# Miscellaneous

 Question 1
Two monoshot multivibrators, one positive edge triggered $(M_{1})$ and another negative edge triggered $(M_{2})$, are connected as shown in figure

The monoshots $(M_{1})$ and $(M_{2})$ when triggered produce pulses of width $(T_{1})$ and $(T_{2})$ respectively, where $(T_{1} \gt T_2)$ . The steady state output voltage $(v_{o})$ of the circuit is
 A A B B C C D D
GATE EE 2014-SET-3   Analog Electronics
Question 1 Explanation:
Not in the GATE-2018 Syllabus.
 Question 2
The following circuit has a source voltage $V_s$ as shown in the graph. The current through the circuit is also shown.
The element connected between a and b could be
 A A B B C C D D
GATE EE 2009   Analog Electronics
Question 2 Explanation:
Diode act as a switch. When forward biased it is short circuited. But when suddenly reverse biased, current does not become zero instantly , initially the same current flow in opposite direction and after some time (turn off time) it will become zero.
 Question 3
The truth table of monoshot shown in the figure is given in the table below :

Two monoshots, one positive edge triggered and other negative edge triggered, are connected shown in the figure, The pulse widths of the two monoshot outputs $Q_1 \; and \; Q_2$ are $T_{ON1}$ and $T_{ON2}$ respectively.

The frequency and the duty cycle of the signal at $Q_1$ will respectively be
 A $f=\frac{1}{T_{ON_{1}}+T_{ON_{2}}},D=\frac{T_{ON_{1}}}{T_{ON_{1}}+T_{ON_{2}}}$ B $f=\frac{1}{T_{ON_{1}}+T_{ON_{2}}},D=\frac{T_{ON_{2}}}{T_{ON_{1}}+T_{ON_{2}}}$ C $f=\frac{1}{T_{ON_{1}}},D=\frac{T_{ON_{1}}}{T_{ON_{1}}+T_{ON_{2}}}$ D $f=\frac{1}{T_{ON_{2}}},D=\frac{T_{ON_{1}}}{T_{ON_{1}}+T_{ON_{2}}}$
GATE EE 2008   Analog Electronics
Question 3 Explanation:
$f=\frac{1}{T_{ON1}+T_{ON2}},$ $D=\frac{T_{ON1}}{T_{ON1}+T_{ON2}}$
 Question 4
IC 555 in the adjacent figure is configured as an astable multi-vibrator. It is enabled to to oscillate at t=0 by applying a high input to pin 4.
The pin description is :
1 and 8-supply;
2-trigger;
4-reset;
6-threshold
7-discharge.
The waveform appearing across the capacitor starting from t=0, as observed on a storage CRO is

 A A B B C C D D
GATE EE 2007   Analog Electronics
Question 4 Explanation:
An astable multi-vibrator is providing pulse as given below

But in this case initial voltage at capacitor is zero. So it starts from zero. Also charging time will be larger (normally) than discharging time but it is made equal by using a diode.
 Question 5
The input signal $v_{in}$ shown in the figure is a 1 kHz square wave voltage that alternates between +7 V and -7 V with a 50% duty cycle. Both transistor have the same current gain which is large. The circuit delivers power to the load resistor $R_{L}$. What is the efficiency of this circuit for the given input ? choose the closest answer.
 A 46% B 55% C 63% D 92%
GATE EE 2007   Analog Electronics
Question 5 Explanation:
It is a class D amplifier, so $\eta$ should be high $\eta$ of class D amplifier is 90% to 100%.
 Question 6
In the Schmitt trigger circuit shown in figure, if $V_{CE\left ( sat \right )}$= 0.1V, the output logic low level ($V_{OL}$) is
 A 1.25 V B 1.35 V C 2.50 V D 5.00 V
GATE EE 2004   Analog Electronics
Question 6 Explanation:
$V_i=0$, then first transistor will be cut-off and current through left resistor will drive the second transistor into saturation

Then, $V_o=V_{CE,sat}+1.25 \times 10^{-3}\times 10^3$
=1.35 V
 Question 7
The circuit of figure shows a 555 Timer IC connected as an astable multivibrator. The value of the capacitor C is 10 nF. The values of the resistors $R_A \; and \; R_B$ for a frequency of 10 kHz and a duty cycle of 0.75 for the output voltage waveform are
 A $R_{A}=3.62 k\Omega , R_{B}=3.62k\Omega$ B $R_{A}=3.62 k\Omega , R_{B}=7.25k\Omega$ C $R_{A}=7.25 k\Omega , R_{B}=3.62k\Omega$ D $R_{A}=7.25 k\Omega , R_{B}=7.25k\Omega$
GATE EE 2003   Analog Electronics
Question 7 Explanation:
Duty cycle $=\alpha =\frac{T_{ON}}{T}$
$\therefore T_{ON}=\alpha T=0.75 \times 10^{-4}=75 \mu sec$
$\frac{75 \times 10^{-6}}{0.7 \times 10^{-8}}=R_A+R_B$
$\therefore R_A+R_B=10.714 \; k\Omega$
$\therefore T_{OFF}=0.7 \; CR_B$
$\therefore R_B=\frac{25 \times 10^{-6}}{0.7 \times 10^{-8}}$
$\;\; R_B=3.57 \; k\Omega , R_A=7.14 \; k\Omega$
There are 7 questions to complete.