Question about reducing battery output

Discussion in 'The Projects Forum' started by N0ctrnl, Oct 12, 2009.

  1. N0ctrnl

    Thread Starter Member

    Aug 14, 2009
    14
    0
    This is kind of a hard question to word, so bear with me...

    I currently have a string of 10 LEDs wired in parallel. I am using 3 - AA rechargeable batteries (1.2v rated) wired in series to power them. The LEDs all draw 30mA each. I have a plug on the end of the LED string that hooks up to a ~5.2v power source to recharge the batteries. The jack disconnects power from the ground side and uses an alternate wire when charging so that the LEDs are not on when the batteries charge, but the switch has to be (since I have it connect inline with the positive side of the battery).

    It's kind of a pain to make sure the circuit is on when I go to plug the batteries in. So, I've run another connection from the positive side of the battery that bypasses the switch. The problem I'm going to have is that when I give a 5.2v charge to the batteries, I get a 4.8v output. I've used the LED calculator just for giggles to see what size resistor it would tell me to use (5.2v input, 3.6v forward current, 300mA draw), but it comes back with 5.6Ω. Since I know that can't be right, and I suspect my methodology is flawed, can anybody shed some light on what I need to make sure I don't fry my 3.6v LEDs when connected to a 5.2v power supply? Keep in mind that I have 3 batteries connected at the same point.

    Thanks a bunch!
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    That's why we prefer having people post schematics of their circuits; they're worth a thousand words!

    Welcome to AAC.

    Do they have individual current limiting resistors wired in series with each LED?
    So, a 3.6v supply.
    What is their Vf (forward voltage) rating at 30mA, and what value resistors are you using for current limiting?

    This is where a schematic would be a big help.

    If you are using LEDs that are rated for 3.6v @ 30 mA without current limiting resistors, you probably have some LEDs that are getting far too much current, others not enough. Power to LEDs needs to be regulated by current instead of by voltage.

    Your battery charger should be providing a constant current of around 120mA to 180mA to charge your batteries, not a constant voltage. If you try powering your LEDs at the same time you're trying to charge the batteries, you'll wind up with dead batteries and dim LEDs.
     
  3. N0ctrnl

    Thread Starter Member

    Aug 14, 2009
    14
    0
    Hey, thanks for the quick reply. I looked for an easy to use program to whip up a schematic without installing Visio, but looks like I'm going to end up going that route. Probably need to anyway. :) Any suggestions on that front? I did a quick forum search, but undoubtedly missed something.

    The LEDs I'm using in this case are rated at 3.6v and 30mA. So, no resistors on them. I have many different LED configurations I use, so some do require resistors.

    I was just trying to come up with a way to charge the battery without the power switching having to be on. I can simply do that by using the direct connection to the battery and using the DC jack that disconnects the ground from the LEDs when plugged in. That way it'll charge either way and not require the switch to be on. All good. I'll post a picture of exactly what I'm working on and it might explain a little better why I wanted them on while charging. :)

    Thanks so much for the reply!
     
  4. tibbles

    Active Member

    Jun 27, 2008
    249
    3
    "I have a plug on the end of the LED string" ?
    batteries at one end- charger at the other?

    like this
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Look in the Electronic Resources forum.
    a couple of good options:
    Linear Technology's LTSpice; freeware schematic capture & SPICE simulations.
    Cadsoft Eagle; freeware version available; does schematic capture & small PCB's (3"x4") in freeware mode, no SPICE simulation.

    Bad juju. You're depending on the battery internal resistance to limit current.
    It is good practice to always use SOME form of current limiting or regulation.

    Well, why not move the LED power switch from the + supply to the - return? The charge jack removes that anyway.

    Using a single resistor to limit current for multiple LEDs in parallel is a bad idea, unless they have individual current limiting resistors on them already.
     
  6. tibbles

    Active Member

    Jun 27, 2008
    249
    3
    hope my input hasnt confused you, i did what i thought noctrnl was describing,

    great name noctrnl im a bit of an insomniac myself
    d
     
  7. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    A pencil and a piece of paper does can do a good, cheap job of making a diagram. You can use a scanner or digital camera to get the drawing to a form so it can be posted. Alternatively, it's not too hard to make a simple schematic with a program like MS Paint, etc. No need to run out and find a fancy drawing tool -- but no harm in doing so.
     
  8. N0ctrnl

    Thread Starter Member

    Aug 14, 2009
    14
    0
    Well, I forgot about dia. I used that for years instead of Visio. Let me whip something up ....ok, dia doesn't have stencils for switches. That kinda sucks. I tried LTspice IV before, but honestly, I always used a pencil and paper for my diagrams and they're way too ugly to scan in. I'll see if I can't clean it up a little.

    Here's a few images of what I'm building to kind of give you an idea of what we've got going on.

    http://www.gilpin.info/gallery/LED/

    The hula hoop and the staffs both have the 3 AA batteries in 'em and have a 2.5mm jack for recharging.

    So, if I need a current limiting resistor for something that's already rated at the output power, what resistor should I use? Just the smallest one I can find that fits watt rating?
     
  9. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    Next time, don't use the "~" mark. Some people could mistake it for a
    -5.2V. We know what ya mean though! ;)
     
  10. N0ctrnl

    Thread Starter Member

    Aug 14, 2009
    14
    0
    Here's a diagram. Better late than never, right?

    Still curious what to do for a current limiting resistor on these LEDs since they're rated at 3.6v/30mA.

    Thanks!

    [​IMG]
     
    Last edited: Oct 13, 2009
  11. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You will need to have the LEDs disconnected when the batteries are charging.

    The charger (if it is a commercially made charger for AA & AAA batteries) probably won't put out enough current to both charge the batteries and light the LEDs.
     
  12. N0ctrnl

    Thread Starter Member

    Aug 14, 2009
    14
    0
    Well, that's what the DC jack does. I pretty much gave up on my original idea of keeping the LEDs on while it charges. It's more of a novelty than a necessity anyway, so no biggie.

    I'm still curious what I should do as far as a resistor on the LEDs that are operating at their rated voltage.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    OK, good.
    This is a problem, because the Vf of LEDs as they come from the manufacturer may vary by as much as 10% at room temperature.

    Most of them (75%-85%) will fall into the "Typical Vf" range, or about halfway between the minimum and maximum Vf specifications. The remainder will be (to varying degrees) closer to the minimum/maximum Vf specifications.

    Those with a higher Vf will appear more dim. Those with a lower Vf will get more current, when wired in parallel with other LEDs that have a higher Vf.

    As an LED (or other bipolar silicon device, such as diodes or transistors) heats up, it's Vf actually decreases. This can quickly lead to a "thermal runaway" condition, where the LED passes more current dissipating power as heat, heating it up more, which causes it to pass yet MORE current, etc. until it turns into a blackened molten blob.

    The safer approach is to use current regulation, or at least current limiting. In order to use either, you will need to have at least some "headroom" over the Vf of the LED.

    For example, let's say you went from 3 batteries @ 1.2v each to 4 batteries @ 1.2v each, for a total of 4.8v. Then you could use current limiting resistors. The generic formula for calculating current limiting resistors is:
    Rlimit >= (VoltageSupply - LEDForwardVoltage) / DesiredCurrent
    So plugging in your numbers with the higher Vsupply:
    Rlimit >= (4.8v - 3.6v) / 30mA
    Rlimit >= 1.2V / 0.030A
    Rlimit >= 40 Ohms per LED.
    A table of standard resistance values is on this page (bookmark it):
    http://www.logwell.com/tech/components/resistor_values.html
    43 Ohms is the closest standard E24 value >= 40 Ohms.
     
  14. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    You could buy a few thousand LEDs and test them (because most are different) then pick 10 that are exactly 3.6V. Then you can connect them in parallel and power them from 3.6V.
    That is what flashlight manufacturers do.

    But if the LEDs are actually 3.5V or lower or the battery is 3.7V or higher then the current will be much too high and the LEDs might burn out soon.
     
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