# How can I figure out what the forward current and forward voltage will be for an LED given a resistor size

We're trying to order some boards from our PCBA manufacturer but they're having trouble sourcing our original LEDs (QBLP601-IW). They proposed an alternative LED (LSM0603443V) and I'm trying to be sure this is a suitable drop-in replacement.

The problem is we're significantly limiting the current on the original LED with a 150 ohm resistor. We have a 3.26V supply voltage and we're dropping about 2.7V across the original LED (and about 550 mV across the resistor). If I'm correct, we're pushing 3.67 mA through the LED.

Based on the original datasheet, this puts us right about 20% of 330 mcd.

Now I'm trying to figure out bright the new LED would be with the original 150 ohm resistor. I'm looking at this chart and basically using trial and error to figure out what my Vf and If would be.

Aside from the fact that I don't understand what's happening to the Vf scale between 2.8 and 3.4, here's what I did . . .

### My Naive Attempt

Using the formula below, I started plugging in values for Vf until If landed on the curve.

$$I_{F} = \frac{(V_{CC} - V_{F})}{R}$$

That seems to happen at Vf = 2.45V, If = 5.4mA or so. Is that the right way to figure that out? I'm a little concerned because if I use this approach, the datasheet for QBLP601-IW leads me to believe that Vf should be 2.54V not 2.7V as I measured (maybe because it's not at 25 degrees C). :(

Is this the right approach? Is there a better or more accurate way?

### Sidenote

The reason it matters so much is that if I look at the Luminous Intensity curve for the new LED, it looks like at 5.4mA the intensity would be about 40% of 330 mcd which would be (I think) twice as bright as the original and could be disruptive to people using our device in the dark.

## 3 Answers

Here is the old-school load-line method:

The straight red line represents the resistor current at Vled from 1.6V (11.06mA) to 3.26V (0mA)

Note: 11.06mA = (3.26V-1.6V)/150$$\\Omega\$$ -- if the graph started at 0V the Y axis for the resistor load line would only depend on the 3.26V and the resistor value. But it doesn't so it's a bit more complex.

Where it intersects the LED Vf curve will be the current through the LED. Looks like a whisker over 5mA (typical).

Note also that the scale on the X axis is wrong- it should go up by 200mV steps and it skips over 3.2V! Very sloppy.

My technique for figuring out Vf is much easier than reading charts.

1. Obtain a lab benchtop power supply that can be set in "constant current" mode. Virtually all supplies can do this

2. Short circuit the supply, set the current limit for 20mA (or whatever your target current is).

3. Connect your LED directly to the supply, no resistor. The power supply will be limiting the current to what you set in step #2

4. Vf is directly read. Probably the power supply has a voltage readout and will tell you what it's limiting its output to in order to achieve the target current. If not, a voltmeter will do the same thing. Just measure Vf directly across the LED.

Such a simple method it almost seems "wrong", but it's not...

Repeat ad nauseum with different samples until you are satisfied you have characterized the LED well enough.

Another option, which is even easier (because you don't need to measure ANYTHING) is to drive the LED's with a constant-current supply. Two transistors and a few resistors will allow you to set the drive current to "anything" and it'll stay at that level regardless what the input voltage may do. This works great, and I've literally put it into millions of production pieces.

i.e.

What is the formula for the dependence between modulating input voltage - output current for two transistor constant current sink

• This is super helpful in many ways!! Problem for this particular question is that I don't have the LED to test it. The manufacturer said it'd be a drop-in replacement but we're concerned about the brightness increasing. Thanks!! – D. Patrick Feb 12 at 14:40
• I've released literally 100 or so designs using LED's into the market over the last 20 years. I've directed new LED designs, visited & worked with LED manufacturers, etc... You could say I'm something of an "expert" (Everything being relative of course ;) . One thing I've learned above all else, when it comes to "what will it look like", you can not know until you actually try it and see the effect in person. Numbers on a datasheet never tell the story, and neither do photographs or movies. You gotta get a sample and implement it to know it's what you want. There's no shortcut :( – Kyle B Feb 12 at 16:44
• Thanks Kyle! We've decided to order 10 sample assemblies with the new LED. I also ordered a handful of the LEDs and some potentiometers so I can test their characteristics with different current limiting resistors. Hopefully we'll get a new full order in soon before we run out of stock. :) – D. Patrick Feb 12 at 16:48

Your "naive attempt" is reasonable, but you have to take the curves as "typical" and approximations, so calculating Vf=2.54V and measuring 2.7V would be reasonably close.

When you say you measured Vf=2.7V, was that with the new LED? If so, you can do direct comparison for brightness with the original LED and see if the increase is intrusive.

Or you can take the new Vf and ask what current (and hence value of resistor) would restore the brightness to the original 20%, and ask the PCBA mfg to substitute that resistor. (In an orderly development process, you would want a sample or small batch like that to evaluate the subjective brightness).

Bonus : the reduced current may improve battery life.

• I don't have any measurements on the new LED unfortunately. My measurements were all on the original LED. – D. Patrick Feb 11 at 14:55
• There is actually a standardized textbook method, the resistor load line. You usually see it for transistors (where it's most useful) but you can use it for leds too. See here lednique.com/current-voltage-relationships/… Your approach is not bad too. Just remember that Vf for a led is not strictly controlled, they just give a typical/min/max range, if you are lucky. You could go better with a constant current drive if you are targeting a specific brightness – Lorenzo Marcantonio Feb 11 at 17:15
• @LorenzoMarcantonio Yup ... load line may be worth its own answer. Mind you, I haven't used them since vacuum tube days! – user_1818839 Feb 11 at 17:18
• @LorenzoMarcantonio, that's really clever. I like that. Thank you. – D. Patrick Feb 11 at 21:32