I have a small digital timer designed to replace a wall switch. As such, it runs on the 120VAC mains and has only a hot and a ground in (plus a switched hot out). Because it works without its own neutral wire, I assume it powers itself via passing a small amount of current through the load even when off.

Here is the manual: http://waterheatertimer.org/pdf/GE-15313-Manual.pdf

As you can see, it's rated for:

  • Incandescent (Tungsten): 5A/600W
  • Resistive: 15A/600W

The packaging specifically admonishes against using a CFL or LED load, and an inductive or motor rating is notably absent.

I was hoping to use this to control a small, "bathroom vent"-style fan until I noticed the rating. I'm going to get a different timer, but I still wonder what folks here think is the concern.

  • Would back-EMF damage the electronic guts when the fan turned off?
  • Would the (intentional) leakage current damage the fan?
  • Are they using a semiconductor switch that requires a zero-crossing to turn off and the inductive phase-lag messes with that?

Bonus question:

What is special about tungsten loads that warrants a much lower current rating than general resistive loads? Does it have something to do with filaments changing resistance as they heat?

  • 1
    \$\begingroup\$ Based on my own experience trying to use a (plug in) digital timer with a flourescent load, the load will produce substantial inductive spikes on turn on and off that will reset the timer. \$\endgroup\$ Aug 5, 2015 at 6:46
  • \$\begingroup\$ ^ then it would be tempting to crack it open and add some caps to the low-DC section :) \$\endgroup\$ Aug 5, 2015 at 6:54
  • 1
    \$\begingroup\$ It almost looks like the "Resistive" rating is in error anyway, as it is still rated 600W, which is 5A at 120V. 15 Amps would be 1800W, well over the (rated) capacity of the timer. \$\endgroup\$
    – R Drast
    Aug 5, 2015 at 13:34

1 Answer 1


I will answer the bonus question.

The filament of a lightbulb has a low resistance when cold and normal resistance when hot (so do all metals, but wires etc usually do not heat up enough for the change to be noticeable). Which means that when you turn on the lightbulb, the filament is cold and has low resistance, so the initial curent surge is very high (tens of amps), then the filament heats up quickly and its resistance increases, reducing the current.

Your device probably has an output driver that would not tolerate to inrush current of a big lightbulb (or a CFL or LED power supply).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.