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Old 23rd April 2013, 03:19 AM   #1
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Default Some dumb rookie questions about output power

Ok, the first question is a really basic one that I'm 99% I know the answer to, but none the less.....

Say I have 18v rails. The power transistors at most only ever see a Vce of 18V right, not 36? (In a simplified sense. I realize they can't swing quite to the full rail voltage)

Now for some power output computations. Assuming a full power square wave output into 8 ohms, and the aforementioned 18 volt rails, would I be at that time putting out:

(18*18)/8 = 40.5

40.5/2 = 20.25 watts?

also, does computing r.m.s. with a square wave have any meaning? I would assume not.

With that said, r.m.s. is pretty much only correct with a sine wave right?

In short, is the stated r.m.s. wattage of an amp a sort of fudge since music is not sine waves? Or is there something more.
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Old 23rd April 2013, 04:19 AM   #2
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Power in watts is V/R, so your first formula is correct.
Why the division by 2? Are you making an adjustment for duty cycle of the square wave? I believe the formula for square wave RMS (which is valid) is Vpeak*√duty cycle.
Amplifier wattage is normally tested not with music but at a single (e.g. 1kHz) frequency. It is a sine wave.
The fudging is usually a little more, uh, creative.
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Old 23rd April 2013, 04:30 AM   #3
JMFahey is offline JMFahey  Argentina
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Quote:
Originally Posted by Dan Moos View Post
Ok, the first question is a really basic one that I'm 99% I know the answer to, but none the less.....

Say I have 18v rails. The power transistors at most only ever see a Vce of 18V right, not 36? (In a simplified sense. I realize they can't swing quite to the full rail voltage)
No, both transistors are in series and connected from rail to rail (as a pair).
On either signal peak (+ or -, it's the same) one transistor is saturating, meaning it drops 0 volts, and the other is connected end to end, so it gets the full 36 volts.
Same for the opposite peak.
For simplicity I'm considering perfect transistors with 0V drop when saturated.

Quote:
Now for some power output computations. Assuming a full power square wave output into 8 ohms, and the aforementioned 18 volt rails, would I be at that time putting out:

(18*18)/8 = 40.5

40.5/2 = 20.25 watts?
Yes, that's right.

Quote:
also, does computing r.m.s. with a square wave have any meaning? I would assume not.
Yes it does, as shown above.

Quote:
With that said, r.m.s. is pretty much only correct with a sine wave right?
Iffy question.
Really there's not such a thing as RMS power, but RMS voltage.
In loose talking both are used, but if you try to dig deeper problems start to surface, so there's not a good answer because the question already has issues.

Quote:
In short, is the stated r.m.s. wattage of an amp a sort of fudge since music is not sine waves? Or is there something more.
No, the "RMS" rating, more precisely the "average continuous power calculated using the RMS voltage" is an important tool, because it is simple and can be repeated at will and with precision, while "playing music", although conceptually the best, (no doubt about that), is dificult to repeat and also can lead to endless discussion.
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Old 23rd April 2013, 09:32 AM   #4
DF96 is offline DF96  England
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Assuming +18V and -18V rails, then a full amplitude square wave will have power of 18^2/8 = 40.5W. No division by two. A full amplitude sine wave would have a peak power of 40.5W too, but an average power of 20.25W. For a square wave the power stays constant throughout the cycle. For a sine wave it varies up and down at twice the frequency, with an average of half the peak. That is where the RMS value comes from.

For a square wave, RMS voltage = peak voltage.
For a sine wave, RMS voltage = 0.707*peak voltage.
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Old 23rd April 2013, 03:15 PM   #5
benb is offline benb  United States
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Quote:
Originally Posted by JMFahey View Post
...
Really there's not such a thing as RMS power, but RMS voltage.
In loose talking both are used, ...
I think it's helpful to describe the origin of this - RMS power is wrong and has no meaning or use in electrical engineering or electronics, but it was the phrase used for average power in US Government regulations from 1974 that attempted to enforce truth in advertising of the power ratings of audio amplifiers:
Audio power - Wikipedia, the free encyclopedia
Due to this codified mistake it's often or usually CALLED "RMS power," but what is actually measured or calculated for this value is average power.
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Old 24th April 2013, 12:50 AM   #6
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Let me clarify my rational behind my dividing by 2. With the square wave, I divided by 2 to account for the duty cycle. Is this not the correct thing to do to calculate average power?

Let me state things as I had thought them to be, so you guys can set me straight.

I figured peak power was as simple as peak voltage squared / the load resistance.

I figured a square wave is basically a wave that is one continuous peak. I suppose I was considering the full cycle (0v through both peaks and back to 0v) when I divided by 2. I suppose I would have been more correct to state this was average. Is this correct?

I also figured that multiplying a sine waves peak by .707 gives a sort of average that I have understood to be what r.m.s is. Again I divided by 2 to account for the full cycle. Is this correct?

To simplifiy my question, would I derive the average r.m.s. power capability of an amp by feeding it a sine wave, measuring the peak, multiplying that peak by .707, squaring the result, dividing by the load resistance, and dividing by 2 to account for duty cycle?

So if I want to say I have "a 100 watt" amp, is the above how I get the number, or some other way.
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Old 24th April 2013, 01:38 AM   #7
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I was a little confused in my reply. In this instance there is no duty cycle; that's more of a digital thing where the square wave goes from 0V to Vp and back. With an alternating square wave, DF96 is correct - the signal goes from +Vp to -Vp, and there is no duty cycle (or, probably more correctly, it is 100%).
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Old 24th April 2013, 04:59 AM   #8
JMFahey is offline JMFahey  Argentina
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Yes, it is 100% because "RMS" is a heat related concept and as far as heat is concerned, +V and -V are exactly the same.

How come?

Remember :
+1 by +1=+1
-1 by -1=+1

So positive half heats load as much as negative half, so same heat exists in both halves so heat is always present so heat is "continuous", no matter polarity changes 25/50/60/400 times a second.

It even sounds silly or like a charade ... once it's understood, but in a way it's counterintuitive because we are used to "+" something and "-" something not being the same.

And where/why did the RMS thingie become so important?
Because original power generation and distribution (Edison) was 110V DC , and lamps were designed for that.

Then came AC distribution (Tesla) and people was worried trying lamps to provide exact same light as before.
How would the new AC be specified, and what value should be chosen for that end?
Problem is , AC voltage is constantly changing, but people says "keep the theory but give me *one* number I can use to compare".
An AC waveform can be defined by its peak ... but 110V Peak was not same as 110 DC or to be more precise, it *would* be if it were squarewave ... which it was not.
So 110V peak voltage was "weak".
On the contary, 110V "average" was "too strong".
Then a mathematical operation was performed and it was shown that RMS voltage is the key.
RMS is higher than average but lower than peak.
So they could guarantee that 110V "RMS" would heat the lamp filament exactly the same as 110V DC .... providing exact same light .... which is what the end user was interested in.

Last edited by JMFahey; 24th April 2013 at 05:02 AM.
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Old 24th April 2013, 05:28 AM   #9
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@JMFahey:

That's a pretty neat back story to the root of the 110V standard!

@Everyone:

Ok, I think I'm understanding a little better. Let me put thus another way, so I can see if I'm all straightened out.

Lets again say I have 18 volt rails. (This is roughly the DC I get from a common RadioShack transformer I have a few of)

Into 8 ohms, and presuming a fictional amp that can go from 0 to the full 18V (maybe this isn't fictional?).

Would the r.m.s. power of such an amp when fully driven be:

(18*.707)^2 = 162

162 / 8 = 20.24 watts r.ms. average, yes?
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Old 24th April 2013, 06:33 AM   #10
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Pretty close.
If you have 18V DC rails, there's no need to use the 0.707 multiplier to calculate an RMS voltage; 18V DC is 18V DC.
RMS is mathematically the root mean square, or Vrms = √V1 + V2... + Vn/n, where V1, V2, and Vn are the values to be averaged (i.e. the mean) and n is the number of values.
0.707 and its reciprocal 1.414 show the numerical relationship between RMS and peak of a sine wave.
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Last edited by sofaspud; 24th April 2013 at 06:36 AM.
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