I have an LED which has specified Typical forward voltage of 3.5V and Maximum forward voltage 3.9V.

I applied 3.3V across it with a 300 Ohm resistor in series. Why did it light up?

I am wondering if I can pick this LED as a reliable choice for my design (which as noted runs at 3.3V supply across the board).

My thinking:

The LED datasheet has a curve of Forward voltage vs current (I'm also confused why they put Forward current on the Y-axis instead of X, given that current is what one would vary here). Anyway, the curve shows a decrease in the forward voltage at smaller currents; perhaps this is the explanation?

Here is the downloadable PDF datasheet for this LED (it's a tricolor LED and in this question, I was referring to the specs for Blue and Green).

  • \$\begingroup\$ Can you share the datasheet? \$\endgroup\$
    – jippie
    Sep 17, 2012 at 15:24
  • \$\begingroup\$ @jippie: I added it now. \$\endgroup\$
    – boardbite
    Sep 17, 2012 at 15:33
  • \$\begingroup\$ If you want led to be on above a certain voltage and off below, you need a comparator of sorts. \$\endgroup\$
    – Lenne
    Jul 23, 2017 at 13:59

4 Answers 4


You are correct - the forward voltage depends on the forward current.

The forward voltage you see in the table of typical values is for a current of 20mA, which is too high when all 3 colors are used at the same time (footnote two in the absolute ratings table on page 3 - 15mA is the maximum in that case).

When you look at diagram 2 in the data sheet, you can see the relation between forward voltage and forward current. What you see here is that for a forward voltage of 3.3V, a forward current of 20mA can be expected. With 3V, it would be 8mA. A higher resistor value doesn't make this more reliable, it just makes the LED darker. You want to have the resistor as small as possible.

The resistor should be only large enough to drop the forward voltage to about 3.1V with a current of 15mA - this would mean a value of about 13.3 Ohms (the one for the red LED needs to be larger, though).

Whether this LED is usable for you depends on the brightness you need. If you don't need it to light up fully (or you use a version with higher intensity, see page 4), it would work. If you want to be sure you can use the full intensity, you need to use another one. Olin is right - the variation between batches can also mean that some are brighter than others. To ensure a uniform brightness, you need to control the current flowing through the LEDs.

  • \$\begingroup\$ I certainly do not need the full intensity -- this is a rather bright LED (e.g., 90 mcd Red @ 20 mA). So I think I understand now -- the 3.5 V is only necessary for the 20 mA current / full intensity case. If I supply a smaller voltage, e.g. with a constant current source, it just will glow less brightly (instead of not at all), since the LED does not have discrete behavior like an ideal diode. \$\endgroup\$
    – boardbite
    Sep 17, 2012 at 17:37
  • \$\begingroup\$ Exactly. Note that when using a constant current source, you choose a solution which needs only a minimal amount of current. Even a drop of 0.3V means you get only 10mA to the LED. For example, the TLC5940 can deliver 20mA with a drop of about 0.25V. \$\endgroup\$
    – hli
    Sep 17, 2012 at 19:16
  • \$\begingroup\$ In your last comment, could you clarify what this means: "you choose a solution which needs only a minimal amount of current"? \$\endgroup\$
    – boardbite
    Sep 17, 2012 at 19:31
  • \$\begingroup\$ I should proof-read what I write :( I meant 'minimal amount of voltage' (dropout voltage, to be exact). Otherwise your constant current source won't leave enough forward voltage for the LED to light up. \$\endgroup\$
    – hli
    Sep 17, 2012 at 20:27

"Typical" in datasheets doesn't mean anything useful. Those are mostly marketing numbers and usually vendors trying to make themselves look good.

The min and max specs are what matter. It's not surprising that a LED that typically has 3.6 V accross it at its full operating current would light up somewhat at 3.3 V. The current is probably considerably less than the full current, but some LEDs are so bright that they are still easily visible on your bench at a small fraction of the maximum current.

No, this model LED will not reliably light from 3.3 volts. You found one that did, and the next 1000 you get might also, but the next 10000 after that might be too dim. Unless the spec sheet explicitly tells you what you get at 3.3 V, you have to assume there is no guarantee. In reality you will probably get some light at 3.3 V, but the amount of that light could easily vary widely from part to part.

  • \$\begingroup\$ This clarifies. On that note, if I were to place a large enough resistor in series, then the LED has a smaller voltage drop (forward voltage) across it, bringing it under 3.3V. Would this make a reliable solution to ensure that the LED will be lit, irrespective of part variability? \$\endgroup\$
    – boardbite
    Sep 17, 2012 at 16:59

LEDs are not ideal diodes, so the "turn on" point (Vf) is not a perfectly sharp transition. If we look at the I-V curve for a typical LED, we can see this:

LED I-V Curves

The Vf is often taken at e.g. 20mA (some datasheets will give a couple of Vfs at different currents)

From this we can see that it's hard to control an LED by altering the voltage across it, so for the best control a constant current driver is needed. You can buy lots of ICs dedicated to this task, or you can roll your own simple source.
With a constant current driver, if the LEDs Vf varies (process, temperature, etc) then the driver compensates to keep the current constant, so this is the way to do things if you want the current to be exact irrespective of part variability (note the brightness at XmA may be different though, as this varies too)

Driving LEDs with a supply voltage above, below or above/below your output voltage

There are different types of LED drivers - some are just a basic constant current limiter, and some use a boost (or buck) topology or charge pump to provide a wider compliance range for the constant current.

Simple constant current driver:

A simple constant current driver will lose regulation as the voltage approaches the supply voltage (due to the drop across the limiting element) This will be given in the datasheet (see lowest supply overhead in this example part datasheet, pg.10)

Simple LED Driver

Boost LED Driver

A LED driver that uses a boost topology (just like a switching regulator but set for constant current rather than voltage) will still provide a constant current, but it increases it's voltage above the supply range to enable driving of LEDs in series with a total Vf above the supply voltage:

LED Boost Driver

SEPIC, Buck-Boost, Cuk LED drivers

Okay, so what about the case when your input voltage varies above and below the output voltage? A typical case could be when using a Li-Ion battery which can vary over ~4.3V - ~2.7V and an output of 3V is needed to push the desired current through the LEDs.
In this case we use either a SEPIC, buck-boost or Cuk driver. All can do the same thing here but have different topologies (why you would pick one over the other is further reading you may want to do - plenty of Books/App notes out there...)

Anyway, here's an example of a SEPIC circuit using the LM3410:


And here is a table of the efficiency at input voltage above and below the output voltage, you can see regulation of the LED current is maintained perfectly:

LED SEPIC efficiency

  • \$\begingroup\$ Thank you Oli. I did a quick search and saw that there exist a number of such drivers. But now, if I were to use such a driver and set X current -- let's say 10 mA current -- then since the V_F for Blue on the curve corresponding to 10 mA is 3 Volts, would it be true that I could do with a 3.3V supply then? \$\endgroup\$
    – boardbite
    Sep 17, 2012 at 17:32
  • \$\begingroup\$ @Inga - I added some details about different types of constant current driver. The general ideas are there, but the details depend on the IC so you need to check the datasheets carefully. Most simple drivers (the first type shown) will not be suitable for such a close margin. A SEPIC like the last example would work fine. \$\endgroup\$
    – Oli Glaser
    Sep 17, 2012 at 23:38
  • 1
    \$\begingroup\$ I ended up using the boost driver approach, and have a working project as of now. I truly appreciate the comprehensive detail you provided here. \$\endgroup\$
    – boardbite
    Sep 28, 2012 at 17:52

(I'm also confused why they put Forward current on the Y-axis instead of X, given that current is what one would vary here).

I was told that a diode is conducting depending on the voltage drop accross the pins. That's why the current which is able to flow is directly related to the voltage drop accross the diode(or LED). That is why Voltage is X and current is Y axis.


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