# Resistor power dissipation when driving optocoupler from ac mains

With the exception that I used an LTV-814 that doesn't require a bridge rectifier. The forwards voltage of the optos led is 1.2 (1.4max) and If = 20mA

Give: Power = I^2 * R = (0.020A)2*(390000) = 156 watts

That doesn't make much sense to me. I'm only using a 1 watt resistor so it should have blown up by now, but it doesn't even get hot. What am I missing?

Further, I calculated my own resistor value for the 390k one rather than relying on the schematic I followed in the aforementioned link:

R = (170vpeak - 1.2forwardvoltage) / .020A = 8440 ohm

Is this correct? Should I use this instead of the 390k resistor?

Mains: 120v

• why are you assuming 20 mA flow through the 390 kΩ resistor? (That's pretty certainly not the case) Your calculation for the resistor value seems to make sense, why are you even using someone else's resistor value? – Marcus Müller Oct 4 at 7:28
• What minimal current is required by the opto to operate it and what value resistor does that imply? – Andy aka Oct 4 at 7:31
• @MarcusMüller it was from the schematic of the circuit for which this was based. the opto used in that circuit had the same characteristics. – Alex Oct 4 at 8:05