To design and set up the following rectifiers with and without filters and to determine ripple factor and rectifier efficiency.
CENTRE TAP FULL WAVE RECTIFIER
During positive half cycle of the input voltage, the diode D1 is forward biased and conducts and diode D2 is reverse biased and will not conduct. The current will flow through the direction A-B-C-D-E. During negative half cycle diode D2 is forward biased and conducts. Diode D1 is reverse biased and will not conduct. The current will flow through the direction F-B-C-D-E. There fore during both half cycles i.e., positive and negative half cycles the out put current will be in only one direction.
Fig:- Centre tap full wave rectifier
FULL WAVE BRIDGE RECTIFIER
During positive half cycle of the input voltage, the diode D1 and D3 is forward biased and both diodes will conduct and diode D2 and D4 is reverse biased and will not conduct. The current will flow through the direction A-B-C-D-E-F-G.
During negative half cycle, the diode D2 and D4 is forward biased and both diodes will conduct and diode D1 and D3 is reverse biased and will not conduct. he current will flow through the direction G-F-C-D-E-B-A. There fore during both half cycles i.e., positive and negative half cycles the out put current will be in only one direction i.e., from point C to D.
Fig:-full wave bridge rectifier
DESIGN:- (Common for both rectifiers)
For the transformer, primary voltage VP=230 V Secondary voltage VS=12V Voltage across diode VD = 0.6 V (Silicon transistor) For a diode current of 10mA, i.e., ID = 10mA
\( R_L=[V_S-V_D]/I_D \)= [12 – 0.6] / 10mA = 1.14 KΩ ; choose RL = 1 KΩ With capacitor filter, ripple factor
Allowing 3% ripple, i.e., r=0.03, f = 50 Hz, RL = 1 KΩ we get C = 96.225µF, let C=100 µF Vm = 12×Ö2 = 16.97 V (Maximum value of the sinusoidal voltage applied to the rectifier) Vdc = 2Vm/π = 10.8V (For the rectifier without filter, Theoretical)
Idc = Vdc/RL (DC Current through the load)
Idc = 10.8V/47W = 229.78 mA (Full Load current, min value of RL shown in the Table, with RS= Rf =0),
Pdc= Vdc2/RL (Load Power)
Pac= VS×IS (Power supplied by the transformer) Ripple factor γ = Vac/Vdc (=1.21 Theoretical) %Efficiency = Pdc/Pac ((40.6%, Theoretical)
%Regulation = E = (VNL-VFL)/VFL
CIRCUIT DIAGRAM OF RECTIFIER:-
Fig:-Full wave bridge rectifier with filter (remove C for rectifier without filter
|Sl.no||Apperatus used name||Quantity||Specification||Maker name|
|1||Center tapped transformer||1||12V-0V-12V||–|
- Place the components on bread board and connect them as per circuit diagram.
- Connect minimum output resistance and DRB in series
- Switch on the input ac and note down readings. Tabulate the readings for different values of output resistor and find out ripple factor, load regulation, and efficiency.
- Using CRO measure the output wave form and check whether it matches with required wave form.
- Repeat this experiment for by connecting the suitable capacitor across the load to reduce the ripple to less than 3%
For both rectifiers use similar tabular column separately
1.Tabular column for rectifier without filter
|RL||VM(Use CRO)||Vdc the theoretical Vdc=2vm/π||Vdc practical use multi-meter in dc mode||Vdc practical use multi-meter in ac mode||Vrms=Vm/√2||ϒ=Vac/Vdc||η=V2dc/V2rms|
2.Tabular column for rectifier with filter
|RL||VM(Use CRO)||Vr.pp(Use CRO)||Vrms the theoretical Vr.rms=Vr.pp/2√3||Vdc the theoretical Vdc=vm-(Vr.pp/2)||Vdc Practical use multi-meter in dc mode||γ=Vr.rms/Vdc||η=V2dc/V2rms|