calculating output power of amp

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I have the Klispch ProMedia 5.1's and just hooked up four KG-2.5's to them for music. The main problem is that the speakers are 8-ohm, and the amp is 4-ohm. I know that hooking speakers up in this manner reduces the amps power rating by 1/2. I really don't want the amp to clip, since spare parts for the KG's are becoming hard to find. But, I don't have enough money for an amp for them right now, so this is the only way I can use them.

So, what formula do you use to calculate the output power of the amp? It is rated at 60 watts per channel into 4 ohms, so I am at 8-ohms, making it around 30 wpc. I am running the system with Windows/wave volumes full and the control pod on 68 out of 80. Need help...
"So, what formula do you use to calculate the output power of the amp?"

If you know the supply rail voltage, then you can estimate how far the output can swing, typically within 2-3V of the supply. So, for a +/- 35V supply, the output can swing around 32V either side of ground. For RMS power you need to consider the mean amplitude of a sine wave output, i.e. 32V/sqrt2 = around 22.6V. Now power = V^2/R, so into an 8 ohm load = around 64W. You can use the same process in reverse to calculate minimum rail voltages given the RMS power output for a known load.

Anyway, when you say you only have 30W available, you are probably right. If you have another set of 8 ohm speakers available, you could connect them in parallel and roughly double your output power.

Hope this helps you,

BTW, be careful when driving your amp from a sound card, it may be clipping itself.
Huge debate over at the car audio forum. Many sources
of engineering are saying there is no such thing
as RMS power.

Rane Professional says;

rms power - No such thing. A misnomer, or application of a wrong name. There is no such thing as "rms power." Average or apparent power is calculated using rms values but that does not equal "rms power;" it equals continuous sine wave power output into a resistive load.
Come colleges say;

Pavg = Vrms^/R
scroll down to page 8.

I see two factions on the internet, one stating
RMS power = V(RMS) / R and the other faction
saying RMS power doesn't exist, instead it's
called average power (using rms values).

Formulas found on the internet say;
Instanenous power = V (peak) ^ / R
Average Power = V (RMS) ^ / R
RMS power = doesn't exist

PEAK = 1.414
RMS = .707
AVERAGE = .637

To get average power they say take RMS voltage values.

So, what do I get if I use average voltage in my
power calculations ? in other words, this forumla

P(?) = V(avg)^ / R

Define the question mark.
RMS Power

RMS Watts is an imprecise term but one which has become well appreciated and much used (abused) by manufacturers and customers alike.

The are other, more correct terms but any maker of amps using them takes his life in his hands, vis:

Average Power - suggests "also ran" or "nothing special"

Sine Wave Power - implies that the amp will only reproduce sines

Peak Power - A much loved description attached to the microwatt music machines of some years ago to artificially boost what was effectively flea power ratings.

RMS power is now as enshrined in folk lore as brake horse power. It's widely understood (technically incorrect as it may be) but generally accepted as a reference point by the consumer when gauging the output power of an amp.

In any event, what does it matter as long as we all sing from the same hymnbook..??
The only Klipsch speaker that is 8R is the original Heresy. The Heresy II and all other Klipsch speakers will measure 4R when connected to an ohm meter.

Marketing people are allowed to change specs by a factor of 1:10 (or 10:1, depending upon what suits their needs). It is part of their job description.
Yeah, there really is no such thing as "RMS power". There is just "power", which is calculated in an AC circuit by P = (Vrms^2)/R. However, I still sometimes use the term RMS power to differentiate it from those other (mysterious) forms of power manufacturers have come up with. :D
AX tech editor
Joined 2002
Paid Member
output power

Of course there is something like "RMS power". RMS stands for "root mean square". A signal develops 1watt RMS in a load if the load is heated the same amount as if 1 watt thermal power would be applied. Don't look so unbelievingly, that's how they defined it, it's not my fault. Early (and some contemporairy) RMS detectors contained 2 close-coupled resistive elements, one heated by the unknown signal, the other by a DC signal. The trick is to adjust the DC so both elements are heated equally, then the unknown signal had the same RMS value as the DC signal. So instead of having to calculate the equivalent power of a varying signal with odd waveshapes like music you just read the DC on a DCVmeter. Simple.

The other thing you need to know is that a sine signal in a load develops the same heat as a DC signal with a value of 1/sqr2 of the sine peak value. So, a sine with 2V Pk is equivalent to 2/sqr2 VRMS, which is about 1.414vRMS.

Now back to the original question. Suppose the amp has 50V rails. Accounting for losses, let's say the output can swing up to 47volts. Thus, for max sine output, the output peak voltage is 47 volts which is equivalent to 47/sqr2 or 33.2V RMS.
Put that into 8 Ohms and you get sq(33.2)/8 or about 138Watts.

This has only a loose relation to music of course. The only thing you can say that the amp puts out 47V peak music signals. If you get an amp that has rails of say 75VDC, the peak output level would rise from 47 to say 72 volts, which is an increase of 50% in peak voltage which is 3.4dB, just audible. However, in the folder the second amp is spec'd as 324Watts RMS.....

Cheers, Jan Didden
trick question?

thylantyr wrote,
P(?) = V(avg)^ / R
Define the question mark.

Obviously in ac the average voltage is zero. This is the reason why RMS was invented. It is a way to rectify the voltage mathematically. Square the voltage values, average them and them square root them. In the special case of a sinewave and a real resistance the RMS voltage (or current) can be inserted into Ohm's law to calculate the average power. In the general signal case, integral calculus must be used to calculate average power.
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