1
GATE ECE 1996
MCQ (Single Correct Answer)
+1
-0.3
An npn transistor has a beta cut-off frequency $${f_\beta }$$ of 1MHz and Common Emitter short circuit low frequency current gain $${\beta _o}$$ of 200. The unity gain frequency $${f_T}$$ and the alpha cut-off frequency $${f_\alpha }$$ respectively are
A
200 MHZ, 201 MHz
B
200 MHZ, 199 MHz
C
199 MHZ, 200 MHz
D
201 MHZ, 200 MHz
2
GATE ECE 1996
MCQ (Single Correct Answer)
+2
-0.6
In the circuit shown in Fig. is a finite gain amplifier with a gain of K, a very large input impedance, and a very low output impedance. The input impedance of the feedback amplifier with the feedback impedance Z conducted as shown will be GATE ECE 1996 Analog Circuits - Frequency Response Question 7 English
A
$$Z\left[ {1 - {1 \over K}} \right]$$
B
$$Z\left( {1 - K} \right)$$
C
$$\left[ {{Z \over {K - 1}}} \right]$$
D
$$\left[ {{Z \over {1 - K}}} \right]$$
3
GATE ECE 1996
Subjective
+5
-0
A JEFT with VP = -4V and IDss = 12mA is used in the circuit shown in Fig. Assuming the device to be operating in saturation, determine ID, VDS and VGS GATE ECE 1996 Analog Circuits - FET and MOSFET Question 10 English
4
GATE ECE 1996
MCQ (Single Correct Answer)
+2
-0.6
Value of R in the oscillator shown in the given figure. So chosen that it just oscillates at an angular frequencies of' "$$\omega $$". The value of "$$\omega $$" and the required value of R will respectively be GATE ECE 1996 Analog Circuits - Oscillators Question 6 English
A
$${10^5}$$ rad/ sec, 2 $$ \times $$$${10^4}$$ $$\Omega $$
B
$$2 \times {10^4}$$ rad/ sec, $$2 \times {10^4}$$ $$\Omega $$
C
$$2 \times {10^4}$$ rad/ sec , $${10^5}\,\Omega $$
D
$${10^{5\,\,}}rad/\sec ,{10^5}\,\Omega $$
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