AC and DC power of a full wave rectifier

[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

 

[davenn]

implies the obvious that Pac≠Pdc.

that's correct

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


Dave

 

[medwatt]

that's correct

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


Dave

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

 

[berkeman]

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 : $P_{ac} =V^{2}_{rms}R$.
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 $P_{ac} =V^{2}_{rms}R$ and $P_{dc} =V^{2}_{dc}R$. 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

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

 

[medwatt]

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

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 ?

 

[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 $\langle V \rangle$).

 

[milesyoung]

From basic circuit analysis, I already know that the average power delivered to a resistive load by a sinusoidal signal is : $P_{ac} = \frac{V^{2}_{rms}}{R}$.

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

In the text, the author says $V_{dc}$ is the voltage that appears across the load.

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.

 

[medwatt]

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.

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.

 

[The Electrician]

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

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.

 

[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 ?

 

[davenn]

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

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

 

[jim hardy]

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.

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:
https://www.physicsforums.com/showthread.php?t=767072
You might want to read the whole thread .

RMS stands for a simple arithmetic procedure.
In EE we usually apply it to a repetitive waveform like a sine wave.
Here's the concept:
Slow down your thinking like a slow-motion movie. Slow it all the way down to one single frame at a time, that is instant by instant. At any instant AC has some definite value.

Now imagine yourself doing these steps:
Write down the value of instantaneous voltage at each instant of time for one cycle(choose some increment - microsecond by microsecond? )
That'd be 16,666 values for a single cycle of 60 hz AC, 20,000 for 50 hz.

Now you're ready to do RMS on that series of values.
R stands for square Root
M stands for Mean , which is just the average
S stands for Square .

We work it backwards. S then M then R.

First you'd Square every one of your several thousand voltage readings giving you a series of Squares.
Next you'd average those squares, giving you the Mean of the Squares
Then you'd take the square Root of that mean giving you the square Root of the Mean of the Squares.

Calculus let's you do the calculation in one line not thousands of them.

That process has been worked out already for a lot of common waves .
http://ecee.colorado.edu/copec/book/slides/Ap1slide.pdf
see also wikipedia for Root Mean Square

It so happens that RMS gives the effective heating value of a waveshape, that is a DC current of RMS amps carries same power as the complex wave. That's why they thought it up.

Another neat feature is this - average value of a sine wave is zero because it is symmetrical above and below zero. That's why a DC meter connected to AC reads zero.
When you square each individual reading you make them all positive before taking the average.

Now it's your turn - work out some non sinewave RMS's.

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.

Other comments in the thread
https://www.physicsforums.com/showthread.php?t=767072
state it more elegantly.
But that's the down-in-the-mud-dirt-simple explanation. I hope it helps...

old jim

 

[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

 

[The Electrician]

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.

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.

For any sinusoidal wave $V_{ave}$ and $V_{dc}$ are synonymous, are they not ?

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.

 

[medwatt]

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

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.

 

[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 ?

 

[jim hardy]

The question is WHY ??

Okay, i understand your frustration...



An average responding DC voltmeter connected to unfiltered full wave rectified sinewave
will read 0.636 Vpeak.
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:

So the power delivered should be Pdc=V²dc/R. But what about Pac ?

Did the author say that ? He should have said

"Since for this particular DC waveform Vdc= 0.9Vrms, P_{not subscripted}= V²dc/0.81R. "

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 !

 

[medwatt]

Check this link : http://www.brighthubengineering.com...lectronics/96645-efficiency-of-ac-rectifiers/

 

[The Electrician]

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 ?

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.

 

[The Electrician]

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.

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.

 

[jim hardy]

Check this link : http://www.brighthubengineering.com...lectronics/96645-efficiency-of-ac-rectifiers/

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 we begin the study of any science, we are in a situation, respecting that science, similar to that of children; and the course by which we have to advance is precisely the same which Nature follows in the formation of their ideas. In a child, the idea is merely an effect produced by a sensation; and, in the same manner, in commencing the study of a physical science, we ought to form no idea but what is a necessary consequence, and immediate effect, of an experiment or observation.[4] Besides, he that enters upon the career of science, is in a less advantageous situation than a child who is acquiring his first ideas. To the child, Nature gives various means of rectifying any mistakes he may commit respecting the salutary or hurtful qualities of the objects which surround him. On every occasion his judgments are corrected by experience; want and pain are the necessary consequences arising from false judgment; gratification and pleasure are produced by judging aright. Under such masters, we cannot fail to become well informed; and we soon learn to reason justly, when want and pain are the necessary consequences of a contrary conduct.[5]

In the study and practice of the sciences it is quite different; the false judgments we form neither affect our existence nor our welfare; and we are not forced by any physical necessity to correct them. Imagination, on the contrary, which is ever wandering beyond the bounds of truth, joined to self-love and that self-confidence we are so apt to indulge, prompt us to draw conclusions which are not immediately derived from facts; so that we become in some measure interested in deceiving ourselves. Hence it is by no means to be wondered, that, in the science of physics in general, men have often made suppositions, instead of forming conclusions. These suppositions, handed down from one age to another, acquire additional weight from the authorities by which they are supported, till at last they are received, even by men of genius, as fundamental truths.

The only method of preventing such errors from taking place, and of correcting them when formed, is to restrain and simplify our reasoning as much as possible. This depends entirely upon ourselves, and the neglect of it is the only source of our mistakes. We must trust to nothing but facts: These are presented to us by Nature, and cannot deceive. We ought, in every instance, to submit our reasoning to the test of experiment,** and never to search for truth but by the natural road of experiment and observation. Thus mathematicians obtain the solution of a problem by the mere arrangement of data, and by reducing their reasoning to such simple steps, to conclusions so very obvious, as never to lose sight of the evidence which guides them.[6]

**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.

 

[milesyoung]

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.

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.

 

[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 !

 

[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 ?

...gratification and pleasure are produced by judging aright.

 

[sophiecentaur]

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 ?

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.

 

[jim hardy]

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.

Which initial definition, Sophie ?
I suppose textbook authors have the latitude to define their own terms.

From the link offered in post # 18:
http://www.brighthubengineering.com...lectronics/96645-efficiency-of-ac-rectifiers/

Rectifier efficiency is defined as the ratio of DC power to the applied input AC power.

Rectifier efficiency, η = DC output power/input AC power... <<< that's unambiguous enough

... he then presents then his halfwave case..
wherein he defines Pdc as :

The DC output power is given by: Pdc= I²dc x RL = (Im / π)² x RL <<< note average not RMS. Is that the initial definition you say is not correct? I'd agree.

.. for his fullwave case he says...
..

Idc= 2 I am /π = 2 Vm / π RL <<<< average current

Irms= I am / √2 = 0.707 I am <<<< RMS current

Rectifier Efficiency, η = Pdc / Pac

= I²dcRL/ I²rms (rf + RL) <<< Numerator: If one uses the square of average full wave rectified sine current he will not calculate the correct number for power..

= 0.812 / [1+ (rf / RL)] <<<< to his credit he squared pi and sqrt2...

= 81.2% if the diode resistance rf is negligible as compare to RL.



The efficiency of the center tap full-wave rectifier is twice the value of the half-wave rectifier.

I submit this author's "efficiency" claim is a figment of his imagination concocted to disguise his arithmetic mistake. If that's making its way into textbooks then shame on academia.

 

[sophiecentaur]

Which initial definition, Sophie ?
I suppose textbook authors have the latitude to define their own terms.

From the link offered in post # 18:
http://www.brighthubengineering.com...lectronics/96645-efficiency-of-ac-rectifiers/

... he then presents then his halfwave case..
wherein he defines Pdc as :

.. for his fullwave case he says...
..
I submit this author's "efficiency" claim is a figment of his imagination concocted to disguise his arithmetic mistake. If that's making its way into textbooks then shame on academia.

Absolutely but a term like 'efficiency' is pretty universally accepted as power out / power in and I don't think he's doing that.
What is he actually doing in that calculation to allow him to work out the Power in from the AC voltage source? With perfect diodes and no source resistance, there's nowhere for any power to get lost so, working backwards, surely his "AC Power in" must be calculated wrongly.
It may mean that, to get the equivalent Power into a load, you need different supply volts - but there is a nonlinearity in here so is it surprising? That may not actually involve anything other than the load resistance 'looking' different because it's seen through a rectifier.
If you got less power by feeding a load through an ideal step down transformer, you wouldn't say that constituted an efficiency less than unity so is there any more to it than that?

 

[medwatt]

...there's nowhere for any power to get lost...

That was one of the conundrums I realized and stopped reading because I'm not the sort of guy who assumes he understands what he's reading. If you can't understand what you're reading, what read ?

The webpage I cited is just another source that uses the same notations/terminology as the book I'm using. I actually believe the terminologies used are not restricted to just these few sources.

A convoluted book is hell to read. There's the book on Linear Algebra by Gilbert Strang (MIT professor) that can frustrate anyone unfamiliar with the subject. The guy simply cannot hold on to a though long enough. Check out his Linear Algebra course on MIT's youtube page. His lecturing style is very evident in his writing ! That's why for self study, a book with clear, lucid and unambiguous explanation is critical. First book that comes to mind is engineering electromagnetics by Ida.

 

[jim hardy]

What is he actually doing in that calculation to allow him to work out the Power in from the AC voltage source?

He's squaring the RMS of AC side amps and multiplying by R_L .

With perfect diodes and no source resistance, there's nowhere for any power to get lost so, working backwards, surely his "AC Power in" must be calculated wrongly.

Is that a Devil's Advocate statement ?
Power = (I_RMS)²R looks right to me.

It may mean that, to get the equivalent Power into a load, you need different supply volts

I think you hit it with that one. "different supply volts" measured how ?
My alleged brain is so prone to misunderstanding conversations that i always ask myself this question:
"What would make a rational person say that?"
When i asked myself that question about medwatt's textbook authors , my mind flashed back to high school electronics class where we boys measured bridge rectified output of various waveshapes by both oscilloscope and Simpson 260. Teacher's objective was to make us aware of the relationship between Peak, RMS and average for sines triangles and squares. And to alert us that test equipment will fool you if you ignore such facts.
Simpson 260 is an average responding meter.
When selected to DC it indicates average.
But when selected to AC, it is re-scaled to indicate RMS assuming sinewave, that is to indicate higher than the average it senses by ratio .707/.636 .
That's why when you measure a 90 volt B battery on AC scale it reads impossibly high.
Teacher wanted us boys keenly aware of that fact.


So - i assume the author is a rational man.
His mistake would be explained if as an undergraduate he did actually measure power into and out of an unfiltered bridge, but using an RMS indicating meter on AC side and an average indicating meter on DC side.
He would calculate different powers using the meter readings he observed.
Not questioning the reason for that discrepancy he could well suppose that AC and DC power are simply not equivalent by some unseen property of rectification.
He could give to that supposition the catch-all name "efficiency" ,
and hand it down through the ages, supporting it by the additional weight of his authority, ..

That fits the observation. And i think that's likely what has happened.

It may mean that, to get the equivalent Power into a load, you need different supply volts - but there is a nonlinearity in here so is it surprising?

Yes it is surprising because that's what RMS does, mathematically straightens out a nonlinear waveform into a straight horizontal line DC equivalent.

Surely some educators reading here have a mechanism to provide feedback to authors and editors of textbooks promoting this misconception, which medwatt seems to have encountered elsewhere...

If I've blundered please correct me.
Old guys are often too set in their ways...

http://www.rounds.com/blog/wp-content/uploads/2011/08/King-and-I-258x300.jpg
i cannot change

The horse is plenty dead now so i'll not flog it further.

old jim