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 |
The components in the circuit given below are ideal. If R = 2 k\Omega and C = 1 \mu F, the
-3 dB cut-off frequency of the circuit in Hz is
14.92 | |
34.46 | |
59.68 | |
79.58 |
Question 2 Explanation:
Op-amp active filter (LPF) inverting type 3 dB cut-off frequency,
f_{c}=\frac{1}{2\pi RC}=\frac{1}{2\pi \times 2\times 10^{3}\times 10^{-6}}=\frac{500}{2\pi }=79.58Hz
f_{c}=\frac{1}{2\pi RC}=\frac{1}{2\pi \times 2\times 10^{3}\times 10^{-6}}=\frac{500}{2\pi }=79.58Hz
Question 3 |
Using the incremental low frequency small-signal model of the MOS device, the Norton
equivalent resistance of the following circuit is
r_{ds}+R+g_mr_{ds}R | |
\frac{r_{ds} +R}{1+g_m r_{ds}} | |
r_{ds}+\frac{1}{g_m}+R | |
r_{ds}+R |
Question 3 Explanation:
\begin{aligned}v_{\pi }&=-V_{x} \\ V_{x}&=\left ( I_{x}-g_{m}V_{x} \right )r_{ds}+I_{x}R\\ V_{x}\left ( 1+g_{m} r_{ds}\right )&=\left ( r_{ds}+R \right )I_{x} \\ R_{N}&=\frac{V_{x}}{I_{x}}=\frac{R+r_{ds}}{1+g_{m}r_{ds}}\end{aligned}
Question 4 |
An enhancement MOSFET of threshold voltage 3 V is being used in the sample and
hold circuit given below. Assume that the substrate of the MOS device is connected
to -10 V. If the input voltage V_I lies between \pm 10 V, the minimum and the maximum
values of V_G
required for proper sampling and holding respectively, are
3V and -3 V | |
10 V and -10 V | |
13 V and -7 V | |
10 V and -13 V |
Question 4 Explanation:
for holding MOSFET should be OFF.
\begin{aligned}V_{1\, min}&\rightarrow -10\, V \\ V_{G}-V_{1\, min}& \lt 3 \\ V_{G}& \lt 3-10V\Rightarrow -7V \\ \text{For Sampling, } V_{G}-V_{i\, max}& \gt 3\\ V_{G}& \gt 3+V_{i\, max}\\ V_{G}&\gt 13\end{aligned}
Question 5 |
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 5 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 6 |
In the circuit shown below, all the components are ideal. If V_i is +2 V, the current I_o sourced
by the op-amp is __________ mA.
4 | |
6 | |
2 | |
1 |
Question 6 Explanation:
\begin{aligned} V_o=(1+1) \times 2&=4V \\ \text{KCL at node } V_o & \\ \frac{2-4}{1k\Omega }I_o+\frac{0-4}{1k\Omega }&=0 \\ -2+I_o-4 &=0 \\ I_o&=6mA \end{aligned}
Question 7 |
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.
325.2 | |
650.4 | |
125.45 | |
752.36 |
Question 7 Explanation:
Voltage Doubles, V_{o}=2\: V_{m}=2\times 230\sqrt{2}\cong 650.4V
Question 8 |
The components in the circuit shown below are ideal. If the op-amp is in positive feedback
and the input voltage V_i is a sine wave of amplitude 1 V, the output voltage V_o is
a non-inverted sine wave of 2 V amplitude | |
an inverted sine wave of 1 V amplitude | |
a square wave of 5 V amplitude | |
a constant of either +5 or -5V |
Question 8 Explanation:
Given circuit is a Schmitt trigger of non-inverting type.
V_{o}=\pm 5\, V
V^{+}=\frac{V_{o}\times 1+V_{i}\times 1}{1+1}=\frac{V_{o}+V}{2}
let, V_{o}=-5\, V,\, \, \, \,V^{+}=\frac{-5+V_{i}}{2}
V_{o} can change from -5 V to +5 V if V^{+} \gt 0
i.e. \frac{-5+V_{i}}{2} \gt 0\Rightarrow V_{i} \gt 5\, V
similarly, V_{o} can change from -5 V to +5 V if V_{i} \lt -5\, V
But given input has peak value 1 V. Hence output cannot change from +5 V to -5 V or -5 V to +5 V.
Output remain constant at +5 V or -5 V.
Question 9 |
In the circuit shown, V_1=0 \; and \; V_2=V_{dd}. The other relevant parameters are mentioned in the figure. Ignoring the effect of channel length modulation and the body effect, the value of I_{out} is _____mA(rounded off to 1 decimal place).
4.2 | |
6 | |
8.3 | |
2.4 |
Question 9 Explanation:
\begin{aligned} \frac{I_{2}}{I_{1}} &=\frac{(W / L)_{2}}{(W / L)_{1}}=\frac{3}{2} \\ I_{2} &=\frac{3}{2} \times I_{1}=1.5 \mathrm{mA} \end{aligned}
M_{3} is OFF because V_{1}=0 \Rightarrow I_{3}=0
M_{4} is ON because V_{2}=V_{D D}
\begin{aligned} I_{5} &=I_{4}=I_{2}=1.5 \mathrm{mA} \\ \frac{I_{6}}{I_{5}} &=\frac{(W / L)_{6}}{(W / L)_{5}}=\frac{40}{10}=4 \\ I_{6} &=4 I_{5}=4 \times 1.5=6 \mathrm{mA} \\ \therefore \quad I_{\text {out }} &=I_{6}=6 \mathrm{mA} \end{aligned}
Question 10 |
In the circuit shown, the threshold voltages of the pMOS (|V_{tp}|) and nMOS (V_{tn}) transistors are both equal to 1 V. All the transistors have the same output resistance r_{ds} of 6 M\Omega. The other parameters are listed below:
\mu_n C_{ox}=60\mu A/V^2; \left ( \frac{W}{L} \right )_{nMOS}=5
\mu_p C_{ox}=30\mu A/V^2; \left ( \frac{W}{L} \right )_{pMOS}=10
\mu_n \; and \; \mu_p are the carrier mobilities, and C_{ox} is the oxide capacitance per unit area. Ignoring the effect of channel length modulation and body bias, the gain of the circuit is____ (rounded off to 1 decimal place).
\mu_n C_{ox}=60\mu A/V^2; \left ( \frac{W}{L} \right )_{nMOS}=5
\mu_p C_{ox}=30\mu A/V^2; \left ( \frac{W}{L} \right )_{pMOS}=10
\mu_n \; and \; \mu_p are the carrier mobilities, and C_{ox} is the oxide capacitance per unit area. Ignoring the effect of channel length modulation and body bias, the gain of the circuit is____ (rounded off to 1 decimal place).
-800 | |
-750.6 | |
-900 | |
-652.3 |
Question 10 Explanation:
M_{3} and M_{4} are identical PMOS transistor and they have equal current.
Hence their V_{SG} should be equal.
\begin{aligned} V_{S G 3} &=V_{S G 4}=\frac{V_{D D}}{2}=2 V \\ I_{S D} &=\frac{\mu_{p} C_{o x}}{2}\left(\frac{W}{L}\right)_{p}\left(V_{S G}-\left|V_{T}\right|\right)^{2} \\ &=\frac{30}{2} \times 10(2-1)^{2}=150 \mu \mathrm{A} \end{aligned}
now, by using current mirror property all transistor should have equal current.
\begin{aligned} I_{\mathrm{DSN}}&=I_{\mathrm{SDP}}=150 \mu \mathrm{A} \\ \text { For } M_{1} \quad g_{m 1}&=\sqrt{2 \mu_{n} C_{o x} \frac{W}{L} \times I_{D S}} \\ &=\sqrt{2 \mu_{n} C_{0 x}} \frac{W}{L} \times I_{D S}\\ & =\sqrt{2 \times 60 \times 5 \times 150}=300 \mathrm{m} \end{aligned}
M_{2}, M_{3} and M_{4} from active load for M_{1}. This active load in equivalent to resistance r_{d s 2} i.e. 6 \mathrm{M} \Omega
M_{1} is common source amplifier.
\begin{aligned} \therefore \quad \frac{V_{\text {out }}}{V_{\text {in }}} &=A_{v}=-g_{m 1} \times\left(r_{\text {ds2 }} \| r_{\text {ds1 }}\right) \\ &=-300 \mathrm{M} \mathrm{C} \times 3 \mathrm{M} \Omega=-900 \end{aligned}
Hence their V_{SG} should be equal.
\begin{aligned} V_{S G 3} &=V_{S G 4}=\frac{V_{D D}}{2}=2 V \\ I_{S D} &=\frac{\mu_{p} C_{o x}}{2}\left(\frac{W}{L}\right)_{p}\left(V_{S G}-\left|V_{T}\right|\right)^{2} \\ &=\frac{30}{2} \times 10(2-1)^{2}=150 \mu \mathrm{A} \end{aligned}
now, by using current mirror property all transistor should have equal current.
\begin{aligned} I_{\mathrm{DSN}}&=I_{\mathrm{SDP}}=150 \mu \mathrm{A} \\ \text { For } M_{1} \quad g_{m 1}&=\sqrt{2 \mu_{n} C_{o x} \frac{W}{L} \times I_{D S}} \\ &=\sqrt{2 \mu_{n} C_{0 x}} \frac{W}{L} \times I_{D S}\\ & =\sqrt{2 \times 60 \times 5 \times 150}=300 \mathrm{m} \end{aligned}
M_{2}, M_{3} and M_{4} from active load for M_{1}. This active load in equivalent to resistance r_{d s 2} i.e. 6 \mathrm{M} \Omega
M_{1} is common source amplifier.
\begin{aligned} \therefore \quad \frac{V_{\text {out }}}{V_{\text {in }}} &=A_{v}=-g_{m 1} \times\left(r_{\text {ds2 }} \| r_{\text {ds1 }}\right) \\ &=-300 \mathrm{M} \mathrm{C} \times 3 \mathrm{M} \Omega=-900 \end{aligned}
There are 10 questions to complete.