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I want to know how to choose and calculate the values of input and output capacitors while making a power supply circuit(SAY using 7805)
And got the following answer:
It depends on many factors, such as maximum current load, and peak AC voltage out of the rectifiers. For example, the input to a 7805 has to be always above 7 volts for the regulator to operate. So if the lowest average DC voltage into the regulator is 9 volts, that means the AC ripple component can ve 4 volts peak to peak, that is, vary from 7 to 11 volts. That sets one parameter of the calculation. Actually, the ripple is not symmetrical, but this is a start. If the DC voltage worse case is 7.5 volts, then you can have only 1 volt P-P of ripple before the 7805 drops out. Note I said worse case DC voltage which is a calculation or measurement with highest current and lowest AC power voltage. Once you have those numbers, then you can use CV = It to calculate the capacitor size, where t is 8 ms for a 60 Hz full wave rectifier, and V is the P-P ripple from above. And I is max current load. And this gets back to transformer design/selection and thermal considerations. The lower the DC voltage into the regulator is, the less heat you have to dissipate, but the larger the capacitor value will be. Generally it's better to minimize heat and use a larger capacitor. edit: Note that the min input spec to the 7805 is not always 7 volts, that is a simplification. It depends on current and temperature. Worse case, it is 8 volts.