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I am trying to find out if increasing current increases the brightness of an LED.
I'm pretty sure this is the case now that I have looked into semiconductor band theory physics and p-n junctions.
However, I still don't understand how I can calculate the current through an LED if I have been given the brightness and vice versa.
I'm also not sure if I understand exactly why the brightness increases with an increase in current (if it does), but this is probably too big of an ask for someone to explain.

brhans
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Fabajab
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3 Answers3

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Normally we speak of current through and voltage across a component.

The relationship of brightness to current is device dependent and can be done using the datasheet for your LED. As far as I know, an equation that covered all devices would be too general for practical use. Some manufacturers give no information in the datasheet at all, but mostly you'll see a plot like this of a.u. vs mA: (p3): http://docs-asia.electrocomponents.com/webdocs/151e/0900766b8151e40a.pdf

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For the 2nd part of your question:

The power going into a DC component is the current through it times the voltage drop across it. Due to conservation of energy, the LED has to dissipate increases in power somewhere if the current is increased and the voltage drop stays about the same (or perhaps increases slightly). An LED dissipates some of any normal increases in power by becoming brighter (e.g. emits more photons. At least within its normal operating range. All bets are off if you dump enough power into an LED that it smokes or explodes.) Due to less than 100% lighting efficiency, some of the power will also go into heating the LED and the ambient environment around the LED.

hotpaw2
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Assuming you have a resistor in series with the LED, simply measure the voltage across the resistor and use Ohms law.

Current = voltage / resistance

E.g if you have a 1kOhm resistor and measure 5v across it, that's 5mA

FirstTimeEver
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    OP is asking about brightness or luminosity vs current. Not simply measuring current. – Passerby Mar 28 '17 at 00:28
  • Also, when you have an LED you also need to take into account the voltage drop over the LED. – MCG Mar 28 '17 at 13:34