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QUESTION: 1

Which is the frequency of the given 555 timer?

Solution:

The waveforms will be like

The charge of the capacitor varies between

When capacitor is getting charged, output V_{0} is high

When capacitor is getting discharged, output V_{0} is low

The time period of o/p, T = T_{high }+ T_{low}

In T_{high, }i.e. charging time, both R_{A}& R_{B} are in play.

In T_{low}, i.e. discharging time,

T_{low}, i.e. discharging time,

T_{low} = 0.693 R_{B}C

T = T_{high} + T_{low}

= 0. 693(R_{A} + 2R_{B})C

= 0.693 (8 + (2 × 4) × 10^{3} × 0.1 × 10^{-6})

= 0.693 × 16 × 10^{3} × 10^{-7
}

*Answer can only contain numeric values

QUESTION: 2

In the above monostable multivibrator circuit above given C = 10 nF. If the frequency of output pulse is 100 kHz the resistor value is _________ kΩ

Solution:

The pulse duration in mono-stable multivibrator using 555 timer is T = RC In3

10^{-5} = 1.1 × R × 10 × 10^{-9}

R = 909.1 Ω

≃ 0.91 kΩ

QUESTION: 3

For the 555 timer circuit, if R_{A} = R_{B} then

Solution:

The ON time of 555 timer is T_{ON} = R_{A} + R_{B} and OFF time T_{OFF} = R_{B}

_{}≃ 66% > 50%

QUESTION: 4

A dc voltage of 380 V with a peak ripple voltage not exceeding 7 V is required to supply a 500 Ω load. Find out the inductance required.

Solution:

We know that the ripple factor

Where r = RMS value of AC component/DC component

RMS value of AC component = 7/1.414 = 4.95

L = 28.8 H

*Answer can only contain numeric values

QUESTION: 5

A shunt regular is shown in the figure has a regulated output voltage of 10 V. Given V_{z} = 9.3, V_{BE} = 0 .7 V, β = 49. Neglecting the current through Resistor R_{B} the ratio of maximum power dissipated in the transistor to the maximum power dissipated in the Zener diode is ________

Solution:

Maximum input current

The Base current is equal to current flowing through Zener diode

V_{z} = I_{B}

→ I_{c} = βI_{B} = βI_{Z}

I_{1} = βI_{Z} + I_{z}

1A = (49 + 1) I_{z}

I_{z} = 20 mA

∴ Power dissipated by the Zener diode P_{z} = V_{z}I_{z}

= 9.3 × 0.02 = 186 mW

Power dissipated by the Zener diode P_{c} = I_{c} × V_{c}

I_{C} = 99I_{z}; V_{C} = V_{O}

⇒ P_{C} = 0.98 × 10 = 9.8 W

∴ The ratio =

QUESTION: 6

For the circuit shown, the value of C = 1 nF, the value of R_{A} & R_{B} such that oscillation frequency of 100 kHz and duty cycle frequency of 75% is obtained at the output is

Solution:

T = 0.69 C (R_{A} + 2R_{B})

4R_{B} = 14452.75

R_{B} = 3623

3.6 K

R_{A} = 7.2 K

*Answer can only contain numeric values

QUESTION: 7

The frequency of oscillation of the astable multivibrator circuit given below is ______

Solution:

The time period of astable multivibrator given is

T = 0.693 (R_{A} + 2R_{B}) C

Where R_{A} = 9.5 kΩ R_{B} = 7.5 kΩ

C = 0.1 μF

Frequency of oscillation is given by F = 1/T

QUESTION: 8

For the given circuit, what is the value of power dissipated in zener diode ?

Solution:

The given circuit is of a voltage regulator, hence the output voltage will remain fixed due to zener diode.

V_{0} = V_{Z} – V_{BE} = 12 − 0.7 = 11.3V

V_{CE} = 20 – 11.3 = 8.7V

Hence power dissipated in zener diode is P_{dis }= V_{Z }× I = 477.3 mW

*Answer can only contain numeric values

QUESTION: 9

The output voltage of the voltage regulator is constant at 18.5 V. Assuming the op-amp and Zener diode are ideal, the maximum power dissipated in the transistor is ______ mw. The transistor β is very high.

Solution:

Due to virtual short, the voltage at inverting terminal is V_{z
}

Then the current

≃ 0.545 mA

Since op-amp is ideal no current is drawn by it,

⇒ I_{C} ≃ I_{E} = I_{1}

V_{CE} = (20 ~ 40) V – 18.5 V

V_{CEmax} = 21.5 V

Power dissipated (maximum) = 21.5 × 0.545 mA

≃ 11.72 mW

*Answer can only contain numeric values

QUESTION: 10

For the timer circuit shown below the frequency of oscillation is __________ kHz.

Solution:

In the given circuit R_{A }= 10 kΩ R_{B} = 5 kΩ C = 1 nF

The frequency of oscillation

Normally the duration of 'high' of pulse is T_{H} = C(R_{A} + R_{B})ln 2, but the diode bypasses the lower resistance R_{B}. And the capacitor is charged only through R_{A}.

Hence T_{H} is = 0.69R_{A}C and T_{L} = 0.69R_{B}C

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