# 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.