More directly : passing electricity that is 0.5 times the amperage of the one required will make the LED half bright, right? Is it the same with halving its voltage requirement? Or will it not turn on at all if it doesn't half 100% of its required voltage?

And similarly : What kills an LED? Powering it at 25mA where it demands 20mA ? Is it the same with voltage : Does powering it at 3V when it requiers 2V kill it?

  • \$\begingroup\$ first, note that LED is, electrically, just a diode; as such, you'd be well advised to learn the basics of semiconductor operation (e.g. understand voltage drop properly) before trying to understand LED's specifics. That should cover most of your questions indeed. \$\endgroup\$
    – user20088
    Jan 31, 2016 at 22:18

2 Answers 2


Forward voltage versus current: -

enter image description here

Below maybe 1.5 volts you might not see any residual glow from an LED with your eyes and as expected the current (Y axis) is sub 1mA. If you doubled that voltage to 3V, the current would be off-the-scale in the Y axis and, for a normal standard LED this spells the end of the road.

All LEDs have this knee voltage thing and this is why we choose to control the current through LED rather than try and control the voltage across it.

Luminous intensity versus current typical example: -

enter image description here

On this example (right hand graph) the luminous intensity increases linearly with current but as currents rise beyond the scope of that graph you get less payback. Here's an example showing light output versus current and, quantum efficiency versus current: -

enter image description here

Efficiency reaches a peak at some moderate amount of current but the payback gets less as you rise above this value. Always read the data sheet on a specific LED - the graphs above are very generic.

  • 1
    \$\begingroup\$ Second graph shows light output IS linear with current : 20-30mA is a 50% increase in both current and luminous intensity. (Third and fourth graphs suggest otherwise, presumably for different LEDS - are these white LEDs with a phosphor secondary stage? \$\endgroup\$ Jan 31, 2016 at 12:56
  • \$\begingroup\$ @BrianDrummond oops I tried to show too much on the 2nd picture - I'll reword. \$\endgroup\$
    – Andy aka
    Jan 31, 2016 at 13:04
  • 1
    \$\begingroup\$ Just a small note: although luminous intensity does increase linearly with current, perceived brightness is more of a logarithmic scale. \$\endgroup\$
    – brhans
    Jan 31, 2016 at 14:03

In addition to @Andy aka's excellent answer:

  • Halving the amperage to an led will generally about halve its light output (but this assumes it's currently being run in a point in its current/output chart with a near-linear response curve).
  • In general, changing the voltage to an LED does change the current, but without a series resistor the LED's non-linear resistance makes this very sloppy.
  • Two main things (besides physical/thermal abuse) generally kill LEDs before simple aging:
    • Overcurrent - Whether due to slight overvoltage without a series resistor (VERY easy to do), pushing too much current by overvoltage with a series resistor, or too much current through a current-contol device, any time you push more than rated max current through an LED, you're damaging it.
    • Reverse Polarity Breakdown - LEDs (Light Emitting Diodes) are still diodes (and NOT zener diodes), so hitting them with a reverse-polarity voltage above their reverse-breakdown threshold will generally kill them instantly!

  • \$\begingroup\$ Having received my BEE degree in 1978, I'm surprised that the term 'amperage' is used here instead of 'current'. As I recall, current is measured in amperes. ;) \$\endgroup\$ Jan 31, 2016 at 17:10
  • \$\begingroup\$ @joelgoldstick You're correct; I simply used 'amperage' in the first bullet in following the wording convention of OP's question which that answered. \$\endgroup\$ Jan 31, 2016 at 17:54

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