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Chip Amps Amplifiers based on integrated circuits 

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7th May 2003, 06:15 PM  #11 
diyAudio Member
Join Date: Aug 2001
Location: Bath, UK

It works out as half through the maths although the correct way to calculate is to multiply the swing by 0.707 (the reciprocal of the square root of 2).
If the supplies are +/20v, your peak output swing will be 16.4v whereas the RMS will be 16.4 * 0.707 = 12.6v. This will result in an RMS current of 1.45A into 8R, giving 16.82W RMS output power. Nice one, David. 
20th September 2013, 08:34 PM  #12 
diyAudio Member
Join Date: Mar 2003
Location: St. Catherine, Jamaica

sooo for a +35 power supply i would get
8 ohms 353.6=31.4 (31.4*31.4)/8 = 123watts 123*0.707=87 watts rms correct??? for 4 ohms = 246 watts 174watts rms correct??? Can two OPA541 in parrallel handle this power without plowing up?
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21st September 2013, 05:43 PM  #13 
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Join Date: Jul 2004
Location: Scottish Borders

no.
no.
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22nd September 2013, 09:52 PM  #14 
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Join Date: Mar 2003
Location: St. Catherine, Jamaica

so then how do I calculate the power output?
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23rd September 2013, 01:27 AM  #15 
diyAudio Member
Join Date: Nov 2009
Location: Los Angeles

You take the Peak Voltage and multiply time .707 ( square root of 2 divided by 2). Square the result and divide by the load resistance. So your 31 Volt peak output is 21.9 Vrms. Squared is 480.35 and divided by 8 is 60.04. BTW it's hard to get accurate to 3 digits.
If you're measuring the Peak to Peak Voltage you will need to divide by 2 to get to the peak Voltage or else you'll get an answer 4 times larger than it should be. It gets a little more interesting as the power supply varies with the power output which is why power is specified with continuous sine wave drive. ALSO, the power varies when the line Voltage changes as well. If you measure power with 117 Volt line an the line goes up to 125, a ratio of 1.068 %, the power can go up 1.068 SQUARED or 1.14. This must be done with a scope to stay out of clipping. G² 
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