Test: Rectifiers
10 Questions MCQ Test Analog Electronics | Test: Rectifiers
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All of the mentioned are different types of a rectifier.
For a half wave or full wave rectifier the Peak Inverse Voltage of the rectifier is always
The peak input voltage is smaller than the input voltage due to the presence of diode(s). A single diode reduces the output voltage by approximately 0.7V.
For a half-wave rectifier having diode voltage VD and supply input of VI, the diode conducts for π – 2Θ, where Θ is given by
The diode doesn’t conducts when VD ≥VI . Hence Θ = sin-1 (D/VI).
Bridge rectifier is a better alternative for a full wave rectifier.
All of the given statements are true for a bridge rectifier.
The diode rectifier works well enough if the supply voltage is much than greater than 0.7V. For smaller voltage (of few hundreds of millivolt) input which of the following can be used?
For the supply voltages less than 0.7V super diodes are used.
A simple diode rectifier has ‘ripples’ in the output wave which makes it unsuitable as a DC source. To overcome this one can use
A capacitor is parallel with a resistor can only makes ripples go away. Series connection will become equal to an open circuit once the capacitor is fully charged.
Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value Vp = 100 V. Let the load resistance R = 10 kΩ. Calculate the fraction of the cycle during which the diode is conducting
w Δt ~ √(2Vr/Vp)
Θ = √(2 X 2/100)
Θ = 0.2 rad or 3.18% of the cycle
(Q.9-Q.10) The op amp in the precision rectifier circuit is ideal with output saturation levels of ±12 V. Assume that when conducting the diode exhibits a constant voltage drop of 0.7 V.
Find V– when VI is -1V.
VI = -1v
Vo = 1v
VA = 1.7v
V– = 0v
Virtual gnd as negative feedback is closed through R.
VI > 0 D1 conducts D2 cutoff.
VI < 0 D2 conducts D1 cutoff.
V0 ⁄ VI = -1.
: VI = 2v
Vo = 0v
VA = -0.7v
V– = 0v
Virtual gnd as negative feedback is closed through R.
VI > 0 D1 conducts D2 cutoff.
VI < 0 D2 conducts D1 cutoff.
V0 ⁄ VI = -1.
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