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Old 14th December 2012, 10:02 AM   #1
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Default Output power facts

Hello,
Consider an LM3886 running at a trivial voltaje (in the operating range of course) at +/-35vdc rails. At worst case scenario, as datasheet states, accounting for chipamp losses, you will have +/-32 volts on the output at full power before clipping.
Then, ohm's law: P=V^2/R
So, peak output power at 8ohm should be 32*32/8 which equals 128watts

For a 4 ohm load, voltage should be +/-28vdc, so for a +/-25v output it is 25*25/4= 156watts.

How does it correlate to the IC datasheet, which specs 50watts at 8ohm and 68watts at 4 ohms?


Please try to explain that to me, as this is something simple that I cannot manage to understand

Thank you, best regards!
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Old 14th December 2012, 10:10 AM   #2
joram is offline joram  Belgium
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Default peakvoltage vs RMS

32V is peak voltage.
If you would run a sinus with 32V peak you have to calculate the RMS voltage
32V / 1.414 = 22.63V
(22.63 x 22.63) / 8 = 64 W

25 V / 1.414 = 17.68V
(17.68 x 17.68) / 4 = 78W
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Old 14th December 2012, 10:24 AM   #3
sreten is offline sreten  United Kingdom
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Hi,

As the numbers above indicate, RMS is half your calculated peak power.
The supply in a real circuit will also droop, reducing the voltage swing.

rgds, sreten.
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Old 14th December 2012, 11:27 AM   #4
AndrewT is online now AndrewT  Scotland
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Quote:
Originally Posted by regiregi22 View Post
.............At worst case scenario, as datasheet states, accounting for chipamp losses, you will have +/-32 volts on the output at full power before clipping................
The 3886 datasheet gives guidance on what to expect for voltage drop through the amplifier when driving your 8r or 4r test load.
3Volts of amp drop is not the worst case value I would choose.
As Sreten mentioned, your +-35Vdc supply rail only holds that voltage while the amplifier is not delivering power to the load.
This supply voltage will drop when the amplifier delivers power to the load.
In addition the PSU has limited smoothing capacitance. As the current demanded by the load increases so does the ripple on the DC supply.
The droop in voltage due to the increased ripple must also be accounted for. Reading the average rail voltage while delivering full power does not account for the ripple. You need to measure the ripple separately and then subtract half that peak to peak ripple from the average reading that the DMM gave you.

In summary, the supply minimum voltage, at the power PINs, could drop to +-30Vdc from the unloaded +-35Vdc.
The volts drop through the amplifier could be 4V.
The maximum unclipped output voltage will be ~26Vpk for this example.
26Vpk converted to rms for the test sinewave will be 18.4Vac.
P = Vac*Vac/Rload = 18.4*18.4/8 = 42W into 8r0.
quite different from
Quote:
equals 128watts

For a 4 ohm load, ............= 156watts.
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