Using an SMD LED with 12V

[kristakis]

I'm making a circuit with 12V in and want to put a little LED on the board so show when 12V is present. That should be as simple as a resistor and LED in series. I found this LED on JLCPCB and the datasheet says the forward current should be 25mA.

So, with Ohm's law I want a resistor that is V/I which is 12/0.025 = 480R?

The datasheet says the forward voltage for 25mA is around 2V so that means there'll be 10V on the resistor. So the power disipated by that resistor will be 10*0.025 or 250mW?

Is that all correct? It doesn't feel right. It feels like I'm out by order of magnitude but I can't see it. That's a lot less resistance than I typically use for LEDs on 5V and all the SMD resistors I normally use are (only?) 125mW. I'm missing something really simple?

Part page on JLCPCB: jlcpcb.com/partdetail/YongyuPhotoelectric-SZYY0805R/C434431

update: So I was doing the voltage calculation wrong - it's 12-2/0.025 which is 400R. But it's still 250mW through the resistor?

[liaifat85]

Forward current (I) = 25mA = 0.025A
Forward voltage (Vf) = approximately 2V
Voltage across the resistor = 12V (supply) - 2V (LED) = 10V
Resistor value (R) = Voltage across the resistor / Forward current = 10V / 0.025A = 400Ω

Voltage across the resistor = 10V
Current passing through the resistor = 0.025A

Power dissipated by the resistor (P) = Voltage across the resistor * Current passing through the resistor = 10V * 0.025A = 0.25W

So, the power dissipation across the 400Ω resistor would indeed be 0.25W or 250mW.

[langwadt]

I'm making a circuit with 12V in and want to put a little LED on the board so show when 12V is present. That should be as simple as a resistor and LED in series. I found this LED on JLCPCB and the datasheet says the forward current should be 25mA.

no, it says maximum is 25mA

A modern led at 25mA is stupidly bright, a few mA is usually more than plenty

[Daixiwen]

25mA is the maximum DC current, you don't need it to be that high. 10 mA is enough, and if you don't need ti especially bright, 1mA might even suffice.

[tooki]

The datasheet says the forward voltage for 25mA is around 2V so that means there'll be 10V on the resistor. So the power disipated by that resistor will be 10*0.025 or 250mW?

No, it doesn’t say that. As others said, the “absolute maximum rating” is 25mA, which means that above that will damage it. You never want to be operating right at the limit.

Everything else in the datasheet uses 10mA — every other spec is tested at that current, including the forward voltage. So 10mA is (at most) what you should be targeting.

As others have also said: you may not need to run it at its rated (10mA) current. If it’s just a status LED for diagnostics or development purposes that won’t be used in everyday use, use an even lower current so you don’t waste power in normal operation.

I like using modern emerald green (~525nm) LEDs as low-power indicators because they’re insanely efficient. One of those at 0.5mA is still almost too bright for indoor use. For a diagnostics LED, you can drive them at 0.05mA and still see them well enough indoors. That tiny current also means the dropper resistor from 12V will be dissipating much less power.

[Peabody]

I agree.  You should test a variety of resistors, and pick the largest value that still lets you see the LED.

But in any case, you will be dropping about 2V across the LED and 10V across the resistor.  That means about 16% of the total power consumption will be used generating light, and 84% will be used generating heat.  So if there's some other point in your circuit that already uses a lower voltage, that might be a better place to put the LED.

If power usage really matters, more than parts count, using multiple LEDs in series might be an option since it would let you use much lower current but still produce the same light.  But that's probably not a reasonable option if this is just an indicator LED.

[tooki]

But in any case, you will be dropping about 2V across the LED and 10V across the resistor.  That means about 16% of the total power consumption will be used generating light, and 84% will be used generating heat.  So if there's some other point in your circuit that already uses a lower voltage, that might be a better place to put the LED.

From the original post, it seems like this LED is intended as a diagnostic indicator for the 12V rail. Putting it on a lower-voltage rail would defeat its purpose.

[Wallace Gasiewicz]

I usually use about an 800 ohm resistor in series with a white LED.There are 12 volt LEDs that already have the resistorSearch ebay for 12 volt led in business and industrial, pretty cheap. Made for replacing 12 volt lights on radios etc...

[tooki]

I usually use about an 800 ohm resistor in series with a white LED.There are 12 volt LEDs that already have the resistorSearch ebay for 12 volt led in business and industrial, pretty cheap. Made for replacing 12 volt lights on radios etc...

A 12V LED is simply an LED with a built-in resistor. It doesn’t change anything in terms of power consumption or efficiency. The only exception, more relevant for bright lamps than small indicators, is when they put 3 or4 LEDs in series. Assuming 3.2V for a white LED, 3 in series means dropping only 2.4V across the resistor. With 4 in series, each LED is underdriven at just 3V, which probably gets a If of a milliamp or two.

[kristakis]

I was so far into my own head about the resistor I didn't even consider dropping the current. Thank you.