Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
Consider the circuit with an ideal OPAMP shown in the figure.
Assuming \left | V_{\text{IN}} \right |\ll \left | V_{\text{CC}} \right | and \left | V_{\text{REF}} \right |\ll \left | V_{\text{CC}} \right | , the condition at which V_{\text{OUT}} equals to zero is
Assuming \left | V_{\text{IN}} \right |\ll \left | V_{\text{CC}} \right | and \left | V_{\text{REF}} \right |\ll \left | V_{\text{CC}} \right | , the condition at which V_{\text{OUT}} equals to zero is
V_{\text{IN}}\:=\:V_{\text{REF}} | |
V_{\text{IN}}\:=\:0.5\:V_{\text{REF}} | |
V_{\text{IN}}\:=\:2\:V_{\text{REF}} | |
V_{\text{IN}}\:=\:2\:+\:V_{\text{REF}} |
Question 4 Explanation:
For ideal op-amp, V^{\prime}=V^{+}=0
KCL at node \mathrm{V}^{-}:
\begin{aligned} \frac{V_{\text {IN}}-0}{R}+\frac{\left(-V_{\text {REF }}-0\right)}{R}+\frac{V_{\text {OUT }}-0}{R_{F}} &=0 \\ \frac{V_{\text {OUT }}}{R_{F}} &=\frac{1}{R}\left(V_{\text {REF }}-V_{\text {IN }}\right) \\ V_{\text {OUT }} &=\frac{R_{F}}{R}\left(V_{\text {REF }}-V_{\text {IN }}\right)\\ \text { We want, }\qquad V_{\text {OUT }}&=0 \\ \Rightarrow\qquad V_{\text {REF }}-V_{\text {IN }}&=0 \\ \Rightarrow\qquad V_{\text {IN }}&=V_{\text {REF }} \end{aligned}
KCL at node \mathrm{V}^{-}:
\begin{aligned} \frac{V_{\text {IN}}-0}{R}+\frac{\left(-V_{\text {REF }}-0\right)}{R}+\frac{V_{\text {OUT }}-0}{R_{F}} &=0 \\ \frac{V_{\text {OUT }}}{R_{F}} &=\frac{1}{R}\left(V_{\text {REF }}-V_{\text {IN }}\right) \\ V_{\text {OUT }} &=\frac{R_{F}}{R}\left(V_{\text {REF }}-V_{\text {IN }}\right)\\ \text { We want, }\qquad V_{\text {OUT }}&=0 \\ \Rightarrow\qquad V_{\text {REF }}-V_{\text {IN }}&=0 \\ \Rightarrow\qquad V_{\text {IN }}&=V_{\text {REF }} \end{aligned}
Question 5 |
For the circuit with an ideal OPAMP shown in the figure. V_{\text{REF}} is fixed.
If V_{\text{OUT}}=1 volt for V_{\text{IN}}-0.1 volt and V_{\text{OUT}}=6 volt for V_{\text{IN}}=1 volt, where V_{\text{OUT}} is measured across R_{L} connected at the output of this OPAMP, the value of R_{F}/R_{\text{IN}} is
If V_{\text{OUT}}=1 volt for V_{\text{IN}}-0.1 volt and V_{\text{OUT}}=6 volt for V_{\text{IN}}=1 volt, where V_{\text{OUT}} is measured across R_{L} connected at the output of this OPAMP, the value of R_{F}/R_{\text{IN}} is
3.28 | |
2.86 | |
3.82 | |
5.55 |
Question 5 Explanation:
MARKS TO ALL AS PER IIT ANSWER KEY
\begin{aligned} V &=V^{+} \\ \frac{V_{\text {out }} R_{\text {in }}+V_{\text {in }} R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} \\ \frac{1 \times R_{\text {in }}+0.1 \times R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} &\ldots(i)\\ \frac{6 R_{\text {in }}+1 \times R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} &\ldots(ii) \end{aligned}
Equate equation (i) and (ii),
\begin{aligned} 1 \times R_{\text {in }}+0.1 \times R_{F} &=6 \times R_{\text {in }}+1 \times R_{F} \\ -5 R_{\text {in }} &=0.9 R_{F} \\ \therefore \quad \frac{R_{F}}{R_{\text {in }}} &=\frac{-5}{0.9}=-5.55 \end{aligned}
(According to the given data magnitude is taken)
\begin{aligned} V &=V^{+} \\ \frac{V_{\text {out }} R_{\text {in }}+V_{\text {in }} R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} \\ \frac{1 \times R_{\text {in }}+0.1 \times R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} &\ldots(i)\\ \frac{6 R_{\text {in }}+1 \times R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} &\ldots(ii) \end{aligned}
Equate equation (i) and (ii),
\begin{aligned} 1 \times R_{\text {in }}+0.1 \times R_{F} &=6 \times R_{\text {in }}+1 \times R_{F} \\ -5 R_{\text {in }} &=0.9 R_{F} \\ \therefore \quad \frac{R_{F}}{R_{\text {in }}} &=\frac{-5}{0.9}=-5.55 \end{aligned}
(According to the given data magnitude is taken)
Question 6 |
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 6 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 7 |
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 7 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 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 |
An op-amp based circuit is implemented as shown below.
In the above circuit, assume the op-amp to be ideal. The voltage (in volts, correct to one decimal place) at node A, connected to the negative input of the op-amp as indicated in the figure is _________.
In the above circuit, assume the op-amp to be ideal. The voltage (in volts, correct to one decimal place) at node A, connected to the negative input of the op-amp as indicated in the figure is _________.
0.4 | |
0.8 | |
0.3 | |
1 |
Question 9 Explanation:
Applying the concept of virtual ground, we get
\begin{aligned} V_{o}&=-\frac{R_{2}}{R_{1}} \cdot V_{i n}\\ [\therefore & \text{ non-inverting amplifiel}]\\ \therefore \quad V_{o}&=-\frac{31 \mathrm{k} \Omega}{1 \mathrm{k} \Omega} \times 1 \mathrm{V}\\ V_{0}&=-31 \mathrm{V} \lt -15 \mathrm{V} \end{aligned}
which is not possible
Hence, the output voltage of the op-amp is equal to -15 \mathrm{V}
Now applying KCL of node 'A', we get.
\begin{aligned} \frac{V_{A}-(-15)}{31 \mathrm{k} \Omega}+\frac{V_{A}-1}{1 \mathrm{k} \Omega} &=0 \\ \frac{V_{A}}{31 \mathrm{k} \Omega}+\frac{V_{A}}{1 \mathrm{k} \Omega} &=\frac{-15}{31 \mathrm{k} \Omega}+\frac{1}{1 \mathrm{k} \Omega} \\ V_{A}\left[\frac{1}{31}+\frac{1}{1}\right] &=-\frac{15}{31}+1 \\ V_{A} &=0.5 \mathrm{V} \end{aligned}
\begin{aligned} V_{o}&=-\frac{R_{2}}{R_{1}} \cdot V_{i n}\\ [\therefore & \text{ non-inverting amplifiel}]\\ \therefore \quad V_{o}&=-\frac{31 \mathrm{k} \Omega}{1 \mathrm{k} \Omega} \times 1 \mathrm{V}\\ V_{0}&=-31 \mathrm{V} \lt -15 \mathrm{V} \end{aligned}
which is not possible
Hence, the output voltage of the op-amp is equal to -15 \mathrm{V}
Now applying KCL of node 'A', we get.
\begin{aligned} \frac{V_{A}-(-15)}{31 \mathrm{k} \Omega}+\frac{V_{A}-1}{1 \mathrm{k} \Omega} &=0 \\ \frac{V_{A}}{31 \mathrm{k} \Omega}+\frac{V_{A}}{1 \mathrm{k} \Omega} &=\frac{-15}{31 \mathrm{k} \Omega}+\frac{1}{1 \mathrm{k} \Omega} \\ V_{A}\left[\frac{1}{31}+\frac{1}{1}\right] &=-\frac{15}{31}+1 \\ V_{A} &=0.5 \mathrm{V} \end{aligned}
Question 10 |
In the circuit shown below, the op-amp is ideal and Zener voltage of the diode is 2.5 volts.
At the input, unit step voltage is applied, i.e. v_{IN}(t)= u(t) volts. Also, at t= 0, the
voltage across each of the capacitors is zero.
The time t, in milliseconds, at which the output voltage v_{OUT} crosses -10 V is
The time t, in milliseconds, at which the output voltage v_{OUT} crosses -10 V is2.5 | |
5 | |
7.5 | |
10 |
Question 10 Explanation:
\text{For} \quad t \gt 0,
I=\frac{1 V}{1 \mathrm{k} \Omega}=1 \mathrm{mA}
Till t=2.5 \mathrm{msec}, both V_{1} and V_{2} will increase and after t=2.5 \mathrm{msec}, V_{2}=2.5 \mathrm{V} and V_{1} increases with time.
\begin{aligned} \text { when } v_{\text {out }}(t) &=-10 \mathrm{V} \\ & V_{1}=7.5 \mathrm{V}\\ \text{So,}\\ \frac{1}{1 \mu F} \int_{0}^{t}(1 \mathrm{m} \mathrm{A}) d t &=7.5 \mathrm{V} \\ 10^{3} t &=7.5 \\ t &=7.5 \mathrm{msec} \end{aligned}
I=\frac{1 V}{1 \mathrm{k} \Omega}=1 \mathrm{mA}
Till t=2.5 \mathrm{msec}, both V_{1} and V_{2} will increase and after t=2.5 \mathrm{msec}, V_{2}=2.5 \mathrm{V} and V_{1} increases with time.
\begin{aligned} \text { when } v_{\text {out }}(t) &=-10 \mathrm{V} \\ & V_{1}=7.5 \mathrm{V}\\ \text{So,}\\ \frac{1}{1 \mu F} \int_{0}^{t}(1 \mathrm{m} \mathrm{A}) d t &=7.5 \mathrm{V} \\ 10^{3} t &=7.5 \\ t &=7.5 \mathrm{msec} \end{aligned}
There are 10 questions to complete.