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The datasheet for some LEDs I'm considering lists the forward voltages as 2.7V typical and 4.2V max. (I'm looking at the "warm white" version.)

In the past, I've only ever looked at the typical forward voltage for figuring out the resistor value. But I've never used an LED where the maximum value may be so different from the typical.

What factors affect the actual forward voltage? Manufacturing variation? Temperature? Current? Something else?

Suppose I have a 3.3V supply and that I want to drive one of these LEDs at 20mA. Based on the 2.7V typical forward voltage, I'd use a 30 Ohm resistor. But if I end up with an LED that actually has a forward voltage of 4.2V, it won't light because 4.2V > 3.3V.

How do you design a circuit to accommodate such a wide range of forward voltages, or is there a reason I don't have to worry about the max forward voltage?

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  • \$\begingroup\$ Do you really want to drive these LEDs with maximum luminosity, all 20 mA? What is the purpose for your LEDs? \$\endgroup\$
    – Ale..chenski
    Aug 7, 2018 at 1:50
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    \$\begingroup\$ See my answer to electronics.stackexchange.com/questions/353955/… for some info on this topic. \$\endgroup\$
    – Transistor
    Aug 7, 2018 at 1:55
  • \$\begingroup\$ Note that you will often get product that sits fairly close to the center of the spectrum. The typical Vf is typically much more common that values close to the min and max. Also if you are driving LEDs in series, this problem more or less becomes mitigated just by that. \$\endgroup\$
    – K H
    Aug 7, 2018 at 2:01
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    \$\begingroup\$ Those with enough rep should vote to close this as a duplicate of electronics.stackexchange.com/questions/341267/… which I didn't find until after I submitted this one. \$\endgroup\$
    – Adrian McCarthy
    Aug 7, 2018 at 4:29
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    \$\begingroup\$ Or this one: electronics.stackexchange.com/questions/339458/… \$\endgroup\$
    – Adrian McCarthy
    Aug 7, 2018 at 4:37

5 Answers 5

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Suppose I have a 3.3V supply and that I want to drive one of these LEDs at 20mA.

In a production environment you do not use an LED with a max Vf greater than the supply voltage. In a home project you can give it a try.

The solution is to select an LED with a Vf that does not exceed the supply voltage.


How do you design a circuit to accommodate such a wide range of forward voltages

Many manufactures "bin" LEDs by Vf so you can buy your LEDs based on a range of forwards voltage.

An example below is where Samsung packages (bins) the LEDs in 3 ranges of Vf.

enter image description here


What factors affect the actual forward voltage?

  • Current
  • Temperature
  • Manufacturing variation

Most LED datasheets will have an IV graph showing the typical Vf at a specific temperature.

enter image description here

LEDs have a negative temperature coefficient. For example the Vf of an LED may decrease 0.006V for every degree rise in temperature. LEDs are typically specified to work where the ambient temperature is between -40° and +85° C. With a 6mV coefficient the Vf will fluctuate 0.75V across its operating temperature range.

enter image description here

LEDs are manufactured by growing epitaxial crystals on a substrate. So by the nature of the manufacturing process the Vf can vary even when the LEDs come from the same wafer.

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  • \$\begingroup\$ +1 for data on temperature dependence. It would be useful to add that, if driven at constant-voltage Vf, there could be a thermal runaway situation, and "magic smoke" will follow. \$\endgroup\$
    – Ale..chenski
    Aug 7, 2018 at 22:02
  • \$\begingroup\$ You are Inconsistent on text with specs on tempco you ought to say -3 mV/‘C +\-1 not -6mV. Although Vf max/min tolerances are more consistent around typ. +0.4/-0.2, +0.3/-0.2, 0.2/-0.1 \$\endgroup\$
    – Tony Stewart EE75
    Aug 7, 2018 at 23:47
  • \$\begingroup\$ @TonyEErocketscientist I added the graphic after writing the text. 6mV is a very common temperature coefficient. \$\endgroup\$
    – Misunderstood
    Aug 9, 2018 at 1:07
  • \$\begingroup\$ @AliChen Thanks. I do not understand the thermal runaway with a CV supply. As I understand thermal runaway it refers to LEDs wired in parallel with a CC supply. \$\endgroup\$
    – Misunderstood
    Aug 9, 2018 at 1:10
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    \$\begingroup\$ ESR is simply a good linear expression of behaviour over a limited range. Your logic has a false assumption \$\endgroup\$
    – Tony Stewart EE75
    Aug 10, 2018 at 23:22
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The mean value of the typical voltages are 2.55V and 3.14V for the maximum values. However, plugging in the same resistor value for a 2.2V LED vs a 4.0V LED would be very different currents and brightness levels.

enter image description here Source: https://www.lumex.com/article/led-color-guide

Suppose I have a 3.3V supply and that I want to drive one of these LEDs at 20mA. Based on the 2.7V typical forward voltage, I'd use a 30 Ohm resistor. But if I end up with an LED that actually has a forward voltage of 4.2V, it won't light because 4.2V > 3.3V.

The typical forward voltage means that the LED will be lit, how much depends on the LED itself and even the construction of the LED (viewing angle ECT) (not to mention that the human eye sees some colors better than others, so even two LED's with the same intensity will not be perceived at the same intensity).

The only way I've found to really get LED's with the right intensity are to ball park the resistors then fine tune the values based on the intensity, sometimes I'll get a few people to look at it if it's on a product.

The other way is to use a constant current driver, which is more complicated, but allows you to avoid a voltage drop if you can't afford it:

enter image description here
Source: https://hackaday.com/2012/03/08/led-tutorial-demystifies-several-control-techniques/

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I think you are getting a bit confused because LED's are current driven devices. To some degree the 'typical' forward voltage is at a typical safe drive current, while maximum forward voltage could be the peak pulsed current the LED can handle, but not have a long life.

I buy 5mm round high-efficiency pure white LED's that put out 1 watt of light while consuming about 1/9th watt of power. Forward voltage is 2.9, which is the minimum turn-on voltage, to a maximum of 3.5 volts.

I know I need at least 2.9 volts to make them dim, and limit or fix the current depending on how long I want it to last vs. brightness. The manufacture states that 20 mA continuous or 30 mA pulsed @ 10% duty cycle are typical maximum current values.

So ultimately you have a voltage source high enough to turn the LED on full but you limit the current by using a current sink or clamp or a fixed resistor tied to a fixed voltage source. I run mine at 15 to 16 mA in blocks of 25 so I need a 75 volt source just to turn them on.

But my source is 1/2 wave 120 VAC so my source is about 85 volts. A few K ohms of resistors limits the current to a safe 15 to 16 mA. Future version may include a current sink to lock the LED current even if the source voltage fluctuates. Please read @laptop comment and link to some useful current sinks and clamps.

It is important to understand that typical forward voltage must be combined with a safe drive current, usually 2/3 of the rated maximum value so the LED will run cool and have a long life. If you run it in pulsed mode take 2/3 of the maximum pulsed current limit as a good safe value.

LED's do not care about the source voltage as long as it is above its minimum rated voltage. They do care a lot about the drive current from this voltage source. Try not to get voltage and current confused.

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    \$\begingroup\$ You stated " put out 1 watt of light while consuming 1/9 th of power ". Watts are not a magnitude of light intensity, and getting more power out from a system than what you're putting in violates the 1st law of thermodinamics, you can't get power out of nothing! \$\endgroup\$
    – Juan Manuel López Manzano
    Aug 7, 2018 at 4:03
  • \$\begingroup\$ LEDs have a non-linear resistance profile, which varies even between components made out from the same silicon die, but the voltage across a LED does matter and will determine a given current acording to the earlier mentioned resistance profile. \$\endgroup\$
    – Juan Manuel López Manzano
    Aug 7, 2018 at 4:07
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    \$\begingroup\$ I didn't mean to be rude! Sorry for that. I viewed the details I pointed out as important, maybe not so much to you because you readily knew what you were talking about, but I believe users searching for answers in this posts might find those sentences misleading! \$\endgroup\$
    – Juan Manuel López Manzano
    Aug 7, 2018 at 5:04
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    \$\begingroup\$ @JuanManuelLópezManzano Watts are not a magnitude of light intensity? When Sparky said "1 watt of light" he was being very specific that the watt was a measurement of light and not electrical power. According to the SI system of units Radiant Flux is measured in Watts. Radiance and Radiant Intensity are also measured in watts, watt per steradian and watt per square meter steradian respectively. \$\endgroup\$
    – Misunderstood
    Aug 7, 2018 at 19:42
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    \$\begingroup\$ @Misunderstood If you think you can get 1 watt of light out of an LED by putting 1/9 watt of power in, you are NOT using "watt", the unit of radiant flux, you are using "watt", the colloquial unit of approximate incandescent bulb equivalence, which is a very different unit. \$\endgroup\$
    – Glenn Willen
    Aug 8, 2018 at 6:41
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LEDs should be driven with nominal current. This is how people accommodate for wide Vf differences.

The simplest circuit is to have much higher (than Delta-Vf) voltage supply and use a resistor of corresponding value. Then the smaller variations in Vf will not change the supply current much, and LED emission changes will be negligible for human eye. But this is not very energy-efficient solution.

A better way is to have a CC (constant-current) source, there are several circuits posted on EE.

Linear CC sources are not efficient solutions, so for medium-to high power LEDs a CC driver based on switchers should be used.

If a resistor or a CC source drives a LED, the forward voltage will appear automatically in accord with LED's inherent I-V diagram, so nobody really cares much about what it is.

More, with 20-30 mA and around 3.3 V, the heat dissipation of this LED will be about 100 mW, which will cause substantial rise of die temperature given the tiny plastic case and no heat transfer path. LED's I-V do have substantial dependence on die temperature, and the Vf will change by another 0.2V.

You just can't control the Vf, and the manufacturing variance in Vf is largely a don't care "for information mostly" parameter.

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  • \$\begingroup\$ Really? Vf is largely a don't care? I'm going to have to disagree with you on that. \$\endgroup\$
    – Misunderstood
    Aug 7, 2018 at 19:31
  • \$\begingroup\$ @Misunderstood, of course it is not completely "don't care", you need to account for it and have sufficient room in CC LED driver for Vf, whether for individual LEDs, or in-series. It probably would be better to say "for information mostly". \$\endgroup\$
    – Ale..chenski
    Aug 7, 2018 at 21:10
  • \$\begingroup\$ +1 for sticking to the important issues without a lot of suppositions. \$\endgroup\$
    – user105652
    Aug 7, 2018 at 21:57
  • \$\begingroup\$ I manufacture strips with strings of 16 LEDs and the Vf has a 2-3V spread. I will know if my CC driver has sufficient overhead only if I take the LED's manufacturing Vf variance into consideration. Not for "information mostly" but an absolute necessity for proper design. \$\endgroup\$
    – Misunderstood
    Aug 9, 2018 at 1:05
  • \$\begingroup\$ @Misunderstood, right, but you only need to know the maximum, to build in sufficient overhead. So the "spread" would be still for "information". A good DC-DC CC switcher-driver will take care of the rest. \$\endgroup\$
    – Ale..chenski
    Aug 9, 2018 at 2:10
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  • There is no change in Vf for white colour temperature, that is purely a control of the phosphor layer over blue LED substrate
  • Your document from a 3rd party supplier copied from some unknown source has a typo error in the published specs for Vf

Vf typ should read Vmin

  • LEDs like all diodes, zeners and saturated transistor switches starts from a "knee shaped" threshold Vt (or Vzt in Zeners) is obvious from VI charts.

  • What causes the voltage to rise above this depends on how small the chip is and some fabrication controls.

    • the slope of rising current is steeper for bigger chips ( and also different scales in VI plots)
  • size of the chip and thus the package power rating are inversely related to the slope of the VI curve.
  • the slope increases with chip size suitable increased power dissipation in the package. above the Vf max is only due to the diode "bulk" resistance
  • when saturated all diodes have this effective series R, ESR and ESR=k/Pmax
  • although the datasheet does not specify this I/V slope or "typ" VI plot due to wide manufacturing variances
    • i.e. quality controls, yield, cost of large epi-wafer and many 2nd sources which may be qualified by a cost saving factory and also historically from supplier process improvements
  • The wider spread is on the high side of Vf so a White LED might be
    3.0V +0.4/-0.2V
    or +0.2/0.1V
    or single bin +/-.1V or less
  • or WORSE as in your case with a spread
    of 1.5V from 2.7 to 4.2V. which I believe is
    3.2V +1.0/-0.5V
    But again I believe these are bogus specs. , mislabelled & no test conditions I expect this should be

old info

ESR nom. varies from 1 / Pmax to 0.25 / Pmax for the best higher power LEDs typ.

So a 5mm 65 mW LED = 15 Ohms typ and a 1W LED is 1 Ohm typical while 3W LEDs are <1/3 Ohm typ. while the best power LEDs are 1/4 of that per diode.

For a Vf using a white 5mm LED at 20mA with Vf=Vt(3mA)+ESR*If @25’C

Assuming Vt~2.8 @2mA. (knee threshold voltage, Vt or ~10% of If nominal current.

ESR= (Vf-Vt)/20mA For Vf = 3.1,3.4
ESR = 15, 30 Ohms

15 Ohms is the expected nominal value of a white 5mm 65mW part.

I would never accept this part with a spec of 4.2V even if the supplier said this was actually at 30 mA.

ESR =(4.2-2.8)/30mA=47 Ohms or 3x the nominal “good quality” value. This is equivalent to +300% tolerance unacceptable by today’s standards. And most likely another error in the spec.

The same inverse power to ESR relationship is true for silicon diodes.

Here are some 1/3W 5050 LEDs enter image description here - best sources are few but exist

If you prefer 100 % single bin White,
This is what I used to sell and still have some leftover in 200 pc bags.
3.0-3.1V @20mA
30 deg 16,000~20,000 mcd ( too bright to watch at close range )
4000’K ~4500’K
( also similar in Red, Yellow, Blue, true green , Aqua)

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    \$\begingroup\$ I think he's referring to the "Min", "Typical" and "Max" values for Vf at rated current. IIRC these typically fall on a fairly tight, centered bell curve and often improve with manufacturing methods. If this is the case, these are simply the Vf range a diode has to test rated current within before being accepted as "good" by the factory, and the Vf(typical) would be one of the averages (mean, median or mode) \$\endgroup\$
    – K H
    Aug 7, 2018 at 2:11
  • \$\begingroup\$ LEDs do NOT have an ESR. \$\endgroup\$
    – Misunderstood
    Aug 7, 2018 at 19:43
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    \$\begingroup\$ Yes when saturated above ~10% of Imax the linear bulk resistance dominates over the quadratic curve such that Vf =Vth+ESR* If is a good fit to any diode curve and ESR std deviation is an epitaxial quality figure of merit. \$\endgroup\$
    – Tony Stewart EE75
    Aug 7, 2018 at 19:47
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    \$\begingroup\$ @KH It very tight (1%) within one epitaxial wafer but on some suppliers very loose between batches (+50/-25%) \$\endgroup\$
    – Tony Stewart EE75
    Aug 7, 2018 at 23:19
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    \$\begingroup\$ You can ignore ESR if you prefer but I find it very useful to understand behaviours and spot errors in data sheets such as this . You won’t find Rce in all data-sheets except when Zetex started to. But you always find Zzt which is ESR in zener diodes. I did this even before Zetex (Diodes Inc) started doing this for switching BJT’s. How specs are released has nothing to do with the physics that exists here. \$\endgroup\$
    – Tony Stewart EE75
    Aug 9, 2018 at 1:05

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