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This question implies two assumptions:

  1. The frequency of a switched LED driver is high enough that average power, not instantaneous power, should be used to determine maximum drive current.
  2. The limiting factor which determines maximum current at any duty cycle is average power dissipation.

After these assumptions, it is obvious that the current through an LED at maximum power dissipation is inversely related to the duty cycle.

Is the apparent brightness (not necessarily the luminosity) increased, decreased, or not affected by pulsing an LED at higher current and lower duty cycle?

I do not have any particular LEDs or driver topologies in mind, but would welcome references to real parts, datasheets, or appnotes. I would also be interested in knowing if this varies between low-power (say, 20mA) indicator LEDs and high-power, high-brightness lighting LEDs.

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I wanted to ask this too! :-) –  stevenvh Jul 28 '11 at 14:52
    
Of course you did! You even stated the question in the comments here. (Note: That answer and discussion is about driving with a half-rectified 60 Hz AC signal, I'm more interested in pulsed DC here) –  Kevin Vermeer Jul 28 '11 at 15:07
    
Tough question, I think it might have to do with the Weber-Fechner law... –  NickHalden Jul 28 '11 at 15:39
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@jsolarski: Pulse some LEDs at different duty cycles and brightnesses, and (without telling them which was which) ask people which one is brightest? –  Kevin Vermeer Jul 30 '11 at 1:09
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@jsolarski - aka "blind testing", though for this test it sounds a bit funny :-). In "double blind testing" the test person still doesn't know which of the LEDs she sees, but not even the person operating the switches does. –  stevenvh Jul 30 '11 at 5:28

6 Answers 6

I have looked at this in some detail in the past as I design LED based solar charged lights and am generally interested in LEDs.

Firstly, human perception at constant power and variable duty cycle pulses. A say 10% duty cycle would result in 10 x the current at the same voltage for this to hold. Real LEDs will have somewhat higher forward voltages when current is increased by 10x but not greatly so. A fair test is probably Ipeak x time on = constant.

  • In the distant past it was alleged that the human eye response was such that pulsing LEDs at constant power but at low duty cycles resulted in greater apparent brightness. AFAIR the reference was in an HP document.

  • Quite recently I have read just the opposite from a moderately authoritative but unremembered source.

I can probably find the recent document, but the HP one will be lost in the mists of time. However, I believe that any physiological effect ether way is small. Given that you need about a 2:1 change in LED brightness for it to be noticeable when LEDs are viewed separately (one or other but not both together), small differences will certainly not be noticeable. Where eg two flashlights are shone side by side on a general scene so that direct comparison can be made you may need about 1.5:1+ difference before the difference is noticeable - this depends somewhat on the observer. When two lights are used in "wall washing" on a smooth wall, side by side differences of down to about 20% may be discernible.

Secondly - actual brightness.

Using constant mean current, total light output falls for pulsed operation and is lower for increasingly low duty cycle! The effect is even worse for constant mean power !!

Both of these effects can be clearly seen by examining the data sheets of target LEDs. Luminous output per current curves are close to straight lines but curve towards decreasing output per mA as current increases. ie doubling current does not quite double luminous output. This decreasing rate of return accelerates as current increases. ie an LED operated at well below its rated current produces more lumen/mA than at rated current with increasing efficiency with decreasing mA.

Output (lumen) per Watt is even worse than lumen per mA. As mA increase Vf also increases so the Vf x I product increases at a faster rate per lumen than just I does. So, again, maximum lumen/Watt is achieved at low mA compared to rated mA and lumen/Watt efficiency improves with decreasing current.

Both these effects can be seen in the following graphs.

enter image description here

These curves are for the utterly marvellous [tm] Nichia NSPWR70CSS-K1 LED mentioned below. Even though this LED is rated at 60 mA absolute maximum and 50 mA continuous max Nichia have kindly specified it's performance up to 150 mA. Longevity at these current is "not guaranteed". This is about the most efficient <= 50 mA LED available. If anyone knows of any with a superior l/W at 50 mA and in the same price range, please advise!

I use the Nichia "Raijin" NSPWR70CSS-K1 LED in several products. This started life as an 30 mA LED but was uprated to 50 mA by Nichia after testing (with reduced lifetime of 14,000 hours). At 50 mA it delivers about 120 l/W and at 20 mA about 165 l/W. The latter figure puts it amongst the very best real world products available, although recent offerings are now exceeding this value at well below rated currents.

A complicating factor is that modern high power LEDs are often rated for Iabsolute_max values perhaps 20% above Imax_operating. ie it is not possible to operate them in a pulsed mode at less than about 90% duty cycle and constant mean current without exceeding their rated absolute maximum currents. This does not means that they cannot be pulsed at many times their rated maximum continuous currents (ask me how I know :-) ) just that the manufacturer does not certify the results. The Raijin LED is VERY bright at 100 mA.

Special case.

One area where pulsing at very high currents and low duty cycles may make sense is where the LED is rated for this sort of duty and the instantaneous luminous output (brightness) is of more importance than the mean brightness. A commonly encountered example is in Infra Red (IR) controllers where the brightness of each individual pulse is important as individual pulses are detected and the mean level is irrelevant In such cases pulses of 1 amp plus may be used. The limiting current in such cases may be the bond wire fusing currents. The effect on the LED die will be a shortening of lifetime but thi is (presumably) allowed for by th manufacturer in the specification - and required total operating lifetime is usually low. (eg a TV remote control which is used for 0.1 seconds x say 50 pulses per hour for 4 hours per day gets about2 hours of on time per year.


Effective illuminance improvement of a light source by using pulse modulation and its psychophysical effect on the human eye. EHIME university 2008

Enddolith cited a paper that claimed a substantial true visual gain under certain conditions. Here's a full version of the Jinno Motomura paper cited

They are claiming an up to ~ 2:1 true lumen gain (as lumens relate to eye response) at 5% duty cycle but despite the great care they have taken there are some major uncertainties when translating this to real world applications.

  • They seem to place very high emphasis on fast rise and fall times. Are these met when illuminating real world scenes, does it matter? and are there selected examples where it will work better than others?

  • This is looking at LEDs directly (with remaining good eye?) and comparing apparent brightness. How does this translate to light levels reaching observer after scene reflection.

  • How does this apply when the LEDs are used to illuminate targets. Are the average luminance levels from a target compared to direct LED observation going to affect results? By how much?

  • As modern eg White LEDs have Imax_max ~= 110% of I_max_ continuous, and as this effect seems to depend on ~5% duty cycle, has this got any implications for similar real world LEDs at large percentages of rated current?

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@RusselMcMahon, as you reduce your duty cycle you can increase your current because there will be less time for heating affects to kick in. I have done this with a very very short duty cycle using a pulsed laser power supply at 2kV on a normal 1.9V diode. This was pulsing so often that they eye cannot tell. the brightness of the diode increased significantly. My professor and I built a power supply that had a short duty cycle and only overvoltaged the diode to a little over an amp of current and successfully made a normal diode look brighter then a 1W LED when it was only a cheap LED. –  Kortuk Jul 28 '11 at 16:09
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It would be interesting to know actual current, Voltage, duty cycle figures for pulse and DC there so relative power and mean current comparisons could be made. –  Russell McMahon Jul 28 '11 at 16:15
    
the mean current is increasing, I believe. I would like to rebuild this and carefully monitor it now. The advantage and target was higher light output. The duty cycle causes reduced thermal buildup in the LED so you are supposedly getting higher efficiency, but your graphs make me doubt myself. –  Kortuk Jul 28 '11 at 18:54
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Many of my graphs often make me doubt myself :-). Life tends to be like that. The more I know the less I know. By now I know so much that I know just about nothing. Relatively, anyway. –  Russell McMahon Jul 28 '11 at 19:18
    
Was the HP document Appnote 1113? There's a copy of that here, but it's about using pulsed current to increase transmission distance for IR emitters. If not, hpmemory.org seems to be a good place to look for that. Could you find that more recent document? Thanks! –  Kevin Vermeer Jul 31 '11 at 17:13

There seems to be a lot of misinformation around this area. Some say there is a visual effect that pulsing light is perceived brighter than its average level. As far as I can tell, there is some disagreement about this but it applies to rather slow flashing such that the persistance of vision carries the brightness between pulses. This is in the range of a few Hz to low 10s of Hz. I'm not sure there is a consensus about whether this is really perceived as brighter, or that its only more attention getting.

Fast flashing such that the light looks steady (a few 100 Hz) apparently does not increase perceived brightness. What you perceive is the average brightness. That means that a fast flashing LED is actually less bright at the same average power. LED brightness is roughly proportional to current, but higher current also causes a larger forward voltage drop. 10mA continuous and 20mA for 50% at 1 kHz will look pretty close to the same, but the latter will take more power since the voltage drop at 20mA will be higher than at 10mA.

LEDs brightness is mostly proportional to current, but not completely. It does fall off a bit usually with current, but for most indicator type LEDs this effect is so small as to be unnoticable. Humans perceive light intensity logarithmically. A factor of 2 looks like a small but clearly noticeable step. 10% is impossible to notice except in direct comparison.

High power LEDs used for lighting are pushing limits in a different way and exhibit more falloff with higher current. Maximum efficiency and maximum brightness are not the same thing. This difference is enough to matter in demanding applications. This is where you have to check the LED datasheet carefully. High power lighting LEDs usually have figures for brightness as a function of current, and you will see this tail off a bit at the high end. Also note that for those LEDs the instantaneous max current is closer to the average max than for small indicator LEDs. A lot of this has to do with temperature and heat management.

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We'll have to start dividing up answer space between us. This can get very inefficient :-). Some synergy but also lots of overlap (as would be expected). –  Russell McMahon Jul 28 '11 at 15:58
    
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Here's a full version of the Jinno Motomura paper cited Thy are claiming an up to ~ 2:1 true lumen gain (as lumens relate to eye response) at 5% duty cycle but despite the great care they have taken there are some major uncertainties when translating this to real worl applications. –  Russell McMahon Aug 1 '11 at 2:01

I've always learned and been convinced that LED current above nominal (often around 20mA for a common LED) will cause a higher luminosity, but less than proportional, and that it's not worth the current to do so. If this is the case pulsing won't get you more brightness. Suppose a LED with 0.45mcd @ 10mA, and 0.9mcd @ 40mA. Pulsed at 40mA with 25% duty cycle will give an average current of 10mA, and an average luminosity of 0.225mcd, that's only half the luminosity we'd get at 10mA continuous.
I didn't make up these figures. They can be found in the Panasonic LN222RPX datasheet:

enter image description here

I want to make two notes here:

  1. half the value seems like a big difference, but you'll have to remember that our eye perceives light intensities logarithmically; if doubling the intensity is 1 step then the difference between a dimly lit room (10 lux) and bright sunlight (100 000 lux) is only 13 steps. One step will be less noticeable than the numbers suggest.
  2. There's also the other graph, forward current vs forward voltage. Like for any other diode the voltage increases with increasing current. That means that power dissipation in the LED will increase more than proportionally with rising current.

If we stop here we could conclude that pulsed current is worse than continuous current, luminosity-wise and power-wise. BUT!

Kevin came with this graph from a Kingbright datasheet:

enter image description here

That curve is damn straight! For this LED (and others from Kingbright I checked) luminosity is perfectly linear with current, so pulsing should give the same result as continuous current.

conclusion
Apparently not all LEDs are made equal. While it makes no difference whether you pulse or not for some LEDs pulsing may give worse performance for others. However, I haven't found LEDs where performance increases when pulsing.

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I think it's not so much that some LEDs have linear brightness as a function of current whereas with others it tails off at high current. I think all LEDs do this, but for some the normal operating range is in the low linear region. I think Kingbright dumbed it down to a straight line because at 30mA that LED sees little falloff. If you could run it at higher currents I'm sure there would be falloff, but there are other limitations to prevent you from doing that, or more likely that Kingbright doesn't want to bother specing at. –  Olin Lathrop Jul 29 '11 at 15:53
    
@Olin - I see what you mean, but for the Kingbright it's perfectly directly proportional over the full DC range (30mA is Absolute Maximum DC current), and the Panasonic doesn't show this even at low currents. –  stevenvh Jul 29 '11 at 16:02
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I suspect Kingbright simplified the data. –  Olin Lathrop Jul 29 '11 at 17:05
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@Steven: I agree with how datasheets should be. However, over the past 4 years or so I have had specific occasion to pore over a very large number of Asian data sheets to try to determine both if the products were acceptable and also whether the datasheets were credible. In many cases Asian datasheets from other than the very top players are suspect - especially so with LED products. I'd expect Kingbright to be better behaved than that, as they are reasonably large and well known, but I have about zero doubt that that data sheet is wrong. Here more "we just don't care" than outright fraud. –  Russell McMahon Aug 1 '11 at 1:39
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"our eye perceives light intensities logarithmically" I believe this fact is what causes so much confusion about LED intensity. People PWM an LED at 50% power and notice it does not look "that much" dimmer than 100%, and so assume it is due to persistence of vision. –  bt2 Sep 24 '11 at 23:14

Assuming an LED is on for a constant amount of time, then the brightness is proportional to the current flowing through the diode (either linear or exponential). For the sake of this argument, let us assume it is linear (you need to find the voltage vs. current characteristics from the manufacturer's datasheets to determine what it will be for your particular operating range).

Also, for the sake of this argument, I will assume that the PWM frequency is high enough that you will not notice any visible flicker at any duty cycle.

You can also alter the brightness of an LED at a constant current by varying the duty cycle. A 50% reduction in the duty cycle is a 50% reduction in brightness. This also means that the LED is only on for half the time that it was, and assuming that your current/voltage source is not affected by loading/switching, the average current the LED uses over a given interval will also be directly halved.

Is the brightness increased, decreased, or not affected by pulsing an LED at higher current and lower duty cycle?

That all depends, since there is an inherent contradiction there. By pulsing the LED at a lower duty cycle, you effectively lower the average current. If you just decreased your current limiting resistor to allow more current to flow, and didn't modify the duty cycle, the brightness would increase. Thus, the change in brightness would be a function of both the current and change in duty cycle.

You could calculate the new brightness as:

new_brightness = old_brightness * new_average_current / old_average_current

or in other words

new_brightness = old_brightness * ( new_peak_current * new_PWM_duty_cycle ) / ( old_peak_current * old_PWM_duty_cycle)

Since you are lowering the PWM duty cycle but raising the current, the new PWM Duty Cycle should be less then 1 but greater then 0 (implicitly convert it from a percentage to decimal) , and the ratios in current should be a positive number greater then 1.

Thus, if you halve the duty cycle but maintain the same average current, your brightness remains the same (at the expense of a higher instantaneous current flow through your LED, which may not be desirable).

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"brightness is directly proportional to the current". Well, apparently not for all LEDs, see the first graph in my answer. –  stevenvh Jul 29 '11 at 14:38
    
I suppose that it depends on the construction of the diode, and whether or not you can approximate the operating region as a linear curve (since all diodes are technically exponential devices). –  Breakthrough Jul 29 '11 at 14:39
    
@Break: You are confusing voltage and current. Diodes are exponential when relating their voltage and current. However, this does not inherently related how a LED's brightness changes as a function of current. For small currents, the light output of a LED is pretty linear with current. At high currents, secondary effects become more significant and further increases in current result in less increase in brightness. –  Olin Lathrop Jul 29 '11 at 15:17
    
@Olin Lathrop correct, I forgot about Mr. Shockley's equation. –  Breakthrough Jul 29 '11 at 15:19

A completely subjective analysis:

In trying to maximize the output of an infrared LED at 38 kHz, I experimented with a 5 mm visible red led rated 3500 mcd, 1.85 V @ 20mA (3.7 mW). Switching was done with two 2N7000 MOSFETs in parallel, with a gate voltage of roughly 3.0 V.

1 / Freq = on-time + off-time

I varied the on-time from 10% to 50% while powered first by 3.3 V then 5.0 V. Observed brightness increased with both increasing duty-cycle and voltage.

There was a noticeable brightness increase using two MOSFET over using one, plus two were required at 5.0 V given the amount of heat generated when using only one.

Measured LED voltages and currents at this frequency and duty cycle are unreliable with my DMM, but did manage to get one reading of 2 Volts @ 120 mA (240 mW), though take that with a huge grain of salt.

I feel comfortable running these LEDs continuously indefinitely at 5 Volts and 40% duty cycle @ 38 kHz. At 5 V and 50% duty cycle, they get a bit too toasty for longevity.

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The core issue is if you use a higher voltage and shorter duty cycle so you deliver the same average power to the LED does it become brighter or stay the same? –  Kortuk Sep 27 '12 at 21:53

Yes. Once the LED is far enough away (or apparent image occluded) that the variance falls close to noise, no. (And nevermind Shockley if you happen to have an excellent quantum-mechanical model included in the specsheet!) Hasn't anyone taken a photo of you with their LED flash (i.e. camera vintage 2006 or later) active?

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How is this a answer? I can't even figure out what your second sentence is trying to say. It makes no sense, at least in english. –  Olin Lathrop Sep 24 '11 at 13:13

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