1
GATE ECE 2005
MCQ (Single Correct Answer)
+2
-0.6
For an npn transistor connected as shown in the figure, VBE = 0.7 volts. Given that reverse saturation current of the junction at room temperature $${300^0}$$ K is $${10^{ - 13}}\,{\rm A}$$, the emitter current is $$\left( {\eta \, = \,1} \right)$$ GATE ECE 2005 Analog Circuits - Bipolar Junction Transistor Question 37 English
A
30 mA
B
39 mA
C
49 mA
D
20 mA
2
GATE ECE 2004
MCQ (Single Correct Answer)
+2
-0.6
A bipolar transistor is operating in the active region with a collector current of 1mA. Assuming that the 'β' of the transistor is 100 and the transconductance (gm) and the input resistance ($${r_\pi }$$) of the transistor in the common emitter configuration, are
A
gm = 25 mA/V and $${r_\pi }$$ = $$15.6\,\,k\Omega $$
B
gm = 40 mA/V and $${r_\pi }$$ = $$4.0\,\,k\Omega $$
C
gm = 25 mA/V and $${r_\pi }$$ = $$2.5\,\,k\Omega $$
D
gm = 40 mA/V and $${r_\pi }$$ = $$2.5\,\,k\Omega $$
3
GATE ECE 2003
MCQ (Single Correct Answer)
+2
-0.6
In the amplifier circuit shown in the figure, the values of R1 and R2 are such that the transistor is operating at VCE = 3V and IC = 1.5 mA when its $$\beta $$ is 150. For a transistor with $$\beta $$ of 200, the operating point (VCE, IC) is GATE ECE 2003 Analog Circuits - Bipolar Junction Transistor Question 40 English
A
( 2 V , 2 mA)
B
( 3 V , 2 mA)
C
( 4 V , 2 mA)
D
( 2 V , 1 mA)
4
GATE ECE 2003
MCQ (Single Correct Answer)
+2
-0.6
Three identical amplifiers with each one having a voltage gain of 50, input Resistance of 1 KΩ and output resistance of 250Ω, are cascaded. The open circuit voltage gain of the combined Amplifier is
A
49 dB
B
51 dB
C
98 dB
D
102 dB
GATE ECE Subjects
EXAM MAP
Medical
NEETAIIMS
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
Staff Selection Commission
SSC CGL Tier I
CBSE
Class 12