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 ________.
0.01 | |
0.08 | |
0.04 | |
0.06 |
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}
\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 |
The cross-section of a metal-oxide-semiconductor structure is shown schematically.
Starting from an uncharged condition, a bias of +3V is applied to the gate contact with
respect to the body contact. The charge inside the silicon dioxide layer is then measured
to be +Q. The total charge contained within the dashed box shown, upon application
of bias, expressed as a multiple of Q (absolute value in Coulombs, rounded off to the
nearest integer) is __________ .
0 | |
1 | |
-1 | |
2 |
Question 2 Explanation:
Overall charge in side the box q + q - q - q = 0 charge
Question 3 |
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 3 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 4 |
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 4 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 5 |
The enhancement type MOSFET in the circuit below operates according to the square law. \mu_nC_{ox}=100\mu A/V^2, the threshold voltage (V_T) is 500 mV. Ignore channel length modulation. The output voltage V_{out} is
100 mV | |
500 mV | |
600 mV | |
2 V |
Question 5 Explanation:
As, V_{DS}=V_{GS}
MOSFET is in saturation,
\begin{aligned} I_D&=\frac{1}{2}\mu _nC_{Ox}\left ( \frac{W}{L} \right )(V_{GS}-V_T)^2 \\ 5 \times 10^{-6}&=\frac{1}{2} \times 100 \times 10^{-6} \times 10 (V_{GS}-0.5)^2\\ V_{GS}&=0.6 \\ V_0&=600mV \end{aligned}
Question 6 |
A current controlled current source (CCCS) has an input impedance of 10 \Omega and output impedance of 100 k\Omega. When this CCCS is used in a negative feedback closedloop with a loop gain of 9, the closed loop output impedance is
10\Omega | |
100\Omega | |
100k\Omega | |
1000k\Omega |
Question 6 Explanation:
"CCCS" (Current controlled current source amplifier)
Given, Z_0=100k\Omega
Loop gain, A\beta =9
Z_{0F}=Z_0[1+A\beta ] \;\;\;(\text{High impedance CS})
=100k\Omega [1+9]
=100k\Omega \times 10
=1000k\Omega
Given, Z_0=100k\Omega
Loop gain, A\beta =9
Z_{0F}=Z_0[1+A\beta ] \;\;\;(\text{High impedance CS})
=100k\Omega [1+9]
=100k\Omega \times 10
=1000k\Omega
Question 7 |
Given, V_{gs} is the gate-source voltage, V_{ds} is the drain source voltage, and V_{th} is the threshold voltage of an enhancement type NMOS transistor, the conditions for transistor to be biased in saturation are
V_{gs} \lt V_{th};V_{ds}\geq V_{gs}-V_{th} | |
V_{gs} \gt V_{th};V_{ds}\geq V_{gs}-V_{th} | |
V_{gs} \gt V_{th};V_{ds}\leq V_{gs}-V_{th} | |
V_{gs} \lt V_{th};V_{ds}\leq V_{gs}-V_{th} |
Question 7 Explanation:
For NMOS transistor to be in saturation the condition will be
V_{gs} \gt V_{th}
and V_{ds} \geq V_{gs}-V_{th}
V_{gs} \gt V_{th}
and V_{ds} \geq V_{gs}-V_{th}
Question 8 |
In the circuit shown in the figure, the bipolar junction transistor (BJT) has a current gain \beta=100. The base-emitter voltage drop is a constant, V_{BE}= 0.7 V. The value of the The venin equivalent resistance R_{Th} (in \Omega) as shown in the figure is ______ (up to 2 decimal places).
70.45 | |
85.25 | |
90.09 | |
105.65 |
Question 8 Explanation:
To calculate R_{Th} D.C. volatage should be short circuited.
R_{Th}=1k\Omega ||\frac{10k}{1+\beta }
\; \; =1k\Omega ||99.0099
R_{Th}=90.09\Omega
Question 9 |
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 9 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 10 |
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 10 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
There are 10 questions to complete.