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\def\webworkFirstName{林昱成}%
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\def\webworkOpenDate{June 12, 2026, 3:26:00 AM CST}%
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This PDF is available for convenience. Assignments must be submitted
within {\bfseries{}WeBWorK} for credit.
\par}%

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\begin{multicols*}{2}
\begin{questions}
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\titledquestion{\ifx\webworkProblemNumber\empty\#\else\webworkProblemNumber\fi}[\ifx\webworkProblemWeight\empty\#\else\webworkProblemWeight\fi]
\smallskip

Design a feedback amplifier to have an ideal closed-loop gain of \(10\) and to have an amount of feedback of at least \(42\) dB. \leavevmode\\\relax 
\leavevmode\\\relax 
What is the required value of \(\beta\)? {\answerRule[AnSwEr0001]{10}} (V/V) \leavevmode\\\relax 
\leavevmode\\\relax 
What is the minimum required value of the open-loop gain \(A\)? {\answerRule[AnSwEr0002]{10}} (V/V) \leavevmode\\\relax 
\leavevmode\\\relax 
What is the corresponding realized value of closed-loop gain \(A_f\)? {\answerRule[AnSwEr0003]{10}} (V/V) \leavevmode\\\relax 

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\titledquestion{\ifx\webworkProblemNumber\empty\#\else\webworkProblemNumber\fi}[\ifx\webworkProblemWeight\empty\#\else\webworkProblemWeight\fi]
\smallskip

\settoheight{\strutheight}{\strut}\raisebox{-0.5\height + 0.5\strutheight}{\includegraphics[width=0.583\linewidth]{/opt/webwork/courses/114_ELECTRONICS_2/templates/assignment5/problem2.png}}

\leavevmode\\\relax 
The figure shows a series-shunt feedback amplifier without details of the bias circuit.
\leavevmode\\\relax 
(a) If \(\text{R}_E\) is selected to be \(66 \Omega\), find the value for \(\text{R}_F\) that results in a closed-loop gain with an ideal value of \(23 \text{V/V}\).
\leavevmode\\\relax 
(b) If \(\text{Q}_1\) is biased at \(0.6 \text{mA}\), \(\text{Q}_2\) is at \(0.8 \text{mA}\), \(\text{Q}_3\) is at \(3.3 \text{mA}\), and assuming that the transistors have \(\beta\) = \(100\) and larger \(r_o\), and that \(\text{R}_{C1}\) = \(1.6 \text{k}\Omega\) and \(\text{R}_{C1}\) = \(1 \text{k}\Omega\), find the value of the loop gain \(A\beta\) and hence of the closed-loop gain \(\text{A}_f\).
\par\hrulefill\par 
\(R_F =\) {\answerRule[AnSwEr0001]{10}} \(\Omega\)
\leavevmode\\\relax 
\(A\beta =\) {\answerRule[AnSwEr0002]{10}} \(\text{V/V}\)
\leavevmode\\\relax 
\(\text{A}_f =\) {\answerRule[AnSwEr0003]{10}} \(\text{V/V}\)



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\titledquestion{\ifx\webworkProblemNumber\empty\#\else\webworkProblemNumber\fi}[\ifx\webworkProblemWeight\empty\#\else\webworkProblemWeight\fi]
\smallskip



Below shows a three-stage feedback amplifier:
\leavevmode\\\relax 
\(A_1\) has an \(76\text{k}\Omega\) differential input resistance, a \(20\text{V/V}\) open-circuit differential voltage gain, and a \(2.5\text{k}\Omega\) output resistance.
\leavevmode\\\relax 
\(A_2\) has a \(4\text{k}\Omega\) input resistance, a \(21\text{mA/V}\) short-circuit transconductance, and a \(16\text{k}\Omega\) output resistance.
\leavevmode\\\relax 
\(A_3\) has a \(19\text{k}\Omega\) input resistance, unity open-circuit voltage gain, and a \(0.8\text{k}\Omega\) output resistance.
\leavevmode\\\relax 
The feedback amplifier feeds a \(1.4\text{k}\Omega\) load resistance and is fed by a signal source with a \(12\text{k}\Omega\) resistance.

\settoheight{\strutheight}{\strut}\raisebox{-0.5\height + 0.5\strutheight}{\includegraphics[width=0.833\linewidth]{/opt/webwork/courses/114_ELECTRONICS_2/templates/assignment5/assignment5_problem3.jpg}}

\leavevmode\\\relax 
\leavevmode\\\relax 

(a)If \(R1\) = \(22\text{k}\Omega\), Find the value of \(R_2\) that results in an ideal closed-loop gain of 5 V/V.
\leavevmode\\\relax 
\(R_2\) = {\answerRule[AnSwEr0001]{10}} \(\text{k}\Omega\)

\leavevmode\\\relax 
\leavevmode\\\relax 

(b) Find the value of the open-loop gain \(A \equiv V_o/V_s\).
\leavevmode\\\relax 
\(A\) = {\answerRule[AnSwEr0002]{10}} \(\text{V/V}\)

\leavevmode\\\relax 
\leavevmode\\\relax 

(c) Find the feedback factor \(\beta\) and the amount of feedback.
\leavevmode\\\relax 
\(\beta\) = {\answerRule[AnSwEr0003]{10}} \(\text{V/V}\)
\leavevmode\\\relax 
Amount of feedback (\(1 + \beta A\)) = {\answerRule[AnSwEr0004]{10}}

\leavevmode\\\relax 
\leavevmode\\\relax 

(d) Find the closed-loop gain \(A_f\).
\leavevmode\\\relax 
\(A_f\) = {\answerRule[AnSwEr0005]{10}} \(\text{V/V}\)

\leavevmode\\\relax 
\leavevmode\\\relax 

(e) Find the feedback amplifier's input resistance \(R_{\text{in}}\).
\leavevmode\\\relax 
\(R_{\text{in}}\) = {\answerRule[AnSwEr0006]{10}} \(\text{M}\Omega\)

\leavevmode\\\relax 
\leavevmode\\\relax 

(f) Find the feedback amplifier's output resistance \(R_{\text{out}}\) .
\leavevmode\\\relax 
\(R_{\text{out}}\) = {\answerRule[AnSwEr0007]{10}} \(\Omega\)

\leavevmode\\\relax 
\leavevmode\\\relax 

(g) If the high-frequency response of the open-loop gain A is dominated by a pole at 100Hz, what is the upper 3-dB frequency of the closed-loop gain.
\leavevmode\\\relax 
\(f\) = {\answerRule[AnSwEr0008]{10}} \(\text{kHz}\)

\leavevmode\\\relax 
\leavevmode\\\relax 

(h) If for some reason \(A_1\) is drops to half its nominal value, what is the absolute percentage change in \(A_f\)?
\leavevmode\\\relax 
Absolute Percentage change = {\answerRule[AnSwEr0009]{10}} \(\%\)

\leavevmode\\\relax 


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\titledquestion{\ifx\webworkProblemNumber\empty\#\else\webworkProblemNumber\fi}[\ifx\webworkProblemWeight\empty\#\else\webworkProblemWeight\fi]
\smallskip

For the feedback transconductance amplifier shown in Fig. 4, evaluate the closed-loop
transconductance gain \(A_f\) and the output resistance with feedback \(R_{of}\).

\leavevmode\\\relax 
\[g_{m1}=g_{m2}=4 \text{ mA/V}, \quad
R_D=20 \text{ k}\Omega, \quad
r_{o2}=20 \text{ k}\Omega,\]
\[R_F=100 \ \Omega, \quad
R_L=1 \text{ k}\Omega.\]

\leavevmode\\\relax 
For simplicity, neglect \(r_{o1}\), and take \(r_{o2}\) into account only when calculating
the output resistance.

\leavevmode\\\relax 
\settoheight{\strutheight}{\strut}\raisebox{-0.5\height + 0.5\strutheight}{\includegraphics[width=0.6\linewidth]{/opt/webwork/courses/114_ELECTRONICS_2/templates/assignment5/problem4.png}}

\leavevmode\\\relax 


Find \(A_f\) in mA/V:
\[A_f =\]
{\answerRule[AnSwEr0001]{10}} \(\text{mA/V}\)

\leavevmode\\\relax 
Find \(R_{of}\) in k\(\Omega\):
\[R_{of} =\]
{\answerRule[AnSwEr0002]{10}} \(\text{k}\Omega\)


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\titledquestion{\ifx\webworkProblemNumber\empty\#\else\webworkProblemNumber\fi}[\ifx\webworkProblemWeight\empty\#\else\webworkProblemWeight\fi]
\smallskip

In recent years, AI servers have needed to transmit large amounts of digital data between chips, modules, and systems. As a result, high-speed wireline transmission has become increasingly important. Although digital signals look like square waves composed of 0 and 1 in the time domain, they actually contain many high-frequency components in the frequency domain, as shown by the sinc-shaped spectrum in the figure.
\leavevmode\\\relax \leavevmode\\\relax \settoheight{\strutheight}{\strut}\raisebox{-0.5\height + 0.5\strutheight}{\includegraphics[width=1\linewidth]{/opt/webwork/courses/114_ELECTRONICS_2/templates/assignment5/problem5_1.png}}
\leavevmode\\\relax 
\settoheight{\strutheight}{\strut}\raisebox{-0.5\height + 0.5\strutheight}{\includegraphics[width=1\linewidth]{/opt/webwork/courses/114_ELECTRONICS_2/templates/assignment5/Problem5_2.jpg}}
\leavevmode\\\relax \leavevmode\\\relax \leavevmode\\\relax 
As the transmission speed increases, the symbol period (T) becomes shorter, requiring higher circuit bandwidth. If the bandwidth is insufficient, high-frequency components are attenuated, causing waveform distortion and making data detection harder. Therefore, feedback is important in high-speed receiver circuits because it can extend bandwidth and stabilize the frequency response.

\leavevmode\\\relax \leavevmode\\\relax 
Therefore, in a high-speed wireline transmitter or receiver, the Nyquist frequency is an important bandwidth reference. If the symbol period is \(T\), the Nyquist frequency is


\(f_N=\frac{1}{2T}\)

\leavevmode\\\relax \leavevmode\\\relax 
If we want to transmit data at a rate of 50 Gb/s. This means that a binary signal, either 0 or 1, is transmitted \(50\times10^9\)
 times per second. What is the Nyquist frequency?\leavevmode\\\relax  \leavevmode\\\relax  
\(Nyquist\) \(frequency\) \(=\) {\answerRule[AnSwEr0001]{10}}\(GHz\) \(\)
\leavevmode\\\relax \leavevmode\\\relax \leavevmode\\\relax \leavevmode\\\relax 

Assume all transistor operate in saturation in all of the problems. \leavevmode\\\relax \leavevmode\\\relax 

In a high-speed wireline receiver, the input signal is often a small AC current. A transimpedance amplifier, or TIA, is used to convert this input current into an output voltage.

The circuit contains an amplifier \(A_1\), a feedback resistor \(R_F\), and an input capacitance \(C_X\) at node \(X\). The feedback resistor samples the output voltage and feeds a current back to the input node. Therefore, this is a shunt-shunt feedback circuit.\leavevmode\\\relax \leavevmode\\\relax 
\settoheight{\strutheight}{\strut}\raisebox{-0.5\height + 0.5\strutheight}{\includegraphics[width=0.833\linewidth]{/opt/webwork/courses/114_ELECTRONICS_2/templates/assignment5/Problem5_3.png}}
\leavevmode\\\relax 

Assume:\leavevmode\\\relax 


\(R_F = 1.69\text{kΩ}\)\leavevmode\\\relax 

\(C_X = 137.9\text{fF}\)\leavevmode\\\relax 
The amplifier \(A_1\) is modeled as a ideal amplifier:\leavevmode\\\relax 
Gain is \(20\) \(V/V\)\leavevmode\\\relax 

\(Transfer\) \(Function\) is \(\frac{V_o}{I_{AC}}=\frac{A_0}{1+\frac{s}{\omega_p}}\)

\leavevmode\\\relax 
\(\omega_p\) = {\answerRule[AnSwEr0002]{10}}\(rad/s\) \(\)\leavevmode\\\relax 
\(|A_0|\) = {\answerRule[AnSwEr0003]{10}}\(V/A\) \(\)
\leavevmode\\\relax 
\leavevmode\\\relax \leavevmode\\\relax 
Problem: Source Degeneration as Local Feedback\leavevmode\\\relax 
\leavevmode\\\relax \leavevmode\\\relax 

A common-source amplifier uses a source degeneration impedance \(Z_S\), as shown in the figure. Source degeneration is a form of local negative feedback.\leavevmode\\\relax \leavevmode\\\relax 

When the gate voltage increase then the drain current \(i_d\) increases, the source voltage \(v_s\) also increases. Since

\[v_{gs}=v_{in}-v_s\]

an increase in \(v_s\) reduces \(v_{gs}\). As a result, the drain current is reduced. Therefore, the circuit feeds back a signal that opposes the original current change. This is why source degeneration is called local negative feedback.\leavevmode\\\relax \leavevmode\\\relax 
\settoheight{\strutheight}{\strut}\raisebox{-0.5\height + 0.5\strutheight}{\includegraphics[width=0.833\linewidth]{/opt/webwork/courses/114_ELECTRONICS_2/templates/assignment5/Problem5_4.png}}
\leavevmode\\\relax \leavevmode\\\relax \leavevmode\\\relax 
\(g_m =28 \text{mS}\)\leavevmode\\\relax 
\(Z_s =1.35 \text{kΩ}\)\leavevmode\\\relax \leavevmode\\\relax 
By what factor does source degeneration increase the bandwidth of a CS amplifier compared with the case without source degeneration?\leavevmode\\\relax 
\(factor\) = {\answerRule[AnSwEr0004]{10}}\(\) \(\)\leavevmode\\\relax \leavevmode\\\relax 

A CTLE, or continuous-time linear equalizer, is commonly used in high-speed wireline receivers. In wireline channels, high-frequency signal components are usually attenuated more than low-frequency components. This causes slower signal transitions, waveform distortion.\leavevmode\\\relax \leavevmode\\\relax 

The CTLE helps solve this problem by boosting the high-frequency components of the received signal. In this circuit, \(R_S\) provides source degeneration at low frequency, which reduces the gain. At high frequency, \(C_S\) becomes a short circuit and bypasses (R_S), so the gain increases. Therefore, the CTLE has higher gain at high frequency than at low frequency.\leavevmode\\\relax \leavevmode\\\relax 

As a result, the CTLE can compensate for high frequency component loss,  and improve the signal quality at the receiver.\leavevmode\\\relax \leavevmode\\\relax 

\settoheight{\strutheight}{\strut}\raisebox{-0.5\height + 0.5\strutheight}{\includegraphics[width=0.833\linewidth]{/opt/webwork/courses/114_ELECTRONICS_2/templates/assignment5/Problem5_6.png}}
\leavevmode\\\relax 
\(g_m =28 \text{mS}\)\leavevmode\\\relax 
\(R_s =1.35 \text{kΩ}\)\leavevmode\\\relax 
\(C_s =110.2 \text{fF}\)\leavevmode\\\relax 
\(R_D =0.71 \text{kΩ}\)\leavevmode\\\relax 
\(C_L =200 \text{fF}\)\leavevmode\\\relax 
\leavevmode\\\relax \leavevmode\\\relax 
\(Transfer\) \(Function\) is \(\frac{V_{out}}{V_{in}}=\frac{A_1(1+\frac{\omega}{\omega_z})}{(1+\frac{s}{\omega_{p1}}){(1+\frac{s}{\omega_{p2}}})}\)
\leavevmode\\\relax \leavevmode\\\relax \(\omega_{p1}\) \(<\)\(\omega_{p2}\)\leavevmode\\\relax \leavevmode\\\relax 
\(\omega_{p1}\) = {\answerRule[AnSwEr0005]{10}}\(rad/s\) \(\)\leavevmode\\\relax 
\(\omega_{p2}\) = {\answerRule[AnSwEr0006]{10}}\(rad/s\) \(\)\leavevmode\\\relax 
\(\omega_z\) = {\answerRule[AnSwEr0007]{10}}\(rad/s\) \(\)\leavevmode\\\relax 
\(A_1\) = {\answerRule[AnSwEr0008]{10}}\(V/V\) \(\)\leavevmode\\\relax \leavevmode\\\relax 

As you can see, by suppressing the low-frequency components, we can relatively enhance part of the high-frequency components, which helps increase the overall system bandwidth.



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