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The diagram I'm following for a basic TDA7297 stereo audio amplifier specifies 50 kΩ for its volume control, but I only have 10 kΩ and 100 kΩ dual pots.

If I use 100 kΩ I know I could halve it to 50 kΩ by adding a 100 kΩ resistor, but what happens if I use it as-is, and what happens if I use 10 kΩ instead? Would 10 kΩ limit the total output loudness? Somewhere I read changing values would not limit output but limit some frequencies, but I have no clue.

EDIT: source for input will be a cell phone

This one uses 10kΩ

enter image description here

But this one 50kΩ

enter image description here

From datasheet

enter image description here

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  • \$\begingroup\$ Interesting question that cannot be accurately answered with out more information. You need to post a link to the technical information on your amplifier board, the chip does not show a volume control. I suspect it is a voltage divider on the input of the AMP and should not matter but that is also depended on the signal source and its impedance. \$\endgroup\$
    – Gil
    Dec 19, 2022 at 1:57
  • \$\begingroup\$ @Gil just added the diagrams sir. \$\endgroup\$ Dec 19, 2022 at 2:29
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    \$\begingroup\$ @AramAlvarez The input impedance of the IC is not less than 25 kOhm but in that vicinity. Given the cell phone (at least a few milliwatts of output into a headphone at full volume), you will be fine with a 10 kOhm pot. The cell phone expects a significantly lower load. \$\endgroup\$
    – jonk
    Dec 19, 2022 at 5:31

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In the circuits shown the pot acts as a load to the source signal's output. If it draws to much (lower pot resistance) it will attenuate the output signal. The converse would also be true. You can calculate the rough approximation using a resistor divider calculator. Your load is the pot value or the lower resistor on the divider and the source impedance would be the upper resistor. Change the pot values without changing the source impedance. Changing the source impedance will also have the same effect. Here is a link: https://ohmslawcalculator.com/voltage-divider-calculator Assuming the ampolifier gain is fixed as in this design changeing the input voltage will directly effect the output voltage hence the volume. The pots max output voltage would be the max volume. The ratio determines the maximum voltage available for a given signal.

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