Trying to connect 5 leds parallel

Discussion in 'General Electronics Chat' started by Mav, Nov 6, 2012.

  1. Mav

    Thread Starter New Member

    Nov 5, 2012
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    0
    Hi,

    So I'm super new to the world of leds and electronics generally. I'm working on soldering together 5 leds and sticking them in a miniature ship, but am having a lot of trouble.

    The leds I want to use are these: http://www.tme.eu/ro/Document/dcaac778d507c298085b10bd51b3d2e6/OF-SMD3528G.pdf

    I used the led calculator here: http://led.linear1.org/led.wiz to give me a diagram for wiring, but I have a problem. In the above linked PDF, under forward voltage it mentions min and max, but no typical, so I don't know which to enter. The difference seems big, because with 2.8 I need 12 Ohm resistors and with 3.0 I need 1 Ohm resistors.

    Does anyone have a solution to this?

    Thanks in advance.


    PS: As an additional question, I want to use a CR2012 3V battery due to the small space I have to work with, but do you think for this particular case a 9V battery would be better suited?
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,421
    3,357
    You should not connect LEDs in parallel because of different forward voltages.
    LEDs in parallel will have different brightness.
     
  3. tracecom

    AAC Fanatic!

    Apr 16, 2010
    3,869
    1,393
    Some general comments.

    1. The LEDs may be too bright for your purpose at 20 mA.
    2. A 3 V source is too low to run the LEDs at full brightness.
    3. A CR2012 will be depleted very quickly by these LEDs, especially if you try to run all five from one battery.
    4. A 9 V battery would be better, but still won't last very long.

    How big is the model and what are the LEDs for? Maybe a lower current LED would suffice.
     
  4. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    LEDs are NEVER used in parallel unless they are measured and sorted so that they all have exactly the same forward voltage.

    Some of your LEDs might be 2.8V and others might be 3.4V. Then the 2.8V ones will hog all the current (and quickly burn out) and the 3.4V LEDs will not light.

    Your LED calculator is STUPID like most of them. Calculate your own resistor when the battery is new and again when it has run down a little. Also calculate the resistor when the LEDs are 2.8V and again calculate when they are 3.4V. You will see that you need more voltage across the resistor (a higher battery voltage) so that the LEDs all light with enough brightness and do not burn out, and do not dim too much as the battery runs down a little.

    5 LEDs at 20mA each is a total current of 100mA. Your choice of a tiny CR2012 battery will barely light a single 1.8V red LED for one minute. Look at its datasheet. It is designed for a tiny current less than 1000th of what you need.

    With a 9V alkaline battery then you can connect two LEDs in series and in series with a current-limiting resistor. Use two or three strings like that. With 3 strings then the total current is 60mA when the battery is new and it drops to 7V in about 2 hours when the LEDs might be very dim. You might notice dimming in the 2 hours.
    If both LEDs are 2.8V then their total is 5.6V. With a new 9V battery the resistor will have a voltage of (9V - 5.6V=) 3.4V across it and should be (3.4V/20mA=) 170 ohms.

    But if both LEDs are 3.4V then their total is 6.8V. with a new battery and with the 170 ohm resistor the current is only (9V - 6.8V)/170 ohms= 12.9mA and they will dim and turn off soon as the battery voltage runs down.
     
  5. Mav

    Thread Starter New Member

    Nov 5, 2012
    4
    0
    Thanks all for the replies. First off, to reply, the miniature is a "space ship" from a miniature game called Warhammer 40k that some of you might be familiar with, and the leds are meant to be inserted into the "engines". Space is very limited so it'd be hard to conceal a 9V battery, let alone 2.

    @Audioguru: So if I understand correctly, you're suggesting to have 2 or 3 batteries, each with 2 leds in series, with a resistor calculated by Ohm's law after using a DMM (I guess?) to check the exact Fv of each led?
     
  6. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    No.
    I suggested connecting two LEDs in series and in series with a 170 ohm resistor but 170 is not available so use 180 ohms. That is one string. Connect 2 or 3 strings to a single 9V alkaline battery then it will work for at least 2 hours. You can measure the forward voltage of each LED if you use a 270 ohm resistor in series with one LED and a new 9V battery.
    Replace the battery when the LEDs are too dim.
     
  7. Mav

    Thread Starter New Member

    Nov 5, 2012
    4
    0
    So this would be about right, correct?

    [​IMG]

    Thanks again for all the assistance btw.
     
  8. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Correct. I added the resistor values.

    Your schematic is fuzzy because you saved and posted it as a JPG file type. Schematics should be saved and posted as a very clear PNG file type.
     
    Mav likes this.
  9. Mav

    Thread Starter New Member

    Nov 5, 2012
    4
    0
    Oh ok sorry will remember that in the future.
     
  10. crutschow

    Expert

    Mar 14, 2008
    12,991
    3,227
    Note that if you use high brightness type LEDs you may be able to operate them at a significantly lower current to get better battery life and still get the desired lamp intensity.

    You can experimentally try different higher resistor values to get the brightness you want.
     
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