Question 1 |
The \frac{V_{\text {OUT }}}{V_{\text {IN }}} of the circuit shown below is
-\frac{R_{4}}{R_{3}} | |
\frac{R_{4}}{R_{3}} | |
1+\frac{R_{4}}{R_{3}} | |
1-\frac{R_{4}}{R_{3}} |
Question 1 Explanation:
Here, A_{1} is an inverting amplifier and A_{2} is a non-inverting amplifier.
\begin{aligned} V_{01}&=\frac{-R_{2}}{R_{1}} V_{i n} \\ V_{02}&=\left(1+\frac{R_{2}}{R_{1}}\right) V_{i n} \end{aligned}
Also, A_{3} is an inverting summing amplifier,
\begin{aligned} V_{\text {out }} & =\frac{-R_{4}}{R_{3}} V_{01}-\frac{R_{4}}{R_{3}} V_{02}=\frac{-R_{4}}{R_{3}}\left[\frac{R_{2}}{R_{1}} V_{\text {in }}+\left(1+\frac{R_{2}}{R_{1}}\right) V_{\text {in }}\right] \\ V_{\text {out }} & =\frac{-R_{4}}{R_{3}} V_{\text {in }} \\ \text { Gain, } \frac{V_{\text {out }}}{V_{\text {in }}} & =\frac{-R_{4}}{R_{3}} \end{aligned}
Question 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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}
There are 5 questions to complete.