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.
1000 | |
800 | |
400 | |
200 |
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
-89.42 | |
-128.21 | |
-178.85 | |
-256.42 |
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).
30 | |
40 | |
50 | |
60 |
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}
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 _______.
The value of the collector- to - emitter voltage V_{CE} (in volt) is _______.
4 | |
6 | |
8 | |
10 |
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}
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 _________.
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 _________.0.5 | |
0.9 | |
0.13 | |
0.18 |
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.