Connect with us


Discussion in 'LEDs and Optoelectronics' started by colum, Jan 14, 2017.

Scroll to continue with content
  1. colum


    Jul 25, 2013
    I have many older diagrams on how to hook up an IR Detector and they show from no input Resistor to a Million ohms.
    I want to hook up an Emitter Led and I figure that input Resistor should be 150 ohms at 9 Volts to an IR Detector that is also 9 volts but what determines the Resistor ( if any ) into the Detector.
    The Detector description is Radio Shack
    IR Detector Photo Diode
    Reverse Breakdown 60 Volt
    100 MW 900 Peak wave length
    I would like to have alot of Detector sensitivity .Thanks for reading this and any help is appreciated....colum
  2. davenn

    davenn Moderator

    Sep 5, 2009
    give us some links to these parts you are referring to

    100 MW ?? no you wont find any 100 Mega Watt IR LEDs think you mean 100mW ;)
    and that is HUGE power for a LED maybe it should be 10mW ???
    900 Peak wave length ... that doesn't mean much either

    when doing electronics / science etc ... start to learn the correct ways to describe the units for a given thing
    MW = Megawatt
    kW = kilowatt
    mW = milliwatt
    uW = microwatt

    wavelength for IR through visible to UV light will be in nm = nanometres
    IR remotes are usually in the 900 - 980 nm wavelength

    Last edited: Jan 14, 2017
    Harald Kapp likes this.
  3. Audioguru


    Sep 24, 2016
    A photo diode is easily destroyed by a 9V battery. A photo diode has a very low output level so it usually feeds a high gain amplifier.
  4. hevans1944

    hevans1944 Hop - AC8NS

    Jun 21, 2012
    Photo diodes work best when reverse-biased and connected to the summing junction (inverting input) of an op-amp. With the non-inverting input connected to circuit common, the summing junction becomes a "virtual" ground. You connect a feedback resistor from the output to the summing junction to control the "gain" of the circuit.

    Whatever current the photo diode produces also flows in the feedback resistor. Output U = (Ipd)(Rfb), where U is the output of the op-amp in volts, Ipd is the photo-diode reverse bias current (varies with illumination) in amperes, and Rfb is the feedback resistor value in ohms.

    Depending on diode type and illumination level, you might expect to get 100 μA to perhaps 1 mA of diode current under various illumination conditions. If the value of Rfb is, say, 100 kΩ then the output will be 10 volts for 100 μA input current. Adjust the value of Rfb up or down depending on how much sensitivity you need for a given illumination level.

    The circuit is temperature sensitive, but this can be compensated by using two diodes thermally bonded together and applying a positive reverse bias to one diode and a negative reverse bias to the other diode, with both diodes connected to the summing junction. One diode acts as a "dark current" reference (this is current that is temperature sensitive) so it must be blocked from illumination the other diode "sees". One diode has its anode connected to a negative bias supply and the other diode has its cathode connected to a positive bias supply. Thus both diodes are reverse biased, but only one is exposed to illumination. Bias voltage is not critical... dark current barely varies with reverse bias voltage. Anything from a few volts upward to the maximum reverse voltage rating (60 V in your case) of the diodes will work. I generally use ±15 V DC from the op-amp power supply.

    This circuit is quite linear as a function of illumination intensity over several decades, but it is also susceptible to microphonics when the feedback resistor value is made very large in a effort to maximize sensitivity to low-level illumination conditions. Start out with feedback resistors in the range from 10 kΩ to 1 MΩ and make sure the op-amp output doesn't saturate at maximum illumination. Saturation won't hurt anything, but it does slow down the response of the circuit to rapidly changing illumination conditions, and of course the linearity is destroyed once saturation occurs. If all you need is sensitive on/off switching when the IR source is unblocked/blocked, use the largest value resistor that will produce the desired signal "swing" in the output.
  5. colum


    Jul 25, 2013
    Thanks Heavens 1944 Hop ..That was a well thought out answer and I will try to do it justice...Jeff
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day