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I have a really loud and "angry" doorbell that I would love to be able to adjust its volume. Unfortunately it doesn't seem to have a volume control function so I decided to try and hack it.

I opened the case and found out that the speaker is a 16ohm/0.5watt one, connected to the board with a 2 pin connector (see images) https://photos.app.goo.gl/AXwT877BpwxUTnVY6

I thought that if I increase the resistance of the line that drives the speaker it should make it quieter.

However I have no idea what kind of resistor I need. Would a simple resistor of let's say 50ohms work? Can I somehow calculate the resistor required to make this half as loud? Also can I use a potentiometer in order to have finer control?

Any other suggestions are welcome.

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    \$\begingroup\$ Nobody know, how loud could it be for you? How big is the distance between the two pins in the connector? Simply try a value, maybe 10 Ohm or 100 Ohm. You could buy the same connector and socket to be able to remove it fast. Possibly a simple 2.54mm pin header and for the other side a socket strip could work too. \$\endgroup\$
    – MikroPower
    Commented Apr 11, 2023 at 2:06
  • \$\begingroup\$ The pin header and socket look like a good way to be able to switch/try different resistors easily, thanks! \$\endgroup\$
    – ktsangop
    Commented Apr 11, 2023 at 7:15

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Generally when dealing with audio amplifiers you want to keep the load impedance within a certain range. Since the speaker is 16\$\Omega\$ the amplifier should be happy with that impedance.

To decrease the level without changing the impedance you can use an attenuator circuit. This is a combination of several resistors, usually three resistors in a T or \$\pi\$ configuration but there are other types as well. You can use an online calculator such as this one to find the values. They're not going to be super critical, whatever the closest standard values are should work in this application. For half volume you'll use 3dB attenuation and 16\$\Omega\$ for the impedance.

Your device may not be picky about the impedance, it may just be driving the speaker with a square wave output, in that case you could probably use a resistor around 16\$\Omega\$ in series with the speaker to cut the volume. The attenuator is a bit safer in general, and you can easily pick how much you want to reduce the level by.

As for using a potentiometer, you might be able to use a low value pot, if you can find one maybe 100\$\Omega\$ or less it might work. You have to make sure the wiper of the pot goes to the speaker, not to the output so that it doesn't short the output to ground when turned all the way down.

A pot that acts like an attenuator and keeps the impedance constant is called an L-pad, you can get those to match different speaker impedances.

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  • \$\begingroup\$ That's great information thank you! The attenuator circuit should be the best option, since those L-pad pots seem expensive and hard to find (at least the 16Ω ones) On a side note, let's say I want to try a 16Ω resistor in series, could I just extend the cable that connects the speaker to the board, just to increase the resistance? \$\endgroup\$
    – ktsangop
    Commented Apr 11, 2023 at 7:14
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    \$\begingroup\$ it would take a very long cable to add 16 ohms. buy a bunch of 47 ohm resistors (like 10) they can be combined in different ways to make different resistances. \$\endgroup\$ Commented Apr 11, 2023 at 12:17

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