Q21GATE 2014NAT

For the amplifier shown in the figure, the BJT parameters are $V_{BE}=0.7 V,\beta =200$ , and thermal voltage $V_{T}=25mV.$ . The voltage gain $(v_{0}/v_{i})$ of the amplifier is ____.

🚀 Solve in Practice Mode

📖 Explanation

By applying KVL in input loop, we get

\begin{aligned} 3 &=8.25 \mathrm{k} I_{B}+0.7+1.10 \mathrm{k} I_{E} \\ 2.3 &=I_{E}\left[\frac{8.25 \mathrm{k}}{201}+1.01 \mathrm{k}\right] \\ 2.3 &=I_{E} \times 1.051 \mathrm{k} \\ I_{E} &=2.2 \mathrm{mA} \end{aligned}

AC equivalent model of amplifier is given as where $r_{e}$ is given by

\begin{array}{l} r_{e}=\frac{V_{T}}{I_{E}} \\ r_{e}=\frac{25 \mathrm{mV}}{2.2 \mathrm{mA}}=11.36 \Omega \\ V_{o}=-5 \beta i_{b} \quad\ldots(i)\\ V_{i}=2.27 \mathrm{k} i_{b}+0.01 \mathrm{k}\left(i_{b}+\beta i_{b}\right)\quad\ldots(ii) \end{array}

From equations (i) and (ii), we get $\frac{V_{0}}{V_{i}}=-\frac{5 \mathrm{k} \times 200}{4.28 \mathrm{k}}=-233.6$

Q22GATE 2014NAT

For the common collector amplifier shown in the figure, the BJT has high $\beta$ , negligible $V_{CE}(sat)$ , and $V_{BE}$ = 0.7 V. The maximum undistorted peak to peak output voltage $v_{0}$ (in Volts) is _____.

🚀 Solve in Practice Mode

📖 Explanation

DC analysis

\begin{aligned} \left(V_{B}\right)_{q}&=\frac{V_{c c} \times R_{2}}{R_{1}+R_{2}}=\frac{12 \times 10 \mathrm{k} \Omega}{15 \mathrm{k} \Omega}=8 \mathrm{V}\\ \left(V_{E}\right)_{Q} &=V_{B}-0.7=8-0.7=7.3 \mathrm{V} \\ \left(V_{C E}\right)_{Q} &=\left(V_{C}\right)_{Q}-\left(V_{E}\right)_{O} \\ &=12-7.3=4.7 \mathrm{V}\\ V_{o(p-p)}&=2\times4.7=9.4V \end{aligned}

Q23GATE 2014NAT

In the circuit shown, the PNP transistor has $|V_{BE}|=0.7V \; and \; \beta =50$ . Assume that $R_{B}=100k\Omega$ . For $V_{0}$ to be 5 V, the value of $R_{C}$ (in $k\Omega$ ) is ____.

🚀 Solve in Practice Mode

📖 Explanation

Applying KVL in input loop, we get

\begin{aligned} 10&=0.7+I_{B} R_{B} \\ I_{B}&=\frac{9.3}{100 \mathrm{k} \Omega}=0.93 \mathrm{mA}\\ \text{Hence }I_{C}&=\beta I_{B}=4.65 \mathrm{mA}\\ \text { where } I_{C}&=\frac{V_{0}-0}{R_{C}}=\frac{5}{R_{C}} \\ R_{C}&=\frac{5}{4.65} \\ \therefore R_{C}&=1.07 \mathrm{k} \Omega \end{aligned}

Q24GATE 2014MCQ

Consider the common-collector amplifier in the figure (bias circuitry ensures that the transistor operates in forward active region, but has been omitted forsimplicity). Let $I_{C}$ be the collector current, $V_{BE}$ be the base-emitter voltage and $V_{T}$ be the thermal voltage. Also, $g_{m}$ and $r_{0}$ are the small-signal transconductance and output resistance of the transistor, respectively. Which one of the following conditions ensures a nearly constant small signal voltage gain for a wide range of values of $R_{E}$ ?

🚀 Solve in Practice Mode

📖 Explanation

AC equivalent circuit for given common collector amplifier is

\begin{aligned} V_{\text {\in }}&=\beta r_{e} i_{b}+(1+\beta) i_{b} R_{E} \quad\cdots(i)\\ V_{o}&=(1+\beta) R_{E} i_{b}\quad\cdots(ii)\\ \text{Then}\\ \frac{V_{0}}{V_{\text {\in }}}&=\frac{(1+\beta) R_{E}}{\beta r_{e}+(1+\beta) R_{E}}\\ \frac{V_{o}}{V_{\text {\in }}}&=\frac{\beta R_{E}}{\beta r_{e}+\beta R_{E}}=\frac{R_{E}}{r_{e}+R_{E}} \end{aligned}

The condition for small signal voltage gain to be nearly constant is

\begin{aligned} R_{E} & \gt \gt r_{e} \\ R_{E} & \gt \gt \frac{V_{T}}{I_{C}} \\ I_{C} R_{E} &\gt \gt V_{T} \end{aligned}

Q25GATE 2013MCQ

In the circuit shown below, the silicon npn transistor Q has a very high value of $\beta$ . The required value of $R_{2}$ in k $\Omega$ to produce $I_{c}=1$ mA is

🚀 Solve in Practice Mode

📖 Explanation

Given that, $\beta$ is very large.

\begin{aligned} \text{So}, \quad I_{c}&=I_{E}=1 \mathrm{mA} \\ V_{E} &=I_{E} R_{E}=1 \times 10^{-3} \times 500=0.5 \mathrm{V} \\ V_{B E} &=0.7 \mathrm{V} \\ V_{R 2} &=V_{B E}+V_{E}=0.7+0.5=1.2 \mathrm{V}\\ \text{It is a self bias circuit}\\ \text{So, }\quad V_{R 2}&=V_{c c} \times \frac{R_{2}}{R_{1}+R_{2}} \\ 1.2 &=3 \times \frac{R_{2}}{60+R_{2}} \\ 72+1.2 R_{2} &=3 R_{2} \\ 1.8 R_{2} &=72 \\ R_{2} &=40 \mathrm{k} \Omega \end{aligned}

Q26GATE 2012MCQ

The voltage gain $A_{v}$ of the circuit shown below is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \mathrm{KVL}:-&13.7+12 \times 101 I_{B} \times 100 I_{B}+V_{B E}=0 \\ 1312 I_{B}&=B \\ \Rightarrow I_{B} &\equiv 0.01 \mathrm{mA} \\ I_{C}&= \beta I_{B}=1 \mathrm{mA} \\ g_{m}&=\frac{I_{C}}{V_{T}}=38 \mathrm{mV} \\ r_{\pi}&=\frac{\beta}{g_{m}}=\frac{100}{38 \mathrm{m}}=2.63 \mathrm{k} \Omega \end{aligned}

Apply Miller's theorem to 100 k resistor.

\begin{aligned} R_{m}&=\frac{100}{1-A_{v}} \\ R_{n}&=\frac{100}{1-\frac{1}{A_{v}}} \equiv 100 \mathrm{k}\\ \frac{V_{0}}{V_{1}} &=-g_{m} R_{L}^{\prime}=-38 \times (12 \| 100) \\ &=-407.14 \\ R_{m} &=\frac{100}{1+407.1}=0.245 \\ R_{i}^{\prime} &=r_{n}\left\|R_{m}=2.63\right\| 0.245=0.224 \\ \frac{V_{o}}{V_{i}} &=\frac{V_{o}}{V_{i}} \times \frac{R_{i}^{\prime}}{R_{s}+R_{i}} \\ &=-407.14 \times \frac{0.224}{10+0.224}=-8.92 \\ \left|A_{v}\right| & \cong 10 \end{aligned}

Q27GATE 2012MCQ

The current $i_{b}$ through the base of a silicon npn transistor is $1 + 0.1 cos(10000 \pi t)$ mA. At 300 K, the $r_{\pi}$ in the small signal model of the transistor is

🚀 Solve in Practice Mode

📖 Explanation

We know that

\begin{array}{l} r_{\pi}=(\beta+1) r_{e} \\ r_{\pi}=(\beta+1) \frac{V_{T}}{I_{e}} \\ r_{\pi}=(\beta+1) \frac{V_{T}}{(\beta+1) I_{b}} \\ r_{\pi}=\frac{V_{T}}{I_{b}} \end{array}

Where $I_{b}$ is d.c. current through base so $I_{b}=1 \mathrm{mA}$ $V_{T}=25 \mathrm{mV}$ at room temperature So, $r_{\pi}=\frac{25 \times 10^{-3}}{1 \times 10^{-3}}=25 \Omega$

Q28GATE 2011MCQ

In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency. $v_{i}$ is a small signal input. The gain magnitude $|\frac{v_{0}}{v_{i}}|$ at 10 M rad/s is

🚀 Solve in Practice Mode

📖 Explanation

Whenever we use a bypass capacitor in parallel $R_E$ then it always increases voltage gain. As compare to "CE with $R_E$ without capacitor". Hence only answer (A) is possible. and $\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10 \times 10^{-6} \times 10^{-9}}}$ =10 M rad/sec and gain is maximum ar resonance frequency.

Q29GATE 2011MCQ

For the BJT $Q_1$ in the circuit shown below, $\beta =\infty , V_{BEon} =0.7 V,V_{CEsat}=0.7 V$ . The switch is initially closed. At time t=0, the switch is opened. The time t at which $Q_1$ leaves the active region is

🚀 Solve in Practice Mode

📖 Explanation

In active region

\begin{aligned} -5-0.7-4.3 I_{E} &=-10 \\ I_{E} &=\frac{10-5.7}{4.3}=\frac{4.3}{4.3}=1 \mathrm{mA} \\ I_{C} &=I_{E}=I^{\prime}+0.5 \mathrm{mA}=1 \mathrm{mA} \\ \Rightarrow \quad I^{\prime} &=0.5 \mathrm{mA} \end{aligned}

In saturation region $\Rightarrow$

\begin{aligned} V_{C}-0.7-4.3 \times 1&=-10 \\ V_{C}&=-5 \mathrm{V}\\ q &=C V_{C}=-5 \times 10^{-6} \times 5 \mathrm{V} \\ &=-25 \times 10^{-6} \\ \text {and } q=\mathrm{it} & \\ I^{\prime}(0-t) &=-25 \times 10^{-6} \\ t &=\frac{25 \times 10^{-6}}{0.5 \times 10^{-3}}=50 \mathrm{m} \mathrm{sec} \end{aligned}

Q30GATE 2010MCQ

Consider the common emitter amplifier shown below with the following circuit parameters: $\beta =100$ , $g_{m}=0.3861 A/V$ , $r_0=\infty$ , $r_{a}=259\Omega$ , $R_{s}=1k\Omega$ , $R_{B}=93 k\Omega$ , $R_{C}=250 k\Omega$ , $R_{L}=1 k\Omega$ , $C_{1}=\infty$ and $C_{2}=4.7\mu F$ . The lower cut-off frequency due to $C_2$ is

🚀 Solve in Practice Mode

📖 Explanation

Lower cut-off frequency due to $C_{2}$

\begin{aligned} f_{L} &=\frac{1}{2 \pi\left(R_{C}+R_{L}\right) C_{2}} \\ &=\frac{1}{2 \pi(250+1000) \times 4.7 \times 10^{-6}} \\ &=27.1 \mathrm{Hz} \end{aligned}

Q31GATE 2010MCQ

In the silicon BJT circuit shown below, assume that the emitter area of transistor $Q_{1}$ is half that of transistor $Q_{2}$ . The value of current $I_{0}$ is approximately

🚀 Solve in Practice Mode

📖 Explanation

The given circuit is a current mirror circuit in which the output curren Is a mirror image of the input current 1f both the transistors are identical. To calculate $I_{i},$ $9.3 I_{i}+0.7=0-(-10)=10$ $\Rightarrow \quad I_{i}=1 \mathrm{mA}$ since the emitter area of transistor $Q_{1}$ is half that of transistor $Q_{2}$ $\mathrm{So} \quad I_{i}=I_{0} / 2$ Therefore, $I_{0}=2 \mathrm{mA}$

Q32GATE 2010MCQ

The amplifier circuit shown below uses a silicon transistor. The capacitors $C_{C}$ and $C_{E}$ can be assumed to be short at signal frequency and effect of output resistance $r_{0}$ can be ignored. If $C_{E}$ is disconnected from the circuit, which one of the following statements is true

🚀 Solve in Practice Mode

📖 Explanation

By disconnecting emitter bypass capacitor $C_{E}$ (i) Input impedance increases by a factor of $\left(1+\beta R_{E}\right)$ (ii) Magnitude of voltage gain $A_{V}$ decreases by the same factor.

Q33GATE 2009MCQ

A small signal source $v_{i}(t)= Acos 20t + Bsin10^{6}t$ is applied to a transistor amplifier as shown below. The transistor has $\beta$ = 150 and $h_{ie}$ = 3W. Which expression best approximate $v_{0}$ (t)

🚀 Solve in Practice Mode

📖 Explanation

The best approx answer for output voltage $V_{0}$ is

\begin{aligned} V_{o} &=A_{v} \cdot v_{i} \\ & \approx \frac{-h_{t e} R_{c}}{h_{i e}} \times v_{i} \\ & \approx-150(A \cos 20 t+B \sin 106 t) \end{aligned}

since the coupling capacitor is large $V_{0} \approx-150 B \sin 10^{6} t$

Q34GATE 2008MCQ

In the following transistor circuit $V_{BE}$ =0.7V, $r_{e} =25mV/I_{E}$ , and $\beta$ and all the capacitances are very large The mid-band voltage gain of the amplifier is approximately

🚀 Solve in Practice Mode

📖 Explanation

Mid-band voltage gain,

\begin{aligned} A_{V_{m}} &=-\frac{R_{c} \| R_{L}}{r_{e}} \\ A_{V_{m}} &=-\frac{R_{L}^{\prime}}{r_{\theta}}=-\frac{(3 \| 3)}{0.026}=-60 \\ &=-\frac{3 \times 10^{3} \| 3 \times 10^{3}}{25 \times 10^{-3} / 1 \times 10^{-3}}=-60 \end{aligned}

Q35GATE 2007MCQ

The DC current gain ( $\beta$ ) of a BJT is 50. Assuming that the emitter injection efficiency is 0.995, the base transport factor is

🚀 Solve in Practice Mode

📖 Explanation

$\text { Transport factor }=\frac{I_{\mathrm{PG}}}{I_{\mathrm{PE}}}=\beta^{*}$ Current in emitter is both due to holes and electrons Neglect current due to electrons

\begin{aligned} \alpha&=\frac{I_{P C_{1}}}{I_{E}}=\frac{I_{P C_{1}}}{I_{P E}} \times \frac{I_{P E}}{I_{E}}=\beta^{*} \times \eta \\ &\quad\left(\eta=\frac{I_{P E}}{I_{E}}=\text { emitter efficiency }\right) \\ \beta^{*}&=\frac{\alpha}{\eta}=\frac{50}{51 \times 0.995}=0.9853 \\ & \qquad\left(\alpha=\frac{\beta}{1+\beta}=\frac{50}{51}\right) \end{aligned}

Q36GATE 2007MCQ

For the BJT circuit shown, assume that the $\beta$ of the transistor is very large and $V_{BE}$ = 0.7V. The mode of operation of the BJT is

🚀 Solve in Practice Mode

📖 Explanation

Given $\beta$ is large, so $I_{B}=0$ and $I_{E}=I_{C}$ Assuming BJT is in active Applying KVL in Base emitter loop

\begin{aligned} 2-0.7 &=1 \mathrm{k} \Omega \times I_{E} \\ I_{E} &=1.3 \mathrm{mA} \\ \text { Now, } \quad I_{C_{\text{Sat}}} &=\frac{10-0.2}{1 \mathrm{k}+10 \mathrm{k}}=\frac{9.8}{11 \mathrm{k}} \approx 0.9 \mathrm{mA}\\ \text{As} \quad I_{C_{\text{active}}}&>I_{C_{\text{Sat}}} \end{aligned}

So, BJT is in saturation.

Q37GATE 2006MCQ

In the transistor amplifier circuit shown in the figure below, the transistor has the following parameters: $\beta _{DC}=60$ , $V_{BE}=0.7V$ , $h_{ie}\rightarrow \infty$ , $h_{fe}\rightarrow \infty$ The capacitance $C_{c}$ can be assumed to be infinite. The small-signal gain of the amplifier $\frac{V_{c}}{V_{s}}$ is

🚀 Solve in Practice Mode

📖 Explanation

Given circuit is a voltage-shunt feedback amplifier. So, the approximate voltage gain is $A_{v f}=-\frac{R_{f}}{R_{s}}=-\frac{53 \mathrm{k}}{5.3 \mathrm{k}}=-10$

Q38GATE 2005MCQ

The cascade amplifier is a multistage configuration of

🚀 Solve in Practice Mode

Q39GATE 2005MCQ

The circuit using a BJT with $\beta$ = 50 and VBE = 0.7V is shown in the figure. The base current $I_{B}$ and collector voltage by $V_{C}$ respectively

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} I_{E} &=I_{C}+I_{B} \\ &=\beta I_{B}+I_{B}=(\beta+1) I_{B}\\ \text{KVLin input loop gives,}\\V_{C C}-V_{B E} &=I_{B} R_{B}+I_{E} R_{E} \\ &=I_{B} R_{B}+(\beta+1) I_{B} R_{E} \\ \therefore \quad I_{B} &=\frac{V_{C C}-V_{B E}}{R_{B}+(\beta+1) R_{E}} \\ &=\frac{20-0.7}{430 \mathrm{K}+51 \times 1 \mathrm{K}} \\ I_{B} &=40 \mu \mathrm{A} \\ I_{C} &=\beta I_{B}=50 \times 40 \mathrm{\mu A} \\ &=2000 \mu \mathrm{A}=2 \mathrm{mA} \\ V_{C} &=V_{C C}-I_{C} R_{C}=20-2 \mathrm{mA} \times 2 \mathrm{K} \\ V_{C} &=16 \mathrm{V} \end{aligned}

Q40GATE 2005MCQ

In an ideal differential amplifier shown in the figure, a large value of $R_{E}$ .

🚀 Solve in Practice Mode

📖 Explanation

Only common-mode gain depends on $R_{E}$ and difierential mode gain is independent of $R_{E}$ .