Output power

[lowell]

Something crossed my mind today driving home and figured I'd ask to confirm if I'm right here. Regarding output power and high voltage power supply rails. I get that an amps output voltage cannot exceed it's power supply rails.

Say you have a typical 4ohm load, and you're designing the PA for 100watts from the thing. Lets ignore specifics of CLEAN vs not clean wattage for now.

SqrROOT(100 x 4)= 20volts. So we want an output peak to peak voltage of 20volts? So then we might want 20/2=10volts... so +/-18v rails? (added some headroom)

I must be off here. I was thinking about some solid state bass amps I've worked on, and how some have as much as 70v or more rails. When I put the above math together for this it seems like an outrageous amount of voltage headroom compensation. Say we want 300watts into 4ohms.

SqrROOT(300 x 4)= ~35volts. So then our rails would only need be 35/2=~18volts... plus compensation so lets say +/-25volts rails. Even if we wanted 1000watts it still doesn't add up to 70volt or more rails.

What am I missing, or is this simply a matter of CLEAN headroom?

[Enzo]

20v across 4 ohms is 100 watts. That is 20v RMS or DC, not peak to peak. (20v squared over 4 ohms)So ignoring all the details, 20v RMS means 28v peak (20 x 1.414), so ignoring all the details, you would need rails of +28 and -28v.

70v rails? ignoring details, that means a peak signal in each direction of 70v, which is what you have when you have 49v RMS of sine wave. Into 4 ohms then gives 600 watts.

300 watts, 35v into 4 ohms? OK, 35 x 1.414 is 49v peak, so +/-50v rails.

What you are missing is some reason to use peak to peak numbers.

[lowell]

Ok so if we have 20v RMS (or 28v p-p) then doesn't that sine wave go up 14v, then down 14v? So why the +/-28v rails and not 18v?

[Alan0354]

I am always confuse between rms and peak to peak. 20V rms is 28.28V peak. That is 2X28.28=56.56V peak to peak. Am I getting this right? If so, for 100W, you really need 56V peak to peak signal. Am I right?

Also you need to have head room as the amp will never swing rail to rail. If it is a solid state amp, you are going to need at least 4V head room. So you need something like +32V and -32V supply. Exactly how much depend on the design.

OK, I feel so stupid that I need to go on web to look for this:

http://en.wikipedia.org/wiki/Peak-to...peak_amplitude

It is 56.56V peak to peak!!!! Finally for ones, I got to the bottom of this one!!!

[lowell]

Oh wow I feel dumb. I've been working on amps too many years to not know this. So RMS voltage is ~70% of PEAK voltage. NOT PEAK TO PEAK. Sheesh. Ok now these 70v supply rails are starting to make more sense!!

So RMS voltage gives us our wattage then:

20v RMS = 28.5 PEAK = 57 PEAK TO PEAK (28.5 x 2)

So then +/-32v rails it is.

What happened to the "thank you button"?

thank you!

[Alan0354]

Oh wow I feel dumb. I've been working on amps too many years to not know this. So RMS voltage is ~70% of PEAK voltage. NOT PEAK TO PEAK. Sheesh. Ok now these 70v supply rails are starting to make more sense!!

Hey, you have good company..............ME!!!

I finally looked it up, it always been a question and just speak quieter whenever this come up!!!

[lowell]

hahahaaaa!

[Enzo]

20v RMS is 28v PEAK, not Peak to Peak.

Peak is how far from zero a waveform travels. Doesn;t really matter positive or nagative, since either one results in the same power. A sine wave peaks at some V+, then peaks at the same voltage negative. Peak to peak is the entire width of a signal waveform on a scope, the distance from positive voltage peak to negative voltage peak. Peak to peak is nothing more than twice peak.

In 28v peak, the waveform goes up to 28v, 28v is its peak voltage - the highest point. It also drops to -28v on the other half cycle. SO from +28 to -28 is 56v peak to peak. If the waveform is symmetrical, as a sine wave is, then 28v peak and 56v p-p are the same thing

Alan, you are right about things like an extra few volts to cover junction drops and stuff, but that is what we were ignoring as "details" in the above posts. At this point they serve to cloud the issue at hand - which is simply how to relate output power and voltage into load.


Given a sine wave, RMS is 0.707 times peak, or peak is 1.414 times RMS.

[Enzo]

Oh, you figured it out while i was writing that...

[lowell]

Awesome... got it. I always thought RMS was derived from the entire sine wave (peak to peak). Whoops. Got it for sure now, thanks.

[Jazz P Bass]

A key point point that has been overlooked is that the amplifier must be designed, as a system, to deliver a certain output power.
Simply having a certain power supply voltage available does not imply (V x V / R) that the power is available.
My 2 cents.

[lowell]

Yes I understand that, but we were leaving out the REAL details for now, just simplifying the lesson. Bare bones. You are right though and I do get that a PA is really a PSU first, and a PA second. The power supply must be able to deliver the POWER.

[loudthud]

Since this is in the Theory section, a few comments are in order. If you had 100VDC connected to a 100 Ohm resistor, Ohms law would allow you to calculate the current at 1 Amp. The resistor would dissipate 100 Watts. If you connected 100V peak to peak AC across the same resistor you would have 1 Amp peak to peak so that part of Ohms law still works, but the resistor would only dissipate 12.5 Watts.

To fix this, we calculate how much heat that 100V peak to peak would generate and then assign a number to that 100V peak to peak AC that will allow us to accurately predict how much heat it will produce. Without going into the complex math, for a prefect sine wave you take the peak value (50V in this case) and divide by the square root of 2. So 100V peak to peak sine wave AC is called 35.355V RMS. Now 35.355 Volts RMS causes 0.35355 Amps RMS in the 100 Ohm resistor and the power is 12.5 Watts. 100 Volts RMS connected to a 100 Ohm resistor would cause 1 Amp RMS to flow and the resistor would dissipate 100 Watts.

The number changes if the wave is distorted. A "True RMS" meter takes this into account and still gives you a number equal to the amount of DC that would produce the same amount of heat. For a triangle wave, you divide the peak value by the square root of 3. A 100V peak to peak triangle wave is only 28.866 V RMS and dissipates 8.333 Watts in a 100 ohm resistor.

[Austin]

Since this is in the Theory section, a few comments are in order. If you had 100VDC connected to a 100 Ohm resistor, Ohms law would allow you to calculate the current at 1 Amp. The resistor would dissipate 100 Watts. If you connected 100V peak to peak AC across the same resistor you would have 1 Amp peak to peak so that part of Ohms law still works, but the resistor would only dissipate 12.5 Watts.

To fix this, we calculate how much heat that 100V peak to peak would generate and then assign a number to that 100V peak to peak AC that will allow us to accurately predict how much heat it will produce. Without going into the complex math, for a prefect sine wave you take the peak value (50V in this case) and divide by the square root of 2. So 100V peak to peak sine wave AC is called 35.355V RMS. Now 35.355 Volts RMS causes 0.35355 Amps RMS in the 100 Ohm resistor and the power is 12.5 Watts. 100 Volts RMS connected to a 100 Ohm resistor would cause 1 Amp RMS to flow and the resistor would dissipate 100 Watts.

The number changes if the wave is distorted. A "True RMS" meter takes this into account and still gives you a number equal to the amount of DC that would produce the same amount of heat. For a triangle wave, you divide the peak value by the square root of 3. A 100V peak to peak triangle wave is only 28.866 V RMS and dissipates 8.333 Watts in a 100 ohm resistor.

So does a cheap digital volt meter read rms or peak ac volts?

[frus]

It usually reads the average of the rectified voltage and then multiplies by some factor (I forget which) that's right for sine wave. So basically it assumes it's sine.
If you set it to DC, it just reads the average DC