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I am curious if there are practical differences between a DC power supply based on a half wave rectifier or a full wave rectifier.

I mean I have a few small DC power supply units which should give 12V 0.1A each. They all have a transformer 240V->18V, then 1 diode or 4 diodes, then 78L12 (0.1A regulator) and one or two capacitors (typically 220uF or 470uF).

My question is if the power supply can give a good quality DC voltage with just a half wave rectifier (a single diode) when a 470uF capacitor and 78L12 is added, or if bridge rectifier (4 diodes) is better.

I also have one old 12V 0.2A power supply based on a Zener diode instead of 7812 regulator. It also has 18V going to just a single diode, then 33R resistor which limits current to 0.2 Amp, then Zener diode parallel with a 1000uF capacitor. Again: Would it be better to have 4 diodes there, or is the half wave rectification good enough here thanks to the 1000uF capacitor?

(All my power supplies work well, I am just curious "why" and "how" these things work.)

Update:

I found two more interesting information:

  1. Capacitor should be approximately 500 uF for each .1 Amper of output (or more). This applies to full wave rectifier. Since I saw the same values in half wave rectifiers, it isn't enough and they are bad design.

  2. 4-diode rectification cannot be used when we want to have a combined 5V/12V output (or any other two voltages) with a simple transformer, because it can't provide a common ground for two different circuits. (A more complicated real example: I have got a power supply with four output wires from transformer -7/0/+7/+18 Volt. Then it uses 2-diode rectification to get full wave 7V output, and 1-diode rectification to get half wave 18V output. The 18V line can't be "upgraded" to 4-diode rectification here.)

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  • \$\begingroup\$ full wave rectifiers should be better for the Power Factor on the A/C side, in case that is a consideration. \$\endgroup\$ – highBandWidth Jan 17 '12 at 23:42
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Either can work correctly if designed properly. If you have a dumb rectifier supply feeding a 7805, then all the rectifier part needs to do is guarantee the minimum input voltage to the 7805 is met.

The problem is that such a power supply only charges up the input cap at the line cycle peaks, then the 7805 will drain it between the peaks. This means the cap needs to be big enough to still supply the minimum 7805 input voltage at the worst case current drain for the maximum time between the peaks.

The advantage of a full wave rectifier is that both the positive and negative peaks are used. This means the cap is charged up twice as often. Since the maximum time since the last peak is less, the cap can be less to support the same maximum current draw. The downside of a full wave rectifier is that it takes 4 diodes instead of 1, and one more diode drop of voltage is lost. Diodes are cheap and small, so most of the time a full wave rectifier makes more sense. Another way to make a full wave rectifier is with a center tapped transformer secondary. The center is connected to ground and there is one diode from each end to the raw positive supply. This full wave rectifies with only one diode drop in the path, but requires a heavier and more expensive transformer.

A advantage of a half wave rectifier is that one side of the AC input can be directly connected to the same ground as the DC output. That doesn't matter when the AC input is a transformer secondary, but it can be a issue if the AC is already ground-referenced.

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  • \$\begingroup\$ After reading the last paragraph of Olin Lathrop's reply, I realized why sometime half-wave rectifier is used: When there is combined 5V/12V power supply, they need the common ground and that's why cannot use 4-diode rectifier. 2-diode rectifier can be used there, but needs more complicated transformer. Am I correct? \$\endgroup\$ – Al Kepp Jan 18 '12 at 16:07
  • \$\begingroup\$ @AlKepp there are loads of different ways to design a combined 5V/12V supply but yes half wave may well be simpler/cheaper for small loads. \$\endgroup\$ – Peter Green Feb 9 '16 at 18:12
  • \$\begingroup\$ @Olin Lathrop I created an interface where the logic switched between floating ground and -5 VDC with ground being directly connected to one leg of the 24 VAC. When I replaced the half-wave with a full-wave rectifier, my signals look like a clipped sine wave. I can't understand the difference. Is the voltage at ground on a half wave actually 0, or is there some sort of DC offset thing happening here? \$\endgroup\$ – Justin Manuel Mar 10 '17 at 19:37
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Simplified explanation:

An ideal half-wave rectifier only "uses" half of the AC waveform (hence the name half-wave).

half-wave rectifier from Wikipedia

An ideal full-wave bridge rectifier will use the entire AC waveform.

full-wave rectifier from Wikipedia

An ideal full-wave rectifier (with a center-tapped transformer) will also use the entire AC waveform.

another full-wave rectifier from Wikipedia

You can see that that for the half-wave rectifier, every second AC cycle is skipped leaving a gap in the output waveform. For the full-wave rectifier, since the whole waveform is used, the gap is gone (the effective output frequency is doubled).

If these waveforms are applied to a capacitor, you can see pretty clearly that for the half-wave rectifier, in order to maintain clean DC, the capacitor would need to be large enough to hold up the voltage during that big gap. For the full-wave rectifier, since there are more 'peaks', the capacitor can be smaller than for a half-wave rectifier at the same power level.

To your question, a properly-designed half-wave rectifier should have a sufficiently-large capacitor to maintain regulation despite only using half of the AC waveform, so the regulation should be just fine. There's no need to 'upgrade' the circuit with a bridge.

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  • 2
    \$\begingroup\$ The load acts as a resistor R, limiting supply current. Combined with filter capacitor C, you get a low pass filter. You can actually figure out the minimum capacitance needed to ensure your supply stays above minimum for the voltage regulator. Large value capacitors aren't just more expensive - they also take up a lot of space. \$\endgroup\$ – Alan Campbell Nov 24 '14 at 7:28
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    \$\begingroup\$ That's "half of every AC cycle" not "every second AC cycle". \$\endgroup\$ – user253751 Sep 26 '16 at 0:58
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Just to clarify, the transformer doing the voltage stepdown works only on AC. The rectifier converts the AC to DC, which isn't to say that the voltage isn't varying, it's just that the current isn't going both ways. Full wave rectifiers are definitely more awesome than half wave rectifiers because they give you power during both halves of the cycle. They can even correct DC polarity reversal!

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  • \$\begingroup\$ Fixing of polarity reversal - that's an interesting point. \$\endgroup\$ – Al Kepp Jan 21 '12 at 11:16
  • \$\begingroup\$ The rectifier does not convert AC to DC -- what it does is shift the null average voltage of the input to a positive (or negative) average. In that regard you are converting from AC to AC+DC, not just DC. \$\endgroup\$ – Florian Castellane Oct 5 '16 at 15:16
  • \$\begingroup\$ Actually Florian, since the rectifier output is of a constant polarity it can be considered DC. The voltage may not be steady but it is not alternating in polarity (as in AC). \$\endgroup\$ – Benjamin Wharton Nov 27 '16 at 17:20
  • \$\begingroup\$ In ET school they called this "pulsating DC", which seems pretty accurate. \$\endgroup\$ – Michael Gorsich Dec 11 '17 at 14:37
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The rule of thumb I learned in the late '70s was 2000 uF per amp at 60 Hz. That instructor also explained that historically, with the cost being based on using vacuum tube-mercury rectifiers, and with transformer components of copper and steel being cheap, it was an economic decision whether to engineer one or two rectifiers. And time can change the economic basis for engineering decisions.

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There are a couple of advantages to bridge rectification.

One as other answers have already pointed out is that you can get away with smaller smoothing capacitors.

Another is that if you look at the input current waveform of a half wave rectifier it has a DC component. This DC component contributes to transformer saturation issues.

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  • \$\begingroup\$ By "There are a couple of advantages to bridge rectification" did you mean full-wave? \$\endgroup\$ – JYelton Sep 28 '19 at 20:59

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