# Vripple(p-p) Calculation

Discussion in 'Homework Help' started by DiodeMan, Feb 3, 2013.

1. ### DiodeMan Thread Starter New Member

Feb 3, 2013
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0
I'm currently stuck on a homework question regarding power supplies. I am given the primary frequency, the secondary peak voltage, the capacitor value, and the load current. The values are 60Hz, 80Vp, 10(micro)farads. and 20 mA. I am asked to find Vdc across the load. When I plug into the equation Vripple(p-p) = I/(fC) I come up with a value of 33.333 V. However, according to my book, the peak to peak of the ripple is actually half of what I calculated. Can anybody help me with where I may be going wrong?

2. ### WBahn Moderator

Mar 31, 2012
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4,917
Your post is inconsistent. You say you are asked to find Vdc and then proceed to find Vripple. What is it you are trying to find?

Assuming you are trying to find the peak-to-peak ripple voltage, the problem is that you are grabbing equations and throwing things into them without understanding what the equations, and the parameters in them, mean. So when you see an 'f' in an equation, you throw whatever handy 'f' you have lying around at it without asking if it is the right 'f' for that equation.

So take a step back and DERIVE that equation for ripple voltage. What assumptions are being made about the circuits and waveforms involved? What is the 'f' in that equation and how does it relate to the input waveform?

3. ### DiodeMan Thread Starter New Member

Feb 3, 2013
13
0
I would like to thank you for your reply, as it really got me to think. I'm not 100% sure if I am correct, but I'm assuming that a full-wave rectifier would essentially double the frequency of the waveform, and this is the value of the 'f' in the equation.

As well, the reason I was looking for the ripple p-p was because the average dc voltage with a capacitor across the load is the midpoint of the ripple waveform, so if I knew p-p of ripple I could find the peak, and subtract that off the peak voltage of the secondary, effectively finding Vdc.

Oct 2, 2009
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5. ### WBahn Moderator

Mar 31, 2012
18,080
4,917
Okay, that's more or less what I figured.

Your reasoning on why you use twice the primary frequency is correct. The other assumption is that the capacitor charges up instantaneously to the peak voltage and then discharges at constant rate (the average current) until the next peak. Neither of these are particularly good assumptions, but they are reasonable starting points that yield acceptable results in a large fraction of the situations.

But my big point is that if you make a little sketch of the waveform and show the ripple voltage waveform, you not only can derive this formula almost immediately, but you will naturally do so in such a way that you use the correct frequency. On the other hand, if your skill revolves around picking memorized equations out of a hat, then you will routinely mess up their application.