# Led voltage mind bkock!

Discussion in 'LEDs and Optoelectronics' started by JPU, Aug 29, 2012.

281
1
May 19, 2012
2. ### CocaCola

3,635
4
Apr 7, 2012
R = (VS - VL) / I

Note I is in Amps not mA

The datasheet doesn't give a 'happy' mA drive rating, but they do spec 100mA as being the max and rate the 'brightness' at 50mA... So I will do the math based on 50mA...

R = ( 5 - 1.3 ) / 0.05

or

R = 3.7 / 0.05

Solution

R = 74Ω, 82Ω will be the closest common resistor value

Remember this is driving at 50mA, if you want to drive it lower, say around 20mA use a 220Ω resistor, or tailor it to the design with anything from about 47Ω (90mA drive) to 390Ω (10mA drive)...

Last edited: Aug 30, 2012
3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,175
2,690
Jan 21, 2010
From the datasheet,

Vf = 1.3V (typ)
If = 50mA

using ohms law, you want a resistor that will drop 3.7V at 50mA, so the value is 74 ohms (use 68 or 82 ohms).

Stop here if you want to Let's assume you use 68 ohms. This gives a calculated current of 60 mA for a power dissipation of 78 mW -- well within acceptable values.

Now lets assume the worst case. 5.5V, 61 ohms, 1.15V Vf. This will give a LED current of 71 mA and a power dissipation of 82mW. Still well within spec.

The datasheet also tells you how to de-rate this device (1.33 mA/DegC over 25C), and we have 29mA between the worst case current and the absolute maximum, so a quick calculation tells us that we're OK up to an ambient temperature of around 47 degrees, even assuming the worst case for the power supply and the components. In the real world you would be very unlikely to have such a combination.

However, very often, IR leds are pulsed, and if you're doing that, you probably want to run closer to the maximum ratings and so you may want to repeat similar calculations to these.

4. ### JPU

281
1
May 19, 2012
Thank you both for your helpful and comprehensive answers.

I appreciate it.

JPU

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