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I designed a PCB with the following audio amplifier on it: https://eu.mouser.com/datasheet/2/302/TFA9896_SDS-1381317.pdf

According to the data sheet, the amplifier is able to deliver 2.1W RMS into an 8 Ohm speaker at a battery voltage of 3.6V.

For testing the amplifier, I used a 3W speaker with 8 Ohm and played 16 bit wave files with a maximum amplitude of less than 1. When I play these files, there is strong clipping. This can be observed also for other speakers.

The audio amplifier boosts the input voltage to 6.1V and is able to provide a peak current of 2.5A. In my tests, I measure a peak current of 0.5A and an average current of 0.25A during playback.

As I'm lacking experience with audio amplifiers, I wonder how to fix this issue. When lowering the maximum amplitude of the wave files to less than 0.5, the clipping vanishes. However, I would like not to sacrifice for volume as the audio amplifier should be able to provide enough power according to the data sheet.

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    \$\begingroup\$ How are you measuring current? Do you have an oscilloscope, and can you show us the waveforms at the speaker? \$\endgroup\$ – TimWescott Nov 14 '19 at 20:56
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    \$\begingroup\$ According to the datasheet, the amplifier has some kind off digitally controlled gain. It doesn't go into details, but that's where I'd start looking \$\endgroup\$ – JRE Nov 14 '19 at 21:01
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    \$\begingroup\$ @Hermetica The equation is \$\pm\sqrt{2\cdot P\cdot R}\$ to get the voltage needed. But the IC also needs a little bit of voltage overhead for the switches. So that's probably why they DC to DC convert to 6.1 V. Bridged, this gets them enough to match the expression I just wrote. \$\endgroup\$ – jonk Nov 16 '19 at 12:51
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    \$\begingroup\$ @Hermetica Just use the expression I wrote out. If the supply rails are consistent with it, then bridging isn't needed. If the voltage rails are not consistent, then bridging or something other trick is required. \$\endgroup\$ – jonk Nov 16 '19 at 13:01
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    \$\begingroup\$ @Hermetica Even that may seem simple enough. But it assumes a nice sine wave. Real music tends to not look so nice, so a good rule is to figure that you want an amplifier that is rated about five times (Max or peak) of what you really think you need. So if you want about 2 Watts typically, then design to use voltage rails for 10 Watts. There are a lot of reasons why, distortion being one of them. But in general, good sounds result from some excess capability. It is a sad fact of life. \$\endgroup\$ – jonk Nov 16 '19 at 13:08
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Note: This answer is predicated on the fact that you can verify the clipping by putting a scope across the speaker. And the clipping is happening near the 6.1V mark.

enter image description here

If the output of the amp is clipping, then you only have three choices,

  1. Increase the voltage to the amplifier (if you can)
  2. Reduce the gain of the amplifier if you can.
  3. Reduce the amplitude of the input.

I would like not to sacrifice for volume

After you do one (or more) of the above to regain audio quality, you can add more speakers in parallel, based on your output power description. Two speakers should give you twice the output power 3dB(assuming that the amplifier drives them equally as well as it drives one speaker), then 4 speakers would double the output power of the two speakers to 6dB.

Turns out that for humans that 3dB changes in sound are marginally perceptible, 6dB is pretty noticeable and 10dB is considered double the volume (or half if going down).

Doubling Power vs. Doubling Output

Thanks @ Marcus Müller for your comment.

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    \$\begingroup\$ you can't parallel speakers when your amplifier isn't designed to drive the higher load. This one doesn't seem to be! assuming infinite drive strength, i.e. modelling your amp as voltage source, you'd only get twice the output power with two parallel speakers – that's not twice the volume! Human perception is logarithmic. \$\endgroup\$ – Marcus Müller Nov 14 '19 at 21:12
  • \$\begingroup\$ @MarcusMüller You're right, 10dB is a doubling of volume to a human. I'll update my answer. \$\endgroup\$ – Aaron Nov 14 '19 at 21:17

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