GATE 2026 Power Electronics Practice Questions and Solutions

Among the devices listed below, the one that cannot block a reverse voltage across its main terminals, because a parasitic diode is formed as an inherent part of its structure, is the

1. silicon controlled rectifier (SCR)
2. symmetric-blocking gate turn-off thyristor (GTO)
3. n-channel power MOSFET
4. reverse-blocking IGBT

Solution

A vertical n-channel power MOSFET is built on an \(n^{+}\) substrate (drain) with an \(n^{-}\) drift region, a \(p\) body and an \(n^{+}\) source. The source metallisation deliberately shorts the \(n^{+}\) source to the \(p\) body to suppress the parasitic \(npn\) transistor. That short leaves the \(p\) body and the \(n^{-}\) drift region permanently connected between source and drain as a \(pn\) junction, which is the body diode. When the drain is taken negative with respect to the source this diode is forward biased and conducts, so no reverse voltage can be sustained.

The SCR and the symmetric GTO both contain a reverse-blocking junction by design, and the reverse-blocking IGBT is fabricated with a deep isolation diffusion precisely to give it that capability. Only the MOSFET has an unavoidable antiparallel diode. The characteristics of the switch family are collected in Chapter 5.

A thyristor is connected in series with a load consisting of \(R = 20\,\Omega\) in series with \(L = 0.5\) H, the combination being fed from a 100 V DC supply. The thyristor is triggered by a single gate pulse of duration 50 \(\mu\)s. The anode current at the instant the gate pulse ends is _____ mA.

Solution

With the thyristor in conduction the loop equation is

whose solution is the standard first-order rise

The time constant is \(L/R = 0.5/20 = 25\) ms, so in 50 \(\mu\)s the exponent is very small:

So the anode current reaches only 9.99 mA. The same figure comes out of the short-time approximation \(i \approx (V/L)t = (100/0.5)(50\times10^{-6}) = 10\) mA, which confirms the arithmetic. If the latching current of this device were, say, 50 mA, the thyristor would revert to the blocking state the moment the gate pulse was withdrawn; a highly inductive load therefore demands a pulse train or a sustained gate signal, as discussed in Chapter 4.

A single-phase fully controlled bridge converter is supplied from a 230 V, 50 Hz source of negligible impedance. The load is highly inductive, so the load current is continuous and ripple-free at 10 A. For a firing angle of 45°, the average output voltage is _____ V.

Solution

With continuous conduction each pair of thyristors conducts for a half cycle beginning at \(\omega t = \alpha\), so the output voltage over one repetition interval of \(\pi\) is \(v_o = V_m\sin\omega t\) for \(\alpha \le \omega t \le \alpha + \pi\). Averaging,

With \(V_m = \sqrt{2}\times 230 = 325.27\) V,

The load current does not enter the result at all, provided conduction stays continuous. See Chapter 8.

A single-phase AC voltage controller uses two thyristors in antiparallel to feed a purely resistive load of 10 \(\Omega\) from a 230 V, 50 Hz supply. Both devices are fired symmetrically at a firing angle of 90°. The power delivered to the load is _____ W.

Solution

With a resistive load each thyristor conducts from \(\alpha\) to \(\pi\) in its own half cycle, and the output voltage is the supply voltage over that interval and zero otherwise. Over a half-cycle repetition interval,

At \(\alpha = 90^\circ = \pi/2\) the term \(\sin 2\alpha = \sin 180^\circ = 0\), so

The load being resistive, the power follows directly:

Firing at 90° halves the power relative to full conduction (5290 W), not the voltage. More on this converter in Chapter 21.

A type-A (first-quadrant) chopper operates from a 200 V DC source and feeds a load consisting of a resistance of 0.5 \(\Omega\) in series with a constant counter-emf of 100 V and a large smoothing inductance. The chopping frequency is high enough that the load current may be taken as ripple-free. For a duty ratio of 0.7, the average load current is _____ A.

Solution

In a step-down chopper the load is connected to the source for \(DT\) and freewheels through the diode for \((1-D)T\), so the average output voltage is

Averaging the load loop equation over one chopping period, the inductor voltage averages to zero in steady state, leaving

Note how sensitive the current is to duty ratio when the counter-emf is large: raising \(D\) from 0.7 to 0.75 would take \(V_o\) to 150 V and the current to 100 A. This is the sizing problem that dominates chopper-fed drives, treated in Chapter 11.

A single-phase full-bridge voltage source inverter operates in square-wave mode from a 200 V DC link, producing an output voltage that alternates between \(+200\) V and \(-200\) V with equal half-periods. The total harmonic distortion of the output voltage, referred to the fundamental, is _____ %.

Solution

The Fourier series of a symmetrical square wave of amplitude \(V_{dc}\) contains only odd harmonics:

The rms value of the fundamental is therefore

The total rms value is read straight off the waveform, since the magnitude is \(V_{dc}\) at every instant:

The harmonic content is what is left when the fundamental is removed in the mean-square sense:

So the distortion is 48.34 %, and note that it does not depend on \(V_{dc}\): both numerator and denominator scale with it. This fixed, large distortion is the reason square-wave operation is replaced by modulation whenever the load is a motor. See Chapter 16.

A three-phase voltage source inverter with a DC-link voltage of 600 V is controlled by sinusoidal pulse width modulation with a modulation index (amplitude modulation ratio) of 0.9. The frequency modulation ratio is large and the inverter operates in the linear modulation range. The rms value of the fundamental component of the line-to-line output voltage is _____ V.

Solution

In the linear range of sinusoidal PWM the peak of the fundamental pole voltage, measured with respect to the DC-link midpoint, is

The fundamental line-to-line voltage is built from two pole voltages displaced by 120°, which multiplies the magnitude by \(\sqrt{3}\):

Dividing by \(\sqrt{2}\) to convert peak to rms,

For comparison, six-step operation of the same DC link would give \(V_{ab,1(rms)} = (\sqrt{6}/\pi)V_{dc} = 0.7797 \times 600 = 467.8\) V, so linear sinusoidal PWM reaches at most about 78.5 % of that even at \(m_a = 1\). Recovering the shortfall is the motivation for third-harmonic injection and space vector modulation, covered in Chapter 18.

A buck converter in a switched-mode power supply operates from a 48 V input at a switching frequency of 20 kHz with a duty ratio of 0.5, delivering 24 V to a 12 \(\Omega\) load. The filter inductance is 500 \(\mu\)H and the output capacitor is large enough that the output voltage may be treated as constant. The converter works in continuous conduction mode. The peak-to-peak ripple in the inductor current is _____ A.

Solution

During the on-interval \(DT\) the inductor sees the difference between input and output voltage, and the current ramps up linearly:

With \(T = 1/20\,\text{kHz} = 50\,\mu\text{s}\),

The off-interval gives the same figure, as it must in steady state, since the inductor then carries \(-V_o\) across it for \((1-D)T\):

The average inductor current equals the load current, \(I_L = 24/12 = 2\) A. Since \(\Delta I_L/2 = 0.6\) A is less than 2 A, the inductor current never reaches zero and the assumed continuous conduction is confirmed. Converter waveforms are derived in Chapter 12.

A separately excited DC motor is fed from a single-phase fully controlled bridge converter supplied at 230 V, 50 Hz. The armature resistance is 1.0 \(\Omega\) and the armature current is continuous and ripple-free at 10 A. The field is held constant, giving a back-emf constant of \(K_a\phi = 1.2\;\text{V}\cdot\text{s/rad}\). For a firing angle of 30°, the motor speed is _____ rpm.

Solution

The converter applies an average armature voltage of

Averaging the armature circuit equation over a supply period, the armature inductance contributes nothing in steady state, so

The back-emf fixes the mechanical speed:

The torque developed is \(T = K_a\phi I_a = 1.2 \times 10 = 12\) N·m, unaffected by \(\alpha\); the firing angle sets speed, and the load sets current. Drive behaviour across the quadrants is developed in Chapter 25.

A single-phase AC voltage controller with two antiparallel thyristors feeds a series R–L load whose impedance angle is \(\varphi = \tan^{-1}\!\left(\omega L / R\right)\). The range of firing angle over which the output voltage can actually be varied is

1. \(0 \le \alpha \le \pi\)
2. \(\varphi \le \alpha \le \pi\)
3. \(0 \le \alpha \le \varphi\)
4. \(\pi/2 \le \alpha \le \pi\)

Solution

Fire a thyristor at \(\omega t = \alpha\) into the R–L load. The current is the sum of a steady-state and a transient term,

If \(\alpha = \varphi\) the transient term vanishes and the current is the pure sinusoid \(\left(V_m/Z\right)\sin(\omega t - \varphi)\), which extinguishes exactly at \(\omega t = \alpha + \pi\). Each device then conducts for a full 180° and the next device is fired precisely as the previous one turns off, so the load receives the full supply voltage.

For \(\alpha\) less than \(\varphi\) the extinction angle would exceed \(\alpha + \pi\), meaning one thyristor is still conducting when its partner is gated. The partner cannot turn on with reverse voltage across it, and the outcome is a single device conducting continuously with the load current becoming the full sinusoid again. The output is therefore stuck at the uncontrolled value for every \(\alpha \le \varphi\); control only begins once \(\alpha\) exceeds \(\varphi\), and ends at \(\alpha = \pi\) where the conduction angle shrinks to zero.