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I am trying to read this datasheet (NXP TDF8546 Class AB Power Amplifier). On page 12, the power calculations are mentioned as

Po(max): RL = 4 Ohm, Vp = 14.4V, Vi = 2 V RMS Square wave

As far as I know, while calculating power calculations for any power amplifiers, engineers typically consider the sine wave. The reason I know is, its fundamental signal, and harsh input for power amplifier.

Why NXP considered square wave instead of sine wave in this datasheet?

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    \$\begingroup\$ A warm welcome to the site. Please can you add IC details and a diagram rather than just a link. If your link breaks in future, the question is meaningless. Thanks. \$\endgroup\$ – TonyM Apr 29 '20 at 6:51
  • \$\begingroup\$ I couldn't find (CTRL-F) the term "2RMS" anywhere in the linked document - copy and paste the section you refer to. \$\endgroup\$ – Andy aka Apr 29 '20 at 7:52
  • \$\begingroup\$ Sorry, typo error. "2 V RMS". Corrected it. \$\endgroup\$ – Bahubali Apr 29 '20 at 8:14
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Because a square wave is the waveform with a given amplitude that produces the maximum amount of power.

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    \$\begingroup\$ Hi Jakob, any proof or calculations or article which support the answer? Please help. \$\endgroup\$ – Bahubali Apr 29 '20 at 5:37
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    \$\begingroup\$ You really need proof on that? It's apparent from the waveform itself. Try to think of a waveform that produces more power with same amplitude, and you find none. \$\endgroup\$ – Justme Apr 29 '20 at 5:42
  • \$\begingroup\$ Ok understood. In this case, When I have to use either Sine or Square wave signals as input test signals? \$\endgroup\$ – Bahubali Apr 29 '20 at 6:01
  • \$\begingroup\$ I'd just use a sine wave as it is simpler and more pleasant to listen to. \$\endgroup\$ – Jakob Halskov Apr 29 '20 at 6:24
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    \$\begingroup\$ Please can you expand considerably on this one-line answer, for the benefit of future readers. Thanks. \$\endgroup\$ – TonyM Apr 29 '20 at 6:49

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