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Full Wave Bridge Rectifier
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Posted By: Jose Mathew Member Level: Diamond Points/Cash: 3
TOPIC : FULL WAVE BRIDGE RECTIFIER
“Rectification is the process of converting an alternating voltage or alternating current into direct voltage or direct current”. The device used for rectification is called rectifier. Rectifiers are mainly two types, half wave rectifier and full wave rectifier.
Half wave rectifier is a circuit which rectifies only one of the halves of the ac cycle. During the half cycles when P is positive and N is negative, the diode is forward biased and will conduct. When P is negative and N is positive, the diode is reverse biased and will not conduct. Efficiency of the half wave rectifier will be about 40.6%.
Full wave rectifier is a circuit which rectifies both half cycles of the a.c. when P of 1st diode is positive, the 1st diode is forward biased and will conduct. Now the 2nd diode will not conduct as it is reverse biased. In all the half cycles either of the two diodes will be conducting. The efficiency of a full wave rectifier is about 81.2 %, twice the efficiency of a half wave rectifier.
• To obtain DC voltage form AC supply by the full wave bridge rectifier.
• To find the output voltage of a full wave bridge rectifier.
METHOD OF STUDY
The connections are done as per the circuit diagram. Now connect the 230 V supply the input of the transformer, we get the 6 v as output from the transformer. This is fed into the capacitor to get filtered output. This is taken out from the resistor. Thus, we get rectified and filtered output.
The circuit contains transformer, bridge rectifier and resistor.
The transformer here used is the step down transformer, which converts 230 V supply to 6 V. It isolates the electronic component from the external circuit.
The bridge rectifier is used here; the four diodes are connected in the form of the electronic bridge.
The diode is the semiconductor devices that allow the current to flow only in one direction
(When the diode in the forward biased, the P side is connected to the positive & the N side is connected to the negative). When it is in the reverse direction it will not conduct.
The capacitor here used is the electrolytic capacitor. It is used to filter the ripples, which is present in the output of the diode bridge to an extent.
The resistor is connected in the circuit form which the output is obtained.
The supply is fed into the transformer; the output is a reduced one. From the transformer the reduced voltage is supplied rectifier.
During the positive half cycle the point a in the figure is assumed to be in the positive polarity and point B is of negative polarity. At this moment the diode D2 & D3 is in forward biased condition, they conduct the current. The other diode is the reverse biased condition that is they will not conduct the current. The Path of the current will be a, D2, R, C-RL, L, D3 and B. At this time the one plate of the charges to the positive potential and the other to the negative potential. The charging takes place in a small time period. Just after the point A the capacitor starts discharging due to the voltage difference at its ends. The discharging time is more. Thus the current passes through resistor, which is connected parallel to the capacitor. From that resistor we obtain the filtered output.
Positive Half cycle:-
During the negative half cycle the point A in the figure is assumed to be in the negative polarity and the point B is of positive polarity. At this moment the diode D1 & D3 is in the forward biased condition, they will conduct the current. The other diodes will be in the reverse biased ssscondition that is they will not conduct the current. The path of the current will be B, D4, R, C-RL, L, D1, and A. The capacitor starts charging when the voltage of it is lower than that the output of the rectifier. The Capacitor Charges to a maximum value with a short time. Just after the point B the capacitor starts discharging due to the voltage difference between the ends. Thus the current passes through the resistor, which is connected parallel to the capacitor. From that resistor we obtain the filtered output.
Negative Half cycle:-
According to the charging and discharging of the capacitor we obtain the current flow, which is almost constant (the ripple factor is very much reduced). The direction of the current is the same in both the cycle.
Let V= Vm sin ? be the ac voltage to be rectified. Let rf and RL will be diode resistance and load resistance respectively. The rectifier will conduct current through the load in the same direction for both half cycles of input voltage. The instantaneous current I is given by
I = V / (rf + RL) = Vm sin ?/(rf + RL)
The output current is pulsating direct current. Therefore, in order to find the d.c power average current has to be found out from the elementary knowledge of electrical engineering.
I dc = 2 Im / p
d.c power output = P dc = I2dc x Rl = (2 Im/ p) 2/ x RL
the ac input power is given by
P ac = I2 rms [ rF + R] ]
For full wave rectified wave, we have
I rms = Im /v2
Pac = [ Im/v2 ]2 (rf + RL )
Therefore the full wave rectification efficiency is
? = P dc/Pac = (2Im / p)2RL / (Im /v2)2 (rf + RL) = 8 * RL / p2 (rf+ RL) = 0.812 RL /(rf + RL)
from rectifier we get rf =0 , rf <<< RL
? = 0.812/1 = 0.812
The efficiency will be maximum if rf is negligible as compared to RL
I,e Maximum efficiency = 81.2%.
Calculation for example
Rectification efficiency (M) = 0.812/(1+ rf /RL)
Diode resistance (rf ) = 2.7 ?
M =- 0.812/[1+(2.7/4700)] = 81. 15 %
Ripple factor = Vac / Vdc
diode resistance = 2.7 ?
By doing this project we came to knew that a full ssswave bridge rectifier can be made without centre tapping transformer. And also we came to find the efficiency of full wave rectifier. For bridge type, no centre tap is required on the transformers. Here much smaller transformers are used. Bridge rectifier is used for high voltage application. For the same secondary voltage, the output voltage is twice that of the centre trapped full wave rectifier.
At the same time some disadvantages are there. Here two extra diodes are required and voltage regulation is not satisfactory.
The bridge rectifier with filter section is constructed. A.c and D.C voltages are measured for efficiency is calculated.
Give a diode of small resistance and a high load resistance to make maximum efficiency.
? B. Sc BASIC ELECTRONICS AND ELCTRICITY
? B.Sc (Main) PRACTICAL PHYSICS
ITTIAVIRAH, PREMNATH, ABRAHAM
? NCERT Publication physics class XII
? PRINCIPLES OF ELECTRONICS
|Author: David||Member Level: Bronze||Revenue Score: |
|hey guys thanks for the explanation. you indians rock. can you please send me the picture u are referring to?|
|Author: sridhar.v||Member Level: Silver||Revenue Score: |
|very useful in gaining knowledge to me..........thank you very much.|
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