A half-wave 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 Vs
Voltage across diode
Vp → Vs max
Vn → 12 V
Vp > Vn
Therefore the diode is conducting for the period.
A full-wave 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 VAB = 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
D1 → ON; D2 → OFF: RAB = 10 kΩ
D2 → ON; D1 → OFF: RAB = 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 Vo = 0 volts
When input switch to -V state → diode OFF
Applying Kirchhoff’s voltage law around loop we get
-V -V -Vo = 0
Vo = -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
V0 = 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 double-ended clipper.
D) In this input output follows relation Vout = |Vin|. This is a full-wave 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 Vavg = Vm = √2 × 230
= 325.3 V
The load current is given by
This current discharges the capacitor to the load resistance during the interval (t2 to t3)
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 = ILT = 6.51 × 20 × 10-3
= 0.1302 coulombs
Peak to peak ripple voltage
For the diode network shown below is input with voltage pulse Vi. If the knee voltage of the diode is Vk = 0.7 V then the output voltage pulse Vo is
Given circuit can be written with ideal diode
Is input with
During positive values of Vi < s – Vk i.e. Vi < 4.3 V the diode is in on-state, the diode can be taken as short circuit. The output voltage is simply be voltage across 5V voltage source.
⇒ Vo = s – 0.7 V
= 4.3 V
For Vi > 4.3 V the diode is in off-state the circuit become open circuit hence Vo = Vi
The circuit is in on-state for entire negative cycle. Hence Vo = 4.3 V
The wave form of output voltage is
In Fig. the Input Vi 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 Vo is ______
Time period = 1/f = 1/100 = 10msec
1) During +Ve ½ cycle
Vin = + Ve diode is not forward bias
2) During – Ve cycle
Vin = - Ve diode is Reverse Bias
Average value = 0
In the circuit shown I is a dC current and Vi 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 small-signal circuit (η = 1)
I = 157 uA