Question

Design a DC power supply that will deliver 100mA of current to a load with a ripple of 5%. Use a full wave rectifier and a 17 Peak Voltage Source with F= 60 Hz. Load resistance = 1k

So Basically you just need to find the capacitance of the filtering capacitor. Please tell me how. This is what I did, but I think I'm wrong.

The Vripple I got is (.05)*17 = .85 V.
Then according to my manual, C = (Curr_Load* T)/Vripple. T is the period, so 1/f. ILoad can be replaced with Vdc/RLoad.

Using C = (Curr_Load*T)/Vripple, I get 1960uF.
Using Vdc/RLoad, I get a different number. Typing this I realized maybe Vdc is 100, since Vdc= Rload*ILoad = 1000*.1 = 100. Then Vrip becomes 5V. Even then I will get a capacitance of 333uF.

When I go into multisim, neither of these capacitance values give me the corrent Vripple.

Answer 1

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