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I would like to regulate the light flux output of an LED or laser diode over its entire operating temperature range by varying the current.

What are some different methods to achieve this?

Edit: This is for a sensor application. I'm attempting to measure position by blocking the light over a photodiode. If the LED flux changes however, it will seem like position is changing, when in fact it is not.

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  • \$\begingroup\$ What range do you mean? Current or temperature? \$\endgroup\$ May 26 '12 at 7:27
  • \$\begingroup\$ Over it's temperature range, I would like to vary current automatically as the temperature changes, while keeping light flux constant out of the LED. \$\endgroup\$ May 26 '12 at 7:32
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    \$\begingroup\$ It may be the phototransistor which is sensitive to temperature. \$\endgroup\$ May 26 '12 at 9:22
  • \$\begingroup\$ @Rocketmagnet - Very good point. I made a comment about it in my answer. Hope you don't mind. \$\endgroup\$
    – stevenvh
    May 26 '12 at 10:53
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    \$\begingroup\$ lrc.rpi.edu/programs/nlpip/lightinganswers/led/heat.asp \$\endgroup\$ Sep 17 '19 at 21:29
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Your original question mentioned a sensor, which now seems to be gone. How precise do you want to keep that luminosity constant? If it's for visual it's pointless: your eye's sensitivity follows a logarithmic curve, so that it can cope with both 1 lux moonlight and 100 000 lux sunlight. Therefore it is not very sensitive to small changes in luminosity.

Most LED manufacturers will give a linear relationship between current and luminosity, so to get the latter constant you only have to keep current constant.

enter image description here

Note however that you have to derate current at higher temperatures:

enter image description here

So for this particular LED you can't go higher than 5mA if you want the same luminosity for 25°C and 85°C.

The Supertex CL25 is a temperature compensated constant current source for LEDs.

enter image description here

This has a 0.01%/°C typical temperature coefficient. It is 25mA however, so you'll have to use a LED which doesn't need to be derated below that.

Rocketmagnet makes an important remark: "It may be the phototransistor which is sensitive to temperature." (RM, I hope you don't mind that I copy it here.)
Indeed, it's no use to have a temperature controlled LED driver if the sensor's reading varies highly with temperature. You'll have to look into that too.

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  • \$\begingroup\$ Thank you, sorry I tried to edit the question to clarify, but seemed to have de-clarified it. This is for a sensor application. I'm attempting to measure position by blocking the light over a photodiode. If the LED flux changes however (due to a temperature change), it will seem like position is changing, when in fact it is not. \$\endgroup\$ May 26 '12 at 7:52
  • \$\begingroup\$ @kerblogglobel - (jesus, what kind of nick is that! Nearly broke my fingers typing that) Anyway, my answer still applies: all you need is constant current. I've added a reference to a temperature compensated current source to my answer. \$\endgroup\$
    – stevenvh
    May 26 '12 at 8:02
  • \$\begingroup\$ Awesome thanks! I never even knew such a part existed. \$\endgroup\$ May 26 '12 at 8:53
  • \$\begingroup\$ @stevenvh - no probs. And thanks for informing us about this interesting part. \$\endgroup\$ May 26 '12 at 17:01
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Sense the unblocked LED with a stable sensor and adjust it to maintain constant output and then use that light to drive the sensor proper.

ie

The LED is feedback stabilised.
Provide TWO sensors that at are accurate enough and stable enough for your needs.
One sensor views the LED via the blocking mechanism.
The other sensor veiws the LED directly.
Maintain illumination of unblocked sensor constant by adjusting the LED current.
Blockable sensor is now driven by a light output regulated LED.

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  • \$\begingroup\$ Wow, a very brief answer from Russell! ;-) Only, I read it three times and I'm not sure I understand what you mean. \$\endgroup\$
    – stevenvh
    May 26 '12 at 10:48
  • \$\begingroup\$ @stevenvh - extended \$\endgroup\$
    – Russell McMahon
    May 26 '12 at 12:09
  • \$\begingroup\$ Thank you, that is very helpful. I will have a look into feedback circuits. \$\endgroup\$ May 27 '12 at 2:48
  • \$\begingroup\$ Russel, what role does the blockable sensor play in your scheme? I've seen closed loop circuits with feedback-through-light, but they had only one sensor. \$\endgroup\$ Jun 22 '12 at 5:02
  • \$\begingroup\$ The unblocking sensor allows the light level to be stabilised. The blockable sensor output varies with the position of a target (as per his question description). \$\endgroup\$
    – Russell McMahon
    Jun 22 '12 at 13:09
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I have come across the same problem as you in the past. Analog photo sensors can make excellently sensitive position sensors, but suffer from temperature, as you mentioned. The way I have solved it in the past was to use three IR transceivers:

  • Transceiver 1 was set up as the sensor.
  • Transceiver 2 was set up to always have maximum reflection.
  • Transceiver 3 was set up to always have zero reflection.

I used an ADC which had both positive and negative reference voltage inputs. I connected Transceiver 2 to the positive reference input, and Transceiver 3 to the negative reference input. Transceiver 1 was connected to the ADC input.

Now this gives you perfect automatic temperature compensation across the whole sensor range, and across the whole temperature range too. However, the transceivers must all be placed close enough together to be at the same temperature. You might be using separate LEDs and phototransistors, in which case, all three LEDs should be at the same temperature, and all three phototransistors should be at the same temperature. But the LEDs needn't be at the same temperature as the phototransistors.

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Sadly I think the view that constant current will assure constant light output is probably flawed: Intensity vs Temperature

I'm presently trying to do precisely what's asked here, but using a temperature-compensation on a photodiode for feedback to the constant current source and am still struggling to keep reliably within +/- 1% over the range 30C to 60C.

The following is the arrangement I'm using with compensated photodiode feedback. Don't trust the values, they're from before I needed the compensation.

Compensation with photo-feedback

If you can get the curve suitable, this circuit without the photo-feedback may well be good enough and is simpler (mechanically and adjustably, at least). Again find your own values and OpAmp ... these were just a hack.

Simpler version with just compensated feedback

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    \$\begingroup\$ What would help this answer a lot would be the circuit of your CC diode driver's compensation. Perhaps ±1% is sufficient or the OQ's needs... \$\endgroup\$
    – Techydude
    Mar 5 '18 at 2:00
  • \$\begingroup\$ I'll very happily post a complete reference once I've reached a reasonably close result. At present, the only circuit I have still swings by several percent, which makes me think I have an additional problem. That said, I'll post what I have as a), it may help and b), someone may identify me seeded error :-) It's an old circuit with the TC disabled as at that time the few percent drift was OK. Now it isn't. \$\endgroup\$
    – Gordon
    Mar 5 '18 at 13:07
  • \$\begingroup\$ Incidentally, the LMP2011 has the potential for too much drift and I plan to switch to MCP601. Don't trust the thermistor value ... it was a 'first guess' only. \$\endgroup\$
    – Gordon
    Mar 5 '18 at 13:12
  • \$\begingroup\$ I agree. Even with a constant current, the light output can change with junction temperature. lrc.rpi.edu/programs/nlpip/lightinganswers/led/heat.asp \$\endgroup\$ Sep 17 '19 at 21:30
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TSL230R may be your prefect sensor.

Here is a complete reference project for you.

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  • \$\begingroup\$ why is it the perfect sensor? \$\endgroup\$
    – stevenvh
    May 26 '12 at 7:18
  • \$\begingroup\$ could you suggest a better prefect one? \$\endgroup\$ May 26 '12 at 7:27
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    \$\begingroup\$ I don't know. You haven't said what specific feature makes it perfect. Specify and then I can see if it is perfect indeed. \$\endgroup\$
    – stevenvh
    May 26 '12 at 7:29

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