Question 1 |
Consider the circuit shown with an ideal long channel nMOSFET (enhancementmode, substrate is connected to the source). The transistor is appropriately biased in
the saturation region with V_{GG} and V_{DD} such that it acts as a linear amplifier. v_i is the
small-signal ac input voltage. v_A and v_B represent the small-signal voltages at the
nodes A and B, respectively. The value of \frac{v_A}{v_B} is ________ (rounded off to one
decimal place).
-2 | |
2 | |
-1 | |
1 |
Question 1 Explanation:
For ac analysis
\begin{aligned} V_A &=-i_d\cdot 4k \\ V_B&=i_d\cdot 2k \\ \frac{V_A}{V_B}&=\frac{-4}{2} \\ \frac{V_A}{V_B}&=-2 \end{aligned}
\begin{aligned} V_A &=-i_d\cdot 4k \\ V_B&=i_d\cdot 2k \\ \frac{V_A}{V_B}&=\frac{-4}{2} \\ \frac{V_A}{V_B}&=-2 \end{aligned}
Question 2 |
A circuit and the characteristics of the diode (D) in it are shown. The ratio of the
minimum to the maximum small signal voltage gain \frac{\partial V_{out}}{\partial V_{in}} is ________ (rounded
off to two decimal places)
0.25 | |
0.55 | |
0.75 | |
0.95 |
Question 2 Explanation:
Replacing the given circuit with small signal equivalent.
Case-I when diode is ON
As r_d(ON)=0, the 2k\Omega resistor in parallel to the diode becomes open circuit.
\therefore \; V_{out}=V_{IN}\times \frac{2}{4}=\frac{V_{in}}{2}
\therefore \; \frac{\partial V_{out}}{\partial V_{in}}|_{max}=\frac{1}{2}\;\;\;...(i)
Case-I: When diode is off:
r_d(off)=\infty \Rightarrow total \; R_{eq}=2+2+2=6k\Omega
\therefore \; V_{out}=\frac{V_{in} \times 4}{6}=\frac{2}{3}V_{in}\Rightarrow \frac{\partial V_{out}}{\partial V_{in}}|_{min}=\frac{2}{3}\;\;\;...(ii)
From (i) and (ii)
\frac{\left (\frac{\partial V_{out}}{\partial V_{in}}\right )_{min}}{\left (\frac{\partial V_{out}}{\partial V_{in}}\right )_{max}}=\frac{1/2}{2/3}=0.75
Case-I when diode is ON
As r_d(ON)=0, the 2k\Omega resistor in parallel to the diode becomes open circuit.
\therefore \; V_{out}=V_{IN}\times \frac{2}{4}=\frac{V_{in}}{2}
\therefore \; \frac{\partial V_{out}}{\partial V_{in}}|_{max}=\frac{1}{2}\;\;\;...(i)
Case-I: When diode is off:
r_d(off)=\infty \Rightarrow total \; R_{eq}=2+2+2=6k\Omega
\therefore \; V_{out}=\frac{V_{in} \times 4}{6}=\frac{2}{3}V_{in}\Rightarrow \frac{\partial V_{out}}{\partial V_{in}}|_{min}=\frac{2}{3}\;\;\;...(ii)
From (i) and (ii)
\frac{\left (\frac{\partial V_{out}}{\partial V_{in}}\right )_{min}}{\left (\frac{\partial V_{out}}{\partial V_{in}}\right )_{max}}=\frac{1/2}{2/3}=0.75
Question 3 |
A circuit with an ideal OPAMP is shown. The Bode plot for the magnitude (in dB)
of the gain transfer function (A_V(j\omega )=V_{out}(j\omega )/V_{in}(j\omega )) of the circuit is also
provided (here, \omega is the angular frequency in rad/s). The values of R and C are
R=3k\Omega ,C=1\mu F | |
R=1k\Omega ,C=3\mu F | |
R=4k\Omega ,C=1\mu F | |
R=3k\Omega ,C=2\mu F |
Question 3 Explanation:
\begin{aligned} \text{maximum gain}&=12dB\\ 20 \times \log A_{max}&=12\\ A_{max}&=4\\ 1+\frac{R_2}{R_1}&=4\\ R_2&=3R_1\\ R&=3 \times 1=3k\Omega \\ \log _{10}\omega _c&=3\\ omega _c&=1000rad/sec\\ omega _c&=\frac{1}{R_3C}\\ C&=\frac{1}{R_3 \times \omega _c}\\ C&=\frac{1}{1000 \times 1000}=1\mu F \end{aligned}
Question 4 |
For the following circuit with an ideal OPAMP, the difference between the maximum
and the minimum values of the capacitor voltage (V_c) is __________.
15 V | |
27 V | |
13 V | |
14 V |
Question 4 Explanation:
When,
\begin{aligned} V_0&=+15V\\ V^+&=\frac{R}{3R} \times 15=5V\\ V_{cmax}&=5V \end{aligned}
When,
\begin{aligned} V_0&=-12V\\ V^+&=\frac{2R}{3R} \times (-12)=5V\\ V_{cmin}&=-8V \end{aligned}
V_{cmax}=V_{cmin}=5-(-8)=13V
Question 5 |
Consider an ideal long channel nMOSFET (enhancement-mode) with gate length 10\mu m and width 100\mu m. The product of electron mobility ( \mu _n) and oxide
capacitance per unit area ( C_{OX}) is \mu _n C_{OX}=1mA/V^2. The threshold voltage of the
transistor is 1 V. For a gate-to-source voltage V_{GS}=[2-\sin (2t)]V and drain-tosource voltage V_{DS}=1V (substrate connected to the source), the maximum value
of the drain-to-source current is ________.
40 mA | |
20 mA | |
15 mA | |
5 mA |
Question 5 Explanation:
\mu _n Co_x=1mA/V^2; W=100\mu m;L=10\mu m
V_T= \perp V; V_{GS}=[2-\sin 2t]V;V_{DS}=1V
Let,
\begin{aligned} V_{GS} &=3V(max)\\ \Rightarrow V_{DS} &\lt V_{GS}-V_t\\ \because \; 1 &\lt (3-1) \end{aligned}
MOSFET in triode region
\begin{aligned} I_{Dmax}&=\mu _CO_x\left ( \frac{\omega }{L} \right )\left \{ \left ( V_{as \; max}-V_t \right )V_{DS}-\frac{1}{2}V_{DS}^2 \right \}\\ &=1 \times \left ( \frac{100}{10} \right )\left \{ (3-1) \times 1-\frac{1}{2} \times 1^2 \right \}mA\\ &=10(2-1/2)\\ &=15mA \end{aligned}
V_T= \perp V; V_{GS}=[2-\sin 2t]V;V_{DS}=1V
Let,
\begin{aligned} V_{GS} &=3V(max)\\ \Rightarrow V_{DS} &\lt V_{GS}-V_t\\ \because \; 1 &\lt (3-1) \end{aligned}
MOSFET in triode region
\begin{aligned} I_{Dmax}&=\mu _CO_x\left ( \frac{\omega }{L} \right )\left \{ \left ( V_{as \; max}-V_t \right )V_{DS}-\frac{1}{2}V_{DS}^2 \right \}\\ &=1 \times \left ( \frac{100}{10} \right )\left \{ (3-1) \times 1-\frac{1}{2} \times 1^2 \right \}mA\\ &=10(2-1/2)\\ &=15mA \end{aligned}
Question 6 |
An ideal OPAMP circuit with a sinusoidal input is shown in the figure. The 3 dB
frequency is the frequency at which the magnitude of the voltage gain decreases by
3 dB from the maximum value. Which of the options is/are correct?
The circuit is a low pass filter. | |
The circuit is a high pass filter. | |
The 3 dB frequency is 1000 rad/s. | |
The 3 dB frequency is (1000/3) rad/s. |
Question 6 Explanation:
\begin{aligned}
\frac{V_{out}}{V_{in}}&=\frac{-2000}{1000+\frac{1}{j\omega \times 10^{-6}}}\\
Gain&=\frac{V_{out}}{V_{in}}\frac{-2}{1+\frac{1}{\left (\frac{j\omega }{1000} \right ) }}
\end{aligned}
\omega \rightarrow \infty \Rightarrow gain=-2
\omega \rightarrow 0 \Rightarrow gain=0
\omega _c=1000 rad/sec = cutoff frequency
Hence, it is HPF.
\omega \rightarrow \infty \Rightarrow gain=-2
\omega \rightarrow 0 \Rightarrow gain=0
\omega _c=1000 rad/sec = cutoff frequency
Hence, it is HPF.
Question 7 |
The ideal long channel nMOSFET and pMOSFET devices shown in the circuits
have threshold voltages of 1 V and -1 V, respectively. The MOSFET substrates are
connected to their respective sources. Ignore leakage currents and assume that the
capacitors are initially discharged. For the applied voltages as shown, the steady
state voltages are ______
V_1=5 V, V_2=5 V | |
V_1=5 V, V_2=4 V | |
V_1=4 V, V_2=5 V | |
V_1=4V, V_2=-5 V |
Question 7 Explanation:
Question 8 |
Consider the CMOS circuit shown in the figure (substrates are connected to their
respective sources). The gate width (W) to gate length (L) ratios \frac{W}{L} of the
transistors are as shown. Both the transistors have the same gate oxide capacitance
per unit area. For the pMOSFET, the threshold voltage is -1 V and the mobility of
holes is 40\frac{cm^2}{V.s}. For the nMOSFET, the threshold voltage is 1 V and the mobility
of electrons is 300\frac{cm^2}{V.s}. The steady state output voltage V_o is ________.
equal to 0 V | |
more than 2 V | |
less than 2 V | |
equal to 2 V |
Question 8 Explanation:
\begin{aligned} \mu _PCO_x\left ( \frac{\omega }{L} \right )_1[4-V_0-1]^2&=\mu _nCO_x\left ( \frac{\omega }{L} \right )_2[V_0-0-1]^2\\ \Rightarrow \frac{300}{40}\times \frac{1}{5}(V_0-1)^2&=(3-V_0)^2\\ \Rightarrow \sqrt{1.5} (V_0-1)&=3-V_0\\ \Rightarrow V_0&=\frac{3+\sqrt{1.5}}{\sqrt{1.5}+1} \lt 2V \end{aligned}
Question 9 |
A circuit with an ideal \text{OPAMP}
is shown in the figure. A pulse V_{\text{IN}}
of 20\:ms
duration is applied to the input. The capacitors are initially uncharged.
The output voltage V_{\text{OUT}} of this circuit at t=0^{+} (in integer) is _______ V.
The output voltage V_{\text{OUT}} of this circuit at t=0^{+} (in integer) is _______ V.
15 | |
-15 | |
12 | |
-12 |
Question 9 Explanation:
At, t=0^{+}: Capacitor is short circuit
\begin{aligned} \therefore\quad V^{-}=V_{\text {in }}=5 \mathrm{~V} \\ V^{+}=0 \mathrm{~V} \end{aligned}
\begin{aligned} \text{If}\qquad V^- &>V^{+} \\ V_{\text {out }} &=-V_{\text {sat }}=-12 \mathrm{~V} \end{aligned}
\begin{aligned} \therefore\quad V^{-}=V_{\text {in }}=5 \mathrm{~V} \\ V^{+}=0 \mathrm{~V} \end{aligned}
\begin{aligned} \text{If}\qquad V^- &>V^{+} \\ V_{\text {out }} &=-V_{\text {sat }}=-12 \mathrm{~V} \end{aligned}
Question 10 |
An asymmetrical periodic pulse train v_{in}
of 10\:V
amplitude with on-time T_{\text{ON}}=1\:ms
and off-time T_{\text{OFF}}=1\:\mu s
is applied to the circuit shown in the figure. The diode D_{1} is ideal.
The difference between the maximum voltage and minimum voltage of the output waveform v_{o} (in integer) is ______________ V.
The difference between the maximum voltage and minimum voltage of the output waveform v_{o} (in integer) is ______________ V.
7 | |
5 | |
12 | |
10 |
Question 10 Explanation:
V_{\text{in}} = 10 V: Diode is ON
\therefore Capacitor charges upto 10 \mathrm{~V},
\begin{aligned} \therefore \qquad V_{C}&=10 \mathrm{~V} \\ V_{\text {in }}&=0 ; \text { Diode is OFF } \end{aligned}
Discharging time constant =R \times C
\begin{aligned} &=10 \mathrm{~m} \mathrm{sec} \\ \tau_{\text {discharging }} &>>\tau_{\mathrm{OFF}} \end{aligned}
Capacitor discharges negligibly
\begin{aligned} \therefore\qquad V_{C}&=10 \mathrm{~V}\\ \text{In steady state},\qquad V_{C} &=10 \mathrm{~V} \\ V_{\text {out }} &=V_{\text {in }}-V_{C}=V_{\text {in }}-10 \mathrm{~V}\\ \text{When }\qquad V_{\text {in }}&=10 \mathrm{~V}\\ \Rightarrow\qquad V_{\text {out }} &=10-10=0 \mathrm{~V} \\ V_{\text {in }} &=10 \mathrm{~V} \\ \mathrm{~V}_{\text {out }} &=0-10=-10 \mathrm{~V}\\ \Rightarrow\qquad V_{\text {out(max) }}-V_{\text {out(min) }}&=0-(-10)=10 \mathrm{~V} \end{aligned}
\therefore Capacitor charges upto 10 \mathrm{~V},
\begin{aligned} \therefore \qquad V_{C}&=10 \mathrm{~V} \\ V_{\text {in }}&=0 ; \text { Diode is OFF } \end{aligned}
Discharging time constant =R \times C
\begin{aligned} &=10 \mathrm{~m} \mathrm{sec} \\ \tau_{\text {discharging }} &>>\tau_{\mathrm{OFF}} \end{aligned}
Capacitor discharges negligibly
\begin{aligned} \therefore\qquad V_{C}&=10 \mathrm{~V}\\ \text{In steady state},\qquad V_{C} &=10 \mathrm{~V} \\ V_{\text {out }} &=V_{\text {in }}-V_{C}=V_{\text {in }}-10 \mathrm{~V}\\ \text{When }\qquad V_{\text {in }}&=10 \mathrm{~V}\\ \Rightarrow\qquad V_{\text {out }} &=10-10=0 \mathrm{~V} \\ V_{\text {in }} &=10 \mathrm{~V} \\ \mathrm{~V}_{\text {out }} &=0-10=-10 \mathrm{~V}\\ \Rightarrow\qquad V_{\text {out(max) }}-V_{\text {out(min) }}&=0-(-10)=10 \mathrm{~V} \end{aligned}
There are 10 questions to complete.