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I'm looking to add an audio amplifier to a circuit, in this case the TPA0211 in 5V mode for a full 2W of output.

My rather dumb question is this: I should expect at least 400mA of current draw from the amplifier, given the 2W output at 5V into a 4ohm load, correct? I don't see current draw listed on the DS, other than the supply current of 6mA, so I'm wondering if its amperage draw is implied given the output power and input voltage are stated.

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    \$\begingroup\$ 2 W Into 4Ω From 5V Supply \$\endgroup\$ – Sunnyskyguy EE75 May 14 at 19:18
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    \$\begingroup\$ I didn't look long at the datasheet. But the 6 mA max in BTL mode is the quiescent class-AB current without a signal and without a load and with 2.5 V. At 5 V, the quiescent current will be higher, perhaps 9-10 mA as a typical value. I'd plan for more -- say 100 mW quiescent. (The quiescent current is just what keeps it functioning.) When a load is attached and you supply a signal, it will be directing large swings of voltage (and current compliance) at the output and then it will be dissipating lots more power. Look at Table 2 for dissipation under load. \$\endgroup\$ – jonk May 14 at 19:21
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    \$\begingroup\$ I²=P/R=2/4=0.5 thus I = 0.5 A pk \$\endgroup\$ – Sunnyskyguy EE75 May 14 at 19:24
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    \$\begingroup\$ @SunnyskyguyEE75 - Aha -- Now I get what you meant. Thanks! \$\endgroup\$ – t3ddftw May 14 at 19:27
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    \$\begingroup\$ @t3ddftw Bear in mind that for typical music (there are whole papers analyzing music), you will be using only about an average that is 20% of the peak power capability. This means more dissipation in the IC, less in the speaker. So for a 2 W capable system, you'll probably average 400 mW into the speaker, or so. Stuff that into the efficiency equation to work out the IC dissipation. \$\endgroup\$ – jonk May 14 at 22:43

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