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GATE EE · Analog Electronics

Diodes And Their Applications Practice Questions

Generate GATE-level questions on Diode circuits. Focus on: 1. Diode characteristics and models (Ideal, Practical). 2. Rectifiers (Half-wave, Full-wave, Bridge) and Filter circuits. 3. Clipping and Clamping circuits, Zener diode as voltage regulator.

23 questions · 20 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2024NAT

In the given circuit, the diodes are ideal. The current $I$ through the diode $D_{1}$ in milliamperes is ______ (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

Assume $D_{2}$ OFF Tested: $D_{2}$ ON Assume $D_{1}$ OFF

\begin{aligned} I & =\frac{10}{1.66 \mathrm{~K}}=6.02 \mathrm{~mA} \\ I_{2} & =I \times \frac{1 \mathrm{k} \Omega}{3 \mathrm{k} \Omega}=6.02 \mathrm{~mA} \times \frac{1}{3}=2 \mathrm{~mA} \end{aligned}

Tested: $D_{1}$ ON

\begin{aligned} V_{X} & =2 \mathrm{~mA} \times 1 \mathrm{k} \Omega \\ & =2 \mathrm{~V} \end{aligned}

States: $D_{1}$ and $D_{2}$ ON

\begin{aligned} \frac{V_{X}-10}{1 \mathrm{k} \Omega}+\frac{V_{X}-3}{1 \mathrm{k} \Omega}+\frac{V_{X}}{1 \mathrm{k} \Omega} & =0 \\ 3 V_{X} & =13 \\ V_{X} & =\frac{13}{3}=4.33 \mathrm{~V} \\ I_{1}+I_{D 1} & =I_{2} \\ \frac{4.33-3}{1 \mathrm{k} \Omega}+I_{D 1} & =3 \mathrm{~mA} \\ I_{D 1} & =3 \mathrm{~mA}-1.33 \mathrm{~mA} \\ & =1.67 \mathrm{~mA} \end{aligned}

Explanation Figure 2

Explanation Figure 3

Explanation Figure 4

Explanation Figure 5

Q2GATE 2023NAT

The Zener diode in circuit has a breakdown voltage of $5 \mathrm{~V}$ . The current gain $\beta$ of the transistor in the active region in $99$ . Ignore baseemitter voltage drop $\mathrm{V}_{\mathrm{BE}}$ . The current through the $20 \Omega$ resistance in milliamperes is ____ (Round off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

Redraw the circuit : We know,

\begin{aligned} \mathrm{I}_{\mathrm{E}} & =(1+\beta) \mathrm{I}_{\mathrm{B}} \\ & =100 \mathrm{I}_{\mathrm{B}} \\ \Rightarrow \quad \mathrm{I}_{\mathrm{B}} & =\frac{\mathrm{I}_{\mathrm{E}}}{100} \end{aligned}

Apply KVL in loop,

\begin{aligned} & 25-\frac{\mathrm{I}_{E}}{100} \times 7000-I_{E} \times 10-I_{E} \times 20=0 \\ & \Rightarrow \quad \mathrm{I}_{\mathrm{E}}=0.25 \mathrm{~A} \text { or } 250 \mathrm{~mA} \end{aligned}

Explanation Figure 2

Q3GATE 2022MCQ

For the circuit shown below with ideal diodes, the output will be

🚀 Solve in Practice Mode

📖 Explanation

Case (i): For positive half cycle of $V_i$ Diode $D_1$ & $D_2$ are in forward biased. Redraw the circuit: So, $V_o=V_{\in}$ Case (ii): For negative half cycle of $V_i$ Both diodes $D_1$ & $D_2$ are in reverse biased. Redraw the circuit: Hence, $V_o=V_{\in} \text{ for } V_{\in} \gt 0$

Explanation Figure 2

Explanation Figure 3

Q4GATE 2021NAT

In the circuit shown, a $\text{5 V}$ Zener diode is used to regulate the voltage across load $R_{0}$ . The input is an unregulated DC voltage with a minimum value of $\text{6 V}$ and a maximum value of $\text{8 V}$ . The value of $R_{S}$ is $6\;\Omega$ . The Zener diode has a maximum rated power dissipation of $\text{2.5 W}$ . Assuming the Zener diode to be ideal, the minimum value of $R_{0}$ is _________________ $\Omega$ .

🚀 Solve in Practice Mode

📖 Explanation

To calculate $R_{0}$ min' we must find $I_{L}$ max

\begin{aligned} I_{s \min } &=I_{z} \min +I_{L \max } \\ \frac{V_{i \min }-V_{z}}{R_{s}} &=I_{z} \min +I_{L \max } \end{aligned}

For ideal zener diode, $I_{z \min }=0$

\begin{aligned} \frac{V_{i \min }-V_{z}}{R_{s}} &=I_{L \max } \\ \frac{6-5}{6} &=I_{L \max } \\ I_{L} \max &=\frac{1}{6} \mathrm{~A} \\ R_{0} \min &=\frac{V_{z}}{I_{L \max }}=\frac{5}{1 / 6}=30 \Omega \end{aligned}

Q5GATE 2014MCQ

Assuming the diodes to be ideal in the figure, for the output to be clipped, the input voltage $v_i$ must be outside the range

🚀 Solve in Practice Mode

📖 Explanation

CASE-I: $V_i \leq -4 V$ $D_2\rightarrow conducts$ $D_1\rightarrow OFF$ $So,\;\; V_o=-2V$ CASE-II: $-4V\leq V_i \leq -2V$ Both the diodes will be OFF $V_o=V'_1$ CASE-III: $V_i\geq -2V$ $D_1\rightarrow conducts$ $D_2\rightarrow OFF$ $V_o=-1V$

Q6GATE 2014NAT

The sinusoidal ac source in the figure has an rms value of $\frac{20}{\sqrt{2}}$ V. Considering all possible values of $R_L$ , the minimum value of $R_s \; \in \; \Omega$ to avoid burnout of the Zener diode is _____.

🚀 Solve in Practice Mode

📖 Explanation

$I_s \geq I_Z+I_L$ $\frac{1}{20} \geq \frac{20-5}{R_s} \;\;\;\;\; (Putting \;\;I_L=0)$ $\Rightarrow R_s \geq 300\Omega$ $Hence, R_{s(min)}=300\Omega$

Q7GATE 2013MCQ

In the circuit shown below, the knee current of the ideal Zener dioide 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

🚀 Solve in Practice Mode

📖 Explanation

$I_s=\frac{10-5}{100}=50\; mA$ $I_{load}=50\;mA-10\; mA =40\; mA$ $R_L=\frac{5 V}{40mA}=125\Omega$ $P_{min}=50 mA \times 5V=250\;mW$

Explanation Figure 2

Q8GATE 2013MCQ

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

🚀 Solve in Practice Mode

📖 Explanation

For positive half cycle, $V_{WX}=0$ For negative half cycle, Short circuit condition, $V_{WX}=0$

Explanation Figure 2

Explanation Figure 3

Q9GATE 2012MCQ

The i-v characteristics of the diode in the circuit given below are

i=\left\{\begin{matrix} \frac{v-0.7}{500}A, & v\geq 0.7V\\ 0A & v \lt 0.7 V \end{matrix}\right.

The current in the circuit is

🚀 Solve in Practice Mode

📖 Explanation

By applying kVL in loop, $10=10^3i+V$ $10=10^3\left ( \frac{V-0.7}{500} \right )+V$ $10= 2 V-1.4+V$ $3V=11.4$ $\Rightarrow V=3.8$ $\Rightarrow i=(10-3.8) \times 10^{-3}=6.2\; mA$

Explanation Figure 2

Q10GATE 2011MCQ

A clipper circuit is shown below. Assuming forward voltage drops of the diodes to be 0.7 V, the input-output transfer characteristics of the circuit is

🚀 Solve in Practice Mode

📖 Explanation

For positive voltage the waveform clips at +5.7 V. For negative voltage at -0.7 V, the zener diode conducts and clips out.

Explanation Figure 3

Q11GATE 2011MCQ

The transistor used in the circuit shown below has a $\beta$ of 30 and $I_{CBO}$ is negligible If the forward voltage drop of diode is 0.7 V, then the current through collector will be

🚀 Solve in Practice Mode

📖 Explanation

$I_{c(sat)} =\frac{0-(-12)-0.2}{R_c}$ $\;\; =\frac{11.8}{2.2k}=5.3636 \; mA$ This is the maximum current possible in any case.

Q12GATE 2010MCQ

Assuming that the diodes in the given circuit are ideal, the voltage $V_0$

🚀 Solve in Practice Mode

📖 Explanation

Diode $D_1$ is On and diode $D_2$ is OFF. So, $V_0=10 \times \frac{10}{10+10}=5\; V$

Q13GATE 2008MCQ

In the voltage doubler circuit shown in the figure, the switch 'S' is closed at t=0. Assuming diodes $D_1 \; and \; D_2$ to be ideal, load resistance to be infinite and initial capacitor voltages to be zero. The steady state voltage across capacitor $C_1 \; and \; C_2$ will be

🚀 Solve in Practice Mode

📖 Explanation

Forst capacitor charge through Diode ( $D_1$ ) upto $V_{max} (5 \; V)$ with shown polarities. $\therefore V_{c1}=5\; V$ Now diode $D_1$ will be reversed biased and $D_2$ will be forward biased and capacitor $C_2$ will charge in reverse direction through diode $D_2$ upto $2V_{max}$ $\therefore V_{c2}=-10\; V$

Explanation Figure 2

Q14GATE 2008MCQ

The equivalent circuits of a diode, during forward biased and reverse biased conditions, are shown in the figure. If such a diode is used in clipper circuit of figure given above, the output voltage $v_0$ of the circuit will be

🚀 Solve in Practice Mode

📖 Explanation

$V_P=\frac{10}{10+10}\times 10sin\omega t$ $\;\; =5sin\omega t$ Since maximum voltage across at the point P may be 5 V, hence voltage across the diode always will be less than or equal to zero. So it will be reversed always. $\therefore V_0=\frac{10}{10+10}\times 10sin\omega t=5sin\omega t$

Explanation Figure 3

Q15GATE 2007MCQ

The three-terminal linear voltage regulator is connected to a 10 $\Omega$ load resistor as shown in the figure. If $V_{\in}$ is 10 V, what is the power dissipated in the transistor ?

🚀 Solve in Practice Mode

📖 Explanation

$V_0=V_Z-V_{BE}$ $\;\; =6.6V-0.7V=5.9V$ $I_L=\frac{V_0}{R_L}=\frac{5.9}{10}=0.59A=I_C$ $V_{CE}-V_i-V_0$ $\;\; =10-5.9=4.1V$ $P_Q=V_{CE} \times I_C$ $\;\;=4.1 \times 0.59A=2.4 W$

Explanation Figure 2

Q16GATE 2006MCQ

What are the states of the three ideal diodes of the circuit shown in figure ?

🚀 Solve in Practice Mode

📖 Explanation

Analyzing the circuit, we can see that $D_1$ and $D_2$ are in forward bias and $D_3$ is in reverse bias. But no current flow through $D_2$ , bcause current gets shortest path through $D_1$ .

Q17GATE 2006MCQ

Assuming the diodes $D_{1}$ and $D_{2}$ of the circuit shown in figure to be ideal ones, the transfer characteristics of the circuit will be

🚀 Solve in Practice Mode

📖 Explanation

$V_i \lt 10V \;\;\;\;\; D_1 \;OFF, D_2 \; OFF$ $V_o=10V$ $V_i \gt 10V \;\;\;\;\; D_1 \;ON, D_2 \; OFF$ $V_o=V_{\in}$

Q18GATE 2005MCQ

Assume that $D_{1}$ and $D_{2}$ in figure are ideal diodes. The value of current is

🚀 Solve in Practice Mode

📖 Explanation

Current will pass through its simplest path (or) low resistance path. $D_1$ become forward biased and $D_2$ is reverse biased I = 0 mA

Q19GATE 2004MCQ

Assuming that the diodes are ideal in figure, the current in diode $D_{1}$ is

🚀 Solve in Practice Mode

📖 Explanation

From the circuit, if ' $D_1$ ' is 'ON' Kirchoff's law is not satisfied $\therefore \; \; \; D_1$ is OFF and $D_2$ is ON $\therefore \; \; \; I= 0 mA$

Q20GATE 2004MCQ

The current through the Zener diode in figure is

🚀 Solve in Practice Mode

📖 Explanation

$R_Z=\frac{\Delta V_Z}{\Delta I_Z}=\frac{3.5-3.3}{\Delta I_Z}$ $\therefore \Delta I_Z=\frac{0.2}{0.1k}=2 mA$

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