# Led voltage drop resistor

So I was working on the switch backlighting for A switch plate on my boat. There is a single led mounted in a light bar that mounts behind the engraved legend plates.

I bought a replacement led at RadioShack. It is a red 800 mcd LED. The package says 1.7V - 20mA.

I needed to calculate the voltage drop resistor. So I took incoming voltage of 12v - 1.7v= 10.3v.

10.3v / .02A = 515 ohm resistor. Right?

So I buy a 560 ohm resistor and wire it in series with my led. Luckily I am able to bench test with a variable power supply and a volt meter. As I bring the supply voltage up to 12v, the voltage at the led rises over the led 1.7 volts. Why?

What I ended up doing was wiring 2 - 560 ohm resistors in series and that brought the led voltage to where it needed to be. Why did Ohms law not work? What did I do wrong or not account for? Could the LED current be labeled wrong on the package?

• How much above 1.7 V are you measuring? Is the LED lighting up? Nov 22, 2015 at 19:12
• Resistor have tolerance. Your 560 ohm resistor is already included in the E12 series which has a tolerance of 10%. The actual allowed tolerance is specified by the color lines on the resistor. Nov 22, 2015 at 19:21
• The LED forward voltage does vary slightly with current, and 1.7 volts seems a bit low to me - I'd expect 1.8 - 1.9 volts. How high a voltage did you measure? Nov 22, 2015 at 19:36
• Leds have tolerances as well. The 1.7V @ 20mA stated is an average. It could very well be 1.85V @ 20mA or something. It's a scale. If you want absolute, use a constant current circuit. For the most part, a typical led will be bright enough at 5 or 10 mA for use as an indicator. Nov 22, 2015 at 20:26
• The led does light up. I was probably in the 4-5 volt range with one resistor. I don't think tolerances can make that much of a difference? With the two resistors, I measure about 1.7 at the led. Nov 22, 2015 at 21:21

Assuming your 12 volts is in fact 12 volts exactly (and you did check this, I hope), a 1.9 volt reading on the LED means the voltage across the resistor was less than your expected 10.3 volts, and in fact was 10.1 volts. This in turn means that the current was less than 20 mA, nominally $$i = \frac{10.1}{560} = 18 \text{mA}$$ and you were in good shape. Especially since your resistor may well have had a 10% tolerance, and so your current had the same, and could have been as high as 19.8 mA.