Analog Electronics - 8 - Electronics and Communication Engineering (ECE) MCQ

The waveforms will be like

The charge of the capacitor varies between

When capacitor is getting charged, output V₀ is high

When capacitor is getting discharged, output V₀ 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_BC

T = T_high + T_low

= 0. 693(R_A + 2R_B)C

= 0.693 (8 + (2 × 4) × 10³ × 0.1 × 10^-6)

= 0.693 × 16 × 10³ × 10^-7

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 

The ON time of 555 timer is T_ON = R_A + R_B and OFF time T_OFF = R_B

≃ 66% > 50%

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

Maximum input current

The Base current is equal to current flowing through Zener diode

V_z = I_B

→ I_c = βI_B = βI_Z

I₁ = βI_Z + I_z

1A = (49 + 1) I_z

I_z = 20 mA

∴ Power dissipated by the Zener diode P_z = V_zI_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 = 

T = 0.69 C (R_A + 2R_B)

 

4R_B = 14452.75

R_B = 3623

3.6 K

R_A = 7.2 K

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

The given circuit is of a voltage regulator, hence the output voltage will remain fixed due to zener diode.
V₀ = 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

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 = I₁

V_CE = (20 ~ 40) V – 18.5 V

V_CEmax = 21.5 V

Power dissipated (maximum) = 21.5 × 0.545 mA

≃ 11.72 mW

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_AC and T_L = 0.69R_BC