need advice for my diagram :)

Discussion in 'General Electronics Chat' started by sp_vegeta, May 19, 2011.

  1. sp_vegeta

    Thread Starter New Member

    May 19, 2011
    Hi all,

    i'm new here and a newbie in electronic. right now i want to try make a sensor bar for wii. after googling and reading some article, i finally end up make a diagram like this.

    the component i used is listed below :
    - source power is 4.8V - 6V (4 X AAA Battery / USB)
    - 6 IR LED 1.35V @ 100mA
    - 2 LED 2.1V @ 20mA
    - 2 27Ω 1/2W Resistor
    - 1 100Ω 1/4W Resistor

    and the diagram is look like this :


    all the calculation is based on here

    since i'm a newbie and no electronic background so i'd like ask the expert here about my diagram :). i'd like to know if my diagram is already correct? is there any problem might happen ? is there any adjustment needed ?

    thanks a lot ;)
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
    I cannot see your diagram. Are you sure that you have posted it correctly?
  3. sp_vegeta

    Thread Starter New Member

    May 19, 2011
  4. Wendy


    Mar 24, 2008
    Do you have specs on those IR LEDs? 100ma seems way too much.


    The image is not showing for me either, you are not using the correct address. The link on the above post is what you put between the img /img commands to get an image like this.

    Once you hit 10 posts you can use the local hosting at this side, which is much preferred.


    Found the problem. You are using this address...

    Remove the part in read and it would have worked.
    Last edited: May 19, 2011
  5. sp_vegeta

    Thread Starter New Member

    May 19, 2011
    here is the specs of the IR LED link

    btw is there any IR LED that is less than 100mA? :p

    yes, that is the link i use. i click the image button and put that link in the address. have no idea it will be like that :p.

    ok will try use the local hosting :D.

    i can not edit the first post anymore :(. is there any limitation on editing a post ?
  6. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    FYI: when using that host site if you click "Show Codes" you can copy "Hotlink for Forums" and you can put your pic in the post:

    image uploading

    Since a wizard is nice but knowing how to calculate something is better here's how I would calculate the currents.

    For the IR LEDs, you have 6V driving 3 LEDs, so the resistor sees:

    6 V - 3 * 1.35 = 6 V - 4.05 V = 1.95 V

    This voltage is across the resistor, so the current is:

    1.95 V / 27 Ohms = 0.07 A = 70 mA

    Similarly, for the RED LEDs you have:

    6 V - 2 * 2.1 V = 4.2 V

    4.2 V / 100 Ohms = 0.02 A = 20 mA

    The total current is found by just adding all the strings:

    70 mA + 70 mA + 20 mA = 160 mA

    Seems here you changes the resistor you got from the wizard.

    Here's something you should know: when a LED is rated for some voltage at some current this is just one point where they work, and it is usually near the maximum current they can take. You need not (and I never do) run then near the max; generally any part appreciates some "cushion."

    The red LEDs you can check just by looking at them, the IR LEDs are tougher as you can't see em. But I would first try a current half the rating, 10 mA for the reds and 50 mA for the IR LEDs.

    Otherwise everything looks fine.

  7. SgtWookie


    Jul 17, 2007
    Don't forget to calculate the power requirement for the resistors.
    P=EI, or Power in Watts = Voltage x Current(Amperes)
    For the IR LEDs:
    P=1.95v x 0.07A = 0.1365 Watts, but for reliability, we multiply that result by 1.6 and get 0.2184 Watts.
    You can use a 1/4 Watt resistor for the IR LEDs.

    For the green LEDs:
    P=(6v-(2.1x2)) x 0.02A x 1.6 = (6-4.2) x 0.02A x 1.6 = 1.8v x 0.02A x 1.6 = 0.0576 Watts; you can use a resistor rated for 1/10 Watt or more.
  8. Wendy


    Mar 24, 2008
    Yes. Too many people deleted their questions or what they said to hide things from competitors and professors, a definite abuse. So the rules were changed to make 10 posts a requirement for full permissions at this site. Other than that we are probably the most liberal forums out there, once you are in as a full member you can modify your posts forever (and more). However, it is extremely bad form to do so without a good reason, if this is abused the rules may change again.

    I use it to write articles, treating my posts as a online word processor.

    Bill's Index

    100ma is a fairly odd number, but no big deal, as LEDs are extremely variable in their specs. Radio Shack publishes specs like these, but their old generation IR LEDs are not always up to it, so they regularly burn out, and people come over here asking why.
  9. sp_vegeta

    Thread Starter New Member

    May 19, 2011
    since i'm newbie in electronic so it's a little hard to understand here :p

    so from the calculation above i must change the resistor value ?

    here another question :
    - from the diagram above, should i put the resistor before the LED or after? is there any difference putting before or after ?

    -btw the power i plan to use is 4 X AAA Battery or power from USB. for an AAA Chargable battery the voltage is 1.2V. 4 X 1.2V = 4.8V. but if i use the alkaline AAA battery the voltage is 1.5V. 4 X 1.5 = 6V. so the power i assume will be 4.8V - 6V. is it okay in this circuit ?

    - from my diagram, based on the component spec can i say like this ?

    +----|>|---|>|---|>|----- the current draw for this series are 100mA max

    +----|>|---|>|----- the current draw for this series are 20mA max

    so total current draw from this diagram is 100mA + 100mA + 20mA = 220mA max ? right ?

    - i just bougth the IR LED and LED. can i check the Vf & If using multitester? for IR can i check the wavelength? cause the seller don't know the spec when i ask him :(. i'll recalculate the resistor value after get the actual value of th Vf and If :D.
  10. Wendy


    Mar 24, 2008
    You can measure the Vf yourself, and it will be more accurate. The If is dependent on the datasheet.

    Just bias the IR LED with 20ma (give or take) and measure the voltage drop across the LED, that is Vf.

    Among the articles I've written is a tutorial on LEDs.

    LEDs, 555s, Flashers, and Light Chasers
  11. bertus


    Apr 5, 2008

    You say you want to power the IR led with 100 mA.
    In the datasheet is said that the 100 mA is used with a pulse of 20 mS.
    (unfortunately there is not given waht the repetition time may be).


    I would keep the current lower as 50 mA to be on the safe side.
    Also is stated that the forward voltage is 1.35 Volts Typ and 1.6 Volts max.
    Keep tis in mind when powering theleds.

  12. sp_vegeta

    Thread Starter New Member

    May 19, 2011

    a couple days ago i bought 8 IR LED, 2 Blue LED and a digital Multitester. i think i can not used the ID LED datasheet i present earlier cause the component is not the same. there is no datasheet for this IR LED.

    i test 2 of the IR LED and get the Vf is 1.58V . dunno how to test the current so i assume IR LED current will be 50mA

    for the Blue LED, i can not get the Vf value. on the multitester it shown 1. so i asume the Vf is 3.5V and the current is 20mA

    here is my new diagram, please advice :D :


    oh yes, i have a question about current. current in electronic is it drawn or supplied from the power source ? i mean for example i have a power source with voltage 9V and current 500mA, is this mean the power source will always supply 500mA all the time? or the curren will flow depend on the component in a circuit that connect to the power source ?

    and i found an interesting post from SgtWookie (link). it seem with 555 IC, with 1 NiMH AA Battery can handle up to 7 LED. can this be done to IR LED also?