# BJT Analysis

 Question 1
In the voltage regulator shown below, $V_l$ is the unregulated at 15 V. Assume $V_{BE}$= 0.7 V and the base current is negligible for both the BJTs. If the regulated output $V_O$ is 9 V, the value of $R_2$ is ________ $\Omega$.
 A 1000 B 800 C 400 D 200
GATE EC 2020   Analog Circuits
Question 1 Explanation:
\begin{aligned}9\times \frac{R_{2}}{R_{2}+1K\Omega }&=4 \\ 9R_{2}&=4R_{2}+4K\Omega\\ 5R_{2}&=4K \\ R_{2}&=\frac{4000}{5}=800\Omega \end{aligned}
 Question 2
For the BJT in the amplifier shown below, $V_{BE} = 0.7 V, kT/q = 26 mV$. Assume the BJT output resistance ($r_o$) is very high and the base current is negligible. The capacitors are also assumed to be short circuited at signal frequencies. The input $v_i$ is direct coupled. The low frequency gain $v_o/v_i$ of the amplifier is
 A -89.42 B -128.21 C -178.85 D -256.42
GATE EC 2020   Analog Circuits
Question 2 Explanation:
\begin{aligned}I_{EQ}&=\frac{10-0.7}{20}=0.465mA\\ g_{m}&=\frac{I_{EQ}}{V_{T}}=\frac{0.465}{26}A/V \\ \frac{V_{out}}{V_{in}}&=-g_{m}(R_{e}\parallel R_{L})\\ &=\frac{0.465}{26}\times 5000=-89.423\end{aligned}

 Question 3
In the circuit shown, transistors $Q_{1}$ and $Q_{2}$ are biased at a collector current of 2.6mA.Assuming that transistor current gains are sufficiently large to assume collector current equal to emitter current and thermal voltage of 26 mV, the magnitude of voltage gain $V_{0}/V_{s}$ in the mid-band frequency range is _____________ (up to second decimal place).
 A 30 B 40 C 50 D 60
GATE EC 2017-SET-2   Analog Circuits
Question 3 Explanation:
In AC equivalent circuit, $0_{2}$ becomes diode connected-transistor because collector and base get shorted.

Computer AC equivalent circuit is as shown below:

It is a CE amplifier with unbypassed $R_{E}$
\begin{aligned} A_{v} &=\frac{-g_{m} R_{L}}{1+g_{m} R_{E}} \\ \text { where, } \quad g_{m} &=\frac{I_{C}}{V_{T}}=\frac{2.6}{26}=100 \mathrm{m}^{2} \\ R_{L}^{\prime} &=11 \Omega \\ R_{E} &=\frac{1}{g_{m}} \\ A_{v} &=\frac{-100 \times 1}{1+1}=-50 \\ \left|A_{v}\right| &=50 \end{aligned}
 Question 4
Consider the circuit shown in the figure. Assume base-to- emitter voltage $V_{BE}$=0.8 V and common base current gain $(\alpha)$ of the transistor is unity.

The value of the collector- to - emitter voltage $V_{CE}$ (in volt) is _______.
 A 4 B 6 C 8 D 10
GATE EC 2017-SET-2   Analog Circuits
Question 4 Explanation:
By taking the Thevenin's equivalent between Base and Ground nodes, the given circuit can be
reduced as follows:

\begin{aligned} V_{T h} &=\frac{16}{16+44} \times 18 \mathrm{V}=4.8 \mathrm{V} \\ I_{E} R_{E} &=V_{T h}-V_{B E}-I_{B} R_{T h} \\ I_{B} &=0 \mathrm{A} \qquad \; \; \; \because \alpha=1\\ \text { So, } \quad I_{E} R_{E}&=4.8-0.8=4 \mathrm{V}\\ &I_{E}=\frac{4}{2} m A=2 m A\\ I_{C} &=I_{E}=2 \mathrm{m} \mathrm{A} \; \; \; \because \alpha=1 \\ V_{C E} &=V_{C C}-I_{C} R_{C}-I_{E} R_{E} \\ &=18-(2 \times 4)-(2 \times 2)=6 \mathrm{V} \end{aligned}
 Question 5
In the figure shown, the npn transistor acts as a switch
For the input $V_{in}(t)$ as shown in the figure, the transistor switches between the cut-off and saturation regions of operation, when T is large. Assume collector-to-emitter voltage saturation $V_{CE(sat)}=0.2V$ and base-to-emitter voltage $V_{BE}=0.7V$. The minimum value of the common-base current gain ($\alpha$) of the transistor for the switching should be _________.
 A 0.5 B 0.9 C 0.13 D 0.18
GATE EC 2017-SET-1   Analog Circuits
Question 5 Explanation:

\begin{aligned} I_{B} &=\frac{2-0.7}{12}=0.10833 \mathrm{mA} \\ I_{\mathrm{Q(sat)}} &=\frac{5-0.2}{4.8}=1 \mathrm{mA} \\ I_{B} & \geq I_{B(\min )}=\frac{I_{C(\mathrm{sat})}}{\beta} \\ I_{B} & \geq \frac{1 \mathrm{mA}}{\beta} \\ \beta & \geq \frac{1}{0.10833} \text { and } \beta_{\min }=9.23 \\ \alpha_{\min } &=\frac{\beta_{\min }}{1+\beta_{\min }}=0.902 \end{aligned}

There are 5 questions to complete.