\batchmode \documentclass[10pt]{exam} \usepackage{webwork2} \usepackage[text={7in,9in},centering]{geometry} \setlength{\columnsep}{.25in} \setlength{\columnseprule}{.4pt} \extraheadheight[2pc]{0pt} % This removes the margin for questions from the exam class. \renewcommand{\questionshook}{\leftmargin=0pt\labelwidth=-\labelsep} \makeatletter \qformat{{\bfseries \webworkLocalizeProblem~\thequestiontitle.} \if@placepoints{\bfseries\footnotesize(\thepoints)}\fi \hfill} \makeatother \parindent=0pt \pagestyle{headandfoot} \begin{document} \def\webworkCourseName{114\_ELECTRONICS\_2}% \def\webworkCourseTitle{}% \def\webworkCourseURL{https://wicswebwork.ee.ncku.edu.tw//webwork2/114\_ELECTRONICS\_2}% \def\webworkUserId{E24136178}% \def\webworkStudentId{36}% \def\webworkFirstName{陳易安}% \def\webworkEmailAddress{e24136178@gs.ncku.edu.tw}% \def\webworkSection{1142\_E227020\_2}% \def\webworkSetId{assignment5}% \def\webworkPrettySetId{assignment5}% \def\webworkOpenDate{June 12, 2026, 3:26:00 AM CST}% \def\webworkDueDate{June 21, 2026, 11:59:00 PM CST}% \def\webworkAnswerDate{June 22, 2026, 12:00:00 AM CST}% \def\webworkReducedScoringDate{June 18, 2026, 11:59:00 PM CST}% \firstpageheader{\webworkLeftHeader}{\webworkCenterHeader}{\webworkRightHeader} \runningheader{}{}{} \footer{\webworkLeftFooter}{\webworkCenterFooter}{\webworkRightFooter} {\pgmlSetup This PDF is available for convenience. Assignments must be submitted within {\bfseries{}WeBWorK} for credit. \par}% \smallskip \hrule \begin{multicols*}{2} \begin{questions} \def\webworkProblemId{1}% \def\webworkProblemNumber{1}% \def\webworkProblemWeight{1}% \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 \(18\) and to have an amount of feedback of at least \(52\) 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 \vfill \goodbreak \hrule \nobreak \smallskip \def\webworkProblemId{2}% \def\webworkProblemNumber{2}% \def\webworkProblemWeight{1}% \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 \(68 \Omega\), find the value for \(\text{R}_F\) that results in a closed-loop gain with an ideal value of \(12 \text{V/V}\). \leavevmode\\\relax (b) If \(\text{Q}_1\) is biased at \(0.4 \text{mA}\), \(\text{Q}_2\) is at \(1.5 \text{mA}\), \(\text{Q}_3\) is at \(3.9 \text{mA}\), and assuming that the transistors have \(\beta\) = \(100\) and larger \(r_o\), and that \(\text{R}_{C1}\) = \(2.3 \text{k}\Omega\) and \(\text{R}_{C1}\) = \(0.5 \text{k}\Omega\), 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}} \(\text{V/V}\) \vfill \goodbreak \hrule \nobreak \smallskip \def\webworkProblemId{3}% \def\webworkProblemNumber{3}% \def\webworkProblemWeight{1}% \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 \(82\text{k}\Omega\) differential input resistance, a \(20\text{V/V}\) open-circuit differential voltage gain, and a \(3.6\text{k}\Omega\) output resistance. \leavevmode\\\relax \(A_2\) has a \(4\text{k}\Omega\) input resistance, a \(24\text{mA/V}\) short-circuit transconductance, and a \(16\text{k}\Omega\) output resistance. \leavevmode\\\relax \(A_3\) has a \(25\text{k}\Omega\) input resistance, unity open-circuit voltage gain, and a \(1.1\text{k}\Omega\) output resistance. \leavevmode\\\relax The feedback amplifier feeds a \(1\text{k}\Omega\) 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 \(R1\) = \(18\text{k}\Omega\), 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 \vfill \goodbreak \hrule \nobreak \smallskip \def\webworkProblemId{4}% \def\webworkProblemNumber{4}% \def\webworkProblemWeight{1}% \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}} \(\text{k}\Omega\) \vfill \goodbreak \hrule \nobreak \smallskip \def\webworkProblemId{5}% \def\webworkProblemNumber{5}% \def\webworkProblemWeight{1}% \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 \(R_F = 1.81\text{kΩ}\)\leavevmode\\\relax \(C_X = 112.7\text{fF}\)\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 \(g_m =35.3 \text{mS}\)\leavevmode\\\relax \(Z_s =1.57 \text{kΩ}\)\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 \(g_m =35.3 \text{mS}\)\leavevmode\\\relax \(R_s =1.57 \text{kΩ}\)\leavevmode\\\relax \(C_s =197.8 \text{fF}\)\leavevmode\\\relax \(R_D =0.68 \text{kΩ}\)\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. \webworkSetCopyrightFooter \end{questions} \end{multicols*} \end{document}