# Full wave rectifier...

Maybe this shouldn't be in the arduino forum but it's for my arduino!

I have an AC voltage of up to about 18v about 2.5W max, that I would like to use to power my arduino. I am thinking of a 4 diode full wave rectifier circuit with a capacitor to change it to DC (roughly) and power my arduino.
Can anyone see any issues with this?
The DC voltage will vary slightly, not so much because of the capacitor, but the AC voltage will also fall lower occasionally. There should always be at least 5v DC coming from the circuit to power the arduino and I presume that 2.5W would be enough to power most things. Would the arduino/voltage regulator cope with that?

Is this the best way to do what I want to? This should be simple and it should be a very small board...

Mowcius

18V is probably RMS, so peak might be 1.4x higher.

I'd whack a 7-12V voltage regulator in there, you can probably find a circuit diagram in the regulator datasheet.

Yes the problem with that is that it will give 25.5V DC minimum and probably a lot more as those sorts of power supplys are often on the high side when unloaded. That means the regulator will be burning a low of power, also the voltage on the input capacitors might be exceeded, as they are only 16v on some models and 24v on others.

So in short it's a bit too high for comfort.

It Is max 18.(something)V AC.

I have an AC voltage of up to about 18v

I have phsically measured it and it should be averaging at about 15v. Doesn't the arduino accept up to about 20v on the regulator? As I understand it AC voltage is measured as the difference between the +v and -v/gnd... It should only be max of about 9v after the rectifier because of the 4 diode design then.

Surely this would be ok...

Mowcius

As I understand it AC current is measured as the difference between the +v and -v/gnd

That might be AC voltage...

Ok, sorry, I hadn't realised that I had put that, should it be ok then?

Mowcius

It can work but as Grumpy Mike said you are going to have a lot of voltage sitting across the regulator and it could get pretty warm.

You could drop the voltage a bit with a voltage divider and say a 470uF capacitor on the output of the voltage divider, but be careful if you load a voltage divider too much the voltage drops rapidly. You will also need higher power resistors, I have done this before as a hack and it works but its not the best solution, but protected the regulator from getting too hot.

Fortunately the brigde rectifier will drop some of the voltage for you. Use a nice big capacitor as well, like at least 2200uF or something, on the output of the recitifier.

Give it go and see how it works, the more current you draw the hotter the regulator gets.

The power dissipated by the regulator is approximately [Vin - Vout] * Iin

Vin is the DC input voltage to the regulator
Vout is the output of the regulator namely 5V in this case
Iin is the current flowing from the power source.

I have phsically measured it and it should be averaging at about 15v.

What you measure on an AC volt meter is the RMS value of the voltage. That is the equivalent to a DC voltage of the same value to give you the same heat if you put it through a resistor or load.

AC voltage is measured as the difference between the +v and -v/gnd

No that is the peak voltage: difference between the +v and gnd or the peak to peak voltage difference between the +v and -v.

When you put it through a rectifier and capacitor you will get the peak voltage. This is bigger than the RMS voltage by a factor of root two, or 1.414. So if you have measured 15V, after full (or half) wave rectification you will get 15 * 1.414 = 21.21V.

Incidentally the peak to peak voltage you have is 15 * 1.414 * 2 = 42.42V

This will be too much.

Using a rectifier like this one:

Cuts off the bottom and brings it to the top, halving the peak to peak voltage measured from the AC supply so if I had 20v AC peak then it would become ~10v DC peak. Correct me if I'm wrong.

I would also include a smoothing capacitor...

Therefore I would receive half the peak voltage max yes?

Mowcius

so if I had 20v AC peak then it would become ~10v DC peak

Correction:-
so if I had 20v AC peak to peak then it would become ~10v DC peak.

Therefore I would receive half the peak voltage max yes?

Therefore I would receive half the peak to peak voltage max yes?

A half wave rectifier gives you exactly the same peak voltage only half as often as a full wave rectifier.

As I said meters don't measure peak or peak to peak they measure RMS.

(OK I know about audio peak meters but let's not be silly)

The output of the rectifier is called pulsating DC.

The smoothing capacitor smooths out the pulsating DC.

As long as the circuit is unloaded the capacitor doesnt discharge and the output stays constant at the DC peak voltage output of the rectifier.

As you load the output, the capacitor discharges into the load between the "pulses" on the output. The more you load it (ie lower resistance of the load) the more current you pull and therefore the quicker the cap discharges and the output voltage ramps down. This causes the smoothed output not to be so smooth anymore. The output ramps down and then rises up and ramps down over and over and this is called a ripple.

The bigger the capacitor the more charge it must loose before the voltage drops and the less the voltage drops for a given load. This means it doesnt ramp down so much and the ripple is better. Ideally you want as big a capacitor as possible and practical.

A full wave recitfier has 2 pulses in every cycle of AC so the ripple frequency is double that of a halfwave and there is less time for the capacitor to discharge or ramp down, and as such the ripple is inherently smaller with a full wave rectifier vs a halfwave rectifier.

When you have load connected to the circuit you will therefore have a ripple. If you measure the output and you don't have a true rms meter you may measure the average. Some cheapo meters measure average output voltage.

The Vdc output of the rectifier needs to be enough to power the regulator, in otherwords it needs to be greater than the dropout voltage of the regulator. A 5V regulator might actually need to have 6 or 7 volts at the very least at its input in order to "turn it on". I noticed that the Arduino uses a low dropout regulator so that you can use the lowest possible DC input voltage above 5V to power the board.

If your have too high an input voltage to the regulator its going to cook as mentioned earlier don't forget that.

Correction:-
so if I had 20v AC peak to peak then it would become ~10v DC peak.

Sorry yes that's what I meant.

As I said meters don't measure peak or peak to peak they measure RMS.

Ok, I kinda get it now, I was pretty sure my meter measured peak to peak but now I'm not sure. If I get the chance then I'll test the voltage on an oscilliscope...

The Vdc output of the rectifier needs to be enough to power the regulator, in otherwords it needs to be greater than the dropout voltage of the regulator. A 5V regulator might actually need to have 6 or 7 volts at the very least at its input in order to "turn it on". I noticed that the Arduino uses a low dropout regulator so that you can use the lowest possible DC input voltage above 5V to power the board.

Yes I have read up a bit on that...

Thanks,

Mowcius