Halfwave rectifier with and without filter

					                                        EXPERIMENT NO. 5
      HALF WAVE RECTIFIER WITH AND WITHOUT FILTER
AIM:
       To study the half wave rectifier with and without filter to calculate ripple
factor and percentage of regulation.
APPARATUS:
             1. Transformer (0-12) V            -       1
             2. PN junction Diode (IN4007)      -       1
             3. Capacitor- 1000 F / 25V          -      1
             4 .Decade Resister Box              -      1
             5. CRO                              -      1
             6. Bread Board                      -      1
             7. Connecting Wires                 -      10
             8. Digital multi meter              -      1
FORMULAE:
           1. Ripple Factor = Vr,rms / Vdc
           2. % Regulation = VNL- VFL/ VFL x 100
THEORY:
      In half wave rectifier, rectifying element conducts only during
positive half cycle of input a.c supply. The negative half cycle of a.c
supply are eliminated form the output.
      This rectifier circuit consists of resistive load, rectifying element,
i.e. p-n junction diode, and the source of a.c voltage, all connected in
series. The circuit diagram is shown in fig usually, the rectifier circuits
are operated from ac mains supply. To obtain the desired dc voltage
across the load, the ac voltage is applied to rectifier circuit using suitable
step-up or step-down transformer, mostly a step-down one, with
necessary turns ratio.
           The input voltage to the half-wave rectifier circuit shown in
the fig is a sinusoidal ac voltage, having a frequency which is the supply
frequency, 50 Hz.
             The transformer decides the peak value of the secondary
voltage. If N1 are the primary number of turns and N2 are the secondary
number of turns and Epm is the peak value of the primary voltage then,
                N2 =Esm
                N1_=_Epm
          Where
              Esm = the peak value of the secondary ac voltage
As the nature of Esm is sinusoidal the instantaneous value will be,
              es = Esm sin wt_______________________________________________
              w=2 f
     Let R1 represents the forward resistance of the diode. Assume that,
under reverse biased condition, the diode acts almost as an open circuit,
conducting no current.
PROCEDURE:
          1. Connect the Wire in circuit as shown in the figure
          2. Switch on the supply and note down the reading of AC input voltage
             in secondary of transformer and rectified DC voltage without
             connecting the load resister
          3. Vary the load in convenient steps and note down DC&AC voltages
          4. Connect the capacitor filter and repeat the steps 3 for different
             values of load resistance
          5. Calculate the ripple factor and % of regulation from the obtained
             readings
          6. You can draw the input and output waveforms of rectifier (with &
             without filter ) by the using of CRO
          7. Draw the graph between
                a. load resistance and ripple factor by using without filter
                b. load resistance and ripple factor by using with filter


RESULT:
            Thus the half Wave Rectifier has done and the following
      Parameters are calculated.
      1. Ripple factor Without filter is   =
      2. Ripple factor With filter is      =
      3. % of Regulation without filter is =
      4. % of Regulation with filter is    =
CIRCUIT DIAGRAM :-
WITH OUT FILTER :-

                                                                          +
                       12v      1N4007GP                                   CRO o/p
                 T1
                                           Vdc D1                         DRB
 AC 250V,                                              Vac
    50Hz
                        0                                                 _




WITH FILTER :-
                                    D1
                                                                           +
                         12V     1N4007GP                                  CRO o/p
                  T1
     250V                                                  (0-20V)
                                   (0-20) Vdc        Vac              100uF DRB
  AC 50Hz                                                            -
                            0                                               _


MODAL GRAPH :-



                                  Input Waveform


A.C input
Voltage



                                 Rectified out put
Output
Voltage
Without
Filter


                                    Filter Output
Output
Voltage
With Filter
 TABULAR COLUMN:

 WITH FILTER:
                                     VNL (No Load Voltage) =

        Vdc in        Vr,rms in     RL in K    Ripple Factor % Regulation =
S.No    Volts         Volts                    Vr,rms/ Vdc   VNL- VFL/ VFL




 WITHOUT FILTER:
                                     VNL (No Load Voltage) =

       Vdc in Volts   Vr,rms in   RL in K     Ripple Factor    % Regulation =
S.No                   Volts                  Vr,rms / Vdc     VNL–VFL / VFL
HALFWAVE RECTIFIER:-
MODEL GRAPH :-
WITH FILTER:-




Ripple                          %
Factor                       Regulation




                RL (KΩ)                     RL (KΩ)



WITHOUT FILTER:-




Ripple                     %
Factor                    Regulation



                   RL (KΩ)                RL (KΩ)

				
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posted:11/9/2011
language:English
pages:5