RMS, P-P etc...

[lowell]

Need to confirm some math here...

If I have +/-32v power rails my peak to peak AC output max voltage must be less than 64vac?

(64vac/2).707 = 22.6Vrms?

If I have a jfet and 1 9v battery, ignoring SOME specifics, my max output could be 9v p-p?

I am aware the AC voltage will need to be a bit less than the supply voltage due to biasing and Rdrain...I think

[g1]

Correct. Assuming no step up transformers are involved at the outputs.

Edit: to be picky, if I see "Vac", I assume it is RMS. 64V p-p I know means ac, so to call it 64Vac p-p is a bit confusing.

[J M Fahey]

Yes, please use either AC(RMS or Average) or PP depending on what you are actually referring to.

As of available VPP in a real world amp, you'll lose 3 or 4V (DC or Peak) on either rail with regular bipolar transistors.

With MosFets and *clever* driving you might lose less than 1V .

Notice that if you use a bridged power output, you can have twice that across speaker terminals (which are both floating) .

Now 3 or 4V loss in >30V rails can be despised, but in a single rail 12V amp (car amp) that loss is HUGE.

[R.G.]

There is a little bit of a swamp here.

When I see an AC voltage listed purely as "Vac", I too assume it's a sine wave RMS. This is not necessarily true. "Volts AC" is very ambiguous. It only means that the thing has some measured voltage and alternates between positive and negative. It really needs a qualifier for both the type of measurement (RMS, peak, peak to peak, average, etc.) and the waveform.

An amplifier with +/-32V rails and no transformer to step it up or down can, as noted, only put out a waveform that's less than +/-32V peak because the output devices can't go to really zero volts across them when they drive any output current to speak of. It takes a couple of volts off the peak output to drive the output devices.

If you actually get +/-30Vpeak, 60V p-p, out of an amplifier, and you drive an 8 ohm speaker with a sine wave from it, you get a power of 30V pk*0.707 = 21.21Vrms into the 8 ohms or 56Wrms. If you drive the amp with a square wave things change. The RMS value of a 30V p-p square wave is 30Vrms, which comes out of the math of figuring the root of the mean square. The power out from a 30Vrms square wave into 8 ohms is 112.5W. This is the origin of the "peak watts" power race that started back in the ... um, 50s?

"Wrms" is a misnomer. Watts implies heating, and the RMS math was developed at least in part to describe the heating power of different AC voltages compared to the heating power of a DC voltage. A square wave with 30V peaks obviously has a heating power the same as a 30V dc supply because the voltage across the load is always 30V, it just changes polarity. Not so with a 30V peak sine wave, whose voltage is always changing, and almost never the full 30V peak. IIRC, "watts rms" was dreamed up by the FTC to rein in some of the spurious peak power claims, and it has a definition of watts with a sine wave signal. If I remember right...

Triangle waveforms have yet a different RMS. And then there is music.

Music has typically a "crest factor" of 20db or so. That is, the peak signal voltages are about 10x the average signal level. So amplifier makers started off talking about "music power", which is ill-defined, but often referred to the peak power that would result from an amplifier being turned up until the average music level was clipping, not the level where the peaks would be undistorted. So the "music power" was often advertised as about 10X the "Wrms" level that the FTC came up with to cut down on the abuse.

[Chuck H]

The RMS value of a 30V p-p square wave is 30Vrms, which comes out of the math of figuring the root of the mean square. The power out from a 30Vrms square wave into 8 ohms is 112.5W.

Ho! Hold on!.. So when I look at my scope and see, say, a 30V sine wave it's 56 watts (ok, I knew that part), but as the wave form flattens, still at 30V p-p, the closer it gets to a square wave, the closer it gets to putting out 112 watts? (this I didn't know, if it's true). Because if that's the case there is a lot to consider about clipping power tubes and dissipation limits relative to common practices.

[J M Fahey]

Yes, it's exactly that.
With same V peak, a squarewave has *twice* the RMS power of a sinewave, going by the classic definition of RMS which applies to "heating power" , that's why squarewave overdriven amps burn speaker voice coils.

Only (slightly) redeeming factors are that:
a) higher average current demands from the power supply make it drop even more, so the 56W RMS (clean) amp will probably put out no more than, say, 90 or 100W RMS squarewave instead of expected 118W
b) the copper in the voice coil will heat up a lot and rise DC resistance considerably, pulling less power from the amplifier, that's called "Power Compression".
That is a widely known phenomenon among speaker makers but not mentioned aloud because it "sounds bad" and hurts sales, but consider typical values at around 1 or 2 dB, with 3dB not unheard of.

Otherwise speakers would blow all day long, all over the place.

almost forgot:
c) speaker impedance is way above nominal one at most frequencies, except a narrow band usually around 250 to 400 Hz
Since the impedance rise is reactive, not resistive, it does not dissipate watts as heat, simply stops them getting to the VC ..

[Dave H]

The RMS value of a 30V p-p square wave is 30Vrms

That's +/-30V peak isn't it? I'd call that 30V peak or 60V p-p because that's the voltage I see between the positive and negative peaks on a scope.

[Chuck H]

That's +/-30V peak isn't it? I'd call that 30V peak or 60V p-p because that's the voltage I see between the positive and negative peaks on a scope.

See, now, that makes sense to me. With a theoretical 0 rise time, equal duty cycle and undistorted peak clip (which probably wouldn't sound very good ) a 30V p-p voltage is present 50% of the time at either peak. So isn't that 15Vrms since 15V is present 100% of the time?

[R.G.]

That's +/-30V peak isn't it? I'd call that 30V peak or 60V p-p because that's the voltage I see between the positive and negative peaks on a scope.

Doh! Yes, you're right. It's 60V peak to peak, 30V peak. I can only plead temporary insanity.

I fixed it (I think). Changed +/-30V peak to peak to "+/-30V peak, 60V peak to peak".

[loudthud]

See, now, that makes sense to me. With a theoretical 0 rise time, equal duty cycle and undistorted peak clip (which probably wouldn't sound very good ) a 30V p-p voltage is present 50% of the time at either peak. So isn't that 15Vrms since 15V is present 100% of the time?

Yes, a 30V p-p squarewave would be 15VRMS because the load sees 15V all the time.

Edit: the above was corrected. I got confused if we were talking about 30V p-p or 60Vp-p.

[Dave H]

I fixed it (I think). Changed +/-30V peak to peak to "+/-30V peak, 60V peak to peak".

But I'm still having trouble with the paragraph below

The RMS value of a 30V p-p square wave is 30Vrms, which comes out of the math of figuring the root of the mean square. The power out from a 30Vrms square wave into 8 ohms is 112.5W.

Loudthud seems to be saying the same but I see a 30V p-p square wave as 15V rms which makes the power 28W. Am I looking at it incorrectly or is everything bigger in Texas?

[R.G.]

But I'm still having trouble with the paragraph below

The RMS value of a 30V p-p square wave is 30Vrms, which comes out of the math of figuring the root of the mean square. The power out from a 30Vrms square wave into 8 ohms is 112.5W.

Loudthud seems to be saying the same but I see a 30V p-p square wave as 15V rms which makes the power 28W. Am I looking at it incorrectly or is everything bigger in Texas?

Well, everything SOUNDS bigger in Texas, all right.

There are two ways to look at this. One is the intuitive way, one is the math way. By the way, I mis-described that. I meant a 30V *peak* square wave, 60V p-p. I corrected that, I thought.

Intuitively, a square wave that peaks at +30v and -30V, alternating between them, puts 30V across a load resistor all the time, jumping between +30V and -30V, so it has the same heating power as a DC 30V source that never changes.

The math approach can also be simplified. "RMS" is the root of the mean square. To calculate it, you take many small slices, square the voltage at each incremental slice, add up all these slices and divide by the number of slices to get the mean (or average) square value; then take the square root of that. This is a bit complex with waveforms that aren't square, but you can see by inspection that the square of +30V and -30V is always the same answer (900 volts squared) no matter how many you add up then divide by the number of addends. When you then take the square root of that mean/average square, you get 30 volts.

This is good, because Mother Nature insists that the answers come out the same...

Doing integral over a cycle with a sine wave is almost as simple after you've been extruded through integrals, but it gets complicated with any signal that's not simple and easily defined for integrating.

[loudthud]

I corrected my post.

[Gingertube]

One of my buttons got pressed.

Watts are Watts - a formally defined Standard International Scientific Unit for Power, unlike a lot of other Standard International (SI) Units even Americans use Watts.

For a Laser Beam, Watts it is calulated as Joules/Second, the rate of delivery of ENERGY.
For Electrical Equipment, Power is still the rate of delivery of ENERGY, it's units are WATTS, and it is calculated from the vector product of the RMS value of the voltage and the RMS value of the current (sometimes called TRUE Power).
Apparent Power is calculate by the scalar product of the RMS value of the voltage and the RMS value of the Current and it's units are VA (VoltAmps) not Watts.

Watts RMS has no meaning. Watts Peak has no meaning, Watts Peak Music Power has no meaning.
Being undefined their units are also undefined but whatever they are they are NOT Watts.

Cheers,
Ian

[Dave H]

Watts Peak Music Power has no meaning.

Of course it has! It means I can make more profit by marketing my puny 10W amp to the man in the street as 100W and he won't know the difference because "Watts are Watts"