LEDs have varying forward voltages, and the range can be significant.

Example: OSRAM LRTBGVSR RGB LED OSRAM LRTBGVSR : V.f @ I.f = 20 [mA] See the bottom line.

What is not provided is the statistical distribution of these tolerances.

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How it is related

I am designing a battery-powered indicator LED matrix. Due to the battery, efficiency is critical; however, as there is only one LED per indicator (no redundancies), all LEDs functioning is paramount.

This becomes an optimization problem:

Do I want to design my drive with respect to the maximum possible forward voltage of the part, with the understanding that the great majority of parts will consistently burn off the overvoltage?

(Remember: These are not simply losses of usable battery power — these losses also become heat generated within the driver IC. Ick.)

Do I instead make the bus voltage variable and form tests which determine whether the full current is going out to every LED (increasing the bus voltage until it does)?

Here, if the forward voltage tolerance distribution was known, I could at least try to select an initial driving voltage which, at some cost to efficiency (but no longer maximum cost), drove all LEDs for most units, leaving a statistically manageable number of outliers afterward.

Actual Question

Do LED forward voltage tolerance distributions tend to be equivalent, independent of the part series?

  • If so, where could these distributions be found?
  • If not, would manufacturers tend to know for specific parts?

Additional Information

The minimum LED driving voltage for my driver is equal to the forward voltage of the LED \$V_{led.f}\$ plus a specified value, \$V_{driver.channel.knee}\$.

Example : TI TLC6983 (Constant current, PWM-capable LED driver)

TI TLC6983 : V.knee (table)

TI TLC6983 : V.knee (figure)

  • \$\begingroup\$ you also lose power changing voltages around to feed the LED, so there's no free lunch. It is best to feed something close the the vf max, and resistor from there. if you want to save batt, drive it at 1ma instead of 20, for 20x more efficiency. \$\endgroup\$
    – dandavis
    Nov 14, 2023 at 21:10
  • \$\begingroup\$ You are misunderstanding that specification. That is the range of values the chip can have and be considered "working". Lower voltage is a better quality chip (higher efficiency), but you do not get to "select" that value (aside from buying a lot and testing for the higher quality chips). \$\endgroup\$ Nov 14, 2023 at 21:12
  • \$\begingroup\$ @dandavis : I'm driving optimal current and then using PWM for brightness control. \$\endgroup\$
    – kando
    Nov 14, 2023 at 22:58

1 Answer 1


is it generally recommended to design according to the maximum forward voltage of the LED such that all LEDs will operate?

Not at all.

LEDs need to be driven by a constant current. The drive voltage is variable and highly depends on the LED type, production batch, and temperature.

Driving LEDs with a fixed voltage is a nice way of blowing up the LED. Don't do that.

The forward voltage (Vf) in the datasheet is there to help you know what compliance voltage range the current source needs. That is, Vf spec is a voltages range you can expect to measure across the LED when driving it with a specified constant current. In the case of the datasheet you referenced, they specify the forward voltage at a 20mA constant current through the LED.

That does not mean that you have to drive the LED with a 20mA current. Modern LEDs will be very bright at such a current, and plenty of LEDs provide plenty of light in the 1mA-5mA current range. Try it for yourself.

A common way of converting constant voltage into roughly constant current is using a resistor in series with the LED.

the losses can be significant for a portable (battery-powered) device

When ultra-high efficiency is desirable, and a series resistor is too wasteful (is it? calculate how much power it dissipates!), the LED should be driven by a switching constant current driver. Those are available in single chip, and are easy to apply.

  • \$\begingroup\$ I intend to use a constant current driver IC (TI TLC6983) for which I intend to use the datasheet's recommended current and PWM the brightness down as needed from there; however, this requires selection of a bus voltage roughly equivalent to V.LED.f plus some margin defined by the driver V.driver.knee. (In my specific case, it will be a large array of indicators, so losses will be on a large multiplier.) \$\endgroup\$
    – kando
    Nov 14, 2023 at 21:10
  • \$\begingroup\$ @kando Well, what is your question then (I don't understand what you really need)? You've said exactly what you need to do :) Use the highest value of Vf given in the datasheet, since that's the most critical one when determining the minimum bus voltage. Please also edit the missing details into the question itself - they don't belong in the comments. \$\endgroup\$ Nov 14, 2023 at 21:11
  • \$\begingroup\$ @kando Any supply between 3.3V and 5.0V will work for your application. \$\endgroup\$ Nov 14, 2023 at 21:14
  • 1
    \$\begingroup\$ @Kubahasn'tforgottenMonica The question asks if the design should take maximum Vf into account, your first answer is "No" but then your bulk text basically says "Yes" because driving LED with say constant current does need to take max Vf into account. \$\endgroup\$
    – Justme
    Nov 14, 2023 at 21:15

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