Half wave rectifier

[ronaldrc]

I remember one time we where discussing rectifier circuits on one of the electrical forums.

Someone came up with this question.

What is the effective power output of the half wave rectifier circuit? It was determined that it was more than 50 percent of the input. Would someone please refresh my memory and tell me how this can be and how? I did a search using the search tool here and didn't come up with the thread.

Not sure but I think Dsparks was on here then, remember him?

 

[Ed MacLaren]

Re: Half wave rectifier

Hi Ronald,

The DC output from a half-wave rectifier is called the average value, and is the RMS (aka effective) value multiplied by .45, or you could multiply the maximum (peak) value by .318.

Example - 120 x .45 = 54 volts
- 170 x .318 = 54 volts

Ed

[ August 30, 2004, 06:30 PM: Message edited by: Ed MacLaren ]

 

[ronaldrc]

Re: Half wave rectifier

Hi Ed good to hear from you.

I don't really uderstand what you said.

I know you don't mean the peak voltage would be 54 volts you mean the effective value of the voltage of 54 volts but the peak voltage would be 120 if we rectify 120 ac with a half wave rectifier, right?

And if so wouldn't that be equal to about 50 percent?

 

[roger]

Re: Half wave rectifier

Hello Ronald, I hope all is well. I'm not sure if this thread is what you were looking for, but this was when Andre was still a participant.

Roger

 

[ronaldrc]

Re: Half wave rectifier

Hi Roger I think that is the thread I guess I should have did my search with rectifired instaed of rectified.


I didn't take time to read the whole thread just now but I thought everyone came to the conclusion the half wave result would give more power output than 50 percent and I argued it couldn't.

But I went to my shop put a light bulb on a 120 full ac wave circuit it pulled something like .9 of a amp.

And then I rectified it with a diode or rectifier and it pull a little over half that which proved everyone right.

 

[charlie b]

Re: Half wave rectifier

Actually, there is an easy way to understand the situation, without getting too deeply into the math. All you need to do is focus on what the letters stand for, in the term, ?RMS.? The words are Root, and Mean, and Square, but you carry out the math in the opposite order. First you take every point in the cycle, and square it (?S?). Then you take an average, or mean value (?M?). Then you extract the square root (?R?).

If you start with a sine wave, you will see a curve above the zero axis followed by a curve below the zero axis. When you square each value, then all values become positive (i.e., a negative number times itself gives you a positive number).

The difference that comes into play with a half-wave rectifier is that the positive half-cycle passes through, and the negative half-cycle is cut off (i.e., held to a zero value). So when you square this thing, the curved part gets squared to the same value it had with the original sine wave, and the zero part gets squared to (of course) zero. You can easily guess that you will come up with an average value that is one half of the former average value. That is where you are getting confused. The 50% value that you are visualizing comes after the ?S? and ?M? parts of the RMS have been calculated, but you still have to do the ?R? part. What is the square root of one half? It is about 0.7. So the RMS value of the half wave rectified should be about 70%, and not 50%, of the RMS value of the full wave.

 

[ronaldrc]

Re: Half wave rectifier

Hi Charlie

I no you are right although I still don't understand it.I know you tried to explain it without math and I respect that.

The reading I got in my experiment was around 70 percent of the .9 amps.

I used a clamp on amp meter could the pulsing dc have anything to do with it?

Ronald

 

[charlie b]

Re: Half wave rectifier

The pulsing dc has everything to do with it. OK, simpler version (no math whatsoever): For a half-wave rectified signal, the RMS value is supposed to be 71% of the RMS value of the original sine wave.

 

[caosesvida]

Re: Half wave rectifier

thanks for the info charlie.

 

[steve66]

Re: Half wave rectifier

For a halfwave circuit, we need to be clear if we are talking about voltage or power. Since the voltage for a 1/2 wave is less (than a standard sine wave), and the current is less, the power will be less BY MORE. I forget the actual numbers.

Also, inductive or capacitive loads add a lot of complexity. Edit: Since the current is not in phase with the voltage, the current at each point in the current wave gets multiplied by the voltage at another point in the voltage wave. It would take calculus (or a method called convolution) to calculate the power.

Your clamp-on AC meter probably won't read the correct current for a half-wave circuit unless it is a true-RMS meter. The easiest way to make a meter is to assume it will be used with either DC or a sine wave. A true RMS meter has to perform the calculations Charlie talked about.

Steve

[ August 31, 2004, 09:08 AM: Message edited by: steve66 ]

 

[ronaldrc]

Re: Half wave rectifier

Steve and Charlie

Thanks for the input the reading I got with my clamp on amp meter swayed me to believe contrary to what logic tells me, that is if you take half of something away it leaves half. I know its not that cut and dry with electrical circuits because of the impedance but with straight resistance I don't see why not.

I will do the same experiment using a series ammeter and see what I come up with.The way I see it it should be a little less than half if nothing else because of the .7 volt drop across the diode.

Ronald

 

[jimwalker]

Re: Half wave rectifier

any capacitors across it will cause you to get a higher voltage reading and throw the numbers off as it will try to charge to the peak voltage.

 

[charlie b]

Re: Half wave rectifier

Originally posted by ronaldrc:. . . if you take half of something away it leaves half.

Now I think I see where your difficulty lies. Let me re-write this statement as follows, ?. . . if you take half of something away it leaves half of something else.? Would you accept this version of the statement? I would not.

Think of it this way: If you have a bowl of fruit salad, and you take away half of the bananas, how much fruit salad is left? Would you expect there to be exactly half the bowl left, or perhaps more than half? Similarly, the ?current waveform? and the ?RMS value? are not the same things. If you take away half of the one, there is no reason to expect that you will lose exactly half of the other.

 

[ronaldrc]

Re: Half wave rectifier

Charlie

Thats a good point Charlie but I'm assuming that the source that generated that positive side of the first half of the sine wave will generate an identical negative half.I assuming they have the same amount of bananas.

 

[steve66]

Re: Half wave rectifier

Lets forget about the inductors and capacitors for a minuite and assume the load is a resistor. Lets also start with voltage:

RMS or effective is basically the DC voltage that would produce the same amount of heating in a resistive load. You calculate it with the method Charlie talked about. The RMS voltage of a sine wave is Vmax/sqrt(2). If you take this sine wave, and half wave rectify it, the new RMS voltage will be Vmax/2. So the RMS voltage is only reduced by sqrt(2), and not cut in half.

Same discussion applies to current. Half wave rectify it and it gets reduced by sqrt(2).

What about the power? It is the product of current and voltage. So it is reduced by sqrt(2) * sqrt(2). So the power is cut in half!

An average value of voltage or current is something different. It is a simple average, and if you half wave rectify a sine wave, the average voltage or current will be cut in half.

Steve

 

[dereckbc]

Re: Half wave rectifier

Originally posted by steve66:
Lets forget about the inductors and capacitors for a minuite and assume the load is a resistor. What about the power? It is the product of current and voltage. So it is reduced by sqrt(2) * sqrt(2). So the power is cut in half!

Incorrect.Assuming the resistance of the load is the same the power will be reduced by more than 75%. Assume a load of 10 ohm's, and supply voltages of 120 volts, then 54 volts. You get 1440 watts and 292 watts respectively.

 

[charlie b]

Re: Half wave rectifier

I believe that Steve 66 was right. If you start with a sine wave that has an RMS value of 120 volts, and you cut off the negative half-cycles, the RMS value of the remaining signal is 84.85 volts (not 54). Power can be calculated by voltage squared divided by resistance. 84.85*2 / 10 = 720, and that is indeed half of 1440.

Ed: I had not noticed the difference between your value of 54 volts and my result of 71% (i.e., 71% of 120 is about 85 volts). Where did you get the 54 volts? Or specifically, where did you get the 0.45 factor that you used to calculate the RMS value of the half-wave rectified signal?

 

[dereckbc]

Re: Half wave rectifier

Charlie, or anyone else who wants to dive into the math go here:
http://myweb.tiscali.co.uk/robert.booth/uni/docs/Power%20Electronics%20Assignemt%201.pdf

Scroll down to half wave rectifiers.

Have wave rectifeid 120 VAC = 54 VDC.

 

[Ed MacLaren]

Re: Half wave rectifier

Where did you get the 54 volts? Or specifically, where did you get the 0.45 factor that you used to calculate the RMS value of the half-wave rectified signal?

From textbooks originally, and I have confirmed them by actual measurement in school lab projects.

The 0.45 is one way of expressing the "form factor" for a single-phase, half-wave rectifier circuit. One could also divide the RMS value by the reciprocal of 0.45 which gives - 120/2.22 = 54 volts


The FF for a single-phase, full-wave rectifier is 1.11 or 0.9, for example - 120/1.11 = 108 volts
120 x 0.9 = 108 volts


Three-phase, half-wave FF is 1.17, and
Three-phase, full-wave FF is 1.346.

For example, an un-filtered full-wave bridge rectifier supplied with 208 volt three phase will output 280 volts DC

 

[steve66]

Re: Half wave rectifier

The 54 volts is the AVERAGE value of the voltage (120 * sqrt(2)/pi). It is NOT the RMS value. You can't calculate the power using average voltages and currents.

The link Dereckbc provided shows 5.76W for the half wave, and double that 11.52W for the full wave.

Steve