Oscillators and Feedback Amplifiers

Question 1
A current controlled current source (CCCS) has an input impedance of 10 \Omega and output impedance of 100 k\Omega. When this CCCS is used in a negative feedback closedloop with a loop gain of 9, the closed loop output impedance is
A
10\Omega
B
100\Omega
C
100k\Omega
D
1000k\Omega
GATE EE 2019   Analog Electronics
Question 1 Explanation: 
"CCCS" (Current controlled current source amplifier)
Given, Z_0=100k\Omega
Loop gain, A\beta =9
Z_{0F}=Z_0[1+A\beta ] \;\;\;(\text{High impedance CS})
=100k\Omega [1+9]
=100k\Omega \times 10
=1000k\Omega
Question 2
A hysteresis type TTL inverter is used to realize an oscillator in the circuit shown in the figure.

If the lower and upper trigger level voltages are 0.9 V and 1.7 V, the period (in ms), for which output is LOW, is _____.
A
0.32
B
0.63
C
0.82
D
0.98
GATE EE 2014-SET-3   Analog Electronics
Question 2 Explanation: 


Discharging curve,
V_c(t)=0-(0-1.7)e^{-t/RC}
At, t=T_2
V_c(t)=0.9V
0.9=1.7e^{-t/RC}
0.63=T_2
T_2=0.63\;ms
Question 3
An oscillator circuit using ideal op-amp and diodes is shown in the figure.

The time duration for +ve part of the cycle is \Delta t_{1} and for -ve part is \Delta t_{2}. The value of e^{\frac{\Delta t_{1}-\Delta t_{2}}{RC}} will be ______.
A
0.6
B
0.8
C
2
D
2.4
GATE EE 2014-SET-2   Analog Electronics
Question 3 Explanation: 


This circuit is a stable multivibrator (or) free running oscillator.
When V_0=+V_{sat}
V_{UPT}=V_{sat}\times \frac{1k\Omega }{1k\Omega +3k\Omega } = \frac{V_{sat}}{4}
When, V_0=-V_{sat}
V_{LTP}=V_{sat}\times \frac{1k\Omega }{1k\Omega +1k\Omega }=\frac{-V_{sat}}{2}

V_c=V_{final}+(V_{inital}-V_{final})e^{-t/RC}
In time t=\Delta t_1
V_c=V_{UTP}; V_{initial}=V_{LTP}; V_{final}=+V_{sat}
V_{UTP}=V_{sat}+(V_{LTP}-V_{sat})e^{-\Delta t_1/RC}
\frac{V_{sat}}{4}=V_{sat}+\left ( -\frac{V_{sat}}{2}-V_{sat} \right )e^{-\Delta t_1/RC}
V_{sat}\left ( -1+\frac{1}{4} \right )=-V_{sat}\left ( \frac{1}{2}+1 \right )e^{-\Delta t_1/RC}
\left (1-\frac{1}{4} \right )=\left ( \frac{1}{2}+1 \right )e^{-\Delta t_1/RC}
\frac{3}{4}=\frac{3}{2}e^{-\Delta t_1/RC}
e^{\Delta t_1/RC}=2\;\;....(i)

In time t=\Delta t_2
V_c=V_{LTP}; V_{initial}=V_{UTP}; V_{final}=-V_{sat}
V_{LTP}=-V_{sat}+(V_{UTP}+V_{sat})e^{-\Delta t_2/RC}
-\frac{V_{sat}}{2}=-V_{sat}+\left ( \frac{V_{sat}}{4}+V_{sat} \right )e^{-\Delta t_2/RC}
\left (1-\frac{1}{2} \right )V_{sat}=V_{sat}\left (1+ \frac{1}{4} \right )e^{-\Delta t_2/RC}
\frac{1}{2}= \frac{5}{4}e^{-\Delta t_2/RC}
e^{\Delta t_2/RC}=\frac{5}{2}\;\;....(ii)
From equation (i) and (ii),
\frac{e^{\Delta t_1/RC}}{e^{\Delta t_2/RC}}
=e^{(\Delta t_1-\Delta t_2)/RC}
=\frac{2}{5/2}=\frac{4}{5}=0.8
Question 4
In the Wien Bridge oscillator circuit shown in figure, the bridge is balanced when
A
\frac{R_{3}}{R_{4}}=\frac{R_{1}}{R_{2}},\; \; \omega =\frac{1}{\sqrt{R_{1}C_{1}R_{2}C_{2}}}
B
\frac{R_{2}}{R_{1}}=\frac{C_{2}}{C_{1}},\; \; \omega =\frac{1}{R_{1}C_{1}R_{2}C_{2}}
C
\frac{R_{3}}{R_{4}}=\frac{R_{1}}{R_{2}}+\frac{C_{2}}{C_{1}},\; \; \omega =\frac{1}{\sqrt{R_{1}C_{1}R_{2}C_{2}}}
D
\frac{R_{3}}{R_{4}}+\frac{R_{1}}{R_{2}}=\frac{C_{2}}{C_{1}},\; \; \omega =\frac{1}{R_{1}C_{1}R_{2}C_{2}}
GATE EE 2014-SET-1   Analog Electronics
Question 4 Explanation: 
\frac{R_3}{R_4}=\frac{R_1}{R_2}+\frac{C_2}{C_1}
\omega =\frac{1}{\sqrt{R_1R_2C_1C_2}}
Question 5
In the feedback network shown below, if the feedback factore k is increased, then the
A
The input impedance increases and output impedance decreases
B
The input impedance increases and output impedance also increases
C
The input impedance decreases and output impedance also decreases
D
The input impedance decreases and output impedance increases
GATE EE 2013   Analog Electronics
Question 5 Explanation: 
The given configuration is a voltage-series feedback configuration.
So, the input impedance increases
R_{if}=R_i(1+A_o k)
So, the output impedance decreases
R_{of}=\frac{R_o}{1+A_o k}
Question 6
The typical frequency response of a two-stage direct coupled voltage amplifier is as shown in figure
A
a
B
b
C
c
D
d
GATE EE 2005   Analog Electronics
Question 6 Explanation: 


Question 7
The feedback used in the circuit shown in figure can be classified as
A
shunt-series feedback
B
shunt-shunt feedback
C
series-shunt feedback
D
series-series feedback
GATE EE 2004   Analog Electronics
Question 7 Explanation: 
Equivalent circuit can be drawn with input voltage comparison and current feedback. It is shunt-shunt feedback.
Question 8
The output voltage (V_{0}) of the Schmitt trigger shown in figure swings between +15V and -15V. Assume that the operational amplifier is ideal. The output will change from +15V to -15V when the instantaneous value of the input sine wave is
A
5 V in the positive slope only
B
5 V in the negative slope only
C
5 V in the positive and negative slopes
D
3 V in the positive and negative slopes
GATE EE 2002   Analog Electronics
Question 9
For the oscillator circuit shown in figure, the expression for the time period of oscillation can be given by (where \tau=RC)
A
\tau ln 3
B
2\tau ln 3
C
\tau ln 2
D
2\tau ln 2
GATE EE 2001   Analog Electronics
Question 9 Explanation: 
T=2RC \ln 3=2\tau \ln 3
Question 10
An op-amp has an open-loop gain of 10^5 and an open-loop upper cutoff frequency of 10 Hz. If this op-amp is connected as an amplifier with a closed-loop gain of 100, then the new upper cutoff frequency is
A
10 Hz
B
100 Hz
C
10 kHz
D
100 kHz
GATE EE 2001   Analog Electronics
Question 10 Explanation: 
A_{OL}=10^5
f=10Hz
A_{CL}=\frac{A_{OL}}{1+\beta A_{OL}}
(1+\beta A_{OL})=\frac{10^5}{10^2}=10^3
f'_2=f_2(1+\beta A_{OL})
\;\;=10 \times 10^3 Hz
\;\; =10kHz
There are 10 questions to complete.
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