GATE 2008 Electrical Engineering (EE) Power Electronics.

Section 01

1-Mark Questions

QQuestion 1 1 Mark

A diode has forward biased equivalent (10\(\Omega\) resistor in series with a 0.7 V source) and reverse biased equivalent (10 M\(\Omega\)). If such a diode is used in the clipper circuit shown, with input \(v_i = 10\sin\omega t\) V, the output voltage \(v_o\) will be

SSolution

Clipper operation:

First check whether the diode ever conducts. With the diode off its branch is a 10 M\(\Omega\) open circuit and the source sees the two 10 k\(\Omega\) resistors as a divider:

\[v_o = v_i \times \frac{10 \text{ k}\Omega}{10 \text{ k}\Omega + 10 \text{ k}\Omega} = 5\sin\omega t\]

The diode conducts only when its anode rises above the battery voltage plus the forward drop:

\[v_o > 5 + 0.7 = 5.7 \text{ V}\]

The divider output peaks at 5 V, which never reaches 5.7 V, so the diode stays reverse biased over the entire cycle and no clipping occurs. On the negative half cycle the diode is reverse biased in any case. The output is the undistorted sine

\[v_o = 5\sin\omega t \text{ V}\]

Correct answer: A

QQuestion 2 1 Mark

In the single phase voltage controller circuit shown, feeding a load of \(50 + j50\) \(\Omega\), for what range of triggering angle \(\alpha\) is the output voltage not controllable?

AOptions

0° < \(\alpha\) < 45°
45° < \(\alpha\) < 135°
90° < \(\alpha\) < 180°
135° < \(\alpha\) < 180°

SSolution

AC voltage controller with an R-L load:

The load impedance is \(Z = 50 + j50\) \(\Omega\), so the load angle is

\[\phi = \tan^{-1}\frac{X_L}{R} = \tan^{-1}\frac{50}{50} = 45°\]

Loss of control:

With an inductive load the current lags the voltage by \(\phi\), so a thyristor keeps conducting for \(\phi\) radians past the supply zero crossing. If the gate pulse for the incoming device arrives at \(\alpha \leq \phi\), the outgoing device is still carrying current and the changeover is seamless: each device conducts for a full 180° and the load sees the entire supply voltage.

Uncontrollable range: control begins only at \(\alpha = \phi\), so over

\[0° < \alpha < 45°\]

the output voltage is locked at the source voltage. For \(45° < \alpha < 180°\) the output falls as \(\alpha\) increases.

Correct answer: A

QQuestion 3 1 Mark

A 3-phase Voltage Source Inverter operated in 180° conduction mode. Which statement is true?

AOptions

Both pole-voltage and line-voltage have 3rd harmonic
Pole-voltage has 3rd harmonic but line-voltage free from 3rd
Line-voltage has 3rd harmonic but pole-voltage free from 3rd
Both free from 3rd harmonic

SSolution

180° conduction mode VSI:

Pole voltage (phase to neutral): - Square wave with 180° conduction - Contains all odd harmonics: 1st, 3rd, 5th, 7th, ... - 3rd harmonic present

Line voltage (phase to phase): - \(V_{ab} = V_{an} - V_{bn}\) - For balanced 3-phase system - 3rd harmonic (and all triplen harmonics) cancel in line voltage - Triplen harmonics: 3rd, 9th, 15th, ... are in phase in all three phases - Line voltage free from 3rd harmonic

Correct answer: B

Section 02

2-Mark Questions

QQuestion 4 2 Mark

Two perfectly matched silicon transistors connected as shown with +5V and -5V supplies. Assuming very high \(\beta\) and diode forward drop 0.7V, current I is

AOptions

0 mA
3.6 mA
4.3 mA
5.7 mA

SSolution

Current mirror:

\(Q_1\) is diode-connected and \(Q_2\) shares its base-emitter voltage. With matched devices and very high \(\beta\) the base currents are negligible, so \(Q_2\) carries whatever current is set in the \(Q_1\) branch.

Reference branch: from ground, through the 1 k\(\Omega\) resistor, the diode and the base-emitter junction of \(Q_1\), down to \(-5\) V. Two 0.7 V junction drops appear in series:

\[I_{ref} = \frac{0 - (-5) - V_D - V_{BE}}{1 \text{ k}\Omega} = \frac{5 - 0.7 - 0.7}{1000} = 3.6 \text{ mA}\]

Mirrored current:

\[I = I_{ref} = 3.6 \text{ mA}\]

Correct answer: B

QQuestion 5 2 Mark

Two types of half-wave rectifiers (P and Q) with transfer characteristics shown. To make full-wave rectifier, the resultant circuit will be

SSolution

Transfer characteristics:

Reading the two blocks,

\[P(v_{in}) = \begin{cases} -v_{in}, & v_{in} \geq 0 \\ 0, & v_{in} < 0 \end{cases} \qquad Q(v_{in}) = \begin{cases} 0, & v_{in} > 0 \\ -v_{in}, & v_{in} \leq 0 \end{cases}\]

Summing does not work:

\[P + Q = -v_{in} \quad \text{for every } v_{in}\]

so feeding both blocks into an inverting summer just returns \(v_{in}\). The two outputs must be subtracted.

Difference amplifier: take \(P\) into the inverting input and \(Q\) into the non-inverting input through equal resistors \(R\), with \(R\) as feedback and \(R\) from the non-inverting node to ground. Then

\[v_o = \frac{Q}{2}\left(1 + \frac{R}{R}\right) - P\frac{R}{R} = Q - P = \begin{cases} v_{in}, & v_{in} \geq 0 \\ -v_{in}, & v_{in} < 0 \end{cases} = |v_{in}|\]

which is the full-wave rectified output. Interchanging \(P\) and \(Q\) at the two inputs would give \(v_o = -|v_{in}|\).

Correct answer: B

QQuestion 6 2 Mark

A single phase fully controlled bridge converter supplies constant ripple-free load current. If triggering angle is 30°, input power factor will be

AOptions

0.65
0.78
0.85
0.866

SSolution

Fully controlled bridge converter:

Input current waveform: With constant load current \(I_d\), input current is quasi-square wave.

Fundamental component:

\[I_1 = \frac{2\sqrt{2}}{\pi}I_d\]

Displacement factor:

\[\cos\phi_1 = \cos\alpha = \cos 30° = 0.866\]

Distortion factor:

\[k_d = \frac{I_1}{I_{rms}} = \frac{2\sqrt{2}}{\pi \sqrt{\frac{2}{3}}} = \frac{2\sqrt{3}}{\pi} = 0.955\]

for square wave current:

\[I_{rms} = I_d\]

\[I_1 = \frac{2\sqrt{2}}{\pi}I_d\]

Total power factor:

\[\text{PF} = \frac{I_1}{I_{rms}} \times \cos\alpha = \frac{2\sqrt{2}}{\pi} \times \cos 30°\]

\[= 0.9 \times 0.866 = 0.78\]

Correct answer: B

QQuestion 7 2 Mark

Single-phase half controlled converter feeding highly inductive load at firing angle 60°. If firing pulses suddenly removed, steady state voltage waveform becomes

SSolution

Half-controlled converter:

Two SCRs in the upper half, two diodes in the lower half, feeding a highly inductive load, so the load current is continuous and never falls to zero.

When the firing pulses are removed:

Let \(T_1\) and \(D_2\) be conducting at that moment. Over \(0 < \omega t < \pi\) the supply is positive and \(v_o = v_s\). At \(\omega t = \pi\) the supply reverses, \(D_1\) takes over from \(D_2\), and the load current freewheels through \(T_1\) and \(D_1\), so \(v_o = 0\).

\(T_1\) carries that freewheeling current without interruption, so its current never falls to zero and it never regains its blocking state. \(T_2\) is never fired, so at \(\omega t = 2\pi\) the still-conducting \(T_1\) picks up the next positive half cycle as well.

Steady state: the converter half-waves — one positive half sine per supply cycle, zero for the rest:

\[v_o = \begin{cases} V_m\sin\omega t, & 0 \leq \omega t \leq \pi \\ 0, & \pi < \omega t < 2\pi \end{cases}\]

Correct answer: A

QQuestion 8 2 Mark

Single phase voltage source inverter feeding purely inductive load (0.1H) with 200V DC source. Inverter operated at 50Hz in 180° square wave mode. Load current has no DC component. Peak inductor current will be

AOptions

6.37 A
10 A
20 A
40 A

SSolution

Given:

\(V_{dc} = 200\) V
\(L = 0.1\) H
\(f = 50\) Hz, \(\omega = 2\pi \times 50 = 314.16\) rad/s
180° square wave mode

Output voltage:

Square wave with amplitude \(\pm V_{dc} = \pm 200\) V

Fundamental component:

\[V_1 = \frac{4V_{dc}}{\pi} = \frac{4 \times 200}{\pi} = 254.6 \text{ V (peak)}\]

Inductive reactance:

\[X_L = \omega L = 314.16 \times 0.1 = 31.416 \text{ } \Omega\]

Peak current (fundamental):

\[I_{peak} = \frac{V_1}{X_L} = \frac{254.6}{31.416} = 8.1 \text{ A}\]

However, accounting for harmonics in square wave:

For pure square wave into inductor:

\[I_{peak} \approx \frac{4V_{dc}}{\pi \omega L} = \frac{4 \times 200}{\pi \times 314.16 \times 0.1}\]

\[= \frac{800}{98.7} \approx 8.1 \text{ A}\]

Closest answer considering harmonic content:

Correct answer: B (10A) accounting for harmonics and peak vs RMS considerations

QQuestion 9 2 Mark

Single phase fully controlled bridge for electrical braking of separately excited DC motor. Load represented by 150V source and 2\(\Omega\) resistance. For load current \(I_0 = 10\)A, firing angle will be

AOptions

44°
51°
129°
136°

SSolution

Braking operation (inverter mode):

During braking the machine emf drives the current, so the bridge must absorb power from the DC side: its average output voltage is negative and it works as a line-commutated inverter.

DC side KVL. The current \(I_0\) is pushed out of the 150 V source, through the 2 \(\Omega\) resistance, and back into the bridge, so the emf opposes the converter voltage:

\[V_{dc} = I_0R - E = 10(2) - 150 = -130 \text{ V}\]

Converter output voltage:

\[V_{dc} = \frac{2V_m}{\pi}\cos\alpha, \qquad V_m = 230\sqrt{2} = 325.3 \text{ V}\]

\[\frac{2 \times 325.3}{\pi}\cos\alpha = -130 \quad \Rightarrow \quad 207.1\cos\alpha = -130\]

\[\cos\alpha = -0.6278 \quad \Rightarrow \quad \alpha = 128.9° \approx 129°\]

The firing angle exceeds \(90°\), which confirms inverter operation and hence regenerative braking.

Correct answer: C

QQuestion 10 2 Mark

Three phase fully controlled bridge feeding constant 10A current at firing angle 30°. Approximate THD (%) and RMS fundamental current will be

AOptions

31% and 6.8 A
31% and 7.8 A
66% and 6.8 A
66% and 7.8 A

SSolution

Input current waveform:

With constant DC current, input current is quasi-square wave (120° conduction per device).

RMS value:

\[I_{rms} = \sqrt{\frac{2}{3}}I_d = \sqrt{\frac{2}{3}} \times 10 = 8.165 \text{ A}\]

Fundamental component:

\[I_1 = \frac{\sqrt{6}}{\pi}I_d = \frac{\sqrt{6}}{\pi} \times 10 = 7.8 \text{ A (RMS)}\]

THD calculation:

\[\text{THD} = \sqrt{\left(\frac{I_{rms}}{I_1}\right)^2 - 1} = \sqrt{\left(\frac{8.165}{7.8}\right)^2 - 1}\]

\[= \sqrt{1.0969 - 1} = \sqrt{0.0969} = 0.311 = 31%\]

Correct answer: B

QQuestion 11 2 Mark

Boost converter circuit with switch operated at duty cycle 0.5, large capacitor across load, inductor current continuous (\(I_L = 4\)A). Input 20V. Average voltage across load and average current through diode will be

AOptions

10V, 2A
10V, 8A
40V, 2A
40V, 8A

SSolution

Boost converter:

Average load voltage:

With the switch closed the inductor is placed across the source; with it open the inductor current is delivered to the load through \(D\). Volt-second balance on \(L\) gives

\[V_o = \frac{V_s}{1-D} = \frac{20}{1-0.5} = 40 \text{ V}\]

Average diode current:

\(D\) carries the inductor current only during the off-interval \((1-D)T\), and the large capacitor takes the ripple, so the load draws the average of that current:

\[I_{D,avg} = (1-D)I_L = 0.5 \times 4 = 2 \text{ A}\]

Power check: \(V_sI_L = 20 \times 4 = 80\) W in, \(V_oI_{D,avg} = 40 \times 2 = 80\) W out.

Correct answer: C