I keep reading different sources and have little understanding of the vocabulary of resistors and LEDS. Let's say I have a LED that uses 3v and 30mA, if I use a resistor to give it 1.5V at 30mA it will be half as bright; what if I give it 3v, but give it 15mA. Will it be half as bright? Please explain, and maybe add a little meaning to the vocalbulary like 'load' etc. Thanks!
\$\begingroup\$
\$\endgroup\$
2
-
6\$\begingroup\$ Your title has the idea completely swapped - LED current dictates brightness. \$\endgroup\$– W5VOCommented Jan 21, 2017 at 21:42
-
5\$\begingroup\$ You can't independently vary the input current and voltage of an LED or any other device. The two (for an LED or other diode) are related by the Shockley diode equation. \$\endgroup\$– The PhotonCommented Jan 21, 2017 at 21:58
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
5
You cannot control voltage and current independently. Here's the first image result I got for "LED voltage-current curve": 
If your LED is at 3V when current is 30mA, it'll probably be somewhere around 2.9V at 15mA.
Now that this is cleared up, LED brightness is roughly proportional (in photon count) to the current, so with 15mA you'd produce about half the brightness than if your LED were at 30mA. It won't look that way though, since our vision is kinda logarithmic, so half the photons doesn't look "half as bright".
-
3\$\begingroup\$ > Here's the first image result I got for "LED voltage-current curve": that chart is likely just for illustration purposes. the If/Vfwd relationship is monotonic. The green curve on that chart isn't. \$\endgroup\$– dannyfCommented Jan 21, 2017 at 22:15
-
\$\begingroup\$ > Now that this is cleared up, LED brightness is roughly proportional (in photon count) to the current. the If/Vfwd non-linearity, for lighting leds, is actually overplayed by most people. if you look at lighting LEDs' datasheets, you will find that If / Vfwd is actually fairly linear, as is (relative) lumen / If. \$\endgroup\$– dannyfCommented Jan 21, 2017 at 22:17
-
\$\begingroup\$ so for those lighting leds working around their target range, they behave surprisingly like resistors. \$\endgroup\$– dannyfCommented Jan 21, 2017 at 22:17
-
\$\begingroup\$ And the "white" curve does not exist. There are no "white" LEDs, just blue or UV LEDs with luminescent coating. \$\endgroup\$– JankaCommented Jan 21, 2017 at 22:18
-
1\$\begingroup\$ minor errors with curves due to ESR vs Pd of the device. As we know white LED's are actually same as blue but with yellow red phosphor which latter part has no effect on Vf. green negative ESR is an error due to measurement method as vf drops with rising current due to temp and all V/I curves are done at 25'C. Also Red GaAs vs AlGaAs is 1.2 vs 2V. where ESR is approx 1/Pd in the DeltaV/Delta I at rated I. \$\endgroup\$ Commented Jan 22, 2017 at 3:33
\$\begingroup\$
\$\endgroup\$
1
if the led has a forward voltage drop of 3v @ 30ma, it will have a lower voltage drop if you apply 15ma to it; or it will have a lower current if you apply 1.5v to it.
it is that simple.
-
\$\begingroup\$ It's worth adding that an LED with a nominal 3 V forward voltage is going to take very very little current at 1.5 V, and not produce any noticeable light. \$\endgroup\$ Commented Jan 21, 2017 at 21:55