Q21GATE 2007MCQ

In the circuit of adjacent figure the diode connects the ac source to a pure inductance L. The diode conducts for

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📖 Explanation

$i_0=\frac{V_m}{\omega L}(1- \cos \omega t)$ Diode conducts for $360^{\circ}$ .

Q22GATE 2007MCQ

The circuit in the figure is a current commutated dc-dc chopper where, $Th_M$ is the main SCR and $Th_{AUX}$ is the auxiliary SCR. The load current is constant at 10 A. $Th_M$ is ON. $Th_{AUX}$ is trigged at t = 0. $Th_M$ is turned OFF between.

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📖 Explanation

At $t=0^-,v_c=V_s, i_c=0\; and \; i_{T1}=I_0$ At t=0, $Th_{aux}$ is triggered, a resonant current $i_c$ designs to flow from C through $Th_{aux}$ , L and back to C. This resonant current is given by $i_c=-V_s\sqrt{\frac{C}{L}} \sin \omega _0 t$ $i_c=-I_P \sin \omega _0 t$ after half a cycle of $i_c\left \{ t_1=\frac{\pi}{\omega _0} \right \}$ ; $i_c=0, v_c=-V_s \; and \; i_{T1}=I_0$ . As $i_c$ tends to reverse, $Th_{aux}$ is turned off. When $v_c=-V_s$ right hand plate has positive polarity, resonant current $i_c$ now builds up through C, L, D and $Th_m$ . As this current $i_c$ grows opposite to forward thyristor current of $Th_m$ , net forward current $i_m=I_0-i_c$ begins to decrease. Finalyy when $i_c$ in the reversed direction attains the value $I_0, i_m$ is reduced to zero and $Th_m$ is turned off. $i_m=I_0-i_c=I_0-I_p \sin \omega _0\Delta t=0$ $\Delta t=\frac{1}{\omega _0}\sin ^{-1}\left ( \frac{I_0}{I_p} \right )$ So, $Th_m$ is turned off between $t_1 \lt t \lt t_1+\Delta t$ $t_1 =\frac{\pi}{\omega _0}= \pi \sqrt{LC}=\pi \times \sqrt{10 \times 25.28}=50\mu sec$ Since commutation of $Th_m$ starts from $t_1=50\mu sec$ .

Q23GATE 2007MCQ

A 1:1 Pulse Transformer (PT) is used to trigger the SCR in the below figure. The SCR is rated at 1.5 kV, 250 A with $I_L$ =250 mA, $I_H$ =150 mA, and $I_{Gmax}$ =150 mA, $I_{Gmin}$ =100 mA. The SCR is connected to an inductive load, where L = 150 mH in series with a small resistance and the supply voltage is 200 V dc. The forward drops of all transistors/diodes and gate-cathode junction during ON state are 1.0 V The minimum approximate volt-second rating of pulse transformer suitable for triggering the SCR should be : (volt-second rating is the maximum of product of the voltage and the width of the pulse that may applied)

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📖 Explanation

Forward voltage drop of SCR during ON-state

\begin{aligned} V_T &=1V \\ E-\frac{Ldi_a}{dt}-Ri_a-V_T&=0 \\ \Rightarrow \; 200-0.15\frac{di_a}{dt}-i_a-1 &=0 \\ \Rightarrow \;\; i_a &=199(1-e^{-t/0.15}) \end{aligned}

Gate pulse width required = time taken by i_a to rise up to $I_L=T$

\begin{aligned} I_L&= i_a\\ 250 \times 10^{-3}&=199(1-e^{-T/0.15}) \\ T&=188.56\mu s \\ \text{Width of the pulse } &=T= 188.56\mu s\\ \text{Magnitude of voltage } &=V=10V \\ \text{Voltage } & \text{ second rating of PT} \\ VT&=T=10 \times 188.56 \\ &= 1885.6 \approx 2000\mu s \end{aligned}

Q24GATE 2006MCQ

An SCR having a turn ON times of 5 $\mu$ sec, latching current of 50 mA and holding current of 40 mA is triggered by a short duration pulse and is used in the circuit shown in figure. The minimum pulse width required to turn the SCR ON will be

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📖 Explanation

Current through $5 k\Omega$ resistor, $i_R=\frac{V}{R_2}=\frac{100}{5 \times 10^3}=20mA=0.02A$ Current through inductor,

\begin{aligned} i_L&=\frac{V}{R_1}(1-e^{-R_1/Lt}) \\ &= \frac{100}{20}(1-e^{20/0.5t})\\ &= 5(1-e^{-40t}) \end{aligned}

Anode current, $I_a=i_R+i_L=0.02+5(1-e^{-40t})$ Let minimum pulse width is T To turn on $i_a \geq$ latching current $\Rightarrow \;\;0.02+5(1-e^{-40t})=50mA=0.5$ $T=150\mu sec$

Q25GATE 2006MCQ

A voltage commutation circuit is shown in figure. If the turn-off time of the SCR is 50 $\mu$ sec and a safety margin of 2 is considered, then what will be the approximate minimum value of capacitor required for proper commutation ?

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📖 Explanation

In this type of commutation, athyristor carrying load current is commutated by transferring its load current to another incoming thyristor. Firing of SCR $Th_1$ commutates $Th_2$ and subsequently, firing of SCR $Th_2$ would turn-off $Th_1$ . Circuit turn-off time $t_{c1}$ for $Th_1$ $t_{c1}=R_1C \ln 2$ and circuit turn-off time $t_{c2}$ for $Th_2$ $t_{c2}=R_2C \ln 2$ as $R_1=R_2=50 \Omega$ $t_{c2}=t_{c1}=R_1 C \ln 2$ Safety margin =2 So, $R_1C \ln 2=2 t_{c1}$ $C=\frac{2 \times 50 \times 10^{-6}}{50 \times \ln 2}=2.88 \mu F$

Q26GATE 2005MCQ

An electronics switch S is required to block voltage of either polarity during its OFF state as shown in the figure (a). This switch is required to conduct in only one direction its ON state as shown in the figure (b)

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Q27GATE 2005MCQ

The conduction loss versus device current characteristic of a power MOSFET is best approximated by

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📖 Explanation

Let, I= device current $R_{ON}=$ ON state resistance of power MOSFET Conduction loss $=P=I^2R_{ON}$ Therefore, condition losses versus device current characteristics can be best approximated by a parabola.

Q28GATE 2005MCQ

The figure shows the voltage across a power semiconductor device and the current through the device during a switching transitions. Is the transition a turn ON transition or a turn OFF transition ? What is the energy lost during the transition ?

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📖 Explanation

During interval $t_2$ , voltage starts decreasing and becomes zero and current starts increasing and becomes constant (I), so transition is turn on. During $t_1$ interval: Power loss $=vi$ $E_1=\text{Energy loss}=\int vi\; dt=V\int i\; dt$ V is constant during this period v=V $\int i\; dt$ represents area under i-t curve $\int i\; dt=\frac{1}{2} \times I \times t_1$ $E_1=V\int i\; dt=\frac{1}{2} V I t_1\;\;...(i)$ During $t_2$ interval: Power loss = vi $E_2=\text{Energy loss}=\int vi\; dt=I\int v\; dt$ i is constant during this period i=I $\int v\; dt$ represents area under v-t curve $\int v\; dt=\frac{1}{2} VIt_2$ $E_2=I\int v\; dt=\frac{1}{2} V I t_2\;\;...(ii)$ Total energy lost during the transition $E=E_1+E_2$ $E=\frac{1}{2}VIt_1+\frac{1}{2}VIt_2$

Q29GATE 2004MCQ

A bipolar junction transistor (BJT) is used as a power control switch by biasing it in the cut-off region (OFF state) or in the saturation region (ON state). In the ON state, for the BJT

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📖 Explanation

$V_{CB}=V_{CE}-V_{DE}\;\;...(i)$ Under saturated state, $V_{BES}$ is greater than $V_{CES}$ this means base -emitter junction (BES) is forward baised. Further eq, (i) shows that $V_{CB}$ is negative under saturated conditions, therefore base-collector junction (CBJ) is also forward baised.

Q30GATE 2004MCQ

A MOSFET rated for 15 A, carries a periodic current as shown in figure. The ON state resistance of the MOSFET is 0.15 $\Omega$ . The average ON state loss in the MOSFET is

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📖 Explanation

Rated current during on state $I=10$ ON state resistance $R_{ON}=0.15\Omega$ MOSFET is ON

\begin{aligned} 0\lt \omega t \lt \pi \\ 0 \lt t \lt \pi/ \omega \end{aligned}

MOSFET is OFF

\begin{aligned} \pi \lt \omega t \lt 2\pi \\ \pi/ \omega \lt t \lt 2\pi/ \omega \end{aligned}

Average ON state Loss,

\begin{aligned} P_{avg}&=\frac{1}{2 \pi/\omega }\int_{0}^{\pi/\omega }I^2R_{ON}dt\\ &=\frac{\omega }{2 \pi}\times I^2 R_{ON}\times \frac{\pi}{\omega }\\ &=\frac{I^2R_{ON}}{2}=\frac{10^2 \times 0.15}{2}=7.5W \end{aligned}

Q31GATE 2003MCQ

Figure shows a MOSFET with an integral body diode. It is employed as a power switching device in the ON and OFF states through appropriate control. The ON and OFF states of the switch are given on the $V_{DS}$ - $I_{S}$ plane by

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📖 Explanation

when reverse current flow through diode D. So $I_S \lt 0$ and $V_{DS} = 0$ When MOSFET is in ON state $I_S \gt 0$ and $V_{DS} = 0$ When MOSFET is in OFF state $I_S = 0$ and $V_{DS} \gt 0$

Q32GATE 2003MCQ

Figure shows a thyristor with the standard terminations of anode (A), cathode (K), gate (G) and the different junctions named $J_{1}$ , $J_{2}$ and $J_{3}$ . When the thyristor is turned on and conducting

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Q33GATE 2001MCQ

The main reason for connecting a pulse transformer at the output stage of a thyristor triggering circuit is to

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