What value of filter capacitor is required to produce 1% ripple factor for a full wave rectifier having load resistance of 1.5kohm? Assume rectifier produces peak output of 18v.
Answers
Answer:
Very good. Outstanding. You’ve got all the components necessary. I’ll do you one better. Rather than give you the answer, I’ll tell you how to solve it. I’ll give you 1% of 18 volts is 0.18volts or 180mv. That’s the target. To make it easy, we’ll say the ripple contributes an insignificant change in current, and assume the current is, using ohm’s law, 18/1.5k or 12ma.
Explanation:
Now ask your self how much can I allow this capacitor charge to drop between charge cycles. How much time is there for the capacitor to discharge between charge cycles? What’s the period of a 60hz sine wave after it gets through a full wave rectifier? It’s 120 per second with a period of what? What’s the time, in seconds, between each 120hz peek? Not very many.
Now if a Capacitor looses a charge at the rate of 12ma, what is the size of the minimum capacitor value necessary to prevent a voltage drop greater than 0.18 volts in the time period between incoming power peaks?
What is the math, three components, that gives you a voltage on a given size capacitor for a given charge? That’s what you need to know to answer the question. What will be the charge on that capacitor at the end of the period at the moment of the next incoming charge cycle begins such that the existing charge on the capacitor is equal to 18–0.18 = 17.82v? What fraction of a represents 12ma of current over the period between charge cycles of that capacitor? If you know that, you know the minimum value capacitor you need.