Connect with us

LED array

Discussion in 'Electronic Basics' started by Jonathan Leppert, Nov 18, 2003.

Scroll to continue with content
  1. Hi,

    I am creating a 144-element LED array, wired in parallel. My supply voltage
    is 5.0 volts at around 5-6 amps. The LEDs take a forward current of 3.5
    volts, being driven at around 30 ma. My question is should I assume the
    array acts as one gigantic LED with its current consumption summed, with the
    same forward voltage so hence I will need to limit voltage with a resistor,
    right? I am thinking a 1.0 Ohm 10 W resistor.

    Would this work or am I going to destroy 144 LEDs? Any explanations or
    insight would be greatly appreciated.

    Thanks,

    Jon
     

  2. I think you will run into problems with this setup. The 144 LEDs will not
    have identical forward voltages, due of manufacturing tolerances, etc. The
    LED (or group of LEDs) with the lowest Vf will light up first, will draw
    most of the current, and will probably burn shortly, followed by the others
    in a chain-reaction, until all LEDs are gone. This occurs because the V-I
    curve of a LED is steep around its Vf, so small changes in Vf cause large
    changes in If. I don't know what the best configuration would be, but don't
    connect them in parallel.

    Costas
    _________________________________________________
    Costas Vlachos Email:
    SPAM-TRAPPED: Please remove "-X-" before replying
     
  3. Ben Weaver

    Ben Weaver Guest

    Hi Jon...

    I think you have the right idea. But I get about 0.3 ohms. (I did
    144x30mA = 4.32A, and then 1.5V/4.32A = 0.347 ohms. That'll dissipate
    6.48W so 10W would be a good choice. It'll get really hot though.)

    Anyway, I did a similar thing some time back and wired the LEDs up in
    series to give a total voltage drop equal to the power supply. That way
    you don't need the series resistor (which will effectively be wasting
    power).

    You might try this: Get two of your LEDs and wire them up in series. Put
    them across the 5V power supply. If they're bright enough (remember this
    time they're only getting 2.5V each) then you're in luck. Get 72 of your
    LEDs and wire them in parallel to form a gigantic LED, and do the same
    for the other 72. Now wire those two big "cells" in series. Ta da! No
    resistor getting hot!

    But of course, you may think that the LEDs aren't bright enough and so
    you'll have to go back to your original idea.

    HTH

    Ben
    ~~~
     
  4. Costas is right. You need a series resistor for each LED.
    At the day job we manufacture 180 element LED arrays (from a single
    die.) Each has its own series resistor. And, if you want very good
    uniformity you have to select the resistors.


    Steve J. Noll | Ventura California |
    | The Used High-Tech Equipment Dealer Directory
    | http://www.big-list.com
    | The Peltier Device Information Site:
    | http://www.peltier-info.com
     
  5. Ben Weaver

    Ben Weaver Guest

    I didn't find this too much of a problem. But worth bearing in mind.

    Ben
    ~~~
     
  6. John Fields

    John Fields Guest

    ---
    The LEDs will have forward voltages which can lie anywhere in the range
    specified by the data sheet and a negative temperature coefficient of
    forward voltage, so the LED with the lowest Vf will hog more current
    than the rest, which will cause it to heat up more than the rest which
    will cause it to allow more current to flow through itself which will
    cause it to heat up more, which will eventually lead to its destruction.
    If it fails open, then its fate awaits some of the other LEDs in the
    array, so you need to bite the bullet, use a series current limiting
    resistor for _each_ LED, and connect all of the series resistor-LED
    circuits in parallel across the supply. Since your LEDs have a Vf of
    3.5V at 30mA (?) you'll need to drop 1.5V at 30mA in each series
    resistor, so since R = E/I, R = 1.5V/0.035A ~ 42.9 ohms. The closest
    standard 5% value is 43 ohms, and each resistor will dissipate P = IE =
    0.035A*1.5V ~ 53 milliwatts, so you could use 1/4 watters no problem.
     
  7. John Fields

    John Fields Guest

    ^^^^^^ ^^
    0.030A 45

    Oops...
     
  8. Bill Bowden

    Bill Bowden Guest

    Depends on how well the LEDs are matched. LEDs in parallel
    tend to draw different currents due to differences in forward
    voltage drop. It might work, you could try 3 or 4 in parallel
    to see how well the brightness matches. Why not use a higher supply
    voltage so you can arrange the LEDs in series strings, each
    with their own resistor?

    -Bill
     
  9. will cause it to allow more current to flow through itself which will
    How much current are we talking about here? The LEDs I am using are "under
    rated" according to the distributor, and can be run at higher currents (with
    a decreasing life, of course). I'm not too concerned with making sure the
    brightness is the same for each LED, and the array will only be on for 2
    hours a day. Do you really think I'll need resistors (not worried about the
    price, just the logistics of wiring up 144).

    Thanks,

    Jon
     
  10. I agree that LEDs in parallel is a bad idea. Try it with just a few, and you
    will see how bad it can be.
    If you want to use less series resistors, then increase the supply voltage
    and string the LEDs in series, one resistor per string and as many strings
    as you need.
     
  11. John Fields

    John Fields Guest

    ---
    most of the "standard" white LEDs I've seen are rated for 20mA nominal
    and 30mA absolute max continuous, so I'd be leery about what the
    distributor says unless he can back it up with a data sheet. He may be
    referring to the pulse rating which is probably around 100mA with a 10%
    duty cycle.
    ---
     
  12. JeffM

    JeffM Guest

    Jonathan Leppert
    I'm with Bill on this one. Less energy wasted in resistors.
     
  13. Ok. Assuming I up the voltage to 12 VDC, and use 36 series strings of 4
    LEDs, what value/type of series resistor should I get for each string?

    Vf = 3V (2.8-3.8V range)
    Forward Current: 30 ma

    I know this isn't the most efficient way, but its better than 144 resistors,
    and I only have a choice of 5V, 9V, and 12V for power. I'm not familiar with
    a regulator or DC-DC converter that would efficiently up 12V to something
    like 24V or higher. I'm open to better configurations if anyone has any
    ideas.

    Thanks for all your help!

    Jon



     
  14. You should measure the voltage across a set of the LEDs at 30mA. Average the
    value, and call it Vf. Then your resistor value should be a standard value
    close to

    R = (12 - 4 * Vf)/0.030

    Negative R means you won't get 30mA out of it.

    Since you are running the LEDs at 30mA, I'm guessing they will get hot.
    Since this is the case, and since Vf is dependent on temperature, you should
    let them heat up before taking the measurements.

    You can build a 30mA current source as follows:

    VCC
    +
    +---+---+
    | |
    | .-.
    | | |100
    | | |
    | '-'
    | |
    | |
    .-. |<
    1k | |<---| PNP
    | | |\
    '-' |
    | |
    | |
    | LED
    | |
    | |
    | |
    +---+---+
    |
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    Hook the + lead of your multimeter to the collector of the PNP transistor,
    and the - lead to ground, and put the multimeter into current measurement
    mode. Adjust the pot to get 30mA. Then, hook up the LED between the
    collector and ground, and measure the voltage at the collector (remember to
    wait till it heats up). Thats Vf(30mA). It won't be exact, since the current
    will vary depending somewhat depending on the voltage at the collector, but
    it should be close enough for your measurement.

    Regards,
    Bob Monsen
     


  15. If you arrange them in strings of 4 LEDs in series, then you have 4 * 3 =
    12V total Vf (i.e., you don't need a resistor). This might work, but the
    resistor is generally a good thing to have, as it limits the current in
    overvoltage situations, etc., and makes the overall V-I curve less steep
    which is a good thing. So you could string two or three LEDs in series, and
    work out the resistor. For 2 series LEDs at 30mA, R = 6V/30mA = 200 Ohm. For
    3 LEDs, R = 3V/30mA = 100 Ohm. And 144 is divisible by both 2 and 3, which
    is a good thing.

    cheers,
    Costas
    _________________________________________________
    Costas Vlachos Email:
    SPAM-TRAPPED: Please remove "-X-" before replying
     
  16. John Fields

    John Fields Guest

    ---
    If you wanted to run it off the 120V (?) mains and you could assure that
    no one could come in contact with any of the wirng you could do this
    twice, which would get you down to two resistors:


    ACIN>---[R]--+--[36LED>]---+
    | |
    +--[<36LED]---+
    |
    ACIN>----------------------+


    36LED is a string of 36 LEDs wired in series, with two strings wired in
    anti-parallel so that each string lights alternately as the polarity of
    the input voltage reverses. Determining R analytically is tricky
    because the LED array isn't ohmic, so I just scaled the array and ACIN
    down by a factor of 9, built it, and adjusted R until I got about 30mA
    RMS through it:


    -->| |<---1.7VRMS

    13.3AC>---[47R]---+--[4LED>]---+
    | |
    +--[<4LED]---+
    |
    13.3AC>------------------------+
    <--28.5mA RMS-->



    Scaled back up it should look like this:


    120VAC>---[430R]---+--[36LED>]--+
    | |
    +--[<36LED]--+
    |
    120VAC>-------------------------+

    With approximately 28mA flowing through the 430 ohm resistor it should
    dissipate about 337mW, so I'd use a standard 5% 430 ohm 1/2 watt
    resistor. There is one other problem to consider, and that's line
    transients. While they'll be attenuated by the 430 ohm resistor, a sure
    way to get rid of them would be to split the series R and use a couple
    of Zeners in series opposition to soak it up, like this:

    R1 R2
    120VAC>---[220R]-+-[220R]--+--[36LED>]--+
    | | |
    | +--[<36LED]--+
    [TVS] |
    | |
    | |
    120VAC>----------+----------------------+

    Since you'll have 36 LED's in each string they'll be dropping 126V peak,
    and since 120VRMS ~ 170VP, the peak voltage at the junction of the
    resistors will be 170 - ((170-126)/2) = 148VP with nominal (120V) mains.

    For high line (132V) the voltage at the junction will be 187 -
    ((187-126)/2) ~ 163V, so you'll want to get something with a breakdown
    voltage higher than 163V at the low end of its breakdown voltage
    tolerance.

    If you get, say, a 1000V spike across the mains and the TVS is rated for
    180V, R1 will be dissipating (1000V-180V)²/220R ~ 3000 watts (YOW!!!!)

    That's not as bad as it sounds, though... Since 1 watt is 1 joule per
    second, if the spike lasts for 1 millisecond the resistor will only be
    as hot (after 1 second) as if was dissipating 3 watts, _but_ it needs to
    be something that can handle the pulse. Carbon composition or some sort
    of bulk resitor (like carborundum) would work. Also, for a 1000V spike
    the current which would be flowing through the TVS would be
    (1000V-180V)/220R ~ 3.7A and it would be dissipating 670W during the
    time of the pulse, so you'd need to get something that could handle
    that.

    If you're interested in following this approach but you need isolation
    from the mains, you could use a little isolation transformer to do the
    job.


    http://www.premmagnetics.com/pdf/Page9.pdf
     
  17. You must use a resistor for each individual LED. 1.5V / .03 = 50 ohms.
    Use the next higher common value, 51 ohms. You can also put two 100
    ohm resistors in parallel.


    --
    @@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
    ###Got a Question about ELECTRONICS? Check HERE First:###
    http://users.pandora.be/educypedia/electronics/databank.htm
    My email address is whitelisted. *All* email sent to it
    goes directly to the trash unless you add NOSPAM in the
    Subject: line with other stuff. alondra101 <at> hotmail.com
    Don't be ripped off by the big book dealers. Go to the URL
    that will give you a choice and save you money(up to half).
    http://www.everybookstore.com You'll be glad you did!
    Just when you thought you had all this figured out, the gov't
    changed it: http://physics.nist.gov/cuu/Units/binary.html
    @@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
    F
    o
    d
    d
    e
    r

    f
    o
    r

    s
    t
    u
    p
    i
    d

    n
    o
    t

    e
    n
    o
    u
    g
    h

    i
    n
    c
    l
    u
    d
    e
    d

    t
    e
    x
    t

    m
    s
    g
     
  18. Above is really bad advice.

    --
    @@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
    ###Got a Question about ELECTRONICS? Check HERE First:###
    http://users.pandora.be/educypedia/electronics/databank.htm
    My email address is whitelisted. *All* email sent to it
    goes directly to the trash unless you add NOSPAM in the
    Subject: line with other stuff. alondra101 <at> hotmail.com
    Don't be ripped off by the big book dealers. Go to the URL
    that will give you a choice and save you money(up to half).
    http://www.everybookstore.com You'll be glad you did!
    Just when you thought you had all this figured out, the gov't
    changed it: http://physics.nist.gov/cuu/Units/binary.html
    @@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
    F
    o
    d
    d
    e
    r

    f
    o
    r

    s
    t
    u
    p
    i
    d

    n
    o
    t

    e
    n
    o
    u
    g
    h

    i
    n
    c
    l
    u
    d
    e
    d

    t
    e
    x
    t

    m
    s
    g
     
  19. How'd it change from 30 mA to .035A?


    --
    @@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
    ###Got a Question about ELECTRONICS? Check HERE First:###
    http://users.pandora.be/educypedia/electronics/databank.htm
    My email address is whitelisted. *All* email sent to it
    goes directly to the trash unless you add NOSPAM in the
    Subject: line with other stuff. alondra101 <at> hotmail.com
    Don't be ripped off by the big book dealers. Go to the URL
    that will give you a choice and save you money(up to half).
    http://www.everybookstore.com You'll be glad you did!
    Just when you thought you had all this figured out, the gov't
    changed it: http://physics.nist.gov/cuu/Units/binary.html
    @@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
    F
    o
    d
    d
    e
    r

    f
    o
    r

    s
    t
    u
    p
    i
    d

    n
    o
    t

    e
    n
    o
    u
    g
    h

    i
    n
    c
    l
    u
    d
    e
    d

    t
    e
    x
    t

    m
    s
    g
     

  20. Whoa! As if 144 resistors weren't enough!

    Costas
     
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

-