# Confusion between a half-wave and a centre-tapped full-wave rectifier

What voltages would I get if I use a half or full-wave rectifier (the standard ones made by diode)?

I am a high schooler and our textbook just gave a very brief definition about both.

Suppose I have an AC supply of 24 V and I convert it to DC by both a half-wave rectifier and a full-wave rectifier; then the half wave rectifier (one diode) should give (under ideal conditions) peak voltage of 24 V, right?

Now I centre tap the transformer (which how I understand it should make the centre tap have 0 V and the ends have +12 V and -12 V (relatively of course)).

Now the full-wave rectifier (with two diodes) would give (under all ideal conditions) a peak voltage of 12 V, right?

Maybe there is some other formula, but is the peak voltage for a half-wave rectifier double than that of a full-wave rectifier?

Note that I am talking of the peak and not average or RMS.

Suppose I have an AC supply of 24 V

The default assumption for any stated voltage or current quantity is that of RMS so, if you say 24 volts, that means 24 volts RMS. If you mean a peak of 24 volts then write it like this: 24 volts peak or 24 Vpk. So, I assume you mean 24 volts RMS.

Suppose I have an AC supply of 24 V and I convert it to DC by both a half wave rectifier and a full wave rectifier; then the half wave rectifier (1 diode) should give (under ideal conditions) peak voltage of 24 V, right?

The peak output voltage of a rectifier circuit (assuming ideal diodes) is the peak voltage of the AC input. If your AC input is stated as being 24 volts then this is the RMS voltage. The peak voltage of 24 volts RMS is $$\\sqrt2\$$ higher at 33.9 volts. Peak vs RMS: -

Image from here.

If you account for diode losses, the peak voltage will be closer to 33.2 volts for a half wave rectifier and 32.5 volts for a full wave rectifier. And, if you have a smoothing capacitor, the DC output will be close to the peak output under light loading conditions: -

Image from here. Note that it assumes diodes are ideal i.e. don't drop any forward voltage when conducting.

Maybe there is some other formula but I want to know is the peak voltage for a half wave rectifier double than that of a full wave rectifier?

No, it's only a little bit higher (one diode drop or circa 0.7 volts).

Some more RMS and peak identities: -

Image from True RMS value.

• i asked specifically for a centre tapped 2 diode rectfier. Commented Dec 23, 2022 at 8:47

Suppose I have an AC supply of 24 V and I convert it to DC by both a half wave rectifier and a full wave rectifier; then the half wave rectifier (1 diode) should give (under ideal conditions) peak voltage of 24 V, right?

No! When someone says "this AC supply is $$\X\$$ V", they mean that the effective voltage is $$\X\$$ V. Effective voltage is the RMS voltage, not the peak voltage of the AC.

Sine voltages. (1) peak voltage (amplitude), (2) peak-to-peak voltage, (3) effective voltage (RMS voltage) (4) period.
Image from wikipedia article on Alternating Current

Luckily, we know that when someone says "AC", they mean "sine wave" and for sines of any frequency, we know that the effective voltages is $$\1/\sqrt2\$$ of the peak voltage. So, when we multiply 24 V by $$\\sqrt 2\$$, we get the peak voltage – ca. 34 V.

Look at the figure above: what the half-wave rectifier does is simply "cut off" the half of the wave where the voltage is negative. The positive half is left unchanged.

Since the ideal diode (not the one you mention – the one that you can buy from Diodes Inc. is not an ideal diode, which only exists in theory, but it's close enough here) has no voltage drop over it, the peak of the half-wave rectified voltage also happens to be exactly the same peak voltage – so, $$\\sqrt 2\$$ times the effective voltage as you find on the output of your transformer (24 V), so in our case, ca 34 V.

Now I centre tap the transformer (which how I understand it should make the centre tap have 0 V and the ends have +12 V and -12 V (relatively of course)).

Now the full wave rectifier (with 2 diodes) would give (under all ideal conditions) a peak voltage of 12 V, right?

The thing about center tapping your transformer is this: Center-tapping means now you have three wires coming out of the transformer. But you haven't said how you want to connect the full-bridge rectifier. I'll guess you mean this (because if you ignored the center tap, then you could have just not center-tapped it):

and you measure your peak voltage over a load between +VDC and -VDC.

Then, again, the effective voltage of the AC is still 24 V. The peaks are still $$\\sqrt 2\cdot 24\$$ V high. Having both half waves did not change anything regarding the peaks, and all the center tap does is provide you a reference

• Could you refer a source to me to read about centre tapped transformers. Commented Dec 11, 2022 at 13:58
• Not really, no, sorry! It's a transformer, exactly as you say, where on one side there's a cable that runs from the middle of the winding "out", pretty much as the picture of that center-tapped transformer T1 in my answer suggests, graphically. Commented Dec 11, 2022 at 14:03
• I still don't get it. What is +V dc and -V dc(why not a particular value).i also don't understand the circuit.in my book the wires leaving the diodes are connected and then reach a load(resistor) and then join to the central tap.we aren't taught much about AC and i didn't know whether stack exchange physics or this was a better place for me Commented Dec 11, 2022 at 14:07
• The school doesnt care whether anybody understands anything.there is no qauntitative discussion regarding this and you just have to spit out the circuit design in exams Commented Dec 11, 2022 at 14:08
• This is the wrong answer to Karan's question, which is explicitly about center-tapping a 24v transformer secondary to get two 12v phase-locked secondaries. The text of the answer assumes two 24v phase-locked secondaries. Commented Dec 11, 2022 at 22:23