Human eye, peak or averaging detector?

weirdjim

weirdjim

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weirdjim
Away back in the days when dinosaurs roamed the earth, HP came out with an optics guide for their new LED digits. They said that you could run them at DC and get a certain brightness, or run them on a fast repeating pulse width modulation to keep the average current and hence the thermals from destroying the device but get more perceived light out of them.

I've never had to look to see if the bibliography HP published (I think there was one) showed any proof that this is true, but a little experimentation shows me that it is so. However, I've been challenged by something I wrote a few months ago and need a source to show whether or not this is true.

Anybody got a source?

Thanks,

JIm
 
I am no expert on it, but human vision is not exactly linear. If it were, then a pulsing light will appear to have the same brightness as its DC average. You can search for "nonlinear response of the eye" and find many papers on it.

Additionally, if you pulse an LED hard, it will also exhibit its own nonlinearity. At high currents, an LED can produce amplified spontaneous emission (similar to a laser), which increases its output efficiency.
 
Vision does average according to (roughly) a root mean square law. Vision is also roughly logarithmic, meaning that apparent "brightness" and actual illumination aren't the same.

Same for hearing, by the way, which is definitely log.

If I wasn't in the car I'd get you a hard cite.

Edit: Google "weber-fechner law."
 
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Also look up "critical fusion frequency of flicker" or "flicker fusion threshold". If you moved those old calculators fast enough, you'd see the digits "repeating" across whatever you were looking at. It's related to what made TV possible with cathode ray tubes, you'll know the phosphors are "activated" by a single electron beam. The phosphors had to "decay" fast enough that the picture wouldn't lag. The connection is that we have a persistence of vision which hides the pulses of light from us.
 
The eye perceives everything that happens more than 24 "frames" a second as continuous motion. What I'm asking is that if I light an LED with a specific DC current and then (for example) doubling the current with a square wave MUCH faster than 24pps (identical power) if the second LED is perceived to be brighter than the first. Theoretically.

Jim
 
weirdjim said:
The eye perceives everything that happens more than 24 "frames" a second as continuous motion. What I'm asking is that if I light an LED with a specific DC current and then (for example) doubling the current with a square wave MUCH faster than 24pps (identical power) if the second LED is perceived to be brighter than the first. Theoretically.

Jim
Click to expand...
Let me try to rephrase the question, because I see things which seem to be contradictions, which may simply be that I'm not reading it correctly.
Lets say I have a pulse train of one amp, where each pulse occurs 24 times a second, for one millisecond for each pulse. My pulses are square waves of one volt each.
I then keep the voltage and pulse width the same, but now I do 48 pulses per second. My current doubles. Isn't my power (voltage x current) now doubled?
The question is whether the LED running at 48 pulses per second appears brighter than one running at 24 pulses per second?

You may be asking about the Broca-Sulzer effect.
Here are possibly some newer references:
https://www.sciencedaily.com/releases/2012/11/121115152659.htm
https://www.ledsmagazine.com/home/article/16698062/pulsedriven-leds-have-higher-apparent-brightness

This one seems to cover many different aspects of optimizing pulse width modulation for maximum brightness:
http://www2.isu.edu.tw/upload/223/9/MOE/paper/wang-bing-chien.pdf

I see nothing that contradicts your original post, but it is a bit more complex than simply increasing the number of pulses- perhaps that's why the "challenge" you mentioned.
 
What Cap'n Jack said. You can vary apparent brightness by varying pulse width, and yes, short-but-strong pulses can result in a "brighter" looking LED. Heck, I didn't know the name of the effect. I larn't somethin. :)

As far as a specific source for HP's claims, I don't know of one. Cap'n Jack's links should have what you want, though. The LEDS Magazine article looked pretty good. Go to Dialight's Website and see if they don't have some white papers that are just filled with geek math and stuff.

As for the current (I assume you're referring to power usage/dissipation) ... again, that's a root mean square function. Easy way is to use a True RMS meter to measure the averaged current over several seconds to know how much power you're using. Then go to pure DC and you'll probably need more power to get the same perceived brightness.
 
weirdjim said:
What I'm asking is that if I light an LED with a specific DC current and then (for example) doubling the current with a square wave MUCH faster than 24pps (identical power) if the second LED is perceived to be brighter than the first. Theoretically.

Jim
Click to expand...

Except the power will not be identical. If the current is doubled, the voltage drop will also increase. Thus the average power applied to the pulsed LED will be greater than the DC LED. I'm guessing this is the real reason it looks brighter.
 
flyingbrit said:
Except the power will not be identical. If the current is doubled, the voltage drop will also increase. Thus the average power applied to the pulsed LED will be greater than the DC LED. I'm guessing this is the real reason it looks brighter.
Click to expand...
Are you sure about that? In a simple circuit with a single component, that one component drops all the voltage, regardless of the current. In my example, I feed the LED 1 volt- it should always drop that one volt (I'm assuming it can take whatever current flows through it).
 
Somewhat off base, but it's best to consider the eye for what it is: a part of the brain, rather than an external sensor.
 
Cap'n Jack said:
Are you sure about that? In a simple circuit with a single component, that one component drops all the voltage, regardless of the current. In my example, I feed the LED 1 volt- it should always drop that one volt (I'm assuming it can take whatever current flows through it).
Click to expand...

Absolutely sure!

First, an LED with one volt applied won't do much of anything; they typically need about 3 volts to produce light.

See data sheet: http://www1.futureelectronics.com/doc/EVERLIGHT /334-15__T1C1-4WYA.pdf

Look at the graph of forward current vs. forward voltage on p7. With 3 volts applied, the current is about 4mA. But the current will double if the voltage is increased slightly to about 3.2V. This would make the current/brightness somewhat unpredictable. So the usual method of powering a simple LED circuit is with a 5V supply and a series dropping resistor. This a) makes the current less sensitive to the supply voltage and b) allows the current to be adjusted by changing the resistor value.

weirdjim postulated doubling the current by halving the duty cycle (square wave). Using the above example, with 3VDC, the current is 4mA so the power is 12mW continuously. To double the current to 8mA, 3.2V must be applied. This makes the power 25.6mW when the pulse is "on", and zero when it is "off". Average power is thus about 12.8mW, or about 7% higher than the DC case.
 
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@ weirdjim, this guy seems to have a good explanation of the issues:

http://donklipstein.com/ledp.html

Per your original question he states:

Human vision is nonlinear, but that nonlinearity is after a surprisingly accurate time-integration process. When a light is flashing rapidly enough to appear continuously on without flicker, what you see has a good correlation (although nonlinear) with average brightness and is surprisingly independent of peak brightness.
 
flyingbrit said:
Absolutely sure!

First, an LED with one volt applied won't do much of anything; they typically need about 3 volts to produce light.

See data sheet: http://www1.futureelectronics.com/doc/EVERLIGHT /334-15__T1C1-4WYA.pdf

Look at the graph of forward current vs. forward voltage on p7. With 3 volts applied, the current is about 4mA. But the current will double if the voltage is increased slightly to about 3.2V. This would make the current/brightness somewhat unpredictable. So the usual method of powering a simple LED circuit is with a 5V supply and a series dropping resistor. This a) makes the current less sensitive to the supply voltage and b) allows the current to be adjusted by changing the resistor value.

weirdjim postulated doubling the current but halving the duty cycle (square wave). Using the above example, with 3VDC, the current is 4mA so the power is 12mW continuously. To double the current to 8mA, 3.2V must be applied. This makes the power 25.6mW when the pulse is "on", and zero when it is "off". Average power is thus about 12.8mW, or about 7% higher than the DC case.
Click to expand...
I see where you were coming from. I was just keeping it simple with just a single simple component, going back to basics. Let's call it a black box. It just drops the voltage. More current means it uses more power, but it still drops the same voltage.
 
OK, let me break this down into monosyllabic terms. (Ever wonder why monosyllabic is a four-syllable word?)

I have a DC 13.5 volt power source. I have a red LED that lights up at 1.5 volts. I have a 470 ohm resistor that allows 25 mA into the LED. LED watts = 1.5 x 0.025 = 38 milliwatts

I change the DC 13.5 volt power supply into a square wave and drop the resistor to 240 ohms. Current rises to 50 mA on the positive half of the pulse which is 75 milliwatts, but the average power is half of that, as there is no conduction on the zero half of the pulse. Therefore, the LED is consuming 37 milliwatts. For all intents and purposes, one miliwatt difference between the two is negligible.

Which one appears to be brighter in average room lighting? In total darkness?

Dunno. Gonna find out with an experiment if nobody knows the answer for sure.

Jim



flyingbrit said:
Absolutely sure!

First, an LED with one volt applied won't do much of anything; they typically need about 3 volts to produce light.
Click to expand...

That's just not true. Red, 0range, and Yellow need much less.

Color Wavelength Voltage Material Efficacy
Red 685 1.8 Aluminum Gallium Arsenide 85
Orange 615 2.1 Gallium Arsenide Phosphide 90
Yellow 575 2.2 Aluminum Gallium Indium Phosphide 90
Green 535 2.9 Gallium Phosphide 100
Blue 475 3.1 Indium Gallium Nitride / Gallium Nitride 150
Indigo 445 3.2 Indium Gallium Nitride / Gallium Nitride 150
Violet 415 3.4 Indium Gallium Nitride / Gallium Nitride 150
White 390-700 3.1 Blue + Yttrium-Aluminum-Garnet Phosphor (YAG) 175-200
 
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The human vision is good for somewhere around 30 Hz. Above that it perceives things as smooth, below that you will perceive flicker. Your traditional movie is around 24 fps. However to decrease the apparent flicker on modern stuff, they actually flash each frame twice. This dodge works well for most scenes, but if there is a definite motion in the frame, your brain will actually interpolate in a double image.
 
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