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ECE3161 Analogue Electronics - MUM S2 2025

ECE3161 Analogue Electronics - MUM S2 2025

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Q2. Find the emitter resistance

Find emitter resistance R_E of the PNP bias circuit below.

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$R_E = 0.97 \text{m}\Omega$ R_E = 0.97 \text{m}\Omega

$R_E = 4.47 \text{k}\Omega$ R_E = 4.47 \text{k}\Omega

$R_E = 1.87 \text{k}\Omega$ R_E = 1.87 \text{k}\Omega

$R_E = 287 \Omega$ R_E = 287 \Omega

$R_E = 1.37 \text{k}\Omega$ R_E = 1.37 \text{k}\Omega

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Q1. Find bias voltage and emitter current

Find V_c and I_e in the circuit below assuming transistor $\beta$ \beta is very high.

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V_c = 1.994 V, I_e = 0.5 mA

V_c = -2.994 V, I_e = 0.5 mA

V_c = 2.994 V, I_e = 0.7 mA

V_c = -0.994 V, I_e = 0.5 mA

V_c = -1.994 V, I_e = 0.7 mA

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Q5.3: What is the modified AC output voltage ( $V_{o}(s)$ V_{o}(s) ) for the circuit shown in Figure 5.2 in L(s) domain? Choose the correct representation from the choices provided below:

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$V_{o}(s) = \dfrac{sR_{1}C_{f}}{R_{f}V_{in}(s)}$ V_{o}(s) = \dfrac{sR_{1}C_{f}}{R_{f}V_{in}(s)}

$V_{o}(s) = -\dfrac{R_{f}}{R_{1}}[\dfrac{1}{1+sC_{f}R_{f}}]V_{in}(s)$ V_{o}(s) = -\dfrac{R_{f}}{R_{1}}[\dfrac{1}{1+sC_{f}R_{f}}]V_{in}(s)

$V_{o}(s) = sR_{1}C_{f}R_{f}V_{in}(s)$ V_{o}(s) = sR_{1}C_{f}R_{f}V_{in}(s)

$V_{o}(s) = -\dfrac{R_{1}}{sR_{f}C_{f}}V_{in}(s)$ V_{o}(s) = -\dfrac{R_{1}}{sR_{f}C_{f}}V_{in}(s)

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Q5.1: What is the transfer function for the circuit shown in Figure 5.1 in L(s) domain? Choose the correct representation in the Laplace domain from the choices provided below:

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$V_{o}(s) = R_{1}C_{f}V_{in}(s)$ V_{o}(s) = R_{1}C_{f}V_{in}(s)

$V_{o}(s) = \dfrac{sR_{1}C_{f}}{V_{in}(s)}$ V_{o}(s) = \dfrac{sR_{1}C_{f}}{V_{in}(s)}

$V_{o}(s) = sR_{1}C_{f}V_{in}(s)$ V_{o}(s) = sR_{1}C_{f}V_{in}(s)

$V_{o}(s) = -\dfrac{V_{in}(s)}{sR_{1}C_{f}}$ V_{o}(s) = -\dfrac{V_{in}(s)}{sR_{1}C_{f}}

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PL2.7: Consider the circuit shown in Figure 2.4 below and answer the following question.

Image failed to load: Circuit for pre-lab question PL2.7 - PL2.8

Figure 2.4: Circuit for pre-lab question PL2.7 - PL2.8

What is the voltage gain $A_{V}$ A_{V} transfer function of the inverting amplifier circuit shown in Figure 2.4?

$A_{V} = -\dfrac{R_{f}}{R_{1}}$ A_{V} = -\dfrac{R_{f}}{R_{1}}

$A_{V} = 1- \dfrac{R_{f}}{R_{1}}$ A_{V} = 1- \dfrac{R_{f}}{R_{1}}

$A_{V} = -1- \dfrac{R_{f}}{R_{1}}$ A_{V} = -1- \dfrac{R_{f}}{R_{1}}

$A_{V} = \dfrac{R_{1}}{R_{f}}$ A_{V} = \dfrac{R_{1}}{R_{f}}

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Q9. Find the Input Impedance

For the circuit given below, find the input impedance, $R_{in}$ R_{in} as shown assuming both op-amps are ideal and identical.

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$\frac{R_1R_3R_5}{R_2R_4}$ \frac{R_1R_3R_5}{R_2R_4}

$\frac{R_1R_2R_5}{R_3R_4}$ \frac{R_1R_2R_5}{R_3R_4}

$\frac{R_1R_3R_4}{R_2R_5}$ \frac{R_1R_3R_4}{R_2R_5}

$\frac{R_3R_5}{R_4}$ \frac{R_3R_5}{R_4}

$\frac{R_1R_3}{R_2}$ \frac{R_1R_3}{R_2}

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Q8. Output Voltage in Terms of Input Signals

Consider the following circuit with identical op-amps. Find the voltage $V_{out}$ V_{out} in terms of $V_{in1}$ V_{in1} and $V_{in2}$ V_{in2}.

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$- \frac{(2R_2R_3 + R_2R_4)(V_{in1} + V_{in2})}{R_3R_4}$ - \frac{(2R_2R_3 + R_2R_4)(V_{in1} + V_{in2})}{R_3R_4}

$- \frac{(2R_2R_3 + R_2R_4)(V_{in1} - V_{in2})}{R_1R_4}$ - \frac{(2R_2R_3 + R_2R_4)(V_{in1} - V_{in2})}{R_1R_4}

$- \frac{(2R_2R_3 + R_2R_4)(V_{in1} - V_{in2})}{R_3R_4}$ - \frac{(2R_2R_3 + R_2R_4)(V_{in1} - V_{in2})}{R_3R_4}

$- \frac{(R_1R_2 + R_1R_3)(V_{in1} - V_{in2})}{R_3R_4}$ - \frac{(R_1R_2 + R_1R_3)(V_{in1} - V_{in2})}{R_3R_4}

$- \frac{(2R_2R_3 + R_2R_4)(V_{in1} + V_{in2})}{R_1R_4}$ - \frac{(2R_2R_3 + R_2R_4)(V_{in1} + V_{in2})}{R_1R_4}

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