GATE 2024 Analog Electronics Questions and Solutions

As shown in the circuit, the initial voltage across the capacitor is 10 V with the switch open. The switch is then closed at \( t = 0 \). The total energy dissipated in the ideal Zener diode (\( V_Z = 5\,V \)) after the switch is closed in mJ, rounded to three decimal places, is ___.

Solution

[Image of RC circuit discharge graph]

Once the switch closes the capacitor discharges through the 10 k\(\Omega\) resistor and the Zener, which holds a constant 5 V while current flows. The capacitor voltage therefore relaxes towards 5 V, not towards zero, with \( RC = 0.1\,s \):

Conduction continues until the current dies away, by which time the capacitor has fallen from 10 V to 5 V. The charge delivered through the Zener is therefore

The Zener holds 5 V throughout, so

In the circuit shown, the \( n:1 \) step-down transformer and diodes are ideal (no forward drop). If the input voltage is \( V_s(t) = 10\sin t \) and the average load voltage \( V_L(t) \) is \( \frac{5}{\pi}\,V \), the value of \( n \) is ___.

1. 4
2. 8
3. 12
4. 16

Solution

The centre-tapped secondary with \(D_1\) and \(D_2\) gives a full-wave rectified load voltage, so with \(V_M\) the peak of each half-secondary:

The primary peak is 10 V, so the turns ratio follows directly:

In the circuit shown below, transistors M1 and M2 are in saturation. Their small-signal transconductances are \( g_{m1} \) and \( g_{m2} \) respectively. Neglect body effect, channel-length modulation, and capacitances. Assume C1 is AC short. The exact magnitude of small-signal voltage gain \( \frac{v_{out}}{v_{in}} \) is ___.

1. \( g_{m2} R_D \)
2. \( \frac{g_{m2} R_D R_B}{1+g_{m1} R_S + R_B} \)
3. \( \frac{g_{m2} R_D R_B}{1+g_{m1} R_S + R_B} \)
4. \( g_{m2} R_D \frac{1}{1+g_{m1} R_S + R_B} \)

Solution

From small-signal model:

Gate node gives:

where \( v = v_{in} \) (since \( C_1 \) is AC short).

Therefore,

For the circuit shown, long-channel NMOS is biased in saturation with small signal transconductance \( g_m \). Neglect body effect, channel-length modulation, and intrinsic capacitances. The small-signal input impedance \( Z_{in} \) is ___.

1. \( g_m C_1 C_L \frac{2}{1/jC_1 + 1/jC_L} \)
2. \( g_m C_1 C_L \frac{2}{1/jC_1 + 1/jC_L} \)
3. \( \frac{1}{jC_1 + 1/jC_L} \)
4. \( g_m C_1 C_L \frac{2}{1/jC_1 jC_L} \)

Solution

The drain sits at \( V_{DD} \), a small-signal ground, so the device is a source follower. Let \( v_i \) be the gate voltage and \( v_s \) the source voltage; the only path for input current is \( C_1 \), which bridges gate and source.

KCL at the source node, with \( v_{gs} = v_i - v_s \):

so \( v_s = v_i \dfrac{g_m + j\omega C_1}{g_m + j\omega C_1 + j\omega C_L} \), and the input current is

For the closed-loop amplifier circuit, open-loop small-signal gain \( A_{OL} = 40 \). All transistors in saturation and current source is ideal. Neglect body effect, channel-length modulation, and capacitances. The closed-loop small-signal gain \( \frac{v_{out}}{v_{in}} \), rounded to three decimal places, is ___.

1. 0.976
2. 1
3. 1.025
4. 0.488

Solution

Differential amplifier active load, unity feedback: \( V_f = V_{out} \).

Closed-loop gain:

In the op-amp circuit below, if the circuit is to show sustained oscillations, the respective values of \( R_1 \) and the frequency of oscillation are ___ and ___.

1. \( 29R \) and \( \frac{1}{2\pi\sqrt{6}RC} \)
2. \( 2R \) and \( \frac{1}{2\pi RC} \)
3. \( 29R \) and \( \frac{1}{2\pi RC} \)
4. \( 2R \) and \( \frac{1}{2\pi\sqrt{6}RC} \)

Solution

The feedback network is a series \(C\) with shunt \(R\) followed by a series \(R\) with shunt \(C\). Writing \(x = \omega RC\), nodal analysis gives

The Barkhausen phase condition needs \(\beta\) real, so \(1 - x^2 = 0\), that is \(x = 1\):

The magnitude condition \(|A\beta| = 1\) then requires a non-inverting gain of 3:

So the correct values are \( R_1 = 2R \) and \( f_0 = \frac{1}{2\pi RC} \).

Two ideal op-amps saturate at 10 V. Initial inductor current 0 A. Input \( V_i(t) \) is a triangle wave (2V peak, period 8s). Which statement is true?

1. \( V_{01} \) delayed by 2s relative to Vi, \( V_{02} \) is triangular waveform.
2. \( V_{01} \) not delayed relative to Vi, \( V_{02} \) is trapezoidal waveform.
3. \( V_{01} \) not delayed relative to Vi, \( V_{02} \) is triangular waveform.
4. \( V_{01} \) delayed by 1s relative to Vi, \( V_{02} \) is trapezoidal waveform.

Solution

Op-amp A1 is a non-inverting Schmitt trigger: the inverting input is grounded, while the non-inverting input is fed by \( V_i \) through 10 k\(\Omega\) and by \( V_{01} \) through the 100 k\(\Omega\) feedback resistor. Switching occurs when that node reaches 0 V:

With \( V_{01} = \mp 10\,\text{V} \) the trip points are \( V_{UT} = +1\,\text{V} \) and \( V_{LT} = -1\,\text{V} \).

The triangle swings \( \pm 2 \) V over half a period, so its slope is \( 4\,\text{V} / 4 = 1 \) V per second. Starting from a zero crossing, \( V_i \) therefore takes 1s to reach either trip point.

So \( V_{01} \) (a square wave) switches 1s after the corresponding zero crossing of \( V_i \): it is delayed by 1s.

Op-amp A2 (integrator): saturates quickly, resulting in combination of ramp and flat sections; \( V_{02} \) is a trapezoidal waveform.