Connect 5w 100k ohm resistor between neutral and hot to prevent LED light blinking when switch is off

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I have touch operated light switches operating LED lights throughout my home. The switch manufacturer claims that I must add a device to circuits containing LED bulbs of 5w or less to prevent flickering. They show their device connected between the neutral and hot wires before the driver and after the switch. Some of the devices "popped" – they suggested to replace the device with a 5w 100k ohm resistor bridging the hot and neutral wire before the LED driver and after the switch. Does this make sense? Is this correct? Is this not a short circuit when the switch is on?
photo from light switch company

Best Answer

First, you are not allowed to use electronic components in mains electrical wiring. You must use "equipment" which has been tested and certified (listed) for use in mains electrical wiring. Equipment may simply be an electronic component in a safe housing built to a certifiable spec and run through the independent testing lab to test for fire and health safety. TUV or UL are testing labs, CE is not.

A resistor is a resistor, as long as it is made for this purpose and the ohm values are within 50-70%. If Moon Rune Importers there can't provide a Listed resistor, get a resistor from another manufacturer.

If that doesn't work for you, change the smart switch to one that does not have the resistor requirement. Electronic ballasts/drivers are not a surprise. CFLs have been on the market for over 20 years, governments have been expressing their intent to deter/ban incandescents for almost that long, to warn manufacturers to stop making things that aren't compatible with electronic drivers. So they have no excuse.

A 100k resistor on 110/120V will draw as follows: 

E=IR   120V = I * 100,000ohm    I = 0.0012 amps 
W=EI   W = 120V * 0.0012A       W = 0.144 watts

A 100k resistor on 220/230/240V will draw as follows:

E=IR   230V = I * 100,000ohm    I = 0.0023 amps
W=EI   W = 230V * 0.0023A       W = 0.529 watts

Either approach seems reasonable. I'm not sure where you get a 100k resistor that's listed for use with mains electrical, but 0.5 watts is little enough energy that cooling won't be a big factor.

It will be a short, as you suspect, but it won't be a dead short. Current will be limited to 0.0012 amps or 0.0023 amps, depending on your voltage. This will result in watts of power consumed being under 1 watt.

The only thing that surprises me is that such low current flow would suffice.

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