Need to confirm resistor selection

^^ Not mW, Watts. P = .025W - but that's not the best way to figure the R value. See below.

A low wattage 1K resistor will be great for you.

The SSR uses an LED with a Vf of 1.2 and an If of 2 mA. That If figure - 2 mA - is the maximum current needed to guarantee that the SSR will be energized. You could run the LED current a lot higher than that. What you want to do is ensure that the LED gets _at least_ 2 mA and limit the current to something a lot higher.

With a 5V supply, and ~1.2 volts dropped across the LED, you want to drop ~3.8V across the resistor. At 3.8 mA a 1K resistor will drop 3.8 volts. The LED inside the SSR will happily draw the 3.8 mA - in fact, it will draw as much current as it can get. The 1K resistor limits what it can get to 3.8 mA. The LED could draw a helluva lot more current without damage to itself if the resistor was a lower value, but there is no need to have it do that, as the 1K will give it well over the 2 mA maximum it needs for guaranteed correct operation.

The power dissipation for the resistor is computed with P = I^2*R, or .014 watts in this case, so use .03 watts or higher for the resistor wattage.

Ed

Where did the 25 for V2 come from?

The important values are: LED current to operate: 2 mA (you want to supply somewhat more than that) Absolute Maximum control current: 50 mA (you want to stay well below that)

The resistor has to drop [supply voltage] - [LED "on" voltage], or 5 -

I'd plan for about 10 mA, which would be R = E/I = 3.8V/.010mA = 380 ohms (390 ohms is a standard value)

The suggested 1000 ohms would give 3.8 mA, which is still OK.

The power dissipated in the 1000 ohm resistor is about .014 watts, so the lowest-power resistor available would be much more than adequate.

Ah yes, typo. Good catch.

more

well

standard value)

The 25v came from 5^2 - I had used "v2" because my superscript didn't work, I forgot that I could use the "^" character instead.

Thanks for providing the additional formula and explanation on how to achieve the required current. This is all new to me, I love learning this stuff!