4
\$\begingroup\$

TL;DR - based on the physics of LEDs, does current or power give a better estimation of actual light output of an LED?


I am building a dimmer circuit with a few discrete steps, to operate a COB LED from a constant-current source. I want the steps to appear fairly even to the eye, and am using the "square law" dimming curve relating brightness to dimmer position.

I do not have an accurate way to measure my LED's actual brightness, but can measure current and voltage using my power supply. I initially assumed that current would be the right parameter to use to estimate light output (for the purpose of dividing into equal brightness steps according to the dimming curve), but now I'm wondering if using power would give a more technically accurate result. I see lots of discussion and clarification on using current rather than voltage, but I'm hoping someone familiar with the physics of how LEDs operate can clarify whether power correlates more strongly with lumens than current alone?

At any rate, the results are very similar since the voltage does not vary strongly, and so this is more of an academic question than anything else, but I am curious.

\$\endgroup\$
8
  • 1
    \$\begingroup\$ I think your question may be satisfactorily answered by this other question/answer: electronics.stackexchange.com/q/256336/2028 \$\endgroup\$
    – JYelton
    Commented Aug 3, 2023 at 1:39
  • 1
    \$\begingroup\$ Apparent brightness increases about logarithmically with lumen output. Look at LED data sheets. They show radiant output against current. [I suggest that] It's liable to be a closer fit to eye response than power as Vf rises relatively little for large increases in current. If you have few steps then a discrete step change control may be a good choice. I can make this a more considered answer "sometime" if it doesn't get any others. \$\endgroup\$
    – Russell McMahon
    Commented Aug 3, 2023 at 7:15
  • 1
    \$\begingroup\$ Thanks @user1850479 that is helpful. My LED array drops 10V at 12mA, and 12V at 350mA, so while current is changing much more strongly than voltage, I wasn't sure where to attribute the energy associated with that extra 2V of voltage drop if not the emitted photons (but I'm way over my head on the physics). It makes sense that there are other factors like contact resistance etc. that play a role and turn the extra energy into heat, not light. If I can get access to an accurate light meter I'd like to measure this. \$\endgroup\$
    – JAE
    Commented Aug 3, 2023 at 23:19
  • 1
    \$\begingroup\$ Thanks @RussellMcMahon, it's a good point that the datasheets use current and not power, although some of the datasheets I've been looking at (example) don't include a curve at all! \$\endgroup\$
    – JAE
    Commented Aug 3, 2023 at 23:22
  • 1
    \$\begingroup\$ Fig 13 of your cited datasheet shows relative luminous flux versus drive current. (It's easy to miss such.) \$\endgroup\$
    – Russell McMahon
    Commented Aug 3, 2023 at 23:27

1 Answer 1

1
\$\begingroup\$

After moderator input, I have completely reworked my answer.

Years ago I noticed that my Joule Thief produced the same brightness even after doubling the number of LED's. To me, this proves that brightness is proportional to power. I know datasheets show Luminance proportional to current, but when an LED has multiple dies in it, the 12, 24, or other higher voltage has to be taken into account. So the datasheets are really saying "ignoring voltage, here's the graph of current versus Lumens." And when you're electronically controlling things, current is much better than voltage. But I think power is the best. So I've come to believe that power is more indicative of Lumens, just from my discovery with the Joule Thief.

Here is a simulation that in my mind proves what I am trying to say:

(The Joule Thief is exactly the same, but the LED topology changes).
(The textual summary is the same for both files).

enter image description here

\$\endgroup\$
8
  • 1
    \$\begingroup\$ Saturatio of L1 sets maximum energy per cycle. Cycle time ~ propoprtional to [!/Vin + !/Vout] so 1 Vin dominates so frequency tends to be proportioanl to 1/Vin so Power ~~~~ proportionalto 1/Vin x Isat_L1. || Placing 8 LEDS in 2 x 4 strings or 4 x 2 strings would lower Vout and increase dependence of frequency on load. So also MAYBEpower. Useful to try. \$\endgroup\$
    – Russell McMahon
    Commented Oct 22, 2023 at 6:44
  • \$\begingroup\$ @RussellMcMahon -- Did you actually run my simulations? The truth can easily be found in the results, which anybody can verify with minimal effort, as I have provided the LTSpice simulation file. So did you actually run my simulations? Actually building the circuit and seeing the results would make a believer out of you. \$\endgroup\$ Commented Oct 24, 2023 at 10:44
  • 1
    \$\begingroup\$ Feet almost touching floor. Now only burning candle at both ends :-). ***|| Did YOU try placing LEDs in 2P4S and 4P2S as per my comment above. It may or may not make a believer of you - my simulation was so far only run in my head :-) . (Quite often works well enough. Not always :-)). Note that I was not doubting what you said, but rather suggesting an alternat ive that MAY be useful. || *** My refugee camp latrine lighting proposal was accepted !!! :-) - need to more focus asap. More anon. \$\endgroup\$
    – Russell McMahon
    Commented Oct 30, 2023 at 10:05
  • 1
    \$\begingroup\$ Thanks. Yes - power is not much affected by the change from 4S2P tp 2S4P. I wonder what the magnitude and effects are of changes of frequency and how close to saturation the core is. I'm not asking for more simulations, just interesting questions. When I get into the detail of my lighting project I'll look at if & how this sort of circuit may fit into the operall picture. || My current intention is to try to use single cell (LiIon or more likely LiFePO4) to LED linear or buck drive. Using LiFePO4 and very high efficiency LEDS (so also low Vf) there is enough headroom at V_bat_min with care. - \$\endgroup\$
    – Russell McMahon
    Commented Nov 6, 2023 at 2:33
  • 1
    \$\begingroup\$ ... LEDS are > 200 l/W BUT you only get this is Vsupply approaches Vf - either from battery voltage being ~= Vf or highly efficient buck conversion. At say Vbat = 3.2V and Vf = 2.8V linear loses 4/32 = 12.5% so 200 l/W drops to 175 l/W. A buck converter at say 90% gains you "only" 2.5% and 7.5% at 95% efficiency. That's useful but hopefully not pivotal in a solution that needs to be designed to run for N days without sun. ie more run time is good but overprovisioning of energy may be even better, if affordable. \$\endgroup\$
    – Russell McMahon
    Commented Nov 6, 2023 at 2:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.