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 Resistance seen by the source Vs is

A
$$258\Omega $$
B
$$1258\Omega $$
C
$$93\Omega $$
D
$$100\Omega $$
2
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
3
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
$$\eqalign{ & {f_T} = 1.47 \times {10^{10}}\,\,Hz\,and \cr & {f_\beta } = 1.33 \times {10^{12}}\,\,Hz \cr} $$
4
GATE ECE 1999
MCQ (Single Correct Answer)
+2
-0.6
An NPN transistor (with $${C_\pi }\,\, = \,\,0.3\,\,$$ pf) has a unity gain cutoff frequency $${f_T}$$ of 400 MHz at a DC-bias current Ic = 1mA. The value of its $${C_\mu }$$ (in pf) is approximately $$\left( {{V_T}\, = \,26\,mV} \right)$$.
A
15
B
30
C
50
D
96
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