How to calculate required ripple filter cpacitor value in bridge rectifier circuits?
Answer:
First you have to calculate how much charge the filter capacitor needs to provide between peaks of the rectified waveform. If you've got a full wave rectifier and your line frequency is 50 Hz, the peaks occur at 100 Hz, so the period is 0.01 seconds. By way of example, assume the maximum load is 1 ampere. Then from the formula Q = ∫ i dt, that is, charge equals the integral of current with respect to time, the charge the capacitor will need to provide is 0.01 coulombs.
Now you've got to decide how much ripple is acceptable. If it is a 5 volt power supply, you might accept 0.1 volts. Then you can determine the minimum capacitor from the formula Q = CV, or charge equals capacitance times voltage. Rewriting the formula as C = Q/V and inserting the above values, you've got C = 0.01/0.1 or 0.1 farads. In more familiar terms that is 100,000 microfarads.
This is a large capacitor. That is one of the reasons that higher current power supplies are switching power supplies. If the switching frequency is 20 KHz, the period between peaks for a full wave rectifier is 25 microseconds. Comparing with 10,000 microseconds in the first example, the ratio is 400. That means that the filter capacitor could be 250 microfarads instead of 100,000 microfarads.
Good design practice always leaves a margin, and so larger capacitors should be specified in both cases.
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