Diodes Applications

 Question 1
A circuit and the characteristics of the diode (D) in it are shown. The ratio of the minimum to the maximum small signal voltage gain $\frac{\partial V_{out}}{\partial V_{in}}$ is ________ (rounded off to two decimal places)

 A 0.25 B 0.55 C 0.75 D 0.95
GATE EC 2022   Analog Circuits
Question 1 Explanation:
Replacing the given circuit with small signal equivalent.

Case-I when diode is ON
As $r_d(ON)=0,$ the $2k\Omega$ resistor in parallel to the diode becomes open circuit.
$\therefore \; V_{out}=V_{IN}\times \frac{2}{4}=\frac{V_{in}}{2}$
$\therefore \; \frac{\partial V_{out}}{\partial V_{in}}|_{max}=\frac{1}{2}\;\;\;...(i)$
Case-I: When diode is off:
$r_d(off)=\infty \Rightarrow total \; R_{eq}=2+2+2=6k\Omega$
$\therefore \; V_{out}=\frac{V_{in} \times 4}{6}=\frac{2}{3}V_{in}\Rightarrow \frac{\partial V_{out}}{\partial V_{in}}|_{min}=\frac{2}{3}\;\;\;...(ii)$
From (i) and (ii)
$\frac{\left (\frac{\partial V_{out}}{\partial V_{in}}\right )_{min}}{\left (\frac{\partial V_{out}}{\partial V_{in}}\right )_{max}}=\frac{1/2}{2/3}=0.75$
 Question 2
An asymmetrical periodic pulse train $v_{in}$ of $10\:V$ amplitude with on-time $T_{\text{ON}}=1\:ms$ and off-time $T_{\text{OFF}}=1\:\mu s$ is applied to the circuit shown in the figure. The diode $D_{1}$ is ideal.

The difference between the maximum voltage and minimum voltage of the output waveform $v_{o}$ (in integer) is ______________ V.
 A 7 B 5 C 12 D 10
GATE EC 2021   Analog Circuits
Question 2 Explanation:
$V_{\text{in}} = 10 V:$ Diode is ON

$\therefore$Capacitor charges upto $10 \mathrm{~V},$
\begin{aligned} \therefore \qquad V_{C}&=10 \mathrm{~V} \\ V_{\text {in }}&=0 ; \text { Diode is OFF } \end{aligned}

Discharging time constant $=R \times C$
\begin{aligned} &=10 \mathrm{~m} \mathrm{sec} \\ \tau_{\text {discharging }} &>>\tau_{\mathrm{OFF}} \end{aligned}
Capacitor discharges negligibly
\begin{aligned} \therefore\qquad V_{C}&=10 \mathrm{~V}\\ \text{In steady state},\qquad V_{C} &=10 \mathrm{~V} \\ V_{\text {out }} &=V_{\text {in }}-V_{C}=V_{\text {in }}-10 \mathrm{~V}\\ \text{When }\qquad V_{\text {in }}&=10 \mathrm{~V}\\ \Rightarrow\qquad V_{\text {out }} &=10-10=0 \mathrm{~V} \\ V_{\text {in }} &=10 \mathrm{~V} \\ \mathrm{~V}_{\text {out }} &=0-10=-10 \mathrm{~V}\\ \Rightarrow\qquad V_{\text {out(max) }}-V_{\text {out(min) }}&=0-(-10)=10 \mathrm{~V} \end{aligned}

 Question 3
In the circuit shown below, all the components are ideal and the input voltage is sinusoidal. The magnitude of the steady-state output $V_0$ (rounded off to two decimal places) is _________ V.
 A 325.2 B 650.4 C 125.45 D 752.36
GATE EC 2020   Analog Circuits
Question 3 Explanation:
Voltage Doubles, $V_{o}=2\: V_{m}=2\times 230\sqrt{2}\cong 650.4V$
 Question 4
In the circuit shown, $V_s$ is a 10 V square wave of period, T=4 ms with R= 500$\Omega$ and C = 10 $\mu F$. The capacitor is initially uncharged at t=0, and the diode is assumed to be ideal. The voltage across the capacitor ($V_c$) at 3 ms is equal to ____ volts (rounded off to one decimal place).
 A 2.8 B 3.3 C 4.6 D 5.4
GATE EC 2019   Analog Circuits
Question 4 Explanation:

\begin{aligned} \tau &=R C=500 \times 10 \times 10-6=5 \mathrm{m}_{8} \\ 0 \lt t \lt \frac{t}{2} \; :\; \; v_{\mathrm{cap}} &=v_{f}+\left(v_{i}-v_{f}\right) e^{-t / \tau} \\ &=10+(0-10) e^{-t / \mathrm{Rc}}\\ \text{At }t&=2 \mathrm{msec,} \\ v_{\mathrm{cap}}&=10-10 e^{\frac{-2 \times 10^{-3}}{5 \times 10^{-3}}}\\ &=3.296 \mathrm{V} \simeq 3.3 \mathrm{V} \end{aligned}
For $2 \mathrm{ms} \lt t \lt 4 \mathrm{ms},$ diode is OFF and Capacitor
has no path to discharge. Hence, at t=3 ms $V_{\text {cap }}=3.3 \mathrm{V}$
 Question 5
In the circuit shown, the breakdown voltage and the maximum current of the Zener diode are 20 V and 60 mA, respectively. The values of $R_1 \; and \; R_L$ are $200\Omega \; and \; 1k\Omega$, respectively. What is the range of $V_i$ that will maintain the Zener diode in the 'on' state?
 A 22 V to 34 V B 24 V to 36 V C 18 V to 24 V D 20 V to 28 V
GATE EC 2019   Analog Circuits
Question 5 Explanation:

\begin{aligned} V_{z} &=20 \mathrm{V} \\ I_{z \max } &=60 \mathrm{mA} \end{aligned}
Set zener diode be OFF
$V_{o}=\frac{V_{i} \times 1}{0.2+1}=\frac{V_{i}}{1.2}$
Zener diode can become ON i.e. it goes into
breakdown, when
\begin{aligned} \frac{V_{i}}{1.2} & \gt 20 \mathrm{V} \\ V_{i} & \gt 24 \mathrm{V} \end{aligned}
When Zener diode is in breakdown region,
$\begin{array}{l} I_{1}=\frac{V_{i}-20}{0.2 \mathrm{k} \Omega}=\frac{V_{i}-20}{0.2} \mathrm{mA}\\ I_{L}=\frac{V_{0}}{R_{L}}=\frac{20}{1 \mathrm{k} \Omega}=20 \mathrm{mA} \\ I_{z}=I_{1}-I_{L}=\frac{V_{i}-20}{0.2}-20 \end{array}$
For safe operation, $I_{z} \leq I_{z \max }$
$\frac{V_{i}-20}{0.2}-20 \leq 60$
$\Rightarrow V_{i} \leq 36 \mathrm{V}$
Hence, $24 \lt V_{i} \lt 36 V$
 Question 6
In thecircuit shown, $V_s$ is a square wave of period T with maximum and minimum values of 8 V and -10 V, respectively. Assume that the diode is ideal and $R_1=R_2=50\Omega$. The average value of $V_L$ is____ volts (rounded off to 1 decimal place).
 A -3 B -3.8 C -2.1 D -2.8
GATE EC 2019   Analog Circuits
Question 6 Explanation:
When $V_{s}=8 \mathrm{V} \Rightarrow$diode is in reverse bias

$V_{L}=\frac{8 \times 50}{50+50}=4 \mathrm{V}$
If $V_{s}=-10 \mathrm{V},$ diode is in forward bias

Average value of $V_{L}=\frac{\text { Area }}{\text { Time period }}$
$=\frac{4 \times 0.5 T+(-10) \times 0.5 T}{T}=-3 V$
 Question 7
A DC current of 26$\mu A$ flows through the circuit shown. The diode in the circuit is forward biased and it has an ideality factor of one. At the quiescent point, the diode has a junction capacitance of 0.5 nF . Its neutral region resistances can be neglected. Assume that the room temperature thermal equivalent voltage is 26 mV.

For $\omega =2 \times 10^{6} \; rad/s$, the amplitude of the small-signal component of diode current (in $\mu A$, correct to one decimal place) is _______.
 A 6.4 B 2.4 C 8.7 D 5
GATE EC 2018   Analog Circuits
Question 7 Explanation:
The small-signal equivalent model of the given circuit can be drawn as shown below.

\begin{aligned} \text{Given that, }\omega&=2 \times 10^{6} \mathrm{rad} / \mathrm{sec} \\ C_{j} &=0.5 \mathrm{nF} \\ I_{\mathrm{DC}} &=26 \mu \mathrm{A} \\ V_{T} &=26 \mathrm{mV} \\ \eta &=1\\ \mathrm{So} \quad r_{d} &=\frac{\eta V_{T}}{I_{\mathrm{DC}}}=\frac{26 \mathrm{mV}}{26 \mu \mathrm{A}}=1 \mathrm{k} \Omega \\ \frac{1}{\omega C_{j}} &=\frac{1}{2 \times 10^{6} \times 0.5 \times 10^{-9}} \Omega=1 \mathrm{k} \Omega \end{aligned}
So, total impedance of the circuit will be,
\begin{aligned} Z &=\left(r_{d} \| \frac{1}{j \omega C_{j}}\right)+100 \Omega \\\left(r_{d} \| \frac{1}{j \omega C_{j}}\right) &=\frac{(1000)(-j 1000)}{1000-j 1000} \Omega \\ &=\frac{-j(1+j)}{2} \mathrm{k} \Omega \\ &=\frac{1}{2}(1-j) \mathrm{k} \Omega=(500-j 500) \Omega \\ \therefore \quad Z &=600-j 500 \Omega \\|Z| &=100 \sqrt{36+25}=100 \sqrt{61} \Omega \\ I_{m} &=\frac{V_{m}}{|Z|} \\ &=\frac{5 \mathrm{mV}}{100 \sqrt{61} \Omega}=\frac{50}{\sqrt{61}} \mu \mathrm{A}=6.40 \mu \mathrm{A} \end{aligned}
 Question 8
The circuit shown in the figure is used to provide regulated voltage (5 V) across the 1k$\Omega$ resistor. Assume that the Zener diode has a constant reverse breakdown voltage for a current range, starting from a minimum required Zener current, $I_{Z_{min}} = 2 mA$ to its maximum allowable current. The input voltage $V_{I}$ may vary by 5% from its nominal value of 6 V. The resistance of the diode in the breakdown region is negligible.

The value of R and the minimum required power dissipation rating of the diode, respectively, are
 A 186 $\Omega$ and 10 mW B 100 $\Omega$ and 40 mW C 100 $\Omega$ and 10 mW D 186 $\Omega$ and 40 mW
GATE EC 2018   Analog Circuits
Question 8 Explanation:
\begin{aligned} V_{I} &=6 \mathrm{V} \pm 5 \%=6 \mathrm{V} \pm 0.3 \mathrm{V} \\ &=5.7 \mathrm{V} \text { to } 6.3 \mathrm{V}\\ I_{L} &=\frac{5 \mathrm{V}}{1 \mathrm{k} \Omega}=5 \mathrm{m} \mathrm{A} \\ I_{\mathrm{s}(\mathrm{min})} &=I_{L}+I_{Z_{\mathrm{Tmin}}}=5 \mathrm{m} \mathrm{A}+2 \mathrm{m} \mathrm{A} \\ &=7 \mathrm{m} \mathrm{A} \\ I_{s} &=\frac{V_{I}-V_{z}}{R} \\ I_{\mathrm{s}(\min )} &=\frac{V_{I(\min )}-V_{z}}{R}=7 \mathrm{m} \mathrm{A} \\ \text{So,}\quad R &=\frac{5.7-5}{7} \mathrm{k} \Omega=\frac{700}{7} \Omega=100 \Omega\\ \text{When},\quad R=100 \Omega \\ I_{s(\max )} &=\frac{6.3-5}{100} \mathrm{A}=13 \mathrm{mA} \\ I_{z(\max )} &=I_{s(\max )}-I_{L}=13 \mathrm{mA}-5 \mathrm{mA} \\ &=8 \mathrm{mA} \\ P_{z(\min )} &=V_{z} I_{z(\max )}=(5 \times 8) \mathrm{mW}=40 \mathrm{mW} \end{aligned}
 Question 9
In the figure, D1 is a real silicon pn junction diode with a drop of 0.7V under forward bias condition and D2 is a zener diode with breakdown voltage of -6.8 V. The input $V_{in}(t)$ is a periodic square wave of period T, whose one period is shown in the figure.

Assuming 10$\tau \lt \lt T$, where $\tau$ is the time constant of the circuit, the maximum and minimum values of the output waveform are respectively,
 A 7.5 V and -20.5V B 6.1 V and -21.9V C 7.5 V and -21.9V D 6.1 V and -22.6V
GATE EC 2017-SET-2   Analog Circuits
Question 9 Explanation:
Its a negative clamper with $V_{\text {ref }}=6.8 \mathrm{V}$
In steady state $V_{\text {cap }}=14-0.7-6.8=6.5 \mathrm{V}$
\begin{aligned} \Rightarrow \quad V_{\text {omax }} &=V_{\text {in } \max }-V_{\text {cap }}=14-6.5 =7.5\mathrm{V} \\ V_{\text {omin }} &=V_{\text {in in }}-V_{\text {cap }}=-14-6.5 =-20.5 \mathrm{V} \end{aligned}
 Question 10
The output $V_{0}$ of the diode circuit shown in the figure is connected to an averaging DC voltmeter. The reading on the DC voltmeter in Volts, neglecting the voltage drop across the diode, is ____________.
 A 2.84 B 3.18 C 4.55 D 1.62
GATE EC 2017-SET-2   Analog Circuits
Question 10 Explanation:
The given circuit is a halfwave rectifier.
Voltmeter reads the average value of $V_{o}$
Average value of $V_{o}=\frac{V_{m}}{\pi}$
$V_{m}=$ peak value of the applied sine wave $=10 \mathrm{V}$
So, Reading of meter $=\frac{10}{\pi}V=3.183\mathrm{V}$
There are 10 questions to complete.