Q21GATE 2015NAT

In the circuit shown below, the Zener diode is ideal and the Zener voltage is 6 V. The output voltage $V_{o}$ (in volts) is_______.

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📖 Explanation

Zener is not in breakdown region Hence, $V_{0}=10 \times \frac{1}{2}=5 \mathrm{V}$

Q22GATE 2015MCQ

If the circuit shown has to function as a clamping circuit, then which one of the following conditions should be satisfied for the sinusoidal signal of period T ?

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Q23GATE 2015MCQ

For the circuit with ideal diodes shown in the figure, the shape of the output ( $v_{out}$ ) for the given sine wave input ( $v_{\in}$ ) will be

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📖 Explanation

The circuit can be redrawn as During positive pulse. both diodes are forward biased. During negative pulse, both diodes are reverse biased. So, $V_{o}=0V$

Q24GATE 2015NAT

In the circuit shown, assume that diodes $D_{1} \; and \; D_{2}$ are ideal. In the steady state condition, the average voltage $V_{ab}$ (in Volts) across the 0.5 $\mu$ F capacitor is _____.

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📖 Explanation

The circuits work as a voltage doubler. $\therefore V_{a b}=2 \times 50=100$

Q25GATE 2015NAT

In the circuit shown, assume that the diodes $D_1 \; and \; D_2$ are ideal. The average value of voltage $V_{ab}$ (in Volts), across terminals 'a' and 'b' is _________.

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📖 Explanation

During positive cycles: $V_{1}=\frac{V_{\text {\in }}}{2}$ During negative cycles: $V_{0}=\frac{V_{\text {\in }}}{3}$ $V_{D C}=\frac{V_{m}}{2} \cdot \frac{1}{\pi}+\frac{V_{m}}{3} \cdot \frac{1}{\pi}=\frac{6 \pi}{2 \pi}+\frac{6 \pi}{3 \pi}=5 \mathrm{V}$

Q26GATE 2014MCQ

Two silicon diodes, with a forward voltage drop of 0.7 V, are used in the circuit shown in the figure. The range of input voltage $V_{i}$ for which the output voltage $V_{0}=V_{i}$ , is

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📖 Explanation

\begin{aligned} \text { For case-i } & V_{i} \lt -1.7 \\ & D_{1} \rightarrow \mathrm{ON} \\ & D_{2} \rightarrow \text { OFF }\\ &V_{0}=-1.7 \mathrm{V} \end{aligned}

For case-ii

\begin{aligned} -1.7 &\lt V_{i} \lt 2.7 \\ D_{1} & \rightarrow \text { OFF } \\ D_{2} & \rightarrow \text { OFF } \end{aligned}

Q27GATE 2014NAT

The figure shows a half-wave rectifier. The diode D is ideal. The average steady-state current (in Amperes) through the diode is approximately_____.

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📖 Explanation

The average current through the diode is approximate, $i_{D(\arg )} \approx \frac{V_{p}}{R}=\frac{10}{100}=0.10 \mathrm{A}$

Q28GATE 2014MCQ

The diode in the circuit shown has $V_{on}$ = 0.7 Volts but is ideal otherwise. If $V_{i} = 5 sin(\omega t)$ Volts, the minimum and maximum values of $V_{o}$ (in Volts) are, respectively,

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📖 Explanation

\begin{aligned} \text{If }\quad V_{\text {\in }}&=-5 \mathrm{V} \Rightarrow \text{Diode is off }\\ V_{0}&=V_{\text {\in }}=-5 V \\ \text{If }\quad V_{\text {\in }}&=5 \mathrm{V} \Rightarrow \text{Diode is on }\\ V_{0} &=\frac{V_{\text {\in }}-0.7-2}{1 \mathrm{k} \Omega+1 \mathrm{k} \Omega} \times 1 \mathrm{k} \Omega+0.7+2 \\ &=\frac{5-2.7}{2}+2.7=3.85 \mathrm{V} \end{aligned}

Q29GATE 2014MCQ

In the figure, assume that the forward voltage drops of the PN diode $D_{1}$ and Schottky diode $D_{2}$ are 0.7 V and 0.3 V, respectively. If ON denotes conducting state of the diode and OFF denotes non-conducting state of the diode, then in the circuit,

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📖 Explanation

Consider $D_{1} \rightarrow \text{OFF }$ and $D_{2} \rightarrow \text{ON}$ then Apply KVL

\begin{aligned} 10 &=1000 I+20 I+0.3 \\ I &=\frac{9.7}{1020} \\ I &=9.5 \mathrm{mA} \end{aligned}

Now, we calculate $V_{D_{1}}$ of

\begin{aligned} & 10=9.5+V_{D_{1}} \\ \therefore V_{D_{1}} &=0.5 \mathrm{V} \end{aligned}

since $v_{D_{1}} \lt 0.7 \mathrm{V}, \quad \mathrm{D}_{1}$ is in OFF state i.e. our assumption is correct and hence (D) is the correct option.

Q30GATE 2013MCQ

A voltage $1000\sin \omega t$ Volts is applied across YZ. Assuming ideal diodes, the voltage measured across WX in Volts, is

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📖 Explanation

Case 1: When $V_{Y Z}$ is positive, then all the four diodes are reverse bias

\begin{aligned} \mathrm{So}, V_{w} &=V_{x}=0 \\ V_{w x} &=0 \mathrm{V} \end{aligned}

Case 2: When $V_{Y Z}$ is negative, then all the four diodes are forward bias So, all the diodes will be short circuit

\begin{aligned} V_{w} &=V_{x} \\ V_{w x} &=V_{w}-V_{x}=V_{w}-V_{w}=0 \\ V_{w x} &=0\\ \text{So}, V_{w x}&=0 \;\text{for all }t \end{aligned}

Q31GATE 2013MCQ

In the circuit shown below, the knee current of the ideal Zener diode is 10 mA. To maintain 5 V across $R_{L}$ , the minimum value of $R_{L}$ in $\Omega$ and the minimum power rating of the Zener diode in mW, respectively, are

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📖 Explanation

\begin{aligned} I &=\frac{V_{\text {\in }}-V_{z}}{R}=\frac{10-5}{100}=50 \mathrm{mA} \\ R_{\text {Lmin }} &=\frac{V_{z}}{I_{L \max }}=\frac{V_{z}}{I-I_{z \min }}=125 \Omega \end{aligned}

Minimum power rating of Zener diode

\begin{aligned} P_{z} &=V_{z} I_{z \max } \\ I_{z \max } &=50 \mathrm{mA} \\ P z &=5 \times 50=250 \mathrm{mW} \end{aligned}

Q32GATE 2012MCQ

The diodes and capacitors in the circuit shown are ideal. The voltage $v(t)$ across the diode D1 is

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📖 Explanation

It is voltage doubles circuit in which $C_{1}$ will be charged to maximum value of input that is 1V. So v(t) = $(\cos \omega t- 1)$ according to KVL.

Q33GATE 2011MCQ

In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage $V_t=KT/q=25mV$ . The small signal input $v_i=V_P \cos (\omega t)$ where $V_P=100mV$ . The ac output voltage $V_{ac}$ is

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📖 Explanation

For a.c. analysis diode will be replaced by its dynamic resistance

\begin{aligned} r_{d}&=\frac{\eta V_{T}}{I}=\frac{1 \times 25 \mathrm{mV}}{1 \mathrm{mA}}=25 \Omega \\ \left(v_{0}\right)_{\mathrm{ac}}&=\frac{v_{i}}{9900+100} \times 100\\ \quad&=\frac{v_{i}}{100}=\frac{v_{p}}{100} \cos \omega t\\ \Rightarrow\left(v_{0} \right )_{ac}&=1\cos \omega t \; V \end{aligned}

Q34GATE 2009MCQ

In the circuit below, the diode is ideal. The voltage V is given by

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Q35GATE 2008MCQ

In the following limiter circuit, an input voltage $V_{i}=10sin100\pi t$ is applied. Assume that the diode drop is 0.7 V when it is forward biased. When it is forward biased. The zener breakdown voltage is 6.8 V The maximum and minimum values of the output voltage respectively are

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📖 Explanation

For the positive half of $V_{i}$ , D1 is forward biased and Zener diode is in breakdown stage $V_{0} = 0.7+ 6.8 = 7.5 V$ For the negative half of $V_{i}$ , D2 is forward biased. $V_{0} = -0.7 V$

Q36GATE 2007MCQ

For the Zener diode shown in the figure, the Zener voltage at knee is 7 V, the knee current is negligible and the Zener dynamic resistance is 10 $\Omega$ . If the input voltage ( $V_{i}$ ) range is from 10 to 16 V, the output voltage ( $V_{o}$ ) ranges from

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📖 Explanation

\begin{aligned} V_{0}&=\frac{V_{\text {\in }}-V_{z}}{R+r_{z}} r_{z}+V_{z} \\ \text{If}\;V_{\in}=10\mathrm{V}&\\ V_{0}&=\frac{10-7}{200+10} \times 10+7=7.14 \mathrm{V} \\ \text{If}\;V_{\in}=16\mathrm{V}&\\ V_{0}&=\frac{16-7}{200+10} \times 10+7=7.43 \mathrm{V} \end{aligned}

Q37GATE 2007MCQ

The correct full wave rectifier circuit is

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Q38GATE 2006MCQ

For the circuit shown below, assume that the zener diode is ideal with a breakdown voltage of 6 volts. The waveform observed across R is

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📖 Explanation

Zener diode works as normal diode in FB. So when $V_{\text {\in }} \lt 0, V_{R}=V_{\text {\in }}$ When $0 \lt V_{\text {\in }}\lt 6,$ Diode is OFF and $V_{A}=0$ When $V_{\text {\in }} \gt 6,$ Diode conducts and $V_{R}=V_{\text {\in }}$

Q39GATE 2006MCQ

A regulated power supply, shown in figure below, has an unregulated input (UR) of 15 Volts and generates a regulated output $V_{out}$ . Use the component values shown in the figure. In the figure above, the ground has been shown by the symbol $\bigtriangledown$ If the unregulated voltage increases by 20%, the power dissipation across the transistor Q1

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📖 Explanation

New unregulated voltage $=18 \mathrm{V}$

\begin{aligned} \therefore \quad V_{CE}&=18-9=9V\\ I_{C}&=1 \mathrm{A} \\ \therefore \quad P=9 \times 1&=9 \text{Watts}\\ \therefore \% \text{increase }&=\frac{9-6}{6} \times 100=50 \% \end{aligned}

Q40GATE 2005MCQ

The Zener diode in the regulator circuit shown in the figure has a Zener voltage of 5.8 volts and a zener knee current of 0.5 mA. The maximum load current drawn from this current ensuring proper functioning over the input voltage range between 20 and 30 volts, is

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📖 Explanation

Maximum load current will be drawn, when

\begin{aligned} V_{i} &=V_{\min }=20 \mathrm{V} \\ &=\frac{20-5.8}{1 \mathrm{k} \Omega}=I_{L}+I_{Z} \\ 14.2 \mathrm{mA} &=I_{L}+I_{Z} \\ I_{L} &=14.2 \mathrm{mA}-0.5 \mathrm{mA}=13.7 \mathrm{mA} \end{aligned}