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}
Question 6 |
For the DC analysis of the Common-Emitter amplifier shown, neglect the base current and assume that the emitter and collector current are equal. Given that V_{T}=25mV,V_{BE}=0.7V, and the BJT output resistance r_{0} is practically infinite. Under these conditions, the midband
voltage gain magnitude. A_{c}=|V_{0}/V_{i}|V/V , is _________.
125 | |
126 | |
128 | |
132 |
Question 6 Explanation:
\begin{aligned} V_{T h}&=\frac{12 \times 47}{73+47}=4.7 \mathrm{V} \\ I_{C}&=\frac{V_{\mathrm{Th}}-V_{B E}}{R_{E}}=\frac{4}{2}=2 \mathrm{mA} \\ & &\because \; \; I_{C}&=I_{E}, I_{B}=0 \\ g_{m}&=\frac{I_{C}}{V_{T}}=\frac{2}{25}=80 \mathrm{m}^{2} \mathrm{S} \\ A_{V}&=-g_{m} R_{L}^{\prime}=-80(2 \| 8)=-128 \\ \left|A_{V}\right|&=128 \end{aligned}
Question 7 |
Consider the circuit shown in the figure. Assuming V_{BE1}=V_{EB2} = 0.7 volt, the value of the dc voltage V_{C2}= (in volt) is __________
0 | |
0.5 | |
1 | |
1.5 |
Question 7 Explanation:
\begin{aligned} Q_{1}: \quad V_{C B}&=0\\ \text{We know,}\\ V_{C B 1}+V_{B E 1}&=V_{C E 1} \\ \Rightarrow V_{C E 1}&=V_{B E 1}=0.7 \mathrm{V}\\ \text{KVL to loop 1}\\ -2.5+V_{C E 1}&+0.7+I_{B 2} \times 10 \mathrm{K}+1 \mathrm{V}=0 \\ \rightarrow I_{B 2}&=\frac{2.5-1.4-1}{10 K}=0.01 \mathrm{mA} \\ I_{C 2} &=\beta_{2} I_{B 2}=0.5 \mathrm{mA} \\ V_{C 2} &=I_{C 2} \times 1 K=0.5 \times 10^{-3} \times 1 \times 10^{3} \\ &=0.5 \mathrm{V} \end{aligned}
Question 8 |
Resistor R_1 in the circuit below has been adjusted so that I_{1} = 1 mA. The bipolar transistors Q1 and Q2 are perfectly matched and have very high current gain, so their base currents are negligible. The supply voltage V_{CC} is 6 V. The thermal voltage kT/q is 26 mV.
The value of R_{2} (in \Omega) for which I_{2}=100 \muA is ________
The value of R_{2} (in \Omega) for which I_{2}=100 \muA is ________
340.25 | |
598.67 | |
690.36 | |
740.87 |
Question 8 Explanation:
\begin{aligned} R_{2} &=\frac{V_{T}}{I_{2}} \ln \left(\frac{I_{1}}{I_{2}}\right)=\frac{26 \times 10^{-3}}{100 \times 10^{-6}} \ln \left(\frac{1 \times 10^{-3}}{100 \times 10^{-6}}\right) \\ &=598.67 \Omega \end{aligned}
Question 9 |
The Ebers-Moll model of a BJT is valid
only in active mode | |
only in active and saturation modes | |
only in active and cut-off modes | |
in active, saturation and cut-off modes |
Question 9 Explanation:
Ebers-Moll model is valid for all the region of operation.
Question 10 |
In the circuit shown in the figure, the BJT has a current gain (\beta) of 50. For an emitter-base voltage V_{EB}= 600 mV, the emitter-collector voltage V_{EC} (in Volts) is _______.
0 | |
1 | |
2 | |
3 |
Question 10 Explanation:
\begin{aligned} V_{E B} &=0.7 \mathrm{V} \\ I_{B} &=0.0383 \mathrm{mA} \\ I_{C} &=1.916 \mathrm{mA} \\ V_{E C} &=3-I_{C} R_{C} \\ &=3-(1.916 \times 0.5)=2.04 \mathrm{V} \end{aligned}
There are 10 questions to complete.