Maker Pro
Maker Pro

Good book for Electronics

vivensub

Sep 18, 2015
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I can give you my understanding of half wave rectification until Steve gives you a proper answer.
Half wave only uses one diode and rectifies half the sine wave, so also means that the voltage drops too.
It can be the top half of the wave or the bottom half.
A full wave rectifier uses two diodes and rectifies (changes to DC) both halves of the sine wave.
Another full wave rectifier is the bridge rectifier. It uses four diodes.

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Martin
Then that would mean that we could use a Full-Wave Rectifier all the time, right? Then Half-Wave Rectifiers are obsolete?
Also can you please explain how the efficiency of the Half-Wave Rectifier is 40%? Why isn't it 50? I mean, it does not give any output when the Input Voltage reaches a trough, so, shouldn't it be 50? Or am I just confusing myself?
If there is a derivation, it would be massively helpful.
Thanks,
V
 

Minder

Apr 24, 2015
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With no load, the DC would be RMS x 1.414 (peak), any load will cause a certain % ripple, the percentage would be dependent on amount of load and value of capacitor.
When using certain inductive devices, motors, relays etc, A capacitor is not often required with a full wave supply, as the inductive effect has a similar effect of the capacitor smoothing.
M.
 

vivensub

Sep 18, 2015
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Sorry to be this annoying, but one final doubt for today(Came home just an hour ago): My teacher once told me that for an AC Supply, a combination of an Inductor, Capacitor and a Resistor would be the best, but not why. Does that apply here as well?
 

Minder

Apr 24, 2015
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At one time it was common to see a Pi filter in a power supply, this is a capacitor, series Inductor, and a second capacitor (forming Pi).
Each has the effect of removing ripple.
M.
 

Arouse1973

Adam
Dec 18, 2013
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Sorry to be this annoying, but one final doubt for today(Came home just an hour ago): My teacher once told me that for an AC Supply, a combination of an Inductor, Capacitor and a Resistor would be the best, but not why. Does that apply here as well?

Not the resistor it wastes energy. The inductor also reduces transient responce. Unless you need any of these for noise purposes it is better to use the output direct with an output capacitor to supply transient current to the load. But it does depend on the application and current profile.
Adam
 

Minder

Apr 24, 2015
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Old Steve

Jul 23, 2015
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Thanks a mil.This is easier to understand now. So, from what I gather, using the Capacitor smoothens out the AC Wavefromjust enough for it to resemble a DC Waveform.
After the rectifier, it's actually no longer an AC waveform because it doesn't go below 0V. (Well, not far below.) It's now pretty much a pulsed DC waveform.

So can we add many Capacitors to the circuit in order to completely convert an AC Waveform into a DC Waveform?
Yes, or one larger one. It's far more sensible, though, to use a full-wave rectifier first, which fills in the troughs to a large degree, making smoothing much simpler. See Martin's diagram in post #10 above. A full-wave rectifier inverts the negative half of the AC cycle. Also increases the ripple frequency to twice that of the mains supply. For a 50Hz AC supply, the ripple from a full-wave rectifier is at 100Hz.
 
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