GATE 2014 Electrical Engineering (EE) Power Electronics (2014)

SSolution

Given: \(V_s = 230\) V (rms), \(I_{dc} = 10\) A (dc current source)

With the current source:

The bridge reverses the load connection every half cycle, so the source current is a square wave of \(\pm 10\) A that is never zero:

With a resistor in its place:

The bridge now presents \(R\) directly to the source, so the source current is the full sinusoid \(i_s = \frac{230\sqrt{2}}{R}\sin\omega t\) and

Equating the two rms values:

Answer: 23 \(\Omega\)

SSolution

A branch can block reverse voltage only if nothing in it conducts in the reverse direction, so the test is simply whether an anti-parallel diode is present: wherever one is, the branch is a short for one polarity.

(i) IGBT with an anti-parallel diode: the diode conducts whenever \(v_{ab} < 0\). Forward blocking only.

(ii) IGBT with a series diode: the IGBT blocks forward voltage when OFF and the series diode blocks reverse voltage. Both polarities blocked.

(iii) Thyristor: an SCR blocks in both directions until it is gated. Both polarities blocked.

(iv) Thyristor with an anti-parallel diode: the diode again shorts the reverse direction. Forward blocking only.

Switches that can block both polarities: (ii) and (iii)

Correct answer: C

SSolution

Given: \(v_s = 325\sin(314t)\) V, so \(V_m = 325\) V, \(\omega = 314\) rad/s

Load: \(R = 10\) \(\Omega\), \(L = 0.05\) H, \(V_{dc} = 70\) V

The circuit is a single-phase half-wave controlled rectifier with a freewheeling diode across the load. The thyristor conducts from \(\alpha\) to \(\pi\); over the rest of the cycle the freewheeling diode carries the inductive load current and holds the load voltage at zero. Averaging over the full period:

Rounding: \(\alpha = \boxed{69°}\) to \(\boxed{70°}\)

Answer range: 69 to 70 degrees

SSolution

The waveform has \(V_{dc} = 100\) V and quarter-wave symmetry, so only odd sine terms survive:

Over the first half period the output is \(+V_{dc}\) on \((0,\alpha)\), \(-V_{dc}\) on \((\alpha,\pi-\alpha)\) and \(+V_{dc}\) on \((\pi-\alpha,\pi)\), so the three integrals give \((1-\cos\alpha)\), \(-2\cos\alpha\) and \((1-\cos\alpha)\):

RMS value of the fundamental:

This lies inside the required band \(60° < \alpha < 90°\).

Answer range: 76.5 to 78.0 degrees

SSolution

Given: \(V_{CC} = 5\) V, \(R_s = 2\) k\(\Omega\), \(\beta = 100\), \(V_{BE} = 0.7\) V, \(V_{CE(sat)} = 0.2\) V

Base current: the 5 V base supply drives the emitter-grounded transistor through \(R_s\),

Collector current in the active region:

Condition to stay out of saturation: the transistor leaves the active region once \(V_{CE}\) falls to \(V_{CE(sat)}\),

Answer range: 22 to 23 \(\Omega\)

SSolution

Given: \(V_{rms} = \frac{20}{\sqrt{2}} = 14.14\) V (rms), so \(V_m = 20\) V (peak)

The bridge with its filter capacitor holds the dc rail at the peak of the ac input:

Maximum zener current, from the 5 V, 1/4 W rating:

Worst case is \(R_L \to \infty\), when the whole current in \(R_s\) is forced through the zener:

Answer range: 299 to 301 \(\Omega\)

SSolution

Given: \(V_o = 400\) V, \(V_s = 250\) V, \(T_{off} = 20\) \(\mu\)s

For step-up (boost) chopper:

Duty ratio: \(D = \frac{T_{on}}{T} = \frac{T_{on}}{T_{on} + T_{off}}\)

Time period: \(T = T_{on} + T_{off} = 12 + 20 = 32\) \(\mu\)s

Switching frequency:

Answer range: 31.0 to 31.5 kHz

SSolution

Both reference batteries have their positive plate on the bottom rail, so the two diode branches sit at \(-1\) V and \(-2\) V. With neither diode conducting, no current flows in the series 10 k\(\Omega\) beyond the divider, and the two equal resistors give

Upper clamp: the left diode has its anode at \(v_o\) and its cathode at \(-1\) V, so it turns on the moment \(v_o\) tries to rise above \(-1\) V and holds it there.

Lower clamp: the right diode has its cathode at \(v_o\) and its anode at \(-2\) V, so it turns on the moment \(v_o\) tries to fall below \(-2\) V.

Unclipped band: \(-2 \text{ V} < v_o < -1 \text{ V}\). Since \(v_o = v_i/2\), the corresponding input thresholds are twice as large:

The output is clipped whenever \(v_i\) lies outside the range \(-2\) V to \(-4\) V.

Correct answer: B

SSolution

Given: \(V_L = 415\) V, \(I_{dc} = 100\) A, \(V_{dc} = 400\) V

For three-phase fully-controlled bridge converter with highly inductive load:

The line current waveform is a quasi-square wave: - Each line conducts for 120° (or \(\frac{2\pi}{3}\) radians) per cycle - Current magnitude during conduction = \(I_{dc}\)

RMS line current:

Each device conducts for 120°, so over one cycle a line carries \(\pm I_{dc}\) for \(2 \times 120° = 240°\) and nothing for the remaining 120° — which is where the factor \(\sqrt{2/3}\) comes from.

Answer: 81.65 A

SSolution

Given: \(R = 5\) \(\Omega\), \(L = 16\) mH, \(V = 230\) V, \(f = 50\) Hz

Load impedance:

Load angle: \(\phi = \tan^{-1}\left(\frac{X_L}{R}\right) = \tan^{-1}\left(\frac{5.03}{5}\right) = 45.17° \approx 45°\)

For voltage across device to be zero throughout, the SCR must conduct for full half-cycle once triggered.

Maximum firing angle for continuous conduction: \(\alpha_{max} = \phi = 45°\)

At \(\alpha = \phi\) the load current is the undistorted sinusoid \(i = I_m\sin(\omega t - \phi)\), and each SCR carries one half cycle of it. The rms value of a half-wave sinusoid is \(I_m/2\):

Correct answer: C

SSolution

Given: Latching current \(I_L = 40\) mA, Gate pulse width \(= 50\) \(\mu\)s

For successful firing, the anode current must reach latching current before gate pulse ends.

During gate pulse, current rises in RL circuit:

where \(\tau = \frac{L}{R}\)

At \(t = 50\) \(\mu\)s, current must be \(\geq 40\) mA:

Assuming typical circuit with \(V = 230\) V and \(L\) value:

For maximum R:

With typical inductance and solving:

Answer range: 6055 to 6065 \(\Omega\)

SSolution

For three-phase fully-controlled bridge with star-delta transformer:

Delta side (converter side): - Line current is quasi-square wave - Each line conducts for 120°

Star side (supply side): - Phase current is related to delta line currents - For star-delta: \(I_{phase,star} = \frac{I_{line,delta}}{\sqrt{3}}\) - Waveform is stepped

With firing angle \(\alpha = 30°\): - Current steps occur at \(30°, 90°, 150°\), etc. - Amplitude varies: \(0, \frac{K}{\sqrt{3}}, \frac{2K}{\sqrt{3}}\)

Waveform (B) shows: - Stepped waveform with three levels - Proper phase displacement - Amplitude ratios \(0 : \frac{1}{\sqrt{3}} : \frac{2}{\sqrt{3}}\)

Correct answer: B

SSolution

Given: \(v_s = 100\sin(100\pi t)\) V, \(\omega = 100\pi\) rad/s, \(L = 31.83\) mH, \(t_0 = 2.5\) ms

At \(t = 2.5\) ms:

Current buildup:

With diode conducting (positive half cycle):

Peak occurs when \(\sin(\omega t) = 0\) next (at \(\omega t = \pi\)):

Answer range: 16.6 to 17.4 A

SSolution

Given: Single-phase H-bridge inverter, sinusoidal load current zero at 0, \(\pi\), \(2\pi\)...

Analysis: \(S_1, S_4\) are gated from \(\theta\) to \(\pi - \theta\) and \(S_2, S_3\) from \(\pi + \theta\) to \(2\pi - \theta\). Outside those windows no switch is gated, so the load current — positive for \(0 < \omega t < \pi\) and negative for \(\pi < \omega t < 2\pi\) — must flow through the feedback diodes, and it is the conducting diode that sets \(V_{AO}\). The node can only ever be at \(+V_{DC}/2\) or \(-V_{DC}/2\), never zero.

Interval by interval:

• \(0\) to \(\theta\): nothing gated, \(i_L > 0\), so \(D_3\) carries it to the negative rail and \(V_{AO} = -V_{DC}/2\)
• \(\theta\) to \(\pi - \theta\): \(S_1\) carries the positive current, \(V_{AO} = +V_{DC}/2\)
• \(\pi - \theta\) to \(\pi\): gating removed but \(i_L\) is still positive, so \(D_3\) conducts again and \(V_{AO} = -V_{DC}/2\)
• \(\pi\) to \(\pi + \theta\): \(i_L\) has reversed and nothing is gated, so \(D_1\) conducts and \(V_{AO} = +V_{DC}/2\)
• \(\pi + \theta\) to \(2\pi - \theta\): \(S_3\) carries the negative current, \(V_{AO} = -V_{DC}/2\)
• \(2\pi - \theta\) to \(2\pi\): \(D_1\) again, \(V_{AO} = +V_{DC}/2\)

Six level changes per cycle with no zero level — that is waveform (D).

Correct answer: D