Re: why filter capacitor boost the voltage in full wave bridge rectifier?
Hi,
Input is 24V RMS sine (I assume).
A 24V RMS sinewave
* starts at 0V
* it rises to it's positive maximum of 24V x sqrt(2) = 33.9V
* then it fall diwn to 0V
* it further falls to it's negative maximum of -33.4V
* then it rises to 0V
.... and starts again
The RMS value is an equivalent value for a DC value. This means a resistor connected to 24V RMS (independent of waveform) will generate the same power (heat) as if it is connected to 24V DC.
For a resistor the (voltage) polarity is not important, thus after the rectifier the resistor will generate the same heat as on the true AC.
A little difference comes from the voltage drop caused by the rectifier diodes.
Thus you see a little less RMS voltage after rectifying.
But when you conne t a capacitor, then it becomes charged as long as the absolute voltage voltage before the rectifier is less than the capacitor voltage (ignoring the voltage drop)
This means the capacitor does not get charged to 24V, but to the peak of the sine wave, which is 33.9V.
You see 33V because the voltage drop of the rectifier diodes.
Klaus
Added:
You will find a lot of documents in the internet explaing this. With scope pictures and more details.
Btw: Even with a half wave rectifier you get a capacitor voltage of about 33.9V (minus voltsge drop).