1
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7V. The thermal voltage Vt = kT/q = 25mV. The small signal input vi = VP cos ($$\omega t$$) where VP = 100mV. GATE ECE 2011 Analog Circuits - Diodes Question 6 English The ac output voltage vac is
A
0.25 cos($$\omega t$$)mV
B
1 cos($$\omega t$$)mV
C
2 cos($$\omega t$$)mV
D
22 cos($$\omega t$$)mV
2
GATE ECE 2011
MCQ (Single Correct Answer)
+1
-0.3
In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency. Vi is a small signal input. The gain magnitude $$\left|\frac{{\mathrm V}_0}{{\mathrm V}_\mathrm i}\right|$$ at 10 Mrad/s is GATE ECE 2011 Analog Circuits - Bipolar Junction Transistor Question 54 English
A
maximum
B
minimum
C
unity
D
zero
3
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
For a BJT, the common-base current gain $$\alpha = \,\,0.98$$ and the colector base junction reverse bias saturation current Ico = 0.6 $$\mu {\rm A}$$. This BJT is connected in the common emitter mode and operated in the active region with a base drive current IB = 20$$\mu {\rm A}$$. The collector current Ic for this mode of operation is
A
0.98 mA
B
0.99 mA
C
1.0 mA
D
1.01 mA
4
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
For the BJT Q1 in the circuit shown below, $$\beta = \infty ,\,\,\,\,{V_{BEon\,}}\, = \,0.7V,\,\,\,{V_{CEsat}}\, = \,0.7V.$$. The switch is initially closed. at time t = 0, the switch is opened. The time t at which q1 leaves the active region is GATE ECE 2011 Analog Circuits - Bipolar Junction Transistor Question 26 English
A
10 ms
B
25 ms
C
50 ms
D
100 ms
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