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Resistors for IR LED's not working?

Discussion in 'LEDs and Optoelectronics' started by Bowman, Jun 6, 2018.

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  1. Bowman

    Bowman

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    Feb 1, 2017
    Hi there,

    Been trying to illuminate two 940nm IR LED's in series and am told by this site that a 100Ohm resistor would work with 5v (USB port) to get the correct voltage drop across the diodes, but for some reason the resistor is not lowering the voltage at all.

    The resistor shows a reading of 100.6 to 101.0 on the 200 setting on my multi meter so I'm sure that is correct. (could be wrong).

    I have checked the wiring and i have it like this:
    1. Positive from USB plug to the 100ohm resistor,
    2. Positive wire after the resistor to first LED's Positive pin,
    3. Negative pin of first LED to the Positive pin of the second LED.
    4. Negative pin of second LED to the Negative wire of the USB plug to complete the circuit,
    So the resistor should be lowering the voltage straight out of the USB plug to 3.2v to be split down to 1.6v per LED but it still sits at 5v. (I believe my LED's are toast)

    Hope that makes sense, not sure what I'm doing wrong, probably something obvious but its late. :D

    Also to mention the LED's are Kingbright L-934F3 / L-932P3C according to the ebay listing; Emitter and sensor pair. Link to them Here.
     
    Last edited: Jun 6, 2018
  2. davenn

    davenn Moderator

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    Sep 5, 2009
    where are you measuring ??
     
    Last edited: Jun 6, 2018
  3. Bowman

    Bowman

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    Feb 1, 2017
    I removed the positive and negative from the LED's and put the multi meter in series so negative to negative and positive to the end of the resistor connected to the positive of the USB plug and it showed 5v despite it still going though the resistor.
     
  4. dave9

    dave9

    1,049
    283
    Mar 5, 2017
    ^ That is normal, you need a load (LEDs) attached for it to drop voltage.

    Hook it all back up and measure across one of the LED's positive and negative to see if you're getting a voltage drop around 1.6V, and put the meter in series (anywhere) to measure current, like between USB 5V and the resistor.
     
    davenn likes this.
  5. Bowman

    Bowman

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    Feb 1, 2017
    Can you clarify that the clear ones would be the IR LED's and the black tinted ones are the phototransistors...

    wondering if i have got them wrong as these spike to 100ma according to my multimeter set to 200m and they are supposed to be rated at 20ma each... unless they are way higher current draw... either that or they are just toast going short circuit but not sure what to make of it really.

    Going to note that i havent solder anything yet just twisted the wires and if i press on the wires for the LED's it makes the current spike like that. Otherwise it sits at low 10's to 40ma max
     
    Last edited: Jun 6, 2018
  6. Harald Kapp

    Harald Kapp Moderator Moderator

    11,280
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    Nov 17, 2011
    Wrong way of measuring. Your multimeter has a very high input resistance (probably in the MΩ range), so it will drop the full voltage and negligible current will flow.
    For measuring voltages you put the meter in parallel, not in series.
    Series connection is for current measurements.
    upload_2018-6-6_7-57-44.png
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,482
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    Jan 21, 2010
    It sounds to me that you are measuring current in parallel and voltage in series. This is exactly the opposite way around.

    Let's say we just do voltage measurements.

    With the circuit connected up having the resistor and the LEDs in series and connected to the 5V, measure the voltage across the resistor. Any voltage higher than zero indicated that there is current.

    Use I = V / R to determine the current.

    So, if it's a 100Ω resistor and you measure 2.2V across the resistor, then... I = 2.2/100 = 0.022. so the current is 0.022 Amps or 22mA.

    Try this and see what you get

    If you were measuring current and placed the probes across the resistor then you may have allowed an excessively high current through the LEDs. This may have damaged them. Having said that, if it only went to 100mA then the LEDs may have survived.

    Edit: SNAP!
     
  8. Bowman

    Bowman

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    Feb 1, 2017
    When i put the multimeter in parallel onto the resistors pins i get 0.00v set to VDC 20.

    I don't really know what to say as its not making any sense, I'm doing the things you guys say but not getting the readings you say i should be getting....:confused:

    My meter is a UNI-T UT50A. I also have been using my other meter which is a Sinometer MS8230B but both say 0.00v in parallel across the resistor pins while the circuit is sill complete and connected to the power source (my PC USB)
     
  9. Audioguru

    Audioguru

    3,132
    698
    Sep 24, 2016
    Doesn't your multimeter have a "diode test"? Mine shows the forward voltage of a diode or LED at a very low current. Maybe you burned out your LEDs by connecting the current meter wrong.
     
  10. dave9

    dave9

    1,049
    283
    Mar 5, 2017
    Is it possible you have the LED's anode (positive) and cathode (negative) reversed?
     
  11. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Which indicates there is no current flowing and at least one of the diodes is defect (post #7) or in the wrong orientation (#10). Check the diodes (#9).
     
  12. Bowman

    Bowman

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    Feb 1, 2017
    I have tried the resistor in parallel on the diode setting and am getting a reading of 120 on the UNI-T, and on the Sinometer i am getting a reading ranging from .050 to .090 .

    When i put the meters onto the diodes i get a 1.---- reading on both so no value is being displayed. So i guess they are both toast.

    I have checked the polarity of the LED's, i have the flat side/short leg on the Negative, and long leg/round side on positive for both.
     
  13. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    I assume you mean on the diode range. Did you try the leads both ways? You will only get a reading one way (with the positive probe on the anode and the negative probe on the cathode.

    Bob
     
  14. Bowman

    Bowman

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    0
    Feb 1, 2017
    Yeah its on the diode range, i tried both ways but got the 1.---- value on both meters.

    Here is a picture of my crude setup:

    20180606_184524.jpg
     
  15. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Assuming they are the right way around, that is correct. And you measure 0V across the resistor when plugged in?

    Bob
     
  16. Bowman

    Bowman

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    Feb 1, 2017
    Yes, 0v across the resistor when plugged in, parallel from the resistors pins.
     
  17. Audioguru

    Audioguru

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    Sep 24, 2016
    In the photo I see clear and black LEDs. Are some of them 2-legged photo-transistors or photo-diodes and not IR LEDs?
    The connections should be properly soldered together, not just twisted.
     
  18. davenn

    davenn Moderator

    13,798
    1,939
    Sep 5, 2009

    OK so lets determine where the break/loss of voltage in the circuit is

    step 1 ….. put meter across the positive and negative out of the USB cable before the resistor
    make sure there is 5V
     
  19. Bowman

    Bowman

    23
    0
    Feb 1, 2017
    There are just two LED's in the photo, there is quite a shadow as i had to use a light to illuminate it enough for my phone to get a decent picture. These LED's are clear but did come in a pack with some dark tinted phototrasistors, emitter and sensor pairs.

    And yes they are supposed to be soldered but i wanted to test them first before committing to soldering.
     
  20. Bowman

    Bowman

    23
    0
    Feb 1, 2017
    I have put the meter on the positive and negative coming out of the USB, before the resistor, and there is indeed 5v coming from it.
     
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