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How to Design a Half-wave Rectifier With a Ripple Factor of 2.72

A rectifier is an electronic device that converts alternating current, or AC, into direct current, or DC. A half-wave rectifier does this by clipping off the bottom half of the AC waveform. The half-wave rectifier consists of three stages: A transformer that determines the voltage level of the output, a diode that clips off half of the AC and the filter that smooths out the half-cycle waveforms into something more steady. The ripple factor is the measure of how much the waveform is smoothed out, and it is entirely determined by one of the components in the filter stage.

Things You'll Need

Transformer
Diode
Coil
Capacitor

Instructions

- 1
  Use a transformer in the first stage of the rectifier to convert the voltage level in the input power source to the voltage level you need for the output of your rectifier. Ordinary house current is 120 volts and most electronic devices use 40 volts or less, so the transformer you need is a step-down transformer with a ratio of 3 to 1 or greater.
- 2
  Put a diode in the second stage of the rectifier. Diodes are solid-state devices that allow current to flow in one direction but not in the other direction. The effect a diode has on an AC waveform is to clip off half of the waveform.
- 3
  Combine a coil and a capacitor to make the filter stage of the rectifier. The coil is in series with the output from the previous stage and the capacitor is across the output just before the coil. When electrons start flowing through the coil, they produce a magnetic field that opposes the buildup of current. This makes electrons back up onto the capacitor plates. As the electron flow subsides, the electrons stored on the capacitor plates are released. The effect of these actions smooth out the rippling current -- squashing out the peaks and filling out the valleys between the peaks.
- 4
  Calculate the size of the capacitor that produces a ripple factor of 2.72 using the formula to calculate the ripple factor of half-wave rectifiers: R = I/FC where R is the ripple factor, I is the current, F is the frequency and C is the capacitance. Solving for C, we have C = I/FR. For a ripple factor of 2.72 and using 120 Hertz, we have C = I/326. For a 15 milliamp circuit, C = (15 X 10^-3)/(3.26 X 10^2) = (15 X 3.26) X (10^-3 X 10^-2) = 48.9 X 10^-5 = 489 X 10^-6. A ripple factor of 2.72 requires a 489 microfarad capacitor if the current is 15 milliamps.

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