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

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 | |

B | |

C | |

D |

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

The element connected between a and b could be

A | |

B | |

C | |

D |

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

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

f=\frac{1}{T_{ON_{1}}+T_{ON_{2}}},D=\frac{T_{ON_{1}}}{T_{ON_{1}}+T_{ON_{2}}} | |

f=\frac{1}{T_{ON_{1}}+T_{ON_{2}}},D=\frac{T_{ON_{2}}}{T_{ON_{1}}+T_{ON_{2}}} | |

f=\frac{1}{T_{ON_{1}}},D=\frac{T_{ON_{1}}}{T_{ON_{1}}+T_{ON_{2}}} | |

f=\frac{1}{T_{ON_{2}}},D=\frac{T_{ON_{1}}}{T_{ON_{1}}+T_{ON_{2}}} |

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

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 | |

B | |

C | |

D |

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.

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.

46% | |

55% | |

63% | |

92% |

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

1.25 V | |

1.35 V | |

2.50 V | |

5.00 V |

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

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

R_{A}=3.62 k\Omega , R_{B}=3.62k\Omega | |

R_{A}=3.62 k\Omega , R_{B}=7.25k\Omega | |

R_{A}=7.25 k\Omega , R_{B}=3.62k\Omega | |

R_{A}=7.25 k\Omega , R_{B}=7.25k\Omega |

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

\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.