GATE 2009 Electrical Engineering (EE) Power Electronics.

Section 01

1-Mark Questions

QQuestion 1 1 Mark

A Linear Time Invariant system with an impulse response \(h(t)\) produces output \(y(t)\) when input \(x(t)\) is applied. When the input \(x(t - \tau)\) is applied to a system with impulse response \(h(t - \tau)\), the output will be

AOptions

\(y(t)\)
\(y(2(t - \tau))\)
\(y(t - \tau)\)
\(y(t - 2\tau)\)

SSolution

LTI System properties:

Time-Invariance property:

If input \(x(t)\) produces output \(y(t)\), then:

\[x(t - \tau) \rightarrow y(t - \tau)\]

Given system:

Original: \(x(t) * h(t) = y(t)\)

Modified system:

Input: \(x(t - \tau)\) (delayed input)
Impulse response: \(h(t - \tau)\) (delayed system)

Output calculation:

\[\text{Output} = [x(t - \tau)] * [h(t - \tau)]\]

Using convolution properties:

\[= \int_{-\infty}^{\infty} x(\alpha - \tau)h(t - \alpha - \tau)d\alpha\]

Let \(\beta = \alpha - \tau\):

\[= \int_{-\infty}^{\infty} x(\beta)h(t - 2\tau - \beta)d\beta\]

\[= x(t) * h(t - 2\tau)\]

But we know that \(x(t) * h(t) = y(t)\)

By time-invariance:

\[x(t) * h(t - 2\tau) = y(t - 2\tau)\]

Alternative reasoning:

Step 1: Delay input by \(\tau\)

\[x(t - \tau) * h(t) = y(t - \tau)\]

Step 2: Delay system by \(\tau\)

\[x(t) * h(t - \tau) = y(t - \tau)\]

Step 3: Both delayed by \(\tau\)

\[x(t - \tau) * h(t - \tau) = y(t - 2\tau)\]

The total delay is \(\tau + \tau = 2\tau\).

Correct answer: D

Section 02

2-Mark Questions

QQuestion 2 2 Mark

The circuit shows an ideal diode connected to a pure inductor and is connected to a purely sinusoidal 50 Hz voltage source. \(V_s = 10\sin(100\pi t)\), \(L = (0.1/\pi)\) H. Under ideal conditions, the current waveform through the inductor will look like:

SSolution

Given:

\(V_s = 10\sin(100\pi t)\) V
\(\omega = 100\pi\) rad/s
\(L = 0.1/\pi\) H
Ideal diode in series

Inductive reactance:

\[X_L = \omega L = 100\pi \times \frac{0.1}{\pi} = 10 \text{ } \Omega\]

Circuit behavior:

The diode starts to conduct at \(\omega t = 0\), when \(v_s\) goes positive. While it conducts the whole source voltage appears across the inductor, so with \(i(0) = 0\):

\[i(t) = \frac{1}{L}\int_0^{t} 10\sin(100\pi \tau)\,d\tau = \frac{10}{\omega L}\left[1 - \cos(100\pi t)\right] = 1 - \cos(100\pi t) \text{ A}\]

The inductor current is the integral of the voltage, not a scaled copy of it. This current never goes negative, so the diode is never asked to block during the negative half-cycle:

\(i = 0\) at \(\omega t = 0\)
\(i\) rises to its peak at \(\omega t = \pi\) (\(t = 10\) ms), where the voltage crosses zero going negative
\(i\) falls back to zero only at \(\omega t = 2\pi\) (\(t = 20\) ms) — the conduction angle is a full \(2\pi\)

The cycle then repeats, giving a train of \((1 - \cos\omega t)\) humps of period 20 ms that touch zero at \(t = 0, 20, 40\) ms with no flat gaps between them.

Peak current:

\[I_{peak} = \frac{2V_m}{\omega L} = \frac{2 \times 10}{10} = 2 \text{ A}\]

Only waveform (C) has this shape; the half-wave pattern of (D) would require the inductor current to jump to zero at \(\omega t = \pi\).

Correct answer: C

QQuestion 3 2 Mark

In the chopper circuit shown, the main thyristor (TM) is operated at a duty ratio of 0.8 which is much larger than the commutation interval. If the maximum allowable reapplied dv/dt on TM is 50 V/\(\mu\)s, what should be the theoretical minimum value of \(C_1\)? Assume current ripple through \(L_o\) to be negligible.

AOptions

0.2 \(\mu\)F
2 \(\mu\)F
0.02 \(\mu\)F
20 \(\mu\)F

SSolution

Given:

Source voltage: \(V_s = 100\) V
Duty ratio: \(D = 0.8\)
Maximum dv/dt: \(50\) V/\(\mu\)s
Load current ripple negligible (constant \(I_o\))

Commutation analysis:

During commutation:

When auxiliary thyristor fires to turn off main thyristor:

Capacitor voltage reverses and applies across TM
Rate of voltage rise across TM must be limited

dv/dt calculation:

The capacitor discharges through load current:

\[\frac{dv}{dt} = \frac{I}{C}\]

where \(I\) is the load current.

For constant load current \(I_o\):

\[\frac{dv}{dt} = \frac{I_o}{C_1}\]

Maximum dv/dt constraint:

\[\frac{I_o}{C_1} \leq 50 \times 10^6 \text{ V/s}\]

\[C_1 \geq \frac{I_o}{50 \times 10^6}\]

Load current:

With the ripple in \(L_o\) neglected, the 8 \(\Omega\) load carries the average output voltage:

\[V_o = D \times V_s = 0.8 \times 100 = 80 \text{ V}\]

\[I_o = \frac{V_o}{R_L} = \frac{80}{8} = 10 \text{ A}\]

Minimum \(C_1\):

\[C_1 \geq \frac{I_o}{(dv/dt)_{max}} = \frac{10}{50 \times 10^{6}} = 0.2 \times 10^{-6} \text{ F} = 0.2 \text{ } \mu\text{F}\]

Correct answer: A