# Operational Amplifiers

 Question 1
The temperature of the coolant oil bath for a transformer is monitored using the circuit shown. It contains a thermistor with a temperature-dependent resistance, $R_{thermistor} = 2(1 + \alpha T) k\Omega$. Where T is the temperature in $^{\circ}C$. The temperature coefficient $\alpha$, is $-(4 \pm 0.25) \%/^{\circ}C$. Circuit parameters: $R_1 = 1 k\Omega , R_2 = 1.3 k\Omega , R_3 = 2.6 k\Omega$. The error in the output signal (in V. rounded off lo 2 decimal places) at 150$^{\circ}C$ is ________.
 A 0.01 B 0.08 C 0.04 D 0.06
GATE EE 2020   Analog Electronics
Question 1 Explanation:
As per GATE official answer key MTA (Marks to ALL)
\begin{aligned} \text{Given data,}\\ R_{thermistor}&=R_{th}=2(1+\alpha T)K\Omega \\ \alpha &=-(4+0.25)\%/^{\circ}C=-(0.04\pm 0.0025)^{\circ}C \\ \alpha _{max}&=-0.0424/^{\circ}C, \; \; \alpha _{min}=-0.375/^{\circ}C\\ \text{Temperature, } T&=150^{\circ}C \\ R_{1}&=1 K\Omega ,\; R_{2}=1.3 K\Omega , \; R_{3}=2.6 K\Omega\\ \text{Considering, }\alpha &=-0.04 \\ R_{th}&=2[1-0.04\times 150]=-10 K\Omega \\ V_{0}&=V_{1}\times \frac{R_{1}}{R_{1}+R_{th}}\left [ 1+\frac{R_{2}}{R_{3}} \right ]\\ &=3\times \frac{1k}{1k+10k}\left [ 1+\frac{1.3k}{2.6k} \right ] \\ &=-0.5 V \\ \text{Case-1:} \\ \text{Considering, }\alpha _{max}&=-0.0425/^{\circ}C \\ R_{thmax}&=2[1+(-0.0425)\times 150] =-10.75\, k\Omega \\ V_{0}&=V_{1}\times \frac{R_{1}}{R_{1}+R_{th}}\left [ 1+\frac{R_{2}}{R_{3}} \right ]\\ &=3\times \frac{1k}{1k-10.75k}\left [ 1+\frac{1.3k}{2.6k} \right ] \\ &=-0.46 V \\ \text{Case-2:} \\ \text{Considering, }\alpha _{min}&=-0.0375/^{\circ}C \\ R_{thmin}&=2[1+(-0.0375)\times 150] =-9.25\, k\Omega \\ V_{0} &=3\times \frac{1k}{1k-9.25k}\left [ 1+\frac{1.3k}{2.6k} \right ] \\ &=-0.54 V \\ \text{Output voltage, }\\ V_0&=0.5\pm0.04 \; \Rightarrow \; \text{Error}=0.04 \end{aligned}
 Question 2
A common-source amplifier with a drain resistance, $R_D = 4.7 k\Omega$, is powered using a 10 V power supply. Assuming that the transconductance, $g_m, \;is\; 520 \mu A/V$, the voltage gain of the amplifier is closest to:
 A -2.44 B -1.22 C 1.22 D 2.44
GATE EE 2020   Analog Electronics
Question 2 Explanation:
Given data:
$R_{D}=4.7 K\Omega ,\; g_{m}=520 \mu A/V$
Voltage gain of CS amplifier
$=-g_{m}R_{D}=-520\, \mu A/V\, \times \, 4.7\, k\Omega =-2.44$
 Question 3
In the circuit below, the operational amplifier is ideal. If $V_1$=10 mV and $V_2$=50 mV, the output voltage ($V_{out}$) is
 A 100 mV B 400 mV C 500 mV D 600 mV
GATE EE 2019   Analog Electronics
Question 3 Explanation:

$V_o=\frac{R_2}{R_1}(V_2-V_1)$
$\;\;=\frac{100k}{10k}(50\; mV-10\; mV)$
$\;\;=10(40\; mV)=400\;mV$
 Question 4
The op-amp shown in the figure is ideal. The input impedance $\frac{v_{in}}{i_{in}}$ is given by
 A $Z\frac{R_{1}}{R_{2}}$ B $-Z\frac{R_{1}}{R_{2}}$ C Z D $Z\frac{R_{1}}{R_{1}+R_{2}}$
GATE EE 2018   Analog Electronics
Question 4 Explanation:
According to virtual ground,
\begin{aligned} V_A&=V_B=V_{in} \\ \text{At node A,} \\ \frac{V_{in}-V_o}{Z}&=i_{in}\;\; ...(i) \\ V_{in}&=\left [ \frac{1}{R_2}+\frac{1}{R_1} \right ]=\frac{V_o}{R_1} \\ V_{in}\left [ \frac{R_1+R_2}{R_1R_2} \right ] \times R_1&=V_o \\ V_o&=V_{in} \times \left [ \frac{R_1+R_2}{R_2} \right ] \;\;...(ii)\\ \text{Equation (ii)}& \text{ in euation (i), }\\ \frac{V_{in}-V_o}{Z}&=i_{in} \\ \frac{V_{in}-V_{in}\left [ \frac{R_1+R_2}{R_2} \right ]}{Z}&=i_{in} \\ \frac{V_{in}}{i_{in}}\left [ 1-\frac{R_1+R_2}{R_2} \right ]&=Z \\ \frac{V_{in}}{i_{in}}\left [\frac{R_2-R_1+R_2}{R_2} \right ]&=Z \\ \frac{V_{in}}{i_{in}}&=-Z\cdot \frac{R_2}{R_1} \end{aligned}
 Question 5
For the circuit shown below, assume that the OPAMP is ideal.

Which one of the following is TRUE?
 A $v_o=v_s$ B $v_o=1.5v_s$ C $v_o=2.5v_s$ D $v_o=5v_s$
GATE EE 2017-SET-2   Analog Electronics
Question 5 Explanation:

$V_A=\frac{V_s}{2}$
$I_3=\frac{V_A}{R}=\frac{V_s}{2R}$
$V_B-V_A=I_3R$
$V_B=V_A+I_3R=\frac{V_s}{2}+\frac{V_s}{2}=V_s$
$I_2=\frac{V_B}{R}=\frac{V_s}{R}$
$I_1=I_2+I_3$
$\;\;=\frac{V_s}{R}+\frac{V_s}{2R}=\frac{V_s}{R}[1.5]$
$V_0-V_B=I_1R$
$\Rightarrow V_0=V_B+\frac{V_s}{R}[1.5]R$
$\;\;\;=V_s+1.5V_s=2.5V_s$
 Question 6
The approximate transfer characteristic for the circuit shown below with an ideal operational amplifier and diode will be

 A A B B C C D D
GATE EE 2017-SET-1   Analog Electronics
Question 6 Explanation:

$V_{in} \gt 0\;\; V_A=+V_{55},$ D on, $V_o=V_{in}$
$V_{in} \lt 0\;\; V_A=-V_{55},$ D off, $V_o=0$

 Question 7
For the circuit shown below, taking the opamp as ideal, the output voltage $V_{out}$ in terms of the input voltages $V_{1},V_{2} \; and \; V_{3}$ is
 A $1.8V_{1}+7.2V_{2} -V_{3}$ B $2V_{1}+8V_{2} -9V_{3}$ C $7.2V_{1}+1.8V_{2} -V_{3}$ D $8V_{1}+2V_{2} -9V_{3}$
GATE EE 2016-SET-2   Analog Electronics
Question 7 Explanation:

$V_A=\left ( \frac{4}{5}V_1+\frac{1}{5}V_2 \right )$
$V_{out}=-9V_3+10V_A$
$\;\;=-9V_3+8V_1+2V_2$
 Question 8
The circuit shown below is an example of a
 A low pass filter. B band pass filter C high pass filter. D notch filter.
GATE EE 2016-SET-2   Analog Electronics
Question 8 Explanation:

$\frac{V_{out}}{V_{in}}=-\left [\frac{\frac{ R_2\cdot \frac{1}{j\omega C}}{R_2+\frac{1}{j\omega C}}}{R_1} \right ]$
$\frac{V_{out}}{V_{in}}=-\left [ \frac{R_2}{R_1(R_2j\omega C+1)} \right ]$
So the system is a low pass filter.
 Question 9
The saturation voltage of the ideal op-amp shown below is $\pm$10 V. The output voltage $v_0$ of the following circuit in the steady-state is
 A square wave of period 0.55 ms. B triangular wave of period 0.55 ms C square wave of period 0.25 ms. D triangular wave of period 0.25 ms.
GATE EE 2015-SET-2   Analog Electronics
Question 9 Explanation:
The given circuit in a astable multivibrator, so output will be a periodic square wave and from the circuit $\beta =0.5$.

So, time period will be $2\tau \ln\left ( \frac{1+\beta }{1-\beta } \right )$
$T=2 \times RC \ln \left ( \frac{1+0.5}{1-0.5} \right )=0.55ms$
 Question 10
The filters F1 and F2 having characteristics as shown in Figures (a) and (b) are connected as shown in Figure (c).

The cut-off frequencies of F1 and F2 are $f_1 \; and \; f_2$ respectively. If $f_1 \lt f_2$ , the resultant circuit exhibits the characteristic of a
 A Band-pass filter B Band-stop filter C All pass filter D High-Q filter
GATE EE 2015-SET-2   Analog Electronics
Question 10 Explanation:
To check the type of system:
we apply a delta function at input $V_i=\delta (t)$
$V(f)=1\forall f$

So, voltage at A will be same as voltage at output $V_o$. $V_A$ will be equal to voltage due to $F_1 +$ voltage due to $F_2$. Since, $f_1 \lt f_2$, so $V_A/V_i$ will be

so, the system works as a band stop filter.
There are 10 questions to complete.