In checking out some LED's. Rated at 1.8 volts and max current 20 ma. Using
9 volt regulated power supply and variable resistance. If I run it at 1.8 volts the ma goes to 40ma. If I run at 20ma the voltage drops to about 1.2 volts. So how do you get the max output of the LED ?
You don't want the max output of the LED. They last a very short time. So select a resistor that gives you an output you want.
But that behavior seems a bit unusual. It suggests a dV/dI slope of around 30 ohms. (1.8V - 1.2V) / (40mA - 20mA). That's closer to circa 1980 red leds (20 ohms, or so.) But the starting voltage for them is close to about 1.5V or thereabouts, and at much less current than 20mA. So I'm not sure what you have there, unless it's an IR led (which I have no experiences with.)
Is it possible that it's an IR led and you are trying to get some output from it, thinking you'll see it much? What do you see at 20mA and at 40mA?
1.8 >>volts the ma goes to 40ma. If I run at 20ma the voltage drops to about
1.2 >>volts. So how do you get the max output of the LED ? >
Not IR. They are 660 nm. No difference in brightness I can see. No difference in heat (checked with IR thermometer) after running at 40ma for 5 hours. 10 mm size. W W
I wasn't thinking just about difference, though thanks for that. I would imagine you should see some difference in brightness between
20mA and 40mA. I can't recall not seeing any difference with a doubling, like that.
But what about absolute levels at 20mA? Are they what you are used to seeing with a 660nm?
Finally, I'm almost positive that you cannot get 1.2V at 20mA from of a 660nm. There is a certain threshold voltage required in order to generate the photons. This is all based upon Planck's proposal, in
1904 if I recall, and Einstein's independent development of it then used to explain the photoelectric effect, in 1905 without reading Planck's earlier paper, that atoms absorb and emit radiation in discrete quantities, according to: E = n * h * f, where n is the quantum number and is always an integer, f is the frequency, and h is Planck's constant. E is energy, but can be related to voltage by the simple q*V, where q is the unit of electronic charge. So, q * V = n * h * f, or V = n * h * f / q. There is often some fudging with n, but it certainly cannot go less than 1, so plug that in: V = h*f/q. f= speed of light/wavelength. Call wavelength 'w' and then plugging all this together, we get:
V = h * c / q * w
With consistent mks units, h=6.626068*10^-34, q=1.60217646*10^-19, and c=2.99792458*10^8... so,
V = 1.23984172*10^-6 / w
With 660nm, you get:
V = 1.878548 volts
Thus, I'm having some difficulty with your low voltage figures. I'm still thinking this is an IR LED.
Had ma meter is series with source. Checked voltage across LED with DVM. Will make more tests. Specs were in catalog from seller Marlin P.Jones & assoc, Inc W W
Sounds reasonable, assuming you did both at the same time or else with the same effective setup, done two different times.
The problem is that you can actually do a decent job of calculating Planck's constant using an LED (or two or three of different wavelengths to get a mean line slope out of it.)
There is also a temperature factor based on thermodynamics, kT. As I understand it, temperature adds some jostling around of energy levels which can cause somewhat lower but stable, quiet voltages that aren't themselves adequate to kick off photons to still yield some photons once in a while. So I can grant that the actual measured voltage (at much lower currents, well below 20mA) might be below the figure the equation otherwise calculates. But at those currents, I'd expect a closer match, I think.
(There is a small resistive component that confounds the value a little at higher currents, and you don't want to use really low currents because of the kT effect on the statistics, so going for middling currents around 200uA-500uA or so helps to eliminate both parts as an error when computing Planck's constant from LEDs.)
Anyway, It seems a pretty good way to estimate. And Red LEDs in the
660m/670nm area are usually guessed at 1.9-2.0V, as a rule of thumb for usual operation and that works out pretty good most of the time.
So although I grant the possibility of 1.2V at 20mA, given my overall limited experience here, I do find it difficult to just buy offhand, not knowing more about what you are doing.
Hi warren, You still working on them healing lights I see. How is things progressing?
I found that if you shine the led's through a magnifying glass and project the light on a wall, you can tell the differences in there brightness better. Jim
Jon. I made more tests. This time with a Fluke meter fo volts and a Simson DVM for current. (was using a antilog meter for current). here are results. At 20 ma, volts were 1.97 up to 70 ma. 80 & 90 ma, volts were 1.98. Then went to 160 ma and volts were 2.0 The only part # I have is the catalog # O206OP I don't have mfg #. However there is a difference in brightness. Intensity rating 450 mcd. When running at 20 ma I get .0394 watts. At 160 ma I get .2 watts. I want to use a PWM on them to get a higher watt output than normal.. I have no idea how to determine % to use if I use .2 watts as 100 %. Plantk's law is beyond my knowledge. W W
Even at 1 volt/amp slope, which is pretty low, you'd expect 60mV change from 20mA to 80mA. You are seeing, supposedly, 10mV or thereabouts. That makes it 1/6th ohm or so of dV/dI + R, which seems darned low to me. But then, I'm no expert here.
Which, if the other figures are right, is congruent with your other readings. I'm just having a hard time with them.
Okay. Looks like a cheap one that is pretty basic.
That's good to hear. You can see something, then.
I don't have that much to say about that. You might see:
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That .0394 is a figure, just computing volts*amps for your diode (1.97V*20mA.) That's not the power in the light beam, itself. Lifetime of the LED will decrease if you crank up the average power, significantly. And that itself will vary some depending on your pwm frequency, as well, as slow PWM allows a heating/cooling cycle to flex the wires and epoxy as it cools and warms and cools... Faster PWM rates will make the oscillations be a small % of the average, so less flex going on.
You are trying to run them at .2 watt at 100%. So just do it and see how they fare. Set your resistor based on what you've learned (meaning about 100mA, I'd guess, since the voltage doesn't seem to be changing no matter what you do.) You can size the resistor according to the voltage you need to drop divided by the current of, say, 100mA.
Not that I trust your numbers, just yet. But if you feel they are right, then that seems to be what they suggest you do to get .2 watt input power. (which won't be optical output.) Then just PWM down from there.
And check out Don's web page above and see what you think for your application.
I have a fair bit of experience with 5 mm and 3 mm through-hole LEDs only conducting about, generally somewhat under, 1 amp from brief pulses of 12 volts across the LED. So I think 10 ohms or a bit more is mostly what these do!
Red 5 mm or 3 mm LED conducts that much with less than at leastv 3, good chance 3.5 volts?
I think you need more LEDs and less current apiece.
I somewhat remember from other articles earlier in this thread that at
20 mA this LED was supposed to achieve about 500 mcd, and I know of others with similar wavelength achieving 4-6 times that at 20 mA. I mention some in
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in the red LED section as ones approaching 9 lumens/watt (as in overall luminous efficacy - and I mention some others of shorter but still red achieving over twice that).
160 ma >>>I get .2 watts. I want to use a PWM on them to get a higher watt output than >>>normal.. I have no idea how to determine % to use if I use .2 watts as
100%. >>
Thanks to all for responses. Will continue testing and respond if future after I get PWM built and tested . Warren
-------------------------------------------------------------------------------- Have you considered getting some red laser diode assembly's and making an array of them. They are usually at 660 nm, being that is what the healing wavelength is best at. Maybe even put them in a stand assy. to sit oven the affected area etc. JTT.
cat#LT-100. With this device it is easy to see how small differences in current can effect light output. Uses 9v battery. Has momentary push button switch. $8.95
I keep an LED hooked up to the tester and in this way I also use it as a small flashlight.
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