1
GATE ECE 2003
+2
-0.6
The oscillator circuit shown in the figure has an ideal inverting Amplifier. It's frequency of Oscillation (in Hz) is
A
$${1 \over {2\pi \sqrt {6\,} RC}}$$
B
$${1 \over {2\pi RC}}$$
C
$${1 \over {\sqrt 6 RC}}$$
D
$${1 \over {\sqrt 6 (2\pi RC)}}$$
2
GATE ECE 2002
+2
-0.6
The circuit in the figure employs positive Feedback and is intended to generate sinusoidal oscillation. If at a frequency f$$_0$$, B(f)= $${{\Delta {V_f}(f)} \over {{V_0}(f)}} = {1 \over 6}\,\,\angle {0^0}$$ then to sustain oscillation at this frequency.
A
$${R_2}\, = \,5\,\,{R_1}$$
B
$${R_2}\, = \,6\,\,{R_1}$$
C
$${R_2}\, = {{{R_1}} \over 5}$$
D
$${R_2}\, = {{{R_1}} \over 6}$$
3
GATE ECE 2001
+2
-0.6
The Oscillator circuit shown in the figure is
A
Hartley Oscillator with $${f_{oscillation}}$$= 79.6 MHz
B
Colpitts Oscillator with $${f_{oscillation}}$$ =50.3 MHz
C
Hartley Oscillator with $${f_{oscillation}}$$= 159.2 MHz
D
Colpitts Oscillator with $${f_{oscillation}}$$=159.2 MHz
4
GATE ECE 1996
+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
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
$${10^{5\,\,}}rad/\sec ,{10^5}\,\Omega$$
EXAM MAP
Medical
NEET