Connect with us

Another "string of LEDs" question

Discussion in 'LEDs and Optoelectronics' started by abominableman, Mar 19, 2010.

Scroll to continue with content
  1. abominableman

    abominableman

    3
    0
    Mar 19, 2010
    Hi everyone,

    Check out the pdf attachment. Looks like a perfect, simple solution, but I do not know how to calculate the values for the resistors.

    I'm also building an array of LEDs - 8 rows of 6 - powered by a 24V DC power supply.
    I need 20mA through each leg
    LED Forward voltage = 3.2V

    Can anyone figure out the resistor values for me and for huntxtrm?

    Thanks!!!
    Dan

    I found the diagram at the on Semiconductor website (can't post links yet) Here's what they say:
    When the circuit operates properly and all the LEDs are
    running, the three sense resistors have about 1.25 V across
    them, which turns the transistor switches ‘on’. This connects
    all three sense resistors back to the Vadj pin allowing the
    proper current to go through each leg. If one string opens up,
    the sense resistor for that leg won’t have any voltage across
    it, turning ‘off’ the transistor and disconnecting its sense
    resistor from the Vadj pin. Therefore, the other two LED
    strings are unaffected by the fault. This same scheme can be
    expanded to accommodate as many LED strings as needed.
     

    Attached Files:

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,505
    2,849
    Jan 21, 2010
    Rsense is determined by ohms law to drop 1.25V at the required LED current.

    I would pick something like 4k7 for R4, 5, and 6 (etc). I doubt that it's particularly critical. It should be at least ten times Rsense, but that won't be an issue unless you want really small LED currents (fractions of a mA)

    Note that for this circuit to work, all the strings must have the same number of LEDs in them.
     
  3. abominableman

    abominableman

    3
    0
    Mar 19, 2010
    Makes sense. Thanks!!

    What do you think about adding a Potentiometer between the LED's and ground (see attachment) to be used as a dimmer? With the LED's dropping most of the voltage, a low value pot would do. 250 Ohms seems to do the trick in 5Spice. Anyone see any reason why this wouldn't work?
     

    Attached Files:

  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,505
    2,849
    Jan 21, 2010
    A potentiometer between the LEDs and ground as you have shown will not work correctly for several reasons:

    1) it is in series with the constant current source. Current will not change until the voltage headroom of the regulator is used up, and it ceases to be able to regulate any longer. The current may then decrease, but it won't do so in a regulated manner.

    2) The pot would need to be rated to carry the full current. Because there is (at least in part of the travel) a constant current flowing through the pot, that rating is (I^2)R, where I is the max current, and R is the total resistance of the pot, In your case (for 60mA -- 20mA per LED) it would be 1W. As the LED current increases (let's say it's a total of 300mA) the pot would need higher and higher ratings up to, in this case, 22.5 Watts. At 1A it would have to be rated at 250W! (Yes, 250W, even though the whole circuit might only be consuming 5W maximum)

    OK, the reason the pot needs to be rated like that is that it's wattage rating presumes that the power will be dissipated over the entire resistance of the pot. When it is being expended by less than the full resistance path, correspondingly less of the power can be dissipated. In this case the current at the minimum resistance is the highest, so a suitable wattage must be chosen so that the pot doesn't burn out the instant you start to dim the LEDs.

    In almost all cases, A pot should not be used to carry power.

    Your initial circuit very cleverly ensures that the current to each string is regulated however it does not provide any simple way to adjust that.

    (Oh, and I messed up a little in the calculation of the sense resistor. The voltage drop across the resistor PLUS the Vce(sat) of the 2222 would have to equal 1.25V. (A 2N2222A has a Vce(sat) of around 0.3V at low currents, so the voltage drop across the sense resistor is probably closer to 0.9V)

    Here is an example of a LED dimmer circuit. I would NOT recommend placing the LEDs in parallel like it is shown here, even if they are matched.

    This is a lot more efficient, but also a lot more complex.
     
    Last edited: Mar 23, 2010
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

-