Question 1 |
The current gain (I_{out}/I_{in}) in the circuit with an ideal current amplifier given below
is
\frac{C_f}{C_c} | |
\frac{-C_f}{C_c} | |
\frac{C_c}{C_f} | |
\frac{-C_c}{C_f} |
Question 1 Explanation:
Redraw the circuit:
From circuit,
\begin{aligned} V_o&=V_c \\ &=\frac{1}{C_f}\int I_{in}dt \\ I_{out}&=C_c\frac{dV_o}{dt} \\ &= C_c\frac{d}{dt}\left ( \frac{1}{C_f} \int I_{in}dt\right )\\ &=\frac{C_c}{C_f}I_{in}\\ \Rightarrow \frac{I_{out}}{I_{in}}&=\frac{C_c}{C_f} \end{aligned}
From circuit,
\begin{aligned} V_o&=V_c \\ &=\frac{1}{C_f}\int I_{in}dt \\ I_{out}&=C_c\frac{dV_o}{dt} \\ &= C_c\frac{d}{dt}\left ( \frac{1}{C_f} \int I_{in}dt\right )\\ &=\frac{C_c}{C_f}I_{in}\\ \Rightarrow \frac{I_{out}}{I_{in}}&=\frac{C_c}{C_f} \end{aligned}
Question 2 |
The output impedance of a non-ideal operational amplifier is denoted by Z_{out} . The
variation in the magnitude of Z_{out} with increasing frequency, f , in the circuit shown
below, is best represented by
A | |
B | |
C | |
D |
Question 2 Explanation:
Bode plot of negative feedback amplifier:
Given amplifier is a voltage series feedback amplifier.
Therefore, Output impedance is given by
Z_{out}=\frac{Z_o}{1+A\beta } =\frac{Z_o}{1+A } \;\;\;(\beta=1 \text{ for buffer})
From Bode plot ; at low frequency, the open loop gain (A) is constant.
when \omega \uparrow, A \downarrow,Z_{out}\uparrow
At A=0, Z_{out}=Z_o\rightarrow constant
Therefore, z_{out} with frequency represented by
Given amplifier is a voltage series feedback amplifier.
Therefore, Output impedance is given by
Z_{out}=\frac{Z_o}{1+A\beta } =\frac{Z_o}{1+A } \;\;\;(\beta=1 \text{ for buffer})
From Bode plot ; at low frequency, the open loop gain (A) is constant.
when \omega \uparrow, A \downarrow,Z_{out}\uparrow
At A=0, Z_{out}=Z_o\rightarrow constant
Therefore, z_{out} with frequency represented by
Question 3 |
The steady state output (V_{out}), of the circuit shown below, will
saturate to +V_{DD} | |
saturate to -V_{EE} | |
become equal to 0.1 V | |
become equal to -0.1 V |
Question 3 Explanation:
Redraw the circuit:
From circuit,
\begin{aligned} V_{out} &=-\frac{1}{C_1}\int I\cdot dt \\ &= -\frac{1}{R_1C_1}\int 0 \cdot 1dt \\ \\ &=-\frac{0.1}{R_1C_1}\int dt \\ \\ &= -\frac{0.1}{R_1C_1}t \end{aligned}
Hence, V_{out}=-V_{EE}
From circuit,
\begin{aligned} V_{out} &=-\frac{1}{C_1}\int I\cdot dt \\ &= -\frac{1}{R_1C_1}\int 0 \cdot 1dt \\ \\ &=-\frac{0.1}{R_1C_1}\int dt \\ \\ &= -\frac{0.1}{R_1C_1}t \end{aligned}
Hence, V_{out}=-V_{EE}
Question 4 |
A \text{CMOS} Schmitt-trigger inverter has a low output level of \text{0 V} and a high output level of \text{5 V}. It has input thresholds of \text{1.6 V} and \text{2.4 V}. The input capacitance and output resistance of the Schmitt-trigger are negligible. The frequency of the oscillator shown is ____________\text{Hz}.(Round off to 2 decimal places.)
1245.23 | |
3157.56 | |
258.36 | |
965.24 |
Question 4 Explanation:
\begin{aligned} v_{c}(t)&=v_{c \text { final }}+\left[v_{\text {initial }}-v_{c^{\prime}} \text { final }\right] e^{- t / R C} \\ \text { Charging, } \qquad \qquad v_{c}(t)&=5+(1.6-5) e^{-t / R C}\\ &=5-3.4 e^{-t / R C}\\ t=t_{1}, \qquad \qquad v_{c}(t) &= 2.4 \mathrm{~V} \\ 2.4 &=5-3.4 e^{-t / \mathrm{RC}} \\ 3.4 e^{-11 / \mathrm{RC}} &= 2.6 \\ t_{1} &= \ln \left(\frac{3.4}{2.6}\right) R C=0.268 \times R C \end{aligned}
Discharging,
\begin{aligned} v_{c}(t) &=0+(2.4-0) e^{-t / \mathrm{RC}} \\ &=2.4 e^{-t / \mathrm{RC}} \\ t=t_{2}, \qquad v_{c}\left(t_{a}\right) &=1.6 \mathrm{~V} \\ 1.6 &=2.4 e^{-12 / \mathrm{AC}} \\ t_{2} &=\ln \left(\frac{2.4}{1.6}\right) R C=0.405 \times R C \\ T &=t_{1}+t_{2}=(0.268+0.405) \mathrm{RC} \\ T &=0.673 \mathrm{RC} \\ f &=\frac{1}{0.673 R C}=\frac{1}{0.673 \times 10^{4} \times 47 \times 10^{-9}} \\ f &=3157.46 \mathrm{~Hz} \end{aligned}
Question 5 |
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 ________.
0.01 | |
0.08 | |
0.04 | |
0.06 |
Question 5 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}
\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 6 |
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:
-2.44 | |
-1.22 | |
1.22 | |
2.44 |
Question 6 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
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 7 |
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
100 mV | |
400 mV | |
500 mV | |
600 mV |
Question 7 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 8 |
The op-amp shown in the figure is ideal. The input impedance \frac{v_{in}}{i_{in}} is given by
Z\frac{R_{1}}{R_{2}} | |
-Z\frac{R_{1}}{R_{2}} | |
Z | |
Z\frac{R_{1}}{R_{1}+R_{2}} |
Question 8 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}
\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 9 |
For the circuit shown below, assume that the OPAMP is ideal.
Which one of the following is TRUE?
Which one of the following is TRUE?
v_o=v_s | |
v_o=1.5v_s | |
v_o=2.5v_s | |
v_o=5v_s |
Question 9 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 10 |
The approximate transfer characteristic for the circuit shown below with an ideal operational
amplifier and diode will be
A | |
B | |
C | |
D |
Question 10 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
There are 10 questions to complete.