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Q1. A crystal diode having an internal resistance rt = 10 Ω is used for center tapped full wave rectification. If the applied voltage is V = 50 sin (πt) and the load resistance is RL = 1 kΩ, determine the followings:
Draw the input and output voltage and current waveforms.
The efficiency of the circuit.
The Ripple factor.
Ans:
(Figure: Circuit arrangement of centre-‐tap Fullwave rectifier)
Data given: In a cerrte-‐tap full wave rectifier the internal resistance of diodes D1 & D2 i.e. rf = 10Ω
The applied input voltage, V=50 sin (πt)
As the input applied voltage is V = 50 sin πt which is a sinusoidal signal can be represented as below.
Operation: During +ve half cycle (i.e. o-‐π) the diode D1 will conduct due to forward bias and diode D2 will be OFF due to reverse bias so that the current will flow at the upper half portion of the circuit. As a result a voltage will develop across load RL.
Similarly during – ve half cycle (i.e. π-‐2π) the diode D2 will conduct due to forward bias and diode D1 will conduct due to forward bias and diode D1 will be OFFdue to reverse bias so that the current will flow at the lower half portion of the circuit. As a result a voltage will develop across load RL.
It is observed that the current which will flow through load RL is unidirectional during both half cycle hence an pulsating during both half cycle hence an pulsating dc voltage obtained across load RL.
The input and output waveform of centre trapped full wave rectifier is shown in figure below.
Q.2. (a) Explain the difference between Voltage divider bias and Self bias circuits.
Ans. The difference between voltage divider bias and self bias circuit given below.
Self bias: The self bias circuit doesn’t provide good stabilization for high base voltage. For improving the performances of self bias either increase the base resistor RB or decrease the base bias supply voltage or both.
Voltage divider bias: The voltage divider bias provides a better stabilization than others, for base voltage (VB) where the level of VB depends on resistor R2 and R2 can be change by using a variable resistor so that it divides the supply voltage Vcc among resistors R1 and R2.
The given circuit can be simplified and base current can be calculated by using Thevenin theorem where, Thevenin equivalent voltage VTh can be obtained as below.
In circuit applying KVL at indicated loop, the expression is
18V – I x 510K – I x 510K +18=0
I = 36 / 1020K
So that Vth =-‐18+36/1020K x 510K = 0volt
The Thevenin equivalent resistance Rth is obtained as below
Q.4. (a) What is a clipper circuit ? Explain with an example.
(b) Analyze and draw the output waveform of the following circuit when Vi=5 Sin (100πt).
Ans. Clipper Circuit: The circuit with which the waveform is shaped by removing (or clipping) a portion of the applied wave is known as a clipping circuit/clipper/clipper circuit.
The clipper circuit may be classified into three type depending upon its clipping action i.e.
Consider diode is an ideal diode [i.e it becomes short circuit when it will forward bias i.e VD > 0 volt and open circuit when it will reverse bias VD<0V, where VD is the voltage across diode]
So during the +Ve half cycle of an i/p ac signal (ie from 0 to π) the diode will be forward bias & make it short circuited. So that there will be a current which flow through ‘RL’
So that we can obtain an output voltage which shown in figure (b).
But during negative half cycle of an signal (ie from ‘π’ to 2π) the diode becomes reverse bias so it makes an open circuited. As a result the current flow through diode is zero. Hence the output voltage is zero during this period.
N.B.: From the input & Output waveform it is clear that by help of the above circuit we can obtain the output during +ve cycle only not in negative cycle. Hence this circuit is called as “Series negative clipper” circuit.