#fullwavebridgerectifier #basicelectronics #ALPHALab
The full wave bridge rectifier explaination
The working principle of the full wave bridge rectifier is as follows:
During the positive cycle, the D2, D4 diode is reverse-biased, so they don't conduct. The D1 diode, D3 diode forward-biased, so it conducts current.
The current is from the A point to the D1 diode, to the load, the D3 diode. Finally, it returns to the B point.
Therefore, the load voltage is positive half-wave.
During the negative cycle, the D1, D3 diodes are reverse-biased, so they don't conduct. The D2, D4 diodes are forward-biased, so it conducts current.
The current is from the B point to the D2 diode, to the load, the D4 diode. Finally, it returns to the A point.
Although the input waveform is a negative half-wave, the current through the load doesn't change direction.
Thus, the voltage load is a positive half-wave. It converted from a negative half-wave.
This type of full-wave rectifier, however, needs two diodes for each half-wave. Therefore, the drop voltage is 1.4 volts. In this case, the drop voltage is 0.7 volts for each of the diodes.
Assume the input voltage is a sinewave that has the cycle is two pi.
The output voltage is given by the formula.
Multiply the peak voltage by 2 and divide the result by pi and minus 1.4 volts.
As usual, we need to flatten the output waveform. The capacitor is connected.
Assuming the load is disconnected, the capacitor is fully charged to the maximum voltage. The maximum voltage is the peak voltage minus 1.4 volts.
When the load is connected, the capacitor is still fully charged. After it reaches the maximum voltage, it discharges. Hence, it makes the ripple voltage at the output.
The ripple voltage depends on the load current, the capacitance of the output capacitor, and the input frequency.
Because the output current depends on the load and the input frequency is fix, one often increases capacitance to reduce the ripple voltage.
What happens is with peak inverse voltage and the current through the diodes.
The current through the load comes from 2 flows. Therefore, the current through the diodes which is half of the current load.
The peak inverse voltage.
When the A point is at positive peak voltage, the B point is zero voltage.
The C point is a positive peak voltage minus 0.7 volts. And the D point is about 0.7 volts.
The peak inverse voltage applies to the D2, D4 diodes.
With the D2 diode, the peak inverse voltage is the C point voltage minus the B point voltage.
With the D4 diode, the peak inverse voltage is the A point voltage minus the D point voltage.
To easily calculate, the drop voltage can ignore. In this case, the peak inverse voltage is the peak secondary voltage.
The full-wave rectifier bridge diode has the advantage to compare to the method of center-tapped transformers such as lower cost, low peak inverse voltage, ripple, and current is equal. In reality, it used to replace the full-wave rectifier with the center-tapped transformer.
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