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Old 5th May 2003, 10:51 AM   #1
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Default 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.
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Old 5th May 2003, 04:23 PM   #2
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at max volume, what is the approx. heat output in watts?
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Old 6th May 2003, 08:21 AM   #3
miguel2 is offline miguel2  Portugal
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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
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Old 6th May 2003, 07:44 PM   #4
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i heard somewhere that an amp outputs 4 times as many watts of heat as power output in watts. is this correct?
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Old 6th May 2003, 08:31 PM   #5
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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.

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Old 6th May 2003, 09:39 PM   #6
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a, can u explain how i work out the output swing
b, can u explain how i work out the current

thanks
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Old 6th May 2003, 10:16 PM   #7
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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.
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Old 7th May 2003, 07:20 AM   #8
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thanks
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Old 7th May 2003, 11:28 AM   #9
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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
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Old 7th May 2003, 05:49 PM   #10
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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?
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