You need to be aware of something called the diode conduction angle. The larger the capacitor the less time the rectifier diodes actually conduct.
Remember the diodes only conduct when the voltage on one side is higher than the other, the diodes charge the capacitor to the peak value then as the input side sine wave drops in value the diodes turn off. Now the circuit load draws power from the capacitor causing its voltage to drop. The bigger the capacitor the smaller the drop (less ripple) and the less time the diodes will be "ON" for the next cycle. But, the entire current for the load flows through these diodes so the smaller the "ON" time the larger the peak Amps.
An example: say there's a 1 amp load, say the capacitor size is such that we have 10% ripple. With a full wave rectifier that means the diodes conduct for about 5.9 degrees (sine of 5.9 is .104) out of 180 degrees. So for this example the diodes would conduct 180/5.9 x 1 Amp = 30 Amps peak. They only conduct for about 270uS so the average heat is still equal to the heat from 1 Amp but clearly there is an upper limit to how big a capacitor can be because the diodes have a short term surge limit. All of this is only true for a capacitor input ripple filter.
A linear approximation of capacitor voltage over time with a given load can be worked out from the following equation:
Amps x Seconds = Volts x Farads.
So with a full wave recifier: in .00833 seconds (1 over 120Hz) a 3300uF capacitor will drop 2.5 Volts. A more reasonable example here would be a drop of 0.25V for a 100mA load. Doubing the capacitor size will more or less double the diode peak current.
Remember the diodes only conduct when the voltage on one side is higher than the other, the diodes charge the capacitor to the peak value then as the input side sine wave drops in value the diodes turn off. Now the circuit load draws power from the capacitor causing its voltage to drop. The bigger the capacitor the smaller the drop (less ripple) and the less time the diodes will be "ON" for the next cycle. But, the entire current for the load flows through these diodes so the smaller the "ON" time the larger the peak Amps.
An example: say there's a 1 amp load, say the capacitor size is such that we have 10% ripple. With a full wave rectifier that means the diodes conduct for about 5.9 degrees (sine of 5.9 is .104) out of 180 degrees. So for this example the diodes would conduct 180/5.9 x 1 Amp = 30 Amps peak. They only conduct for about 270uS so the average heat is still equal to the heat from 1 Amp but clearly there is an upper limit to how big a capacitor can be because the diodes have a short term surge limit. All of this is only true for a capacitor input ripple filter.
A linear approximation of capacitor voltage over time with a given load can be worked out from the following equation:
Amps x Seconds = Volts x Farads.
So with a full wave recifier: in .00833 seconds (1 over 120Hz) a 3300uF capacitor will drop 2.5 Volts. A more reasonable example here would be a drop of 0.25V for a 100mA load. Doubing the capacitor size will more or less double the diode peak current.