Full Wave Bridge Rectifier with Capacitor Filter Design Calculation and Formula

In the previous article, we have discussed a center-tapped full-wave rectifier. Which requires a center-tapped transformer and the peak output of the rectifier is always half of the transformer secondary voltage. The Full Wave bridge rectifier with capacitor filter has no such requirement and restriction.

The average output of the bridge rectifier is about 64% of the input voltage. The Bridge type full wave rectifier can convert an AC to DC by the mean of four diodes. The diodes are connected in such a configuration that the output peak voltage remains equal to the secondary of the transformer peak. In each half-cycle, a set of two diodes conduct and block the current alternately. In contrast to the center-tapped rectifier, the Bridge rectifier needs four diodes instead of two diodes, which become expensive.

full bridge rectifier with capacitor filter block diagram

Bridge Rectifier Schematic:

As the name suggests, the configuration of four diodes connection forms a bridge. At the two corners of the bridge, the input AC voltage is applied and at the other two corners of the bridge, the output DC voltage is collected.

full wave rectifier circuit diagram

Working of Full Wave Bridge Rectifier with Capacitor Filter:

Positive Half Cycle:

During the positive cycle of the AC input, the upper corner of the bridge is comparatively positive where diode D1 and D2 are connected. In addition, the lower corner of the bridge is comparatively negative where diodes D3 and D4 are connected.

In this situation, the diode D2 is forward biased as its anode is connected to comparatively higher potential and diode D1 is reversed biased as its cathode is connected to the comparatively higher voltage. Similarly, at the lower corner, the diode D3 is forward biased as its cathode is connected comparatively lower voltage and diode D4 is reversed biased as its anode is connected to the comparatively higher voltage.

For the positive cycle, the current flows from the upper corner of the bridge through diode D2, then through the load resistor from point a towards point b and diode D3, completing its path to the lower corner.

full bridge rectifier current flow during positive half cycle

Negative Half Cycle:

During the negative cycle of the AC input, the upper corner of the bridge is comparatively negative where diode D1 and D2 are connected. In addition, the lower corner of the bridge is comparatively positive where diodes D3 and D4 are connected.

In this situation, the diode D1 is forward biased as its cathode is connected to comparatively lower potential and diode D2 is reversed biased as its anode is connected to the comparatively lower voltage. Similarly, at the lower corner, the diode D4 is forward biased as its anode is connected to a comparatively higher voltage and diode D3 is reversed biased as its cathode is connected to the comparatively higher voltage.

For the negative cycle, the current flows from the lower corner of the bridge through diode D4, then through the load resistor from point a towards point b and diode D1, completing its path to the higher corner.

full bridge rectifier current flow during negative half cycle

Note that during both cycles, the current flow in the load is from point a towards point b and the current is unidirectional like DC rather than AC.

Average Output of the Bridge rectifier:

For a complete input sinusoidal cycle, the output of the full-wave bridge rectifier repeats twice. In other words, the cycle of the time-period of output is $\pi $ instead of $2\pi $. So, the average of the output waveform will be

Full wave bridge rectifier average output waveform

$v_{avg}=\frac{V_{p}}{\pi }(\int_{0}^{\pi }{sin t dt} )$

$v_{avg}=\frac{V_{p}}{\pi }(2)$

$v_{avg}=\frac{2V_{p}}{\pi }$

$v_{avg}=0.637 V_{p}$

Peak Inverse Voltage of the Bridge rectifier :

Consider a positive half cycle, where D2 and D3 are forward biased and D1 and D4 are reversed biased. Peak inverse voltage appears across the diode D1 and D4. The inverse voltage across the diode D4 can be determined by applying KVL at the loop

$v_{p}-PIV_{D4}-0.7v=0$

$PIV_{D4}=v_{p}-0.7v$

Ripple factor of the rectifier:

Ripple factor shows the effectiveness of full wave bridge rectifier with capacitor filter and defined as

$r=\frac{v_{r(pp)}}{v_{dc}}$

Filter Output of Full wave bridge rectifier and ripple factor calculation

Where v$_{r(pp)}$ is the ripple voltage (peak-peak) and v$_{dc}$ value of the filtered output. The formulas for v$_{dc}$ and v$_{r(pp)}$ is given below

$v_{r(pp)}=(\frac{1}{fR_{L}C})(\frac{v_{p(s)}}{2}-0.7)$

$v_{dc}=(1-\frac{1}{2fR_{L}C})(\frac{v_{p(s)}}{2}-0.7)$

Notice the output waveform of the rectifier that the frequency the output voltage is twice the input voltage.

Conclusion :

  • Bridge rectifier converts both halves of the AC input cycle into DC output
  • The rectifier uses four diodes that’s why it is considered expensive
  • The average output of the Bridge rectifier is twice that of a half-wave rectifier
  • The ripple voltage of a full-wave bridge rectifier with capacitor filter is less than that of a half-wave rectifier   

5 thoughts on “Full Wave Bridge Rectifier with Capacitor Filter Design Calculation and Formula”

  1. When negative half cycle potential of top wire is -Vp .Capacitor has potential of Vp . Is the PIV of D2 = 2Vp…?am i right?

    Reply
  2. Good Question.
    As far as, the capacitor and load are connected in parallel, the voltage of point a will be same as the capacitor. Where the anode of diode D2 will be at zero volts, so the PIV will be equal to Vp.

    Reply
  3. In a bridge full wave capacitor filtered rectifier circuit, I thought the available current was less than the available current for a 2-diode full wave system.

    In other words, if I have a 17 VAC winding capable of 1/2 Amp, what would be the available DC current after the capacitor filter, using a full wave bridge rectifier with a large capacitance (which would result in only 5% ripple) ???

    Reply

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