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Analog Circuits
Bipolar Junction Transistor
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1
GATE ECE 2010
MCQ (Single Correct Answer)
+2
-0.6
Consider the common emitter amplifier shown below with the following circuit parameters:


$$\beta = 100,\,{g_m} = 0.3861\,{\rm A}/V,\,{r_0} = \infty ,\,{r_\pi } = 259\,\Omega, $$
$${R_s} = 1\,K\Omega ,{R_B} = 93\,K\Omega ,\,{R_C} = 250\,\Omega, $$
$${R_L} = 1\,K\Omega ,\,{C_1} = \infty \,\,and\,\,{C_2} = 4.7\,\mu F.$$

GATE ECE 2010 Analog Circuits - Frequency Response Question 3 English

The lower cut-off frequency due to C2 is

A
33.9 Hz
B
27.1 Hz
C
13.6 Hz
D
16.9 Hz
2
GATE ECE 2010
MCQ (Single Correct Answer)
+2
-0.6
Consider the common emitter amplifier shown below with the following circuit parameters:


$$\beta = 100,\,{g_m} = 0.3861\,{\rm A}/V,\,{r_0} = \infty ,\,{r_\pi } = 259\,\Omega, $$
$${R_s} = 1\,K\Omega ,{R_B} = 93\,K\Omega ,\,{R_C} = 250\,\Omega, $$
$${R_L} = 1\,K\Omega ,\,{C_1} = \infty \,\,and\,\,{C_2} = 4.7\,\mu F.$$

GATE ECE 2010 Analog Circuits - Frequency Response Question 4 English

The Resistance seen by the source Vs is

A
$$258\Omega $$
B
$$1258\Omega $$
C
$$93\Omega $$
D
$$100\Omega $$
3
GATE ECE 2003
MCQ (Single Correct Answer)
+2
-0.6
An ideal sawtooth voltage waveform of a frequency 500 Hz and Amplitude 3 V is generated by charging a capacitor of 2 $$\mu F$$ in every cycle the charging requires
A
Constant voltage source of 3 V 1 ms
B
Constant voltage source of 3 V 2 ms
C
Constant current source of 3mA for 1 ms
D
Constant current source of 3mA for 2 ms
4
GATE ECE 2001
MCQ (Single Correct Answer)
+2
-0.6
An npn BJT has gm = 38 mA/V, $${C_\mu }\, = {10^{ - 14}}$$ F, $${C_\pi }\, = 4\, \times {10^{ - 13}}\,F$$ and DC current gain $$\beta \, = \,90$$. For this transistor fT and $${f_\beta }$$ are
A
$$\eqalign{ & {f_T} = 1.64 \times {10^8}\,\,Hz\,and \cr & {f_\beta } = 1.47 \times {10^{10}}\,\,Hz \cr} $$
B
$$\eqalign{ & {f_T} = 1.47 \times {10^{10}}\,\,Hz\,and \cr & {f_\beta } = 1.64 \times {10^8}\,\,Hz \cr} $$
C
$$\eqalign{ & {f_T} = 1.33 \times {10^{12}}\,\,Hz\,and \cr & {f_\beta } = 1.47 \times {10^{10}}\,\,Hz \cr} $$
D
$$\eqalign{ & {f_T} = 1.47 \times {10^{10}}\,\,Hz\,and \cr & {f_\beta } = 1.33 \times {10^{12}}\,\,Hz \cr} $$
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Questions Asked from Marks 2
GATE ECE 2010 (2) GATE ECE 2003 (1) GATE ECE 2001 (1) GATE ECE 1999 (2) GATE ECE 1996 (1)
GATE ECE Subjects
Signals and Systems
Network Theory
Control Systems
Digital Circuits
General Aptitude
Electronic Devices and VLSI
Analog Circuits
Engineering Mathematics
Microprocessors
Communications
Electromagnetics