Q41GATE 2004MCQ

In a full-wave rectifier using two ideal diodes, $V_{dc}$ and $V_{m}$ are the dc and peak values of the voltage respectively across a resistive load. If PIV is the peak inverse voltage of the diode, then the appropriate relationships for this rectifier are

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Q42GATE 2004MCQ

In the voltage regulator shown in the figure, the load current can vary from 100 mA to 500 mA. Assuming that the Zener diode is ideal (i.e., the Zener knee current is negligibly small and Zener resistance is zero in the breakdown region), the value of R is

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📖 Explanation

\begin{aligned} I&=I_{z}+I_{L}=I_{2 \min }+I_{L m_{a x}} \\ &=0+500 \mathrm{mA} \\ R&=\frac{V_{\text {\in }}-V_{z}}{I}=\frac{12-5}{500 \mathrm{mA}}=14 \mathrm{a} \end{aligned}

Q43GATE 2003MCQ

The circuit shown in the figure is best described as a

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📖 Explanation

As voltage at non inverting terminal is 3V due to zener diode, voltage at inverting terminal will be 3V because of virtual ground. So, current in 20 k is $\quad \frac{3}{20 k \Omega}=\frac{3}{20} m A$ $\therefore \quad V_{0}=\frac{3}{20 \mathrm{k} \Omega} \times 60 \mathrm{k} \Omega=9 \mathrm{V}$

Q44GATE 2003MCQ

The output voltage of the regulated power supply shown in the figure is

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Q45GATE 2002MCQ

A zener diode regulator in figure is to be designed to meet the specifications: $I_{L}=10mA, \; V_{0}=10V$ and $V_{\in}$ varies from 30 V to 50 V. The zener diode has $V_{Z}$ = 10V and $I_{ZK}$ (knee current) = 1 mA. For satisfactory opertion

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📖 Explanation

\begin{aligned} \frac{V_{\text {\in }}-V_{o}}{R} &\geq I_{Z}+I_{L} \\ \text { when } \quad V_{\text {\in }}&=30 \mathrm{V} \\ \frac{30-10}{R} &\geq(10+1) \mathrm{mA} \\ \frac{20}{R} &\geq 11 \mathrm{mA} \\ \Rightarrow \quad R &\leq 1818 \Omega \end{aligned}

Q46GATE 2001MCQ

The transistor shunt regulator shown in figure has a regulated output voltage of 10 V, when the input varies from 20 V to 30 V. The relevant parameters for the Zener diode and the transistor are: $V_{Z}$ =9.5, $V_{SE}$ =0.3V, $\beta$ =99. Neglect the current through $R_{B}$ . Then the maximum power dissipated in the Zener diode ( $P_{Z}$ ) and the transistor ( $P_{T}$ ) are

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📖 Explanation

\begin{aligned} I_{1 \max }&=\frac{V_{\text {\in } \max }-V_{O}}{20}\\ I_{1} &=\frac{30-10}{20}=1 \mathrm{A} \quad \text { (i.e. when } I_{Z}=0) \\ I_{E} &=I_{C}+I_{Z} &\ldots(i)\\ I_{B} &=I_{Z}\left(\text { as no current flows \in } R_{B}\text { ) } \right.\\ \beta &=\frac{I_{C}}{I_{B}}=\frac{I_{C}}{I_{Z}} \\ \Rightarrow \quad I_{C} &=\beta I_{Z} \\ \text { From (i) }, I_{E} &=\beta I_{Z}+I_{Z}=(99+1) I_{Z} \\ I_{E} &=100 I_{Z} \\ I_{1} &=I_{E}=100 I_{Z} \\ I_{Z} &=\frac{I_{1}}{100}=\frac{1}{100}=0.01 \mathrm{A} \\ P_{Z} &=V_{Z} I_{Z}=9.5 \times 0.01=95 \mathrm{mW} \\ I_{C} &=99 I_{Z}=99 \times 0.01=0.99 \mathrm{A} \\ P_{C} &=V_{C} I_{C}=10 \times 0.99=9.9 \mathrm{W}\\ \end{aligned}