# Amplifier RMS rating



## RongGe (Oct 25, 2010)

Amplifier RMS Rating.

Hello,
let me introduce myself, I am currently majoring in electrical engineer and studying amplifier feedback and design circuits. Car audio have been my hobby for the last 10 years. I rarely post in public forums because most technical response will be based on somebody's "experience" or lack of. As a result, there is a lot of bias answers to every question.

I am writing this article to clarify some possible misleading information about amplifier RMS ratings.

Recently I needed to purchase an amplifier and I looked through various brands and this RMS Rating companies rate their product. I had a budget and a targeted power I wanted from the amplifier. But what caught my attention was during my research i came across a thread where I read about some users doing actual test on amplifiers. For example: X company's amplifier rated at 1800 Watts RMS at 1 OHM. These users who purchased this amplifier did a "Clamp" test and the output numbers @ the rated load were 
"Clipping" tested at about 1.4k wRMS
"Non-Clipping" measured about 1k wRMS

Of course several conditions must be met to get the "true" value (assuming its not over or under rated.. etc)
such as
Constant power supply @ rated
Matching Amp and load Impedance.. etc etc

But the "non-clipping" signal measured too low and it got me thinking that even with the whole RMS vs Peak rating drama, the RMS numbers can still misleading. We have learn to go by the RMS rating and not peak. But nobody ever ask themselves, is the RMS rating in the correct operating region?

A square wave will have a Vrms = Vp. A sine wave will have Vrms = Vp (2)^1/2 = Vp * .707 (do you see any relationship from the numbers above? hint: 1.4k*.707 = 989.94 ; which is ~1k)
a clipped signal can be thought of as a output square wave. This is a result of full saturation on the amplifier.
non-clipping is the sine wave we want to see when setting our gains.

For a amplifier, we can input a square wave and get a output square wave.
We can also input a sine wave, put the amplifier into saturation (too much gain) we will have a output square wave. a Square wave has about 41% (sqrt(2)) more RMS than a sine wave.

Companies do not state whether an amplifiers maximum power output is rated in the linear or non-linear region. Both are legally correct because maximum wRMS can be in both linear and non-linear regions. 

In conclusion
-expect to used 70%*(rated RMS)
-Amplifiers are not underrated nor overrated, RMS power is overrated.

We ditched the PEAK or MAX watts because it was used to appeal to the ignorant and had no real world nor theoretical backing. Now I feel that "rated RMS" is used loosely to sell X amplifier product.
The next time you open a box and it has a sheet of paper that says this amplifier did XXXXwRMS, unless stated otherwise assume that was measured in the non-linear region. Until there is another rating reform where all amplifiers are rated in the linear region its buyer beware.


feel free to post any inputs, thoughts, comments.



RG.


**The amplifier that was tested above was obviously rated in the non-linear region.
** if an amplifier does have a birth sheet XXXXwRMS, you can actually use this number to set gains with a voltmeter. Vcar battery =Vbattery on birthsheet; VT = output terminal with Zload connected;
VT = Vac = (.707*XXXXwRMS*ZLoad)^1/2 +- error.
you can always set gains lower than this, but never higher.
to compensate for error just turn it a pinch back. Gain adjustment is best done with an oscilloscope.


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## Cruzer (Jul 16, 2010)

if we dont go by the rms, what do we go by? or are you simply saying, if we need 500 watts rms, we should look at least at a 1000 watt amp, that way we can get the true 500 non clipped to the speaker/sub?

what was the said amp being tested? i have not searched for it, but i heard someone tested a Hifonics amp, and it clipped well before the rms rating, especially at lower frequencies. well thats just an example of a cheap, crappy built amp.

your disclaimer about the amplifier having a birth sheet. an amp such as a rockford fosgate amp has a birth sheet, and it will prove it does rated or more. this is just proof that its a well built amp, and will do rated, and should have no clipping problems. (maybe im wrong, if so feel free to correct me. i have a RF amp and have had it power subs for years, if its clipping, it hasnt blown them yet. so if it is clipping, who cares, it isnt damaging or not allowing me to crank the volume to desired level)

i look forward to learning about this though for sure.


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## lycan (Dec 20, 2009)

there's no such thing as "RMS Power" (more accurately, it's made up by marketing guys who understand neither RMS ratings, nor Power ... and it has no practical value).

The RMS value of a voltage delivered to a load, multiplied by the RMS value of the current driven through the load, results in AVERAGE power ... not "RMS Power".

RMS values of voltage & current are so defined, in order to facilitate the calculation of AVERAGE power. The reason is simple : for thermal heating, it's the AVERAGE value of the power waveform that is most interesting & useful.

Power is a function of time. It's a waveform, with maximum (peak) values, minimum values, and an _average_ value. One can certainly calculate the "RMS Value" of the power waveform, but it has no practical value ... and it's NOT what you get when you multiply (RMS Voltage)*(RMS Current).


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## RongGe (Oct 25, 2010)

well we will still go by RMS, but I am just sheding light that it can be rated in two operating regions. Linear and non-linear.

If an subwoofer truely needs 500wrms to move then yes you should problaby be looking in the 1k output range.

but similar story with subwoofers, the number that people read is normally the maximum value. there is also a minimum value thats needed to get it moving. Most people just want to use the maximum value. 

Amplifiers have a frequency response. Different designs will yield different response. Poorly built amplifiers will fall apart. Poorly designed amplifiers will burn up. 

Assuming you have your rockford amplifier set to the highest possible gain before clipping correct? you can play a test tone in various frequencies and record the VAC across the terminals. Compare that to what the birth sheet says. I'll do this experiment in the weekend and return with my results.

Also understand output corresponds to your gain setting. You set your gain so you can match an internal Vref. Your amplifier can only produce maximum output if your Vin >= Vref max. This is why if your source produce too weak of a signal, even if the gains are set at maximum the amplifier cannot produce max power. On the contrary, if your source produce a signal too powerful Vin max > Vref max, then you start to saturate the input pushing the amplifier to the non-linear region. If Vin max >> Vref max, much greater then input sine waves output to squares.
We adjust the Vrca that the amplifier sees (Vin) by adjusting the gain circuit.

This is also an example why a good pre-out voltage is desired because noise is also amplified if the gain needs to be set high to match the H/U.

Regardless of this theoretical stuff, you will know when you measure it.


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## RongGe (Oct 25, 2010)

yes power is a function of time. let me get you thinking a little

Vavg of a sine wave with 0 dc is zero. What do you calculate for Average power of this sine wave?

We use Watt RMS because it contains both DC and AC power.


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## lycan (Dec 20, 2009)

RongGe said:


> yes power is a function of time. let me get you thinking a little
> 
> Vavg of a sine wave with 0 dc is zero. What do you calculate for Average power of this sine wave?
> 
> We use Watt RMS because it contains both DC and AC power.


no, that is NOT why marketing guys invented "RMS Power".

Let me get YOU thinking a bit :

Multiply a sinewave that has ZERO average value by another (in-phase, same frequency) sinewave with ZERO average value. Consider the first sinewave to be _voltage_, with peak value of *Vp* ... and the second sinewave to be _current_, with peak value of *Ip*. The product waveform will therefore be _power_.

What is the _average value_ of the resultant product waveform? Is it zero?


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## RongGe (Oct 25, 2010)

I think I see what you are trying to say.
I ended up confusing myself when I misread what you originally wrote. I just reread what you wrote and you are correct. I appreciate the clarification. Whether or not you approve of the term RMS power, it is used to describe average power. It would be weird to be buying a 1/4 watt resistor from radio shack vs 1/4 watt rms.


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## lycan (Dec 20, 2009)

For general reader interest :

A _voltage_ sinewave, with average value=*0* and peak=*Vp*, multiplied by a _current_ sinewave, with average value=*0* and peak=*Ip*, results in a _power_ waveform with the following characteristics :

max (peak) value of power waveform = *Vp*Ip*
min value of power waveform = *0*
average value of power waveform = *(Vp*Ip)/2* = *[Vp/sqrt(2)]*[Ip/sqrt(2)]* = *Vrms*Irms*
rms value of power waveform = who the hell cares

Note that even though the average value of the voltage is zero, and the average value of the current is zero, the average value of the product is NOT zero 

And the key point is that it's the AVERAGE power that is most interesting, because the AVERAGE power will determine resistive heating effects. What makes "rms voltage" and "rms current" interesting, is that they are useful in calculating AVERAGE power.

The concept of "rms power" ... even though it IS something you can calculate ... is physically meaningless.


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## Oliver (Jun 25, 2007)

Nice explanation 

*Thanks Lycan ! *



> Quote:
> Originally Posted by a$$hole View Post
> 
> 
> ...


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## RongGe (Oct 25, 2010)

RongGe said:


> yes power is a function of time. let me get you thinking a little
> 
> Vavg of a sine wave with 0 dc is zero. What do you calculate for Average power of this sine wave?
> 
> We use Watt RMS because it contains both DC and AC power.


I was rereading this and i realize where i confused myself at.

Average total Power = RMS of a DC and AC component.
if sine wave have 0 DC, the avg DC power is 0 however there is still a AC component which is not zero.


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## Cruzer (Jul 16, 2010)

good old lycan at his best


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## lycan (Dec 20, 2009)

lycan said:


> For general reader interest :
> 
> A _voltage_ sinewave, with average value=*0* and peak=*Vp*, multiplied by a _current_ sinewave, with average value=*0* and peak=*Ip*, results in a _power_ waveform with the following characteristics :
> 
> ...


Let's do the same thing for a squarewave, shall we ? 

A _voltage_ squarewave, with average value=*0* and peak=*Vp*, multiplied by a _current_ squarewave, with average value=*0* and peak=*Ip*, results in a _power_ waveform with the following characteristics :

max (peak) value of power waveform = *Vp*Ip*
min value of power waveform = *Vp*Ip*
average value of power waveform = *Vp*Ip* = *Vrms*Irms*
rms value of power waveform = who the hell cares

Note that even though the average value of the voltage is zero, and the average value of the current is zero, the average value of the product is NOT zero 

And the key point is that it's the AVERAGE power that is most interesting, because the AVERAGE power will determine resistive heating effects. What makes "rms voltage" and "rms current" interesting, is that they are useful in calculating AVERAGE power.

The concept of "rms power" ... even though it IS something you can calculate ... is physically meaningless.


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## aphexacid (Oct 24, 2009)

My head hertz...


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## JoeyFiasco (May 2, 2009)

what?


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## t3sn4f2 (Jan 3, 2007)

JoeyFiasco said:


> what?


Who's on first? :laugh:


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## AdamTaylor (Sep 7, 2008)

Here is a question, some of my amplifiers have come with stat sheets that say "to get optimal output at 4 ohms or 1 ohm or whatever, set amplifier to XXX v measured at the outputs by a multimeter using a 50hz test tone" if that info is not given, how would i determine that voltage? i just bought 4 of the ImageDynamics Q series amps. i got two 1200.1's and two 450.4's. the 1200's running 1 ohm and the 450's running 4 ohm.

http://www.imagedynamicsusa.com/pdf/iD Product Spec's/Manual Q-amp.pdf


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## squeak9798 (Apr 20, 2005)

AdamTaylor said:


> Here is a question, some of my amplifiers have come with stat sheets that say "to get optimal output at 4 ohms or 1 ohm or whatever, set amplifier to XXX v measured at the outputs by a multimeter using a 50hz test tone" if that info is not given, how would i determine that voltage? i just bought 4 of the ImageDynamics Q series amps. i got two 1200.1's and two 450.4's. the 1200's running 1 ohm and the 450's running 4 ohm.
> 
> http://www.imagedynamicsusa.com/pdf/iD Product Spec's/Manual Q-amp.pdf


Voltage = sqrt(power * resistance)

For the mono amps;

sqrt(1200*1) = 34.64V

For the 4-channel, assuming it was running at 4ohm per channel, then the voltage per channel would be;

sqrt(75*4) = 17.32V

That said, setting the gain with a DMM is not necessarily going to be the best method and the level of the test tone you use is going to affect results. A lot of people who set their gain w/ a 0db test tone you'll find complaining about low power output from the amp. This is because music is hardly ever at a level of 0db. You can alleviate this problem somewhat by using a test tone recorded at a lower level....-3db, -6db or -10db. The lower the level of the test tone you use, the higher average power you'll get from the amp but more clipping will possibly be introduced on dynamic peaks. 

Easiest thing to do is set the gain by ear. You can go back and check with a test tone afterwords to get an idea of where it's set.


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## AdamTaylor (Sep 7, 2008)

see, i usually set it by multimeter and then play around with it by ear, that way i have a for sure home base to go back to if it gets all wonky. i appreciate your speedy response. maybe i can put it to use when i get the amp rack built in 2 years :worried: (I'm lazy!)


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## Ludemandan (Jul 13, 2005)

lycan said:


> RMS values of voltage & current are so defined, in order to facilitate the calculation of AVERAGE power. The reason is simple : for thermal heating, it's the AVERAGE value of the power waveform that is most interesting & useful.


I would think resistive heating is a function of current, not power. Wires are sized for current, and voltage is not a factor. 



> Power is a function of time. It's a waveform, with maximum (peak) values, minimum values, and an _average_ value. One can certainly calculate the "RMS Value" of the power waveform, but it has no practical value ... and it's NOT what you get when you multiply (RMS Voltage)*(RMS Current).



Why are you saying power is a function of time? Time is not a factor in how you calculate power, it's an instantaneous measurement. When you add time, you get energy.


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## link2009 (Dec 16, 2010)

This is very informative stuff, thank you!


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## lycan (Dec 20, 2009)

Ludemandan said:


> I would think resistive heating is a function of current, not power. Wires are sized for current, and voltage is not a factor.


Resistive heating is a function of dissipated _power_, period. If you happen to know _current_ and _resistance_, you can certainly calculate _power_ ... but this certainly does NOT mean that voltage is "not a factor". If you have current and resistance, then you also have a voltage drop.

Knowing that current and resistance also gives you voltage, you can use a variety of calculations to determine power :

*Power = voltage*current*, or
*Power = (voltage)^2/resistance*, or
*Power = (current)^2*resistance*

Use whichever one you like. But picking number 3 (for example) does NOT mean that voltage is zero. If you have current flow, you must have non-zero resistance to dissipate _power_ as _heat_. And as soon as you have current flowing through non-zero resistance, you also have _voltage_.


> Why are you saying power is a function of time? Time is not a factor in how you calculate power, it's an instantaneous measurement. When you add time, you get energy.


I'm saying it because it's true. POWER is a time-varying waveform, anytime we have voltage & current that are time-varying waveforms. The particular aspect of the _time-varying power waveform_ that we are most often interested in ... for resistive heating effects ... is the AVERAGE value of the _time-varying power waveform_. And indeed, it's the AVERAGE value of the _time-varying power waveform_ that we get, when we multiply rms voltage times rms current.

Don't believe me? Draw a sinewave voltage, and a corresponding sinewave current. Draw them in-phase, if you like, to represent a purely resistive relationship. Now _multiply_ the two waveforms, point-by-point. See if the resulting _product_ ... which is the _power waveform_ ... varies in time.

If so ... how might we determine the AVERAGE value of the _time-varying power waveform_?


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