Worksheetselectronic circuits(module 1 and module 2)
Total questions: 25
Worksheet time: 21mins
The efficiency of a half-wave rectifier is approximately:
40.6%
81.2%
50%
In a full-wave rectifier, the frequency of the output ripple is:
Double the input frequency
Half of the input frequency
Equal to the input frequency
The ripple factor of a half-wave rectifier is:
1.21
0.707
1.414
For a full-wave rectifier with capacitor filter, ripple decreases when:
Capacitance increases
Capacitance decreases
Load resistance decreases
A capacitor filter is more effective at:
High load resistance
High load current
Constant load current
In a zener voltage regulator, zener diode operates in:
Forward bias
Reverse breakdown region
Reverse saturation region
The main purpose of a voltage regulator is to:
Convert AC to DC
Maintain constant output voltage
Reduce ripple
A clipper circuit is used to:
Remove a portion of the input signal
Store charge
Amplify the input signal
A clamper circuit shifts:
The DC level of a signal
The frequency of a signal
The amplitude of a signal
The most stable biasing method for BJT is:
Voltage divider bias
Fixed bias
Base bias
In self-bias, the emitter resistor provides:
Thermal stability
AC gain improvement
Phase shift
In FET biasing, gate current is:
High
Zero (approximately)
Same as drain current
The common emitter amplifier has:
Moderate input impedance, high voltage gain, phase shift of 180°
Low input impedance, unity gain
High input impedance, low output impedance, unity gain
A half-wave rectifier uses a sinusoidal input of peak voltage 20 V. The DC output voltage is approximately:
6.37 V
9 V
12.73 V
A full-wave rectifier with peak voltage 30 V produces average DC output:
19.1 V
15 V
30 V
The RMS value of a half-wave rectifier with Vm = 50 V is:
25 V
35.3 V
50 V
A 12 V zener regulator supplies a load of 1 kΩ. If the current through the zener is 10 mA, the output voltage is:
12 V
10 V
13 V
A BJT amplifier has β = 100. If the collector current Ic = 2 mA, then the base current Ib is:
20 µA
200 µA
2 mA
In a self-bias JFET circuit, IDSS = 10 mA, VGS(off) = –4 V. If VGS = –2 V, the drain current ID is:
2.5 mA
5 mA
10 mA
In designing a DC supply for an IoT sensor node, you want to minimize the filter capacitor size while keeping ripple low. Which factor will help most?
Increase input AC frequency
Reduce load resistance
Use zener diode before rectifier
A telecom device requires 5 V DC regulated from 230 V AC mains. The most practical design is:
Voltage divider → Transistor amplifier → Zener diode
Full-wave bridge rectifier → Capacitor filter → 7805 regulator
Full-wave bridge rectifier → Capacitor filter → 7905 regulator
An industrial amplifier is overheating due to thermal runaway in BJT. The best solution is:
Add emitter resistor with bypass capacitor
Increase Vcc
Remove emitter resistor
In a PLC power supply, excessive ripple is observed under variable loads. The most cost-effective improvement is:
Use larger filter capacitor or LC filter
Increase transformer rating
Replace bridge rectifier with half-wave rectifier
A data acquisition system suffers from signal clipping at input. To restore full waveform swing around 0 V without distortion, the circuit needed is:
Clamper
Positive clipper
Voltage divider
A JFET amplifier is required for better stability across devices. The best biasing method is:
Voltage divider bias
Gate bias
Self-bias
