LED

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HLMP4700 is a common 2mA LED.

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In a series circuit, the supply voltage divides between the elements
in the circuit depending on their resistances, so for Jennie's
circuit, with 3.6V being dropped across the LED with 20mA going
through it, the circuit looks like this:


      |||
      |          |          | 
4.5V>-+--[RLED]--+----[R]---+
                            |
0V>-------------------------+

LEDs are more efficient at lower currents, so you might consider using

2 or 3 of them at lower current rather than just one at higher current. Hook up your LED at 5 mA to see how much dimmer the light is compared to 20mA. I think you will get a lot more light from 4 LEDs at 5 mA as opposed to one LED at 20mA.

-Bill

How do you hook more than one up..in series or parallel?

In a series circuit, the current is the same at all points, so the resistor is calculated with ohms law (E=I*R), so R=E/I or R=4.5/.02 = ~220 ohms.

start with 220 ohms and increase it once or two values at a time until you reach a point where brightness is just insufficient on "half flat" batteries, then go down one or two values. This ensures maximum life, yet sufficient brightness once the batteries begin to reach their mid-file point where're they're no longer fresh, but not yet flat.

Craig

Assuming no empirical information on LEDs generally, accepting your statement as true in some context, and then carrying it to the end, I'd deduce that a million LEDs operating at 20nA each would produce more candelas than one at 20mA. Yet, I'd tend to argue instead that there is an optimizing parabola associated with all this, with an optimum point that is neither too far one way nor too far the other.

But according to an HP document I'm reading here, the relative efficiency for a given LED and type "__increases__ with forward current," it does not __decrease__. That is, prior to the junction reaching saturation.

For example, they say that with high-efficiency red LEDs and using 1.0 as the standard efficiency at 5mA, it reaches a relative efficiency of about 1.59 at about 43mA. This means that if you maintained an average current of 5mA but pulsed it at 43mA while maintaining that average (holding mean die temperature roughly constant) and compared it against a continuous 5mA, the photons produced in both cases would be 1.59 times greater when pulsing but maintaining the 5mA average. (Their chart is a relative guide that can be used to estimate millicandelas for different driving methods against some reference condition.)

The efficiency factor does take into account die temperature changes. But ambient temperature does impact things. For the high efficiency red LED:

ratio = I(Tambient)/I(Tcal) = e^(-.0131*(Tambient - Tcal))

(Temperature affects other LEDs, of course, but the -.0131 factor varies with their type.) So, for example, when comparing a temperature of 60C with some spec given at a calibration point of 25C, the equivalent flux ratio will be e^(-.0131*(60-25)) or about 63.2% of what was specified for 25C.

For example, they suggest the following calculation. If we already know that the LED is supposed to provide 5 millicandelas at 10mA and 25C, and we expect to pulse it at 50mA with a duty cycle of 50%, and the ambient temperature will be

55C, then:

result = 5 mcd * (50mA*50% / 10mA) * (1.61 / 1.24) * e^(-.0131*(55-25))

or about an 11 mcd equivalent output. (The 1.61 comes from the table for 50mA and the 1.24 comes from the table at 10mA.)

I've not made the measurements here, but it appears they disagree with your comments. For example, using their calculations, and assuming the same ambient temperature in all cases:

relative performance = (5mA / 20mA) * (1.00 / 1.43) = about 17.5%

In other words, driving at 1/4th the current (for high-eff red LED) doesn't provide 25% of the brightness, but instead 17.5%. Now, this does NOT consider human logarithmic responses. But if you now use four of these 5mA LEDs, then the total is 4*17.5% or 70%. One doesn't need to understand details about human logarithmic responses to "see" that 70% will appear to be less bright than 100%.

But then, I'm taking this from empirical HP studies I have at hand and I'm not making measurements, myself. So I'd like to see the basis for arguing that (4) LEDs operating at 25% of the current are brighter than (1) LED operating at

100%. (Unless saturation takes place, of course.)

Jon

I do suggest that higher efficiency LEDs will give adequate brightness for use as an indicator lamp at currents around 2 mA or in a really good case as low as .5 or .2 mA. With current 1.3333 to 2 orders of magnitude below the "industry standard" of 20 mA, expect the voltage drop across the LED to be slightly lower than what the datasheet says. As in to an extent that can be significant. For colors ranging from orangish red or "high efficiency red" to yellowish green, expect a voltage drop at 2 mA to be close to 1.75-2 volts. Assuming an "average for average condition" of 3 cells to be 3.75 volts, the difference between battery voltage and LED voltage would be

1.75-2 volts. To achieve 2 mA through the LED, use a 100 ohm or an 82 ohm resistor. Wattage requirement is satisfied by all resistors of wattage 1/10 watt or higher.

For colors from non-yellowish-green to blue, consider a higher voltage drop but also a lower current requirement for decent LEDs.

Most LEDs of color yellow-green to near-infrared have a non-linearity in favor of higher currents of 10's of mA to around 100 mA. Most LEDs of colors from non-yellowish green (wavelength less than 548 nm) to blue (as low as mid-400's of nm) as well as the usual white LEDs which use bue LED chips have a different nonlinearity:

Such LED chips with peak wavelength and dominant wavelength in the sub-550 nm part of the visible spectrum mostly have efficiency maximized at lower currents of a few mA. As for larger size high power LED chips designed for high currents, please consider that blue, blue-green, "pure green" and white ones have maximum efficiency when substantally underpowered.

- Don Klipstein ( snipped-for-privacy@misty.com)

What book are you guys referring to? Some of you guys are pretty smart.......

If you have enough voltage, you would hook them in series, which will increase the efficiency since you get twice the light (for 2 leds) at the same current or 3 times the light for 3 in series. See my LED calculator for working out the resistor needed for various combinations of supply voltage, LEDs and current.

-Bill

Well, I may be misreading this white LED data sheet, but the "current vs luminous intensity" graph is bowed upward, and indicates better efficiency at 10mA verses 20mA. And if you include the extra 0.3 voltage drop at 20mA, the efficiency is downgraded further.

-Bill

May have to do with the efficiency of the fluorescence yielding the 'white' from bluer emissions. White LEDs are different animals, perhaps. I'd been thinking about non-white here.

Jon