v^{\text{Ideal}} $ is given by: \[ A_v^{\text{Ideal}} = \frac{1}{\beta} \] Given $ \beta = 0.125 $: \[ A_v^{\text{Ideal}} = \frac{1}{0.125} = 8 \]

Step 2: Calculate the Loop Gain $ T $

The loop gain $ T $ is given by: \[ T = A \beta \] Given $ A = 72 \text{ dB} $: \[ A = 10^{72/20} = 3981.07 \] \[ T = 3981.07 \times 0.125 = 497.63 \] Rounding to the nearest whole number: \[ T \approx 498 \]

Step 3: Calculate the Actual Closed-Loop Gain $ A_v $

The actual closed-loop gain $ A_v $ is given by: \[ A_v = \frac{A_v^{\text{Ideal}}}{1 + \frac{A_v^{\text{Ideal}}}{T}} \] Given $ A_v^{\text{Ideal}} = 8 $ and $ T = 498 $: \[ A_v = \frac{8}{1 + \frac{8}{498}} = \frac{8}{1 + 0.0161} = \frac{8}{1.0161} \approx 7.87 \] Rounding to the nearest whole number: \[ A_v \approx 8 \]