Some dumb rookie questions about output power

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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.
 
Power in watts is V²/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.
 
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.

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.

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

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.

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.
 
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.
 

benb

Member
2010-04-24 1:52 am
...
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.
 
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.
 
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%).
 
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.
 
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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?
 
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 = √V1² + V2²... + Vn²/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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Mooly

Administrator
Paid Member
2007-09-15 8:14 am
You still don't seem clear on voltages in the way you describe things.

In post #1 you say "the output transistors can only see 18 volts Vce". That's only true under no signal conditions. Under full drive each transistor see the full 36 volts for an amplifier running on -/+18 volts.

An amplifier on an 18 volt rail can develop a voltage of (18/2)/ √2 which is 6.36 volts RMS.

An amplifier on -/+18 volts develops (36/2)/ √2 which is 12.7 volts giving the answer you get above. It can go from 0 to plus 18 volts and 0 to minus 18 volts.
 
I chimed in to say about the same, your way of describing the voltage you have available is confusing.
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?).

Do you mean a single 18V rail (what you get from one "transformer" , which I *think* means Power supply ¿¿¿??? ) or two said transformers, properly connected to have +/-18V rails?

Both Mooly's comments are spot-on ... we just don't know which one applies :(

That said, let's apply some "real World" corrections here.

Math posted above is perfect ... but available voltage is less than what measured without signal, for 3 basic reasons (there are even more :( , but let's consider the worst offenders)

1) transformer wiress have internal resistance.
As soon as you start pulling current, voltage drops.

Said resistive loss is much worse than what would be expected, because windings do not supply current continuously (as into a resistive load such as lamps or heaters) but in narrow pulses, because you are charging PSU capacitors to peak voltage 100 or 120 times a second.

So, say, getting 2 A in 1/2 second is "lossier" than getting 1A in 1 full second , although total energy is same in both cases.

2) said capacitors discharge (under load) between those narrow charging pulses, creating ripple voltage.

This is unusable , or, to be more precise, it's *UGLY* if it gets into your audio, so you should "stop" before reaching it so you must consider peak ripple a "lost" voltage too.

3) transistors need a minimum voltage drop across them to work.

Absolutely minimum drop in any junction is "one diode drop", around 0.7V.

But it also needs to meet some minimum "saturation voltage" , usually around 2 Volts.

MosFets technically don't have saturation voltage but (usually quite low) internal resistance, but they usually need much more voltege at the gates to switch them on (think 4V in cheap "switching" MOS used everywhere) so you lose important voltage one way or the other.

So **in practice** , a 18+18V supply will drop, under full load, to, say, +/-14V and you'll typically lose at least 2V on each side , so realistically you'll only have around +/-12V *peak* available.

Oh well :(
 
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Mooly

Administrator
Paid Member
2007-09-15 8:14 am
I'm confused as to why the fact that the rails are DC means no multiplication by .707. Its not the rails that I am measuring, but the sine wave output.

Work backwards :)

Say you want 12 volts RMS output. What voltage rail/s do you need ?

12 volts RMS is 17 volts peak. (12 * √2 which is the same as 12/.707). That gives the peak voltage. That means from "zero" or the middle of the sine to the top. The peak to peak voltage (from the top to the bottom) is that times two. 34 v pk/pk

So you would need an amplifier running on either a single rail of 34 volts or a spilt rail of -/+17 volts.

Does that help :)
 
Ok, let me clear up some confusion pertaining the the power supply I described.

I have a 12.6v-0-12.6v transformer. I have a very simple rectifier where I just use a diode on each leg of the secondary to get a positive and negative rail. The center tap is grounded. A largish decoupling cap is on each rail.

The small signal portions of my "amp" are built and running. the current DC rail voltage is a shade over 18V plus and minus.

Why would I need two transformers here? I guess in my crude power supply I have 2 half-wave rectifiers. Do proper amps really use 2 transformers?

Also, I now realize that the output transistors see the full 36, but I thought the load could only ever see half of that.
 

Mooly

Administrator
Paid Member
2007-09-15 8:14 am
A lot of amps do use two transformers to create what is really two mono amps (to reduce possibility of interaction)

Your 12-0-12 transformer could usefully work into a 4 diode full wave bridge rather than half wave on each rail. The efficiency would be much better and the ripple lower, and being double the frequency (100/120hz rather than 56/60hz) would be easier to filter.

The load does only see 18 volts in your example as it is connected to the "zero" point or ground. The transistors are not and there "reference" is the two rails. That is why each device can see the full combined supply.
 
ok, could someone using the rails and load I described maybe show the full calculation of the average power of a full power sine wave? That would help a lot because while I'm pretty sure its just me being lame, I feel like sometimes you guys are saying I'm doing it right, and sometimes you say I'm doing it wrong. A fully worked out example would probably help me clear it up!
 
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