I'm trying to troubleshoot a Ring doorbell that doesn't seem to be charging from the doorbell wires. I wanted to use my admittedly cheap multimeter to check the voltage that the wires are getting. My question is, how do I set it up? The transformer says 16V / 10VA on it. This is a picture of how I've set my multimeter. Is this correct?

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2 Answers 2


Yes your selecting an AC voltage setting is correct for a typical doorbell transformer.

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If you can find the transformer itself the first test would be to remove one or both low voltage wires from the transformer output and then measure the output with the meter. This will tell you if the transformer is OK or if there is some problem with the AC line voltage feed to the transformer or the transformer itself has failed. This isolation could save a good amount of time trouble shooting the low voltage wiring at the doorbell and switch.


Your multimeter is set to measure ac voltage in the range of 0-200V as pictured which is fine (your doorbell transformer is probably AC).

First measure voltage at the doorbell. If you get ~16V the problem is with your doorbell. If not you need to determine the root cause.

Disconnect the doorbell wires from the transformer and try measuring the output directly at the transformer. If the output is ~16V then you need to find where the wires are shorted / broken (or simply run new wires).

If the output of the transformer is not ~16V then you should measure the input voltage to the transformer. If it is ~115V you have a bad transformer. If it is 0V then you don't have power to the transformer.

You could have multiple problems such as sorted doorbell wires connected to a now-dead transformer.

Also, if the transformer tests good and you think the problem is in the wires to the doorbell make sure the pre-existing wires don't power anything else before replacing them. You could fix the doorbell but kill a second device (probably the back doorbell).

  • Please correct you answer to point out that the meter is set to measure AC VOLTAGE not AC current.
    – Michael Karas
    Jan 21, 2017 at 22:59

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