Connect with us

Voltage Stabilization?

Discussion in 'General Electronics Discussion' started by icor1031, Apr 27, 2010.

Scroll to continue with content
  1. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    As with cars, when running off of the battery you will hit a max of 12.6v. With the car running, you will hit a max of .. 15.8? and the commonly stated average is 14.4.

    I plan to connect 4 LEDs in series, which require 3.0-3.2v @ 24ma each. If I make an error of say 1.5v, (off vs on, battery condition, etc.) then even if I plan for 3.1v - I am over-powering each LED by about 0.2v..

    So, here's the question: How do I keep the voltage pretty stable?
     
  2. selva

    selva

    25
    0
    Apr 26, 2010
    Hi,

    you can use a dc-dc converter ICs to get the stabilized voltage needed. Selecting dc-dc converter ICs depends upon your requirement. For guidance you can visit TI and linear technology sites.

    comments much appreciated.

    warm regards,
    selva
     
  3. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    The linear site is broken, and I don't know what the other one stands for.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,251
    2,704
    Jan 21, 2010
    You sound awfully like you're planning to just connect the LEDs to the battery. If you do that without any form of current limiting, then their live will be very short, but very brilliant...

    LEDs are not driven by an appropriate voltage, but by an appropriate current. The voltage across them at any particular current is provided as a guide, it's not a specified voltage. Very small changes in voltage (say by 0.1 of a volt) may result in many times more current passing through the diode.

    You don't worry about the voltage, you provide a constant current source.

    There are many ways to do this, here are a couple:

    1) have a string of LEDs with a total voltage significantly less than the battery voltage, and preferably where this difference is small compared to the variation in battery voltage. Use a resistor calculated to drop the excess voltage (calculated at the maximum battery voltage) to achieve the required current.

    2) use a voltage regulator (typically 2 or 3 volts less than the minimum battery voltage) and a resistor appropriate to limit the current at this voltage.

    3) construct a constant current source.


    OK, for option 1, you could put 2 diodes in series (approx 6V drop) and use a 390 ohm resistor in series with them. That will give you between 15mA and 25mA.

    For option 2, use a 9V regulator, and put 2 diodes in series with a 120 ohm resistor (gives about 25mA)

    Option 3 is more complex.

    There are countless more ways to do this, and if you're after options, I'm sure people can give them to you. But none involve connecting LEDs across a voltage source without current limiting.
     
  5. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    No, I was planning to use a resistor. Couldn't nail down the ohm rating without stabilizing the voltage, though..

    With option 1, won't that mean I need high watt resistors? I've been using .. 1/4w? on my other projects, which was plenty.. I'm concerned about project Size, and a 1w resistor for example, will be too large - physically.

    Option 2, what's a regulator cost?

    Thanks!
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,251
    2,704
    Jan 21, 2010
    Once you know the current through the resistor and its value, use I^2*R to determine the wattage. (note that this will give you the power dissipated -- you probably want to select a resistor capable of dissipating about twice this to prevent it from getting too hot -- depending on the environment it's in).

    A cheap 3 terminal regulator will set you back about a dollar for one suited for 100mA. Note that you'll need a couple of capacitors, and probably some protection from voltage spikes.

    The simple resistor may be easiest if the difference in brightness is tolerable.
     
  7. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    ~~OK, for option 1, you could put 2 diodes in series (approx 6V drop) and use a 390 ohm resistor in series with them. That will give you between 15mA and 25mA.

    How did you determine that? :)

    ~~For option 2, use a 9V regulator, and put 2 diodes in series with a 120 ohm resistor (gives about 25mA)

    I've used a regulator before.. Maybe I just hooked it up wrong, but it overheated and died. . I'm looking to hook up about 60 LEDs.. Can they handle that much?
     
    Last edited: Apr 27, 2010
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,251
    2,704
    Jan 21, 2010
    R = V / I

    Where V is the difference between the LED voltage and the battery voltage, and I is the current.
     
  9. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    ~~ A cheap 3 terminal regulator will set you back about a dollar for one suited for 100mA.

    That certainly won't work for 60 LEDs? :)

    Would I be better off just getting a voltmeter reading of my car's voltage when on, and running 4 LEDs in series at a time with a resistor with the proper ohm, as opposed to trying to stabilize the voltage?

    I assume not, since you said dropping 6v first - which I don't understand, to be honest. :D Maybe you can explain?

    The help is appreciated.
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,251
    2,704
    Jan 21, 2010
    2 LEDs (I assume blue or white) in series will drop approx 6V at 24 mA (from your figures).

    so *I* would calculate the resistor value using 15.8V, so

    R = V / I
    = (15.8 - 6) / 0.024
    = 9.8 / 0.024
    = 408 ohms

    the nearest value resistor to this is 390 ohms.

    If you want to power 60 LEDs (I presume you want some fancy lighting for your car?) then a more sophisticated solution may be appropriate.

    Now is when you need to tell me what the LEDs will be used for, whether they're all turned on and off at once, whether you want to be able to dim them, etc., because the problem has just grown from 4 LEDs to 60!
     
  11. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    OK.

    I've done this before, but I killed them. I don't know if I over-volted them (probably), or if it's because I used them when it was -20º outside. .. . It's too bad, too.. they were gorgeous. This is it, if you'd like to see. :D

    What I did before is: take accessory wire (might already be regulated to 12v??)... Attach 3 leds in series with a resistor (they were 3.8v), and .. 10ohm? resistor.
    Attach another 3 leds in series with a resistor, etc.. Always going back to the original power wire, not the one that came off of the LEDs. I suppose that's both parallel and series?

    Now, what I want to do is this:
    Run about 60 LEDs (The thought I had is 4 LEDs per series, with a resistor in each series.) along a wood board.. Run a wire along that board, attaching to each first (positive) LED, and each last (negative) LED in that series.

    They each use 3.0-3.2v and require 24ma.

    I might have to run it from the battery, though.. I'm concerned running over 150 LEDs might blow the fuse quite often.. I said 60, because that's one side of the car.. :)
     
    Last edited: Apr 27, 2010
  12. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    Last edited by a moderator: Apr 27, 2010
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,251
    2,704
    Jan 21, 2010
    The smaller the voltage that the resistor is dropping, the more widely the LED current can vary.

    Did you wire them in series or parallel?

    Please ignore that site's first recommendations about placing LEDs in parallel. There is a very good chance that you will kill LEDs doing it that way. The second recommendation (with a resistor for each LED is perfectly OK)

    150 LEDs arranged with 2 LED per string will draw a total of 1.9 Amps, hardly anything for a car battery.

    Have you calculated the power rating required for the resistor(s)?

    Please get over this. LEDs are not like light bulbs.

    At 24mA the LEDs have a forward voltage drop of between 3.0 and 3.2 volts (your figures).

    At 3,1 volts some led you pick at random may draw 24mA, a second one may draw 16mA, and a third 55mA. If you increase the voltage by 0.1 volt, the current may double. This is why they are NEVER describes as "requiring a particular current at some voltage". In addition, as they heat up, they draw MORE current at the same voltage (and heat up more), until eventually they stop working forever.
     
  14. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    "150 LEDs arranged with 2 LED per string will draw a total of 1.9 Amps, hardly anything for a car battery."

    Yes, my concern was the Accessory (cig) fuse, 10a.

    I wired 3 LEDs in series... pos power to pos led, neg led to pos led, neg led to pos led, neg led to neg wire.
    I did that about 6 times.

    No, I haven't calculated it yet.. I'm asking what to calculate .. Can I do 4 LEDs, or does it have to be 2?

    ~~ Please get over this. LEDs are not like light bulbs.

    Sorry, just quoting the stats from the place I would buy them.. Apparently I miss-understand the specs. :D So, it sounds like 24ma is just an average? :)

    http://cgi.ebay.com/100-x-5mm-White...tem&pt=LH_DefaultDomain_0&hash=item3ca601d6ca (They're actually 5mm, not 8mm.)

    Excuse the cheap LEDs, it's just a cheap/fun project until I get a 'real' car. ;)
     
    Last edited: Apr 27, 2010
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,251
    2,704
    Jan 21, 2010
    That seems fine.

    I would recommend 2 LEDs per string.

    With 4 the current would vary from about 25mA to 2.5mA over your voltage range (160 ohm resistor).
     
  16. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    I don't understand that, but I'll take your word for it ;)

    Thanks, buddy!
     
  17. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    What would my current vary from if I did 3? The reason I ask, is..
    #
    # the wizard thinks 1/2W resistors are needed for your application Help
    That's what I get if I do 2.

    If I do 3, it's 1/4W..

    :)
     
  18. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,251
    2,704
    Jan 21, 2010
    Try connecting 1 in series with a 510 ohm resistor,
    and 2 in series with a 470 ohm resistor,
    and 3 in series with a 270 ohm resistor,
    and 4 in series with a 160 ohm resistor.

    connect each of these to your accessory power and check out what they look like with the engine off, idling, and up to high RPM (preferably whilst NOT driving).

    Decide for yourself which of these show little enough variation for you. I would recommend 2 LEDs in series.
     
  19. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    Appreciated. :D
    Sorry for my ignorance. ;)
     
  20. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,251
    2,704
    Jan 21, 2010
    Calculate it yourself.

    P = I^2 * R

    I = 0.025 (I calculated this at the max voltage)
    R = whatever resistor you want to calculate

    tell me what you get and what power resistor you would use and I'll tell you if you're right :)
     
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.
Similar Threads
There are no similar threads yet.
Loading...
Electronics Point Logo
Continue to site
Quote of the day

-