OPA541 power rating - diyAudio
 OPA541 power rating
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 Chip Amps Amplifiers based on integrated circuits

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 5th May 2003, 10:51 AM #1 diyAudio Member     Join Date: Feb 2003 Location: Bristol OPA541 power rating what is the approx. power output if the OPA541 at different voltages? the datasheet has no watt rating. +/-40v and +/-12v are the voltages im interested in. PS: sorry to post new threads, i can get any answers if i put all my questions in one thread. __________________ If it aint broke, don't fix it. If it is broke, fix it. If you can't fix it, take it apart and see how it "worked".
 5th May 2003, 04:23 PM #2 diyAudio Member     Join Date: Feb 2003 Location: Bristol at max volume, what is the approx. heat output in watts? __________________ If it aint broke, don't fix it. If it is broke, fix it. If you can't fix it, take it apart and see how it "worked".
 6th May 2003, 08:21 AM #3 diyAudio Member   Join Date: Jun 2002 Location: Portugal hi matttcattt, I saw in some other thread a calculation of the power from a chip. From what I remember it is P=V*V/R, being V the voltage swing and R the resistance. If you have 25 V supply, and if the chip can swing up to -4V from the supply, that is 21 V, and if R=8, then the power will be 21*21/8=55W. That´s 2.6 amps, so within the OPA541 ratings. I have a GC based on the OPA541. I mounted the two channels in a L shaped aluminium bar 20 cm long and 5 cm width (20x5x5 cm). I never play it at the highest volumes but even at a loud volume the aluminium bar does not even get warm. Miguel
 6th May 2003, 07:44 PM #4 diyAudio Member     Join Date: Feb 2003 Location: Bristol i heard somewhere that an amp outputs 4 times as many watts of heat as power output in watts. is this correct? __________________ If it aint broke, don't fix it. If it is broke, fix it. If you can't fix it, take it apart and see how it "worked".
 6th May 2003, 08:31 PM #5 diyAudio Member   Join Date: Aug 2001 Location: Bath, UK The chip will dissipate the difference between the supply rails and the output voltage, multiplied by the ouput current. If the supply is +/-18v and the output swing is +/-12v then there's 6v across the device, since only one of the push-pull transistors is on. A load of 8R means the current is 1.5A so the output power is 18W and the dissipation is 9W. Nice one, David.
 6th May 2003, 09:39 PM #6 diyAudio Member     Join Date: Feb 2003 Location: Bristol a, can u explain how i work out the output swing b, can u explain how i work out the current thanks __________________ If it aint broke, don't fix it. If it is broke, fix it. If you can't fix it, take it apart and see how it "worked".
 6th May 2003, 10:16 PM #7 diyAudio Member   Join Date: Aug 2001 Location: Bath, UK The output swing depends on how loud you set the volume control! Your maximum power will be the supply rails minus a few volts - the actual amount is spec'ed in the data sheet (see OUTPUT section in the table on page two). If you're driving 2A, the swing will typically be the supply rails minus 3.6v. The output current roughly is the output volatge swing divided by your load impedance. So, if the supplies are +/-20v, your maximum output swing will be roughly 16.4v. This will give a current of 2.05A into 8R. Power output is V^2/R = (16.4 x 16.4)/8 = 33.62W Power dissipated in the chip is V*I = (20 - 16.4)*2.05 = 7.38W These figures are the maximum you can acheive with a 20v supply and 8R load. You need to recalculate depending on your own situation. Nice one, David.
 7th May 2003, 07:20 AM #8 diyAudio Member     Join Date: Feb 2003 Location: Bristol thanks __________________ If it aint broke, don't fix it. If it is broke, fix it. If you can't fix it, take it apart and see how it "worked".
diyAudio Member

Join Date: May 2002
Location: The great city of Turnhout, BE
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Quote:
 Originally posted by daatkins The output swing depends on how loud you set the volume control! Your maximum power will be the supply rails minus a few volts - the actual amount is spec'ed in the data sheet (see OUTPUT section in the table on page two). If you're driving 2A, the swing will typically be the supply rails minus 3.6v. The output current roughly is the output volatge swing divided by your load impedance. So, if the supplies are +/-20v, your maximum output swing will be roughly 16.4v. This will give a current of 2.05A into 8R. Power output is V^2/R = (16.4 x 16.4)/8 = 33.62W Power dissipated in the chip is V*I = (20 - 16.4)*2.05 = 7.38W These figures are the maximum you can acheive with a 20v supply and 8R load. You need to recalculate depending on your own situation. Nice one, David.
Correct for peak power. For a sinusoidal signal, the RMS power, which would normally be regarded as more meaningfull, of course is 16.81 Watts.

Jan Didden

diyAudio Member

Join Date: Feb 2003
Location: Bristol
Quote:
 Originally posted by janneman Correct for peak power. For a sinusoidal signal, the RMS power, which would normally be regarded as more meaningfull, of course is 16.81 Watts. Jan Didden

half the peak power?
__________________
If it aint broke, don't fix it. If it is broke, fix it.
If you can't fix it, take it apart and see how it "worked".

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