Full wave Rectifier | Properties | Efficeince | Ripples

FULL WAVE RECTIFIER:

In full wave rectifier, current flows through the load in the same direction for both half cycles of ac input voltage .it allows unidirectional (one way) current through the load during the entire 360 degrees of the input cycle, whereas a half-wave rectifier allows current through the load only during one half of the cycle. the result of full-wave rectification is an output voltage with frequency twice the input frequency and that pulsates every half cycle of the input. This can be obtained with two diodes working alternately. For the positive half cycle of input voltage, one diode supplies current to the load and for the negative half cycle, the other diode does so, current flows in the same direction across the load. Therefore a full-wave rectifier utilizes both half cycles of input ac voltage to produce the dc output. The number of positive alternations that make up the full rectified voltage is twice that of the half-wave voltage for the same time interval.

full wave rectifier

                                                            FIG[1]

The average voltage value is the value measured by a dc voltmeter for full-wave rectified sinusoidal voltage is twice that of the half-wave as shown in the following formula:

                                              Vavg=2Vp/π

Vavg is about  63.7% of vp for a full wave rectified voltage  .

The following two circuits are commonly  used for full wave rectification :

  1. center tap full-wave rectifier.
  2. full-wave bridge rectifier.

Center tapped full wave rectifier operation:

A center-tapped rectifier uses two crystal diodes that are connected to the secondary winding of a center-tapped transformer as shown in figure[2]. The input voltage is coupled through a transformer to the center-tapped secondary. Half of the total secondary voltage appears between the center-tapped and each end of the secondary winding.

center tapped full wave rectifier operation

                                                                   FIG[2]

During the positive half cycle of the input AC signal, point A becomes positive, and point B becomes negative and center tap is grounded (zero volts). The positive side A is connected to the anode of the diode d1 and the negative side B is connected to the cathode of the diode d2. So the diode d1 is forward biased during the positive half cycle and passes current through it. whereas diode d2 is reverse biased because the anode of diode d2 is connected to the negative side of polarity and cathode of diode d2 is connected to the positive side of polarity and blocks the current. in this way, only diode d1 supplies the dc current to the resistive load. The DC current produced at the load will return to the secondary winding through a center tap so the current only flows upper part of the circuit as the diode d2 is reverse biased so the lower part of the circuit carry no current to load. Thus, in the half positive cycle of the ac signal, only diode d1 passes the electric current whereas diode d2 blocks the electric current.

During the negative half cycle of the input AC signal, point A becomes negative, and point B becomes positive and the center tap is grounded (zero volts). The negative side of polarity is connected to the anode of the diode d1 and the positive side of polarity is connected to the cathode of the diode d1. So the diode d1 is reverse biased and does not pass the electric.

on the other hand diode, d2 becomes forward biased as the positive side of polarity is connected to the anode of diode d2 and the negative side of polarity is connected to the cathode of diode d2 in this way d2 supplies the current to load. The DC current produced at the load will return to the secondary winding through a center tap, so the current only flows lower part of the circuit as the diode d1 is reverse biased so the upper part of the circuit carry no current to load. Thus, in the half negative cycle of the ac signal, only diode d2 passes the electric current whereas diode d1 blocks the electric current. Because the output current during both the positive and negative portions of the input cycles is the same direction through the load, the output voltage developed across the load resistor is full-wave rectified dc output voltage.

what is the effect of the turn ratio on  output voltage

If the transformer’s  turns ratio is equal to one then the rectified output voltage is equal to half of the peak value of primary input voltage less the barrier potential and the                   

(vp(sec)=vp(vpri))     

vout= vp/2-0.7v

So we will define the forward voltage due to the potential barriers as a diode drop. To order to achieve output voltage peak equal to input voltage peak with less drop of potential barrier we need to use a transformer with turns ratio 1:2 in this case secondary voltage (vs) is twice the primary voltage (2vp). Due to double voltage at the secondary side, the half value of voltage will appear across the half part of the secondary winding and equal to the (Vpri)

  vout=vp(pri)-0.7

 So we concluded one point here that it does not matter what value of turn ratio is the value of rectified voltage output of the center-tapped rectifier is always half of the total secondary voltage minus diode drop or potential barrier voltage.

          Vout =(Vsec /2)- 0.7 V

Peak inverse voltage :

It is maximum voltage appear across diode when diode is in reverse biased.

As shown in figure[3]. During positive half cycle diode D1 is conducting and diode D2 is reverse biased if we apply kvl in the lower part of the circuit the  voltage across the diode D2  is denote by  vD2 is

                                                vin+vo=vD2

if diode D1 is ideal then output voltage vo is equal to vin.

therefore                           

                                                  vD2=2Vin

                                                 PIV=2Vm

when voltage drop across the diode then

                                               PIV=2Vm(out)+0.7

full wave rectifier piv

FIG[3]

Disadvantages

  1. it is difficult to locate the centre tap on the secondary winding .
  2. The dc output is small as each diode uses only one half of the  transformer secondary voltage.
  3. The diode used must have high peak inverse voltage.

full wave bridge rectifier

The bridge rectifier is another circuit to produce the full waveform output. it uses four diodes connected in a bridge configuration to produce the output .The main advantage of this bridge circuit is that it does not need a special centre tapped transformer, so minimize its size and cost. The single secondary winding is attached to one side of the diode bridge circuit and the load to the other side .

Bridge rectifier construction:

 The bridge rectifier is uses of four diodes namely D1, D2, D3, D4 and load resistor RL as shown in figure[4].The four diodes are attached in a closed loop (Bridge) configuration to efficiently change the Alternating Current (AC) into Direct Current (DC).

bridge rectifier

                                                       FIG[4]

The four diodes D1, D2, D3, D4 are set in series with only two diodes passes electric current during each half cycle. For instance, diodes D1 and D3 are considered as one pair which passes electric current during the positive half cycle whereas diodes D2 and D4 are considered as another pair that passes electric current during the negative half cycle of the input AC signal.

During the positive half cycle of the supply the end P of secondary winding becomes positive and end Q negative so this makes diodes D1 and D2  forward biased and diodes D3 and D4 are reverse biased so the diodes D1 and D3 conduct. These two diodes will be in series through load RL  and the current flows through the load as shown below in fig[5](i)

bridge rectifier current passes

                      FIG[5]

During the negative half cycle of the supply, end P becomes negative and ends Q positive. This makes diodes D2 and D 4  forward biased and diodes D1 and D3 reverse biased. so only diodes  D2 and D4 conduct. These two diodes will be in series with load RL as shown in fig[5](ii). Again current flow from A to B through the load i.e in the same direction as for positive half cycle, therefore dc output is obtained across load RL.

PEAK INVERSE VOLTAGE FOR BRIDGE RECTIFIER

Refer to the figure [6] as for the positive half cycle diodes D1 and D3 will be forward biased, considered the diodes are ideal so the diodes are replaced by wire as shown in the figure. The reverse-biased diodes D2 and D4 are parallel with the transformer secondary. therefore  PIV of each diode D2 and D4 is equal to the maximum voltage Vm across the secondary. likewise, during the next half-cycle, D2 and D4 are forward biased while D1 and D3 will be reverse biased. It is easy to see that reverse voltage across D1 and D3 is equal to VM.

fig[6]

 Advantages:

  1. In the bridge rectifier there is no need of centre-tapped transformer .
  2. The output from the bridge rectifier is twice that of the centre-tap circuit for the same secondary voltage.
  3. The peak inverse voltage is one-half that of the centre-tap circuit (for same d.c. output).

Disadvantages:

(i) It consist of four diodes.

Full-Wave Rectifier output frequency:

The output frequency from a full-wave rectifier is double the input frequency. A wave has a complete one cycle when it repeats the same pattern. In Fig[7](i), the input  AC  completes one cycle from 0° – 360°. However, the full-wave rectified wave completes 2 cycles in this period [See Fig[7] (ii). So, output frequency is twice the input frequency i.e.

                                               fout = 2 fin

fig[7]

For instance, if the input frequency to a full-wave rectifier is 50 Hz, then the output frequency will be 100 Hz.     

Efficiency of Full Wave Rectifier:

Below the figure demonstrate the process of full-wave rectification.

the waveform of full-wave rectification

Let v =VmsinѲ be the AC voltage to be rectified. Let RL and rf be the load resistance and diode resistance respectively. clearly, the rectifier will conduct current through the load in the same direction for both half-cycles of input  AC voltage. The instantaneous current i is given by :

instantaneous current

d.c. output power.

The output current from the rectifier is not a perfect dc but it is pulsating direct current. so, in order to get the d.c. power, the average current has to be found out. using the elementary knowledge of electrical engineering,



if rf is negligible as compared to RL then the efficiency will be maximum.

∴ Maximum efficiency = 81.2% . In a full-wave rectifier, the efficiency will be double as compared to the half-wave rectifier.

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