# AC and DC power of a full wave rectifier

1. Sep 16, 2014

### medwatt

Hello,
I am reading a book on Power Electronic and got confused on the actual interpretation of ac and dc power.
From basic circuit analysis, I already know that the average power delivered to a resistive load by a sinusoidal signal is : $P_{ac} = \frac{V^{2}_{rms}}{R}$.
When talking about a pure sine wave, the average voltage $V_{dc}=0$. So it made sense why we didn't use this value to calculate the power delivered to the load by the sine wave since it will erroneously tell us that no power is delivered to the load.

Now comes a full-wave rectified voltage. Here $P_{ac} =\frac{V^{2}_{rms}}{R}$ and $P_{dc} = \frac{V^{2}_{dc}}{R}$. But $V_{rms}=\frac{V_{m}}{\sqrt{2}}$ and $V_{dc}=\frac{2V_{m}}{\pi}$ which implies the obvious that $P_{ac}≠P_{dc}$.

So my question is, what is the actual power that is delivered to the load ? In the text, the author says $V_{dc}$ is the voltage that appears across the load. So the power delivered should be $P_{dc} =V^{2}_{dc}R$. But what about $P_{ac}$ ?

Thanks

Last edited: Sep 16, 2014
2. Sep 16, 2014

### davenn

that's correct

the equivalent DC power = the rms value of the AC power

Dave

3. Sep 16, 2014

### medwatt

I don't understand. Are you implying that $P_{ac}=P_{dc}$ ? If so why ? $V_{rms}≠V_{dc}$. Can you elaborate ?

4. Sep 16, 2014

### Staff: Mentor

I didn't read your whole post, but your starting equation for power is incorrect. P is not = to V^2 R.

http://en.wikipedia.org/wiki/Electric_power

5. Sep 16, 2014

### medwatt

I know. I'm sorry. I wanted to use the Latex feature to properly format the equations and in the process left out the division sign. I was copying and pasting the equation so the error duplicated. But that's not the point of the question! Can you tell me what the different powers mean and which one is delivered to the load ?

6. Sep 16, 2014

### milesyoung

The average power delivered to a resistor is given by $\langle P \rangle = \frac{V_\mathrm{RMS}^2}{R}$.

This:
$$\langle P \rangle = \frac{\langle V \rangle^2}{R}$$
is only true if $V$ is constant (since the RMS value of $V$ is then equal to $\langle V \rangle$).

7. Sep 16, 2014

### milesyoung

Just a reminder that this gives the average power delivered to the resistor regardless of what shape the voltage waveform has.

I would have a look around to see if the author mentions somewhere that $V_{dc}$ is the effective value (the RMS value) and not the average value of the voltage across the resistor. Otherwise, it must be an error.

8. Sep 16, 2014

### medwatt

Actually the author goes into calculating the rectifier efficiency as : $\frac{P_{dc}}{P_{ac}}$ which means that he considered $P_{dc}$ to be the actual average power delivered to the load. But everyone knows that $P_{ac}=\frac{V^{2}_{rms}}{R}$ means by definition/derivation the average power delivered by a sinusoidal wave. So my dilemma !

In fact he went on to calculate from the equations above that a half-wave rectifier has an efficieny of about 45% and a full wave rectifier has a 90% efficiency.

Why is $P_{dc}=\frac{2V^{2}_{m}}{\pi}R$ the power delivered to the load and not $P_{ac}=\frac{V^{2}_{m}}{\sqrt{2}}R$ ? Please help me with that part of my question. Thank you.

Last edited by a moderator: Sep 16, 2014
9. Sep 16, 2014

### The Electrician

If the output of the full wave rectifier is filtered with a capacitor, then the AC component of voltage will be greatly reduced. If your book says that Pdc is the power delivered to the load, then there would seem to be an assumption that there is no Pac.

If the output of the full wave rectifier is not filtered, then there is a large AC component of voltage and you should introduce another term to avoid confusing yourself.

In the initial case of a pure sine wave, you should say Ptotal = Pac + Pdc. In this case Pdc is zero and the expression holds true.

For the output of a full wave rectifier, it's still true that Ptotal = Pac + Pdc, where Ptotal is the power delivered by the full wave rectified, unfiltered, sine wave. Since both sides of this expression are divided by the load R, you can eliminate that and just show that:

$V_{rect} = \sqrt{V^{2}_{ac}+V^{2}_{dc}}$ where $V_{rect}$ is the RMS value of the unfiltered full wave rectified sine wave, $V_{ac}$ is the RMS value of the AC part, and $V_{dc}$ is the RMS value of the DC part.

The RMS value of the unfiltered full wave rectified sine is the same as the RMS value of the sine wave input to the rectifier (assuming an ideal rectifier with no voltage drops in the diodes), so that:

$V_{rect}=V_{rms}=\frac{V_{m}}{\sqrt{2}}$

Here are the calculations:

Thus the total power delivered to the load is the sum of the power delivered by the AC part and the power delivered by the DC part of the waveform.

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10. Sep 16, 2014

### medwatt

Thanks for the explanation. Where does $P_{dc}$ come from ? I already know that an ac source and a dc source in series will have the rectified voltage you mentioned. In my discussion, there's no dc source in the rectifier. Actually, in my original posts, I used the term $V_{dc}$ to be analogous to $V_{ave}$ of the the sinusoidal wave. For any sinusoidal wave $V_{ave}$ and $V_{dc}$ are synonymous, are they not ?

11. Sep 16, 2014

### davenn

you completely reversed what I said

I agreed with your earlier statement that Pac ≠ Pdc

and I said (yet you again reversed it) Pdc = the rms of Pac

12. Sep 16, 2014

### jim hardy

The answer to the confusion seems to me this simple truism we learned around ninth grade:

"The square of the average is not equal to the average of the squares."

You're(or perhaps your author is) trying to equate Vrms^2 and Vaverage^2.

So -- the answer to your question is : power is in proportion to Vrms^2 and not to Vaverage^2.

Elaborating, hope you don't mind:

DC means unidirectional not constant magnitude.
Only for a constant magnitude waveform like steady DC or perfect square wave does Vrms equal Vaverage.
Full wave rectified sinewave is a long way from constant magnitude.

RMS of a sinewave is Vpeak(or Vmax)/√2 , indeed.

Average of a sinewave is indeed zero, so long as it's averaged over a whole cycle.
Average of half a sinewave cycle, from one zero crossing to the next, is indeed 2Vpeak/∏.
If you full wave rectify you flip the negative half cycle up above zero, so you can now take average of a whole cycle and get the same average : 2Vpeak/∏.

So why don't (RMS) and (Average) equal one another? How'd one come out to be 1/√2 and the other 2/∏ ? 0.707 versus 0.636 ?

It's because of how we calculate RMS. In calculating RMS we square before we average.
I'll quote from another thread a few days back:
So -- the answer to your question is : power is in proportion to Vrms^2 not to Vaverage^2.

In taking Vrms we average the squares then unsquare that average. In taking Vaverage we only average.

So Vaverage^2 is NOT equal to Vrms^2 . You can't equate them and when multiplied by R they will give different results.

state it more elegantly.
But that's the down-in-the-mud-dirt-simple explanation. I hope it helps....

old jim

Last edited: Sep 16, 2014
13. Sep 16, 2014

### jim hardy

Now you've discovered an EE "gotcha".
DC voltmeters indicate the average voltage.
There exist true RMS meters that'll indicate RMS. But most of them - only when selected to an AC scale.
So if you want the true RMS value of a "DC" waveform, use the right meter and read its directions carefully.

Glad to see you're questioning these things. It'll help in your career to be aware of such nuances of measurement. Next you'll be investigating "Crest Factor".....

old jim

14. Sep 16, 2014

### The Electrician

Look at the first image of a full wave rectified sine wave in post #9. The DC comes from the operation of the rectifier. The negative going half waves of the input sine wave are flipped over, so the output of the rectifier only goes positive.

The average of a pure sine wave is zero, as you already know. But, after you full wave rectify it, the average is no longer zero. That's where Pdc comes from; that's why rectifiers are used, to make DC from AC.

If by the term "sinusoidal wave", you mean a pure sine wave with no DC offset, then yes, $V_{ave}$ and $V_{dc}$ are the same thing--they are zero.

In fact, $V_{ave}$ and $V_{dc}$ for any waveform would generally be considered to represent the same thing.

The output of a full wave rectifier is not sinusoidal. It is made of pieces of a sinusoidal waveform, but it's no longer a sinusoid after rectification. $V_{ave}$ and $V_{dc}$ of a full wave rectified waveform are still the same, but they're no longer zero. This DC part can deliver power to a load.

Look at the second image in post #9. That shows the full wave rectified sine wave with the DC part removed. That waveform has no DC part, and delivers no Pdc to a load. But, even though its average value is zero, it is not a sinusoidal waveform.

I showed how the output of a full wave rectifier fed with a pure sinusoid can be decomposed into an AC part and a DC part, and how the value of each part can be calculated.

15. Sep 16, 2014

### medwatt

Forgive me for saying this, I don't think you understand my question. I am very aware of the fact that the "square of averages and the average of squares are not equal". No one is saying otherwise. What is confusing me is when the efficiency of the the rectifier was being calculated the author (of the power electronics book) divided $P_{dc}$ by $P_{ac}$ without justifying it where $P_{dc}$ used the average voltage of the sine wave and $P_{ac}$ used the rms value of the sine wave. The question is WHY ?? The prompted me to ask in my first post what is the power delivered to the load ? Is the power delivered calculated from the RMS value of the voltage or the average (dc) value of the voltage. My reason for asking that is because I've always known that the power delivered uses the RMS value.

16. Sep 16, 2014

### medwatt

The Electrician, thanks for making things clear. This is certainly a nuanced concept that the author glossed over. One final question, when calculating the efficiency of a full wave rectified wave (assuming unfiltered) the formula used is :
$\eta=\frac{P_{dc}}{P_{ac}}=\frac{V^{2}_{ave}/R}{V^{2}_{rms}/R}=\frac{2V^{2}_{m}/\pi}{V^{2}_{m}/\sqrt{2}}≈0.9$
So $P_{dc}≈0.9P_{ac}$ . . . where has the extra power gone if we're assuming the only resistance is that of the load and the diodes used are idea ?

17. Sep 16, 2014

### jim hardy

An average responding DC voltmeter connected to unfiltered full wave rectified sinewave
The RMS value of that waveform is 0.707 Vpeak.
Calculating power from the average responding DC meter reading will give a result that's in error by (.636/.707)^2 .

So this statement from your original post, #1, is in error:
Did the author say that ? He shoulda said
Not subscripted because power is power irrespective of waveform; that's what RMS is for .

I don't have any idea what he meant by "efficiency"... an ideal rectifier would lose no energy so would have to be 100% efficient.

are we on same track now?

EDIT Nice job Electrician !

Last edited: Sep 16, 2014
18. Sep 16, 2014

### medwatt

19. Sep 16, 2014

### The Electrician

The Pac in that formula is the AC power before rectification. The Pdc in the formula is the DC power after rectification.

Before rectification, all the power is AC power, of course. After rectification, some of the power is AC power, but it's not $P_{ac.in}=(\frac{V_{m}}{\sqrt{2}})^2/R$, it's $P_{ac.out}=(\frac{V_{m}}{\pi}\sqrt{\frac{\pi^2-8}{2}})^2/R$, a much smaller number, but you wouldn't want to use that in a formula for rectifier efficiency.

What you care about is the ratio of DC output power to the AC power into the rectifier, and that number is expected to be less than 1.

As to where has the extra power gone--if the output of the rectifier is applied to a load resistor, without any filtering, then the extra power is the AC power out of the rectifier, and that power heats the resistor just as does the DC power. The DC power out is less than the AC power input to the rectifier by just the amount of the AC power out.

If there is filtering, the AC power in the rectifier output waveform is wasted heating the ESR of the filter capacitor, the diodes, the copper losses in the transformer, etc.

20. Sep 16, 2014

### The Electrician

When you say "$P_{dc}$ used the average voltage of the sine wave" the problem is that the $P_{dc}$ is not calculated from a sine wave; it's calculated from the output of the rectifier, and that waveform is no longer a sine wave. It's the waveform shown in the first image of post #9, and it has a DC component, unlike a pure sine wave.

$P_{ac}$ is the power delivered to the input of the rectifier and $P_{dc}$ is the DC component of power coming out of the rectifier.

21. Sep 16, 2014

### jim hardy

I'd say Messr's Sunita Sanguri and Lamar Stonecypher made a fundamental mistake trying to calculate DC power from average instead of RMS currents and voltages.

Then they blamed their mistake on "Efficiency".

Do they actually teach to calculate DC power from average current not RMS ?
That gives the wrong answer. As you have already observed, by asking ""where has the extra power gone ...."
Is that mistake actually showing up in bona-fide textbooks nowadays?

sheesh....

Lavoisier:
**When in a lossless system they calculated different input and output powers you'd think they'd have gone back and figured out why they arrived at that impossible result.
Answer is of course in their [DC side] arithmetic they used product of average values not average value of products.

Mediocrity. I blame it on the computers.

old jim

ps you're obviously a thinker. You might enjoy Lavoisier's 1789 essay at http://web.lemoyne.edu/giunta/ea/LAVPREFann.HTML
I've used it before...
Middle part gets a bit long, but don't miss last three paragraphs theyre especially relevant today.

Last edited: Sep 17, 2014
22. Sep 17, 2014

### milesyoung

It's very confusing, as you've discovered, that your author uses the word efficiency to denote the quantity $\frac{P_\mathrm{DC}}{P_\mathrm{AC}}$, where $P_\mathrm{DC}$ is the power supplied to the load by the DC-component, and $P_\mathrm{AC}$ is the total power supplied to the system. That's not standard terminology and, in my opinion, it's wrong to do so.

That definition of efficiency can only result in some very convoluted terminology when you have to consider different loss mechanisms in the rectifier.

23. Sep 17, 2014

### medwatt

Well, I thank you all for taking time to answer my questions. As you can see poor terminology can lead one astray ! One will end up looking for something that is really not there !

24. Sep 17, 2014

### jim hardy

Thanks Mr milesyoung , you said in a half inch of vertical space what took me a foot.

Thanks medwatt and bravo on you for sticking this one out ! Feels good when a mystery resolves, eh ?

25. Sep 17, 2014

### sophiecentaur

I think the basic idea behind this 'efficiency' figure may be flawed - or at least it doesn't accord with the strict definition of the term. If you have a voltage source (i.e. no source resistance) of a given AC value and you full wave rectify it, you will get an equivalent DC value for the voltage. However, comparing the Powers in each case is not the same as comparing Power in against Power out of a machine. No power is 'lost' because the load on the rectifier just takes less from the supply. The paradox only appears to exist because the initial definition is not strictly correct.