Im solving a problem where I have to calculate the ripple voltage for a half-wave rectifier (with a diode and capacitor) and the formula is V=I/(fC). I like to know where the formulas come from, how can I arrive at this result ?
Electrical – Derivation of ripple voltage formula for simple half-wave rectifier
ripplevoltage
Related Solutions
The ripple formula you have is an approximation and just to demonstrate that here's another: -
The formula used here is not too disimilar from yours but it more accurately shows the time and not the frequency as being the important factor. However, the article makes an error in stating the 10msecs should be used at 50Hz. When the diode stops conducting at the top of the cycle and when it restarts is slightly less than 10msecs.
But, in the article's credit, look at the final paragraph - small print indicating where problems with the formula might lie and of course the OP's example falls into this area where all bets are off.
In truth the decay of the voltage is exponential from the top of the peak and not-linear and this will make a difference too.
There is a mass of online help to aid calculating this. This one at Hyperphysics is particularly complete with a calculator provided.
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Best Answer
The formula is an approximation.
A half wave rectifier will recharge your cap on every period, which means every \$ T=1/f \$ seconds.
If the load draws a current \$ i \$, since \$ i = C dv/dt \$ then \$ v \$ will decrease by \$ iT/C = i/(fC) \$ on every period, so you have your answer.
The approximation lies in the fact that the time needed to recharge the cap is not zero, so during a part of the period the load will be supplied directly through the rectifying diode. Real-world ripple voltage will thus be a bit less than predicted. The real value depends on lots of factors, like diode dynamic resistance and other resistances in the circuit (transformer wiring, etc) so it's difficult to model... but \$ i/fC \$ gives a worst case estimation, which in most cases is all we need.
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