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I have a doorbell I'd like to make louder (existing speaker driver is ~2W).

I had an idea of intercepting the signal going into (or out of?) the amplifier chip and feeding those signals through a separate, powered amplifier to driver a higher-powered speaker.

Here is the datasheet for the amplifier: https://datasheetspdf.com/pdf-file/935353/ChipstarMicro-electronics/CS8305E/1

Any help is appreciated!

enter image description here

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  • \$\begingroup\$ The volume is likely set by resistors. Reverse engineer the circuit into a schematic to see if you can simply make it louder by replacing one or two resistors. \$\endgroup\$
    – Justme
    Commented Sep 6, 2022 at 22:45
  • \$\begingroup\$ Use a bigger more efficient 4" speaker and hide it in a plastic coffee container as I did. \$\endgroup\$
    – D.A.S.
    Commented Sep 6, 2022 at 22:48
  • \$\begingroup\$ @TonyStewartEE75 Thanks for the comment. I considered this, but suspected I would be limited by 5W power capability of the amplifier itself. \$\endgroup\$
    – R S
    Commented Sep 7, 2022 at 0:34
  • \$\begingroup\$ I got at least 10 dB gain with a better speaker and same D cells battery power, if you had a 12" speaker in a large cabinet, you probably would want to turn down the volume with 5W \$\endgroup\$
    – D.A.S.
    Commented Sep 7, 2022 at 0:38
  • \$\begingroup\$ @Justme Thanks for the tip. I wish i knew more about how to do this properly! \$\endgroup\$
    – R S
    Commented Sep 7, 2022 at 0:39

1 Answer 1

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The audio signal is available at the bypass capacitor to the inverting input of the amplifier - the node circled in red below.

The top view of the PCB area around the amplifier, with the audio signal node circled in red.

This node is also pointed to by the arrow on the application schematic below.

The single-ended input application schematic with the audio signal node pointed to by the arrow.

Whether you actually need an external amplifier or just an increase in volume is hard to tell.

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