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25th September 2009, 02:43 PM  #1 
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Join Date: Mar 2008

Amplifier theoratical max rated output power based on supply voltage
Been wondering about this topic, found a few (rare) mentions that have different opinions, wondering if anyone could clear this up:
Lets say 10V supply into 10 ohms, to simplify things, amp is ideal, so max output voltage = supply voltage, Instinctively, P=V^2/R so max power = 10W. But, 10V is +5V, so my output voltage is +5V max, wouldn't that mean power = 2.5W? Help me clear this misconception pls. Last edited by wwenze; 25th September 2009 at 02:46 PM. 
25th September 2009, 03:12 PM  #2  
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Join Date: Aug 2005
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Quote:
The formula is: P[RMS] = Ec^2/(8*Rl) where: P is the RMS power in watts Ec is the supply voltage. In your case, is either 10V or +/5V=10V (same thing) Rl=Load resistance in ohms. In your case, the RMS power is 10^2/(8 * 10) = 1.25W 

25th September 2009, 03:15 PM  #3 
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a single polarity 10Vdc is equivalent to a dual polarity +5Vdc supply.
An amplifier with an AC output from a theoretically ideal output stage can develop an output voltage of 5Vpk from the 5Vdc supply. +5Vpk is equivalent to 3.5Vac. The maximum power into 10r0 is V^2 / Rload = 3.5^2 /10 = 1.25W or P = Vpk^2 / 2 / Rload = Ipk^2 / 2 * Rload = Vac / Rload = Iac * Rload = Iac * Vac The real maximum output from an amplifier operating on +5Vdc driving a 10r load will be Pmax ~ 4^2 / 2 / 10 ~= 800mW If the load is a reactive speaker then expect the maximum current that could be demanded ~= three times the equivalent resistor, i.e. a 10ohm speaker driven with 4Vpk may demand <=1.2Apk.
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25th September 2009, 03:21 PM  #4 
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If you are lazy and slightly stupid like myself, you can download a program called RFSim99 that has a handy calculator.
Tools > Calculator > Signal Level 
25th September 2009, 03:30 PM  #5 
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Hi, a useful bit of info here: http://www.bcae1.com/voltages.htm
Bottom line is that you have to convert your theoretical 5V peak sine wave into the equivalent DC heating value called the rms value which would be around 3.5V. You can then calculate what power would be dissipated into the particular load you were using by using the usual formula V (squared) divided by R (Load resistance). Assuming this was an 8 Ohm speaker, power dissipated with a continuous pure sine wave would be 3.5 x 3.5 = 12.25/8 = 1.53 Watts. Les P.S. Sorry if I have repeated some of the above posts which were not there when I started mine LOL! Last edited by Hi_Q; 25th September 2009 at 03:52 PM. Reason: recognising dual info postings at same time 
25th September 2009, 03:32 PM  #6 
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if you can remember that
I = V / R  Ohm's law and that P = I * V then all the others follow exactly from those two equations. No reason to be either lazy nor stupid. But frequent repetition may well be required to commit this to permanent memory, for it to stick. Just like a PC: RAM is volatile and HDD can be almost permanent, if we know where the index to the library/files is kept.
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25th September 2009, 03:34 PM  #7  
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Quote:
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25th September 2009, 03:48 PM  #8 
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Agreed Andrew, was not concentrating or rather too much going on around me at the time. I have changed the values to agree more with the original post.
Best regards Les 
25th September 2009, 11:25 PM  #9 
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Join Date: Mar 2008

Thanks, I'd have to bookmark this.
I understand much of the calculations, especially the electrical formalae, I'm just confused as to what other factors I have to include. So what I understand is: 10V > +5V, so equivalent heating power divide by 4 Then 5V peak sine wave > divide by root 2 to get RMS voltage, equivalent power divide by 2 So P[RMS] = Ec^2/(8*Rl) > the 1/8 in the formula comes from above I notice that everybody (and program) is considering RMS power with regard to sine waves, is that standard practice in this particular industry? (Because an amp can output square waves, like classD, or digital buffer, would P[RMS]=Ec^2/(4*Rl) instead?) 
25th September 2009, 11:32 PM  #10 
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Join Date: Apr 2008
Location: Carlisle, England

The problems start when you consider MOSFET output stages.
On my amp it takes 7 volts bias voltage so I have to subtract 3.5 volts from the peak voltage at least to get near the correct result. Just to complicate things further the harder the MOSFET is driven the more volts it needs on the gate. BJT's arent as bad they just need a volt or two bias. Or at least that is my understanding.......
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