I'm experimenting with LEDs with Arduino and have a question. For example; I have one basic 20mA LED that runs constant at 3.3v. I have another LED with that I PWM with a low duty cycle (6%). I up the voltage of this LED by x%~.

I have learned from this experiment that if the second LED passes the 20mA, it burns out. That's why currently it has to run less bright than the first LED. By my calculations (V/A measurement) it comes at 33%~ of wattage used by LED 1.

My question is: say I replace LED 2 with a LED rated with a higher amperage, will I match the brightness of LED 1 with same wattage used? Or will LED 2 still experience more stress and energy loss to heat?

(I intended the question to be applicable to a broad range of LEDs, not just 20mA LEDs)

Duplicate question

For other readers wondering, this is another post with a similar question: Does pulsing an LED at higher current yield greater apparent brightness?


I found out some new things (also thanks to that link); some LEDs are notably more efficient when current is overdriven. Big LEDs are inefficient regarding this.

Also resistors will burn a lot of heat, and limit the rise of current (so a much higher voltage is needed to force the current through the LED than without resistor in front). I noticed one of my COB LEDs has resistors on it, and at roughly the same brightness, the resistors will seriously heat up and overall the COB LED will be inefficient. Perhaps matching the resistor better for the overdriven scenario would work, but at the moment I am relying on my variable voltage supply and no resistor at all.

I seem to have success with a 3W LED, running at 1 W. By eye, the constant and the overdriven PWM'd seem to have the same brightness per watt. My theory is that using a higher rated LED and running at an average lower wattage increases efficacy per Watt; which is generally true but especially so when overdriven. I had less success with a few COB LED strings in parallel with a resistor each, I assume because of the earlier mentioned resistors burning a lot of energy.

  • \$\begingroup\$ Never exceed rated current by more 25% and never exceed absolute max. This applies to PWM and DC. Efficacy (mcd/mA) drops with rising current above rated 20mA \$\endgroup\$ Nov 14, 2017 at 21:17
  • \$\begingroup\$ Average current is the same for both. \$\endgroup\$
    – LongLog
    Nov 14, 2017 at 21:19
  • 1
    \$\begingroup\$ doesn't matter. PWM with 200% I is a good way to blow diodes \$\endgroup\$ Nov 14, 2017 at 21:20
  • 1
    \$\begingroup\$ 5mm parts are 20mA rated and 30mA absolute max pulsed \$\endgroup\$ Nov 14, 2017 at 21:21
  • 1
    \$\begingroup\$ Longlog take a good look as the gold bond wire. Consider that as a fuse and realize why there is a max current rating in addition to Pd. Those are 100mA rated for 10% duty cycle pulses as long as Pd is not exceeded for entire part and Red is most sensitive to Pd. Not all are the same. The ABSmax spec implies 10% of 1kHz or 100us*100mA = 10A-us in fuse time constants \$\endgroup\$ Nov 14, 2017 at 22:12

2 Answers 2


From the comments:

The peak forward current is what matters in this case. In most documents it is shown with a duty cycle of 10% @ 1KHz. The question remains though; if I drive an LED at peak forward current with PWM, will it be as efficient as a constant LED with same average wattage used?

It's not clear from your original post that that's what you are asking. It shows some confused thinking in your mention of doubling voltage.

enter image description here

Source: Purdue Engineering.

To answer your question you need to look at the relative luminous intensity vs. forward current on the datasheet. If it's linear then using PWM at \$n\$ times continuous reference current for \$ \frac {1}{n} \$ duty cycle will result in the same light output.

For the LED shown in Fig 1. indicates that there will be some reduction in light output if the current is pulsed as described in the previous paragraph.

Check the datasheets for the devices you are using.

Update 1

An article in LEDs Magazine titled Pulse-driven LEDs have higher apparent brightness makes interesting reading.

According to an article by Naoshige Shimizu of Nikkei Electronics, a research group at Ehime University, Japan, has developed a pulse drive control method to make LEDs look twice as bright by leveraging the properties of how people perceive brightness.

When a short-cycle pulse voltage with a frequency of approximately 60Hz is applied to an LED at a duty ratio of about 5%, the LED looks about twice as bright to the human eye in comparison with an LED driven by a direct voltage, the research group said.

And further down ...

There are two principles, the Broca-Sulzer effect and the Talbot-Plateau effect, involved in how human eyes perceive brightness. The Broca-Sulzer effect refers to a phenomenon in which light looks several times brighter to the eye than it actually is when exposed to a spark of light, such as a camera flash.

In addition, the Talbot-Plateau effect is a principle where human eyes repeatedly see flashes and sense the average brightness of the repeated lights. Until now, said Jinno, it has been believed that, due to the Talbot-Plateau effect, the brightness perceived by human eyes would not change even if an LED is pulse driven.

  • \$\begingroup\$ I was just looking at that exact graph after you told to look at the sheets. I think you're right with that answer. Thanks a lot for your help, sorry if I wasn't clear enough from the beginning; the terminology escapes me sometimes! \$\endgroup\$
    – LongLog
    Nov 14, 2017 at 22:02
  • 1
    \$\begingroup\$ See update 1. Pulsing may make the LEDs look brighter despite the fall-off on the flux curve. \$\endgroup\$
    – Transistor
    Nov 15, 2017 at 6:02
  • \$\begingroup\$ Awesome, that's very beneficient and good to know! Can be as high as a factor of 2 (for red), that's a lot. \$\endgroup\$
    – LongLog
    Nov 15, 2017 at 10:19
  • \$\begingroup\$ I have a follow up question if you dont mind; I have tried this in practice with a duty cycle of ~3% @ 60Hz. I am comparing constant source vs. the PWM'ed one (driven over peak current, 16V). Both at 2.5W average. The constant source is definitely a lot brighter. None of my parts emit excessive heat. I'm thinking the premise doesn't hold when driven over peak current of the spec (so I need a higher watt LED) or a resistor is burning a lot of heat, or something else. Any ideas? \$\endgroup\$
    – LongLog
    Dec 5, 2017 at 13:18
  • \$\begingroup\$ This Q&A seems to suggest that brightness will fall off, most notably more with high power LEDs. So I guess my options are using more low power LEDs or higher power LEDs that aren't driven over peak. electronics.stackexchange.com/a/26029/81943 \$\endgroup\$
    – LongLog
    Dec 5, 2017 at 13:33


Dec. 5, 2017

What baffles me is that the current rises a lot with little increase in voltage.

My original answer was about the luminous flux differences comparing one LED to another.

Now I think you are not driving the LEDs correctly. It never had anything to do with the PWM duty cycle. The PWM and duty cycle just lowers the average current.

Based on what you said today, it appears you are driving the LEDs with a constant voltage source with no current limiting resistor rather than a constant current driver. If you were using a constant current driver a higher voltage would only impact efficiency. Current would remain the same.

A small increase in voltage, even if a current limiting resistor were used, the current would only increase proportionately. If the LED is connected directly to the voltage power supply a small increase in voltage over the rated forward voltage of the LED would likely damage the LED. The current would spike dramatically until the LED burned up.

Assuming no PWM dimming, you should be using a 15Ω resistor between the LED and a 3.3V supply. 100Ω if using a 5V supply. Anytime you change the voltage you need to change the resistor. Use a resistance calculator like this to get the proper value of resistance: LED Resistance Calculator

BTW, regarding the brightness, use a Cree XP-G3 and you will see a dramatic increase in brightness are the same current. If a color LED, use an OSRAM Olson SSL.

Big LEDs are inefficient regarding this.

It's not the size, it the technology used in the manufacturing process. Today the just released Samsung LM301B is the leader in efficiency at 218 lumens per watt. Last week it was the Samsung LM561 @ 212 lm/W.

In Hi-Power White LEDs 1-3 watt, the Cree XPG3 is the most efficient at 185 lumens per watt.

Even the CoBs are now getting very efficient. Today the 100W Luminus Devices CXM CoB is the leader at 176 lumens per watt.

resistors will burn a lot of heat, and limit the rise of current (so a much higher voltage is needed to force the current through the LED than without resistor in front)

Not if the value of the resistor is properly selected. You can easily get 99% with tight tolerances on the supply voltage.

By eye, the constant and the overdriven PWM'd seem to have the same brightness per watt.

Highly doubtful use can differentiate the difference in brightness with the naked eye. You need a good spectrometer to measure the number of photons.

My theory is that using a higher rated LED and running at an average lower wattage increases efficacy per Watt;

Almost always an LED performs at its peak efficiency when run at or below its "Test Current". The higher you go above its test current,the less efficient it gets. I have seen 5 Amp LEDs rated with a test current of 1 Amp.

End of Update

Nov. 14, 2017

will I match the brightness of LED 1 with same wattage used?

Not necessarily. Wattage is not the correct criteria.

LEDs are rated by their emitted flux in Candela, Lumens, Photons (µmol), and/or Radiant power (mW) at a specific current and temperature.

For example there are many 1 Watt LEDs. At the same electrical wattage (Vf x current) another LED will have a different emitted flux.

LEDs produce both radiated flux (light) and non-radiated flux (heat). The sum of these two flux types will equal the electrical wattage.

The ratio of the radiated flux to electrical wattage is the LED's efficacy.

The forward current is only one factor in the wattage. Forward voltage is the other. You need to look at both factors.

All LEDs have a range (min typ max) of emitted flux at a specific current.

When comparing LEDs you cannot just look at the Relative Flux vs. Current but also factor in the forward voltage at the target current and temperature.

Also you must be aware the relative flux curves can be Luminous (Photometric) or Radiant (Radiometric).

Forward Voltage and Temperature

enter image description here

Forward Current and Temperature

enter image description here

Luminous and Radiant

enter image description here

  • 1
    \$\begingroup\$ What baffles me is that the current rises a lot with little increase in voltage. Luminous flux vs current is pretty much linear in most cases. I would think that applying a slightly higher voltage, leading a large increase of current, would net a higher luminous flux for total wattage used (since P = V x A). However this doesn't seem the case in practice. \$\endgroup\$
    – LongLog
    Dec 5, 2017 at 13:41
  • \$\begingroup\$ @LongLog See the update to my original answer. \$\endgroup\$ Dec 5, 2017 at 18:33
  • \$\begingroup\$ Thanks Misunderstood. I'm not using a constant current source, but I am monitoring the voltage & current with a variable voltage supply/labsupply. Please also see the referenced link in the edit of the post; I found it very informative regarding the issue. I'll place some extra updates in the original post, to keep it concise and informative to others. I'll address your comments in there as well. \$\endgroup\$
    – LongLog
    Dec 5, 2017 at 19:12
  • \$\begingroup\$ "It's not the size, it the technology used in the manufacturing process." You say that, but what exactly is it then? I obviously know it's not just size, but there is a distinction that higher rated LEDs have a bigger falloff. "Not if the value of the resistor is properly selected." I figured, but haven't gotten around to selecting proper resistors. "Highly doubtful use can differentiate the difference in brightness" To an extent I can yes. "The higher you go above its test current..." Yes, but it depends on the LED how big the fall off is. \$\endgroup\$
    – LongLog
    Dec 5, 2017 at 20:28

Your Answer

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

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