Analog Electronics - 8
10 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Analog Electronics - 8
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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 = Thigh + Tlow
In Thigh, i.e. charging time, both RA& RB are in play.
In Tlow, i.e. discharging time,
Tlow, i.e. discharging time,
Tlow = 0.693 RBC
T = Thigh + Tlow
= 0. 693(RA + 2RB)C
= 0.693 (8 + (2 × 4) × 103 × 0.1 × 10-6)
= 0.693 × 16 × 103 × 10-7
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
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 TON = RA + RB and OFF time TOFF = RB
≃ 66% > 50%
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.
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
A shunt regular is shown in the figure has a regulated output voltage of 10 V. Given Vz = 9.3, VBE = 0 .7 V, β = 49. Neglecting the current through Resistor RB the ratio of maximum power dissipated in the transistor to the maximum power dissipated in the Zener diode is ________
Maximum input current
The Base current is equal to current flowing through Zener diode
Vz = IB
→ Ic = βIB = βIZ
I1 = βIZ + Iz
1A = (49 + 1) Iz
Iz = 20 mA
∴ Power dissipated by the Zener diode Pz = VzIz
= 9.3 × 0.02 = 186 mW
Power dissipated by the Zener diode Pc = Ic × Vc
IC = 99Iz; VC = VO
⇒ PC = 0.98 × 10 = 9.8 W
∴ The ratio = 
For the circuit shown, the value of C = 1 nF, the value of RA & RB such that oscillation frequency of 100 kHz and duty cycle frequency of 75% is obtained at the output is
T = 0.69 C (RA + 2RB)
4RB = 14452.75
RB = 3623
3.6 K
RA = 7.2 K
The frequency of oscillation of the astable multivibrator circuit given below is ______
The time period of astable multivibrator given is
T = 0.693 (RA + 2RB) C
Where RA = 9.5 kΩ RB = 7.5 kΩ
C = 0.1 μF
Frequency of oscillation is given by F = 1/T
For the given circuit, what is the value of power dissipated in zener diode ?
The given circuit is of a voltage regulator, hence the output voltage will remain fixed due to zener diode.
V0 = VZ – VBE = 12 − 0.7 = 11.3V
VCE = 20 – 11.3 = 8.7V
Hence power dissipated in zener diode is Pdis = VZ × I = 477.3 mW
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.
Due to virtual short, the voltage at inverting terminal is Vz
Then the current 
≃ 0.545 mA
Since op-amp is ideal no current is drawn by it,
⇒ IC ≃ IE = I1
VCE = (20 ~ 40) V – 18.5 V
VCEmax = 21.5 V
Power dissipated (maximum) = 21.5 × 0.545 mA
≃ 11.72 mW
For the timer circuit shown below the frequency of oscillation is __________ kHz.
In the given circuit RA = 10 kΩ RB = 5 kΩ C = 1 nF
The frequency of oscillation
Normally the duration of 'high' of pulse is TH = C(RA + RB)ln 2, but the diode bypasses the lower resistance RB. And the capacitor is charged only through RA.
Hence TH is = 0.69RAC and TL = 0.69RBC
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21 docs|263 tests
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21 docs|263 tests
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