capacitorMATLABrectifier
I am currently learning electrical engineering,and I'm having trouble in finding the capacitance of a full wave rectifier so that it would match the graphs below.
From the image below, the input voltage is a 10 V sine wave and the load of the rectifier is 2000 ohms. I am having trouble finding the capacitance and the only idea I have tried is looking when the input voltage is at 0. I would like any help possible.
Thank you.
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.
The problem with analysing it formulaicly is that the diodes are very non-linear devices and this is implied in the diagram below (red curve is output): -
The blue curve shows a full-wave rectified signal with no capacitor on the output, just a load resistor. When you add a capacitor it charges very quickly due to the forward conduction of one of the diodes but as soon as it attains near-peak voltage (and the blue waveform turns back down towards zero volts), the capacitor remains charged thus reverse biasing the diode and no further diode conduction is seen until the next peak.
You have to think along these lines to adequately "model" what happens. The red curve shows a discharging element between peaks and this is the effect of drawing current into the load resistor. It's a reasonable approximation to regard this as a time-linear discharge but in fact it's an exponential discharge. However, for most decently designed (and used) bridge circuits, using a linear approximation is fine and the ripple voltage formula works pretty good.
Best Answer
First of all, it cannot be a full-bridge rectifier. It's a half-bridge with one diode because it passes only a half cycle of AC voltage.
Anyway, to calculate the capacitor, you should consider the discharging rate of the capacitor's voltage between two successive peaks. According to the graph, a discharge happens from 11V to 6V in a time of 180 (all values approximately.)
So, the solution would be achieved by putting these values into the discharging equation of a capacitor with a time-constant of \$T= R_{load}*C\$. Initial value of capacitor is 11V, final value would be zero. So, 180 is the time when the capacitor's voltage reaches 6V. The value of the capacitor would be derived in this way.
Something like this: