A halfwave rectifier is used to charge a 12V battery through a resistance ‘R’. The input transformer is fed with 34 AC with turns ration 2 : 1. The conduction period of the diode is
Voltage at secondary winding of the coil is V_{s
}
Voltage across diode
V_{p} → V_{s max}
V_{n} → 12 V
For conduction:
V_{p} > V_{n
}
Therefore the diode is conducting for the period.
A fullwave rectifier circuit employs inductor filter at its output. If the source frequency is 50 Hz and the inductor value is 10 mH. Find the ripple factor for 100 Ω loads.
For full wave rectifier using inductor filter at output is
The voltage source V_{AB} = 4 sin ωt is applied to the terminal A and B of the circuit shown In Fig. the diode is assumed to be ideal. The impedance by the circuit across the terminals A and B is ________ kΩ
When,
D_{1} → ON; D_{2} → OFF: R_{AB} = 10 kΩ
D_{2} → ON; D_{1} → OFF: R_{AB} = 10 kΩ
∴ The impedance by the circuit across the terminals A and B of 10 kΩ
For the damping network shown below, resulting output for the applied Input will be…
Clamping network “Clamp” a signal to different DC level.
During the Internal O → T/1 → Diode in the ON state during this internal output voltage is directly across the short circuit and V_{o} = 0 volts
When input switch to V state → diode OFF
Applying Kirchhoff’s voltage law around loop we get
V V V_{o} = 0
V_{o} = 2V
So output wave form will be,
Assuming the diode to be ideal, the current I in the circuit shown is ________ mA.
The above circuit can be redrawn as:
Assume diode is off:
Voltage across 20 k resistor connected to left of diode is
This will turn on the diode
the equivalent circuit is now:
Now the two 20k resistors are in parallel whose equivalent is 10k
this 10 k is in series with other 10k resistor
9V divides equally among them
V_{0} = 4.5V
Cuurent I = 4.5/20 mA
= 0.225 mA
The correct option matching List  I to List  II is
A) In graph ‘A’ the output follows input, but above certain positive input, output stays constant. The circuit is called positive clipper.
B) Similarly the graph ‘B’ is of negative clipper.
C) In this path positive and negative side are clipped, it is a doubleended clipper.
D) In this input output follows relation V_{out} = V_{in}. This is a fullwave rectifier circuit.
The figure shows a half wave rectifier with a load resistor of 50 Ohms. The input supply voltage is 230V(runs) at 50 Hz. A 10000 μF filter capacitor is added across the load resistor to reduce ripple. The percentage of ripple voltage at the output is ________
Due to filter capacitor the output voltage is given by waveform
Output DC voltage V_{avg }= V_{m }= √2 × 230
= 325.3 V
The load current is given by
This current discharges the capacitor to the load resistance during the interval (t_{2} to t_{3})
The time period of AC voltage.
= 20 ms (for 50 Hz)
Thus the charge supplied by the capacitor to the load resistance during this interval will be
∆Q = I_{L}T = 6.51 × 20 × 10^{3}
= 0.1302 coulombs
Peak to peak ripple voltage
For the diode network shown below is input with voltage pulse V_{i}. If the knee voltage of the diode is V_{k} = 0.7 V then the output voltage pulse V_{o} is
Given circuit can be written with ideal diode
Is input with
During positive values of V_{i} < s – V_{k} i.e. V_{i} < 4.3 V the diode is in onstate, the diode can be taken as short circuit. The output voltage is simply be voltage across 5V voltage source.
⇒ V_{o} = s – 0.7 V
= 4.3 V
For V_{i} > 4.3 V the diode is in offstate the circuit become open circuit hence V_{o} = V_{i}
The circuit is in onstate for entire negative cycle. Hence V_{o} = 4.3 V
The wave form of output voltage is
In Fig. the Input V_{i }is 100 Hz triangular wave having a peak to peak amplitude of 2 volts and an average value of 0 volts. Given that the diode is ideal, the average value of output V_{o} is ______
Time period = 1/f = 1/100 = 10msec
1) During +Ve ½ cycle
V_{in} = + Ve diode is not forward bias
2) During – Ve cycle
V_{in} =  Ve diode is Reverse Bias
Average value = 0
In the circuit shown I is a dC current and V_{i} is a sinusoidal signal with peak amplitude 10 mV and a frequency 100 kHz. The value of I that will provide a phase shift of 45° at the output is ________. μA (Assume η = 1)
Opening the current source we get the AC smallsignal circuit (η = 1)
I = 157 uA
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