I bought a video doorbell that runs on battery and "trickle-charges" from the doorbell chime's wiring. But my street is so busy that the battery died in under a day. The support forum for this product mentioned upping the transformer from 16V 10VA to 24V 40VA. Trouble is, my current chime box is rated for 16V, so I'm worried that 24V will burn it out.

I've seen some posts online about this exact scenario (24V transformer on a 16V chime box), and people keep mentioning using Ohm's law to calculate the resistance needed to correct for this overvoltage. The chime draws 0.2A when ringing (measured with a multimeter), and we're over by 8V, so that means we would need a 40Ω resistor.

So if I'm understanding this right, the resistance in the circuit is fixed, and we're just upping that number. So the original resistance is 24V/0.2A = 120Ω, and we're moving that to 160Ω, making the new current 24V/160Ω = 0.15A.

But the original transformer with the original resistance gives us 16V/120Ω = 0.133A.

To reach the original current, wouldn't I need a 60Ω resistor?

Also, am I correct in understanding that the battery's rate of charge is controlled by the power in the circuit, and thus adding resistance to decrease current would result in a slower battery charge? Wouldn't that defeat the purpose of having a higher voltage transformer (to some extent)?

Finally, is the thing that's important for the chime box not to burn out the current? i.e. does dropping the current back to the original ~0.13A actually solve the problem I'm trying to solve?

Thanks in advance for your help. It's been 10 years since I my college electrical engineering class and I'm just baffled at this point.

  • \$\begingroup\$ Hard to say. There are so many assumptions and interdependencies in there. \$\endgroup\$
    – Jens
    Jun 23 at 0:37

2 Answers 2


Stepping up your voltage is a mistake.

It won't do what you want, because you misunderstand the problem.

When a smart doorbell is wired up with 2 wires, it powers itself (recharges its battery) by leaking current through the mechanical chime. It is counting on the chime being an old-fashioned solenoid type, drawing fairly high current (like 0.5 amps), and allowing a fair amount of current to pass through it without actuating (e.g. 0.2 amps).

Unfortunately your chime is modern, I would guess electronic, and has a high resistance when it is not being actuated.

As such, the smart doorbell cannot flow enough current to power itself (without triggering the chime continuously, which it does not want to do).

And you can see the problem: if you step up voltage by 50%, the chime will still be too high-impedance to power the doorbell. And you'll be making the impedance even higher by putting a resistor in series with the doorbell!

I get the urge to "throw more power at it" but that will do nothing and only make things worse.

Use the method used with smart switches.

This same problem comes up with "no neutral" smart switches which leak their operating current through the incandescent bulb. Incandescents are great at that, but LED screw-in bulbs are really not.

In lighting, the cure is to put a bypass "resistor" across (shorting across) the bulbs, so a low impedance path is kept alive for the smart switch. However when you are dealing with AC power, you don't use a resistor for that. You use a capacitor.

An example is the LUT-MLC product, aka Lutron Makeup Load Capacitor. This is actually UL Listed product and is legal to use in this capacity, unlike almost any electronic components you might buy that have UR-Recognized and not the needed UL Listed.

The LUT-MLC is made for 120V, but it certainly wouldn't hurt to try it across the chime terminals. Measure the series current with the chime not ringing, and make sure it is reasonable for the transformer.

  • \$\begingroup\$ My chime is actually pretty ancient, and definitely mechanical. I would guess the resistance seen is maybe just from the length of wiring? The transformer is in the basement, and the chime is in the upstairs hallway, so it's a long-ish loop. Interestingly, the chime box on 16V doesn't trigger unless the smart doorbell has it marked as "electronic," but on 24V it works either way. I don't really know what to make of that. \$\endgroup\$ Jun 23 at 13:57
  • \$\begingroup\$ @Dylan well you can figure the resistance, just look up 18 AWG wire resistance per foot (listed as per 1000'). \$\endgroup\$ Jun 23 at 21:00

40 is the correct answer, where you got confused is

24V/0.2A = 120Ω,

which is the finished resistance, not the initial resistance.

The initial resistance is 16V / 0.2A = 80Ω

conveniently this is 40 less than the finished resistance confirming your first calculation.

Go out and buy a 2W 39Ω resistor

39Ω is a standard size, and will be close enough to 40 for your task.

Finally, is the thing that's important for the chime box not to burn out the current? i.e. does dropping the current back to the original ~0.13A actually solve the problem I'm trying to solve?

If the camera circuit can make use of the extra voltage available to it id does help, if it can only use the current it does not help.

If the camera needs more current connecting a resistor in parallel with the chime could help. start at 220 ohms.


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