I read that optocouplers and optotriacs use infrared LEDs with infrared sensitive transistors.

What is the idea behind using IR frequency but not visible light freq. LEDs inside a completely closed dark chip?

  • 7
    \$\begingroup\$ Very short answer: It does not matter but IR parts are cheaper \$\endgroup\$ – caconyrn May 31 '16 at 11:51
  • \$\begingroup\$ The cynic in me says "So you can't tell when it isn't light-tight" \$\endgroup\$ – W5VO May 31 '16 at 18:15

Simple, silicon devices are very sensitive to infrared light. See the wikipedia article on a photodiode, there's a graph showing that silicon photodiodes are most sensitive around 900 nm, and that is near infrared light.

Silicon devices are sensitive to visible light so the package has to block all the light.

  • \$\begingroup\$ Also, as a nice bit of synergy, IR LEDs have a lower forward drop than visible LEDs. \$\endgroup\$ – WhatRoughBeast May 31 '16 at 14:00
  • \$\begingroup\$ "Silicon devices are sensitive to visible light so the package has to block all the light" Aren't they already blocking "all the possible light" since they are black and rigid? Imagine one uses normal LED instead of IR LED and assume no light leaks into the IC, in that case are you saying that the only reason IR LED is better because silicon devices more sensitive to IR than light spectrum freq? But they could also produce a device(fx. type of silicon device) which can be most sensitive to green light or red light as in an LDR. Sorry but I didn't really grasp what you meant. \$\endgroup\$ – user16307 May 31 '16 at 20:05
  • \$\begingroup\$ @user16307 1) The 'black and rigid' part is the 'package' so yes, you are agreeing with OP. 2) It's definitely a design concern. Look at this instance where insufficient visible light shielding left the Raspberry Pi 2 susceptible to crashing from use of a camera flash: technobuffalo.com/2015/02/08/… \$\endgroup\$ – Chris Hatton May 31 '16 at 23:26
  • \$\begingroup\$ @ChrisHatton Is this an answer? I couldnt relate it to the question \$\endgroup\$ – user16307 Jun 1 '16 at 0:10
  • \$\begingroup\$ @user16307 Well, you wrote LEDs inside a completely closed dark chip? That sentence does not make much sense, the LED is usually not on the same chip as the diode/transistor/thyristor because of the difference in manufacturing processes and (big one) making good high-voltage isolation will be more difficult on the same chip. Also you cannot "close" the chip, the chip has to be inside some plastic or ceramic package. Which usually blocks all the light. Unless you make it from transparent plastic of course. \$\endgroup\$ – Bimpelrekkie Jun 1 '16 at 9:18

Price. IR diodes for transmission and phototransistors/photodiodes for reception is cheaper and easy to make.

  • \$\begingroup\$ Cheap? Pls quote how cheap they are compared to the visible light LEDs. What is yr reference? \$\endgroup\$ – soosai steven May 31 '16 at 14:00
  • \$\begingroup\$ IR LEDs were created before red LEDs. IR and red can use the same chemicals, so there never was much of a difference. \$\endgroup\$ – CL. May 31 '16 at 14:04
  • \$\begingroup\$ White LEDs need far more prossessing stages, not to mention the phosphor dye on top of the LED to make it white. \$\endgroup\$ – winny May 31 '16 at 14:04
  • \$\begingroup\$ @CL, yes, red or IR is about the same and you find both types in comercial optocouplers but I have not seen a single one with any other type of color in my ten years in the industry. All other types provide no benefit and requires more processing stages. \$\endgroup\$ – winny Jun 1 '16 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.