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LEDs - Simple Wireing Questions

Discussion in 'Electronic Basics' started by ShadowTek, Nov 3, 2004.

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  1. ShadowTek

    ShadowTek Guest

    I just bought 50 3.2V 20mA white LEDs for about $30US total off of
    EBay.
    http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&category=66952&item=3847009435&rd=1&ssPageName=WDVW
    I tried to read up on electrical stuff but there is a lot I dont
    understand.
    For instance, if I use 2 1.5V AA betteries in a series that are 3 Amps
    each, then is the output 3 Amps or 6 Amps?
    I know voltage doubles when connecting cells in a series but what
    about current?
    Assumeing that the total current of the two batteries were 3 Amps,
    does that mean that these 2 batteries alone could power 150 of these
    3V 20mA LEDs if the LEDs were connected in series?
    That doesn't sound correct.
    Or would that need to be parallel?
    Maybe that would work but only for like a few minutes before the cells
    were completely drained?
    How many of these LEDs could I hook up to 2 AA (wired how?)cells
    without using any other form of resistor other than the LEDs
    themselves?
     
  2. Lord Garth

    Lord Garth Guest

    First problem is the batteries in series are 3 volts and the voltage drop
    across one LED is 3.2 volts. This is with no consideration toward the
    voltage drop across any form of current limiting. If you don't limit the
    current, your LED will overheat and burn out.
     
  3. John G

    John G Guest

    You cannot say the voltage drop across a led is 3.2 volts.
    There are many different leds and some are less than 2 volts.
    The proper limiting resistor is absolutely necessary.
    3 amps is a lot, maybe the short-circuit current for an AA cell.
    A lot of study would seem to be indicated here before short answers are
    meaningful.
     
  4. Saying that the batteries are 3 amps each is a bit nonsensical.
    Batteries have an internal resistance (that consumes some of the
    voltage as current passes through it and limits the short circuit
    current to the value that uses up all of the voltage across the
    internal resistance) and batteries have an ampere hour rating that
    generally tells how much current it can supply for how long before it
    is exhausted. Cut the current in half and the life approximately
    doubles.
    If you are talking about short circuit current, it stays the same, 3
    amps, because each cell drops its entire internal voltage across its
    internal resistance at its short circuit current.
    Connecting cells in series doubles the total internal resistance while
    providing not more capacity (ampere hours).
    Not if the 3 amperes is a short circuit current rating, That is the
    current that reduces the battery voltage to zero, so it can't drive
    anything at that current. If that rating is the ampere hour rating of
    the cells, then, you could power a 20 ma load for 3/.02=150 hours, a 1
    amp load for 3/1=3 hours, etc. Whether the output voltage is good
    enough to force any particular amount of current through an LED is
    another matter, entirely.

    The LED ratings of 3.2 volts and 20 ma implies that it takes
    approximately 3.2 volts across each led before they pass 20 ma of
    current. However, LEDs are not resistors, and their current is very
    strongly related to the applied voltage. A few millivolts extra can
    double the current, and a few millivolts less will cut the current by
    half. Also, a small change in temperature can shift the current
    dramatically, if the voltage is fixed. This makes it very difficult
    to power LEDS directly and in parallel from a stiff (low internal
    resistance) voltage source. Normally some additional stuff is needed
    to waste a bit of extra voltage and regulate the current. For
    instance, if you used 3 cells in series for an open circuit voltage of
    about 4.5 volts, you could add a series resistor to each of those
    parallel LEDs and if you pick the right resistance it will waste the
    extra voltage (4.5-3.2=1.3 volts whenever the current is about 20 ma.
    So 1.3 volts divided by .02 amp predicts that this value is about 65
    ohms. Standard 5% values are 62 and 75. If you want ot get maximum
    brightness you might use the lower resistance, but if you want ot make
    sure that you operate the LED well within its ratings you might use a
    75 to 100 ohm resistor with a small drop in brightness. How many of
    these sets you can connect across the battery depends on the sag in
    voltage caused by the internal resistance and the battery life you
    desire.
    If the LEDs take 3.2 volts and the battery produces 3 volts, even with
    no current load, then zero.
     
  5. Maybe a fire extinguisher could be a good analogy for a battery.

    It can deliver a certain amount of liquid and the pressure is high if it
    newly filled.

    If it contains 10 liter of water it can deliver a current of 1 liter per
    second for 10 seconds. But the flow is not constant through these 10
    seconds, the pressure and the current is a little higher in the beginning
    and slows down as it becomes emptier.

    There is a certain flow resistans in the output tubes and stuff, and that
    limits the flow somewhat.

    A battery has a capacity rating, like the amount of water in the fire
    extinguisher, it is measured in AmpereHours (Ah) or mAh for smaller
    batteries.

    A battery with a capacity of 3000mAh can deliver 3000mA for one hour, or
    1000mA for 3 hours, or 100mA for 30 hours.

    Batteries also have a voltage rating, like the water pressure in the fire
    extinguisher, it is measured in Volt.

    The amount of work a battery can deliver is the capacity times the
    voltage. A 3 Volt battery with 3000mAh capacity can do a work of 3*3 VAh,
    or 9 Watthours.

    That is what you pay for when you pay the electricity bill, the work you
    have bought, measured in VoltAmperehours, or Watthours, or kiloWatthours.

    The effekt is how much work is performed per time unit, like Watthours
    per hour, or simply Watt.
     
  6. dB

    dB Guest

    (ShadowTek) wrote

    You MUST use a series resistor. Otherwise your l.e.ds are likely to
    have a very bright life but a very short one.
     
  7. ShadowTek

    ShadowTek Guest

    OK thanks.
    I thought 3 Amps for a AA battery sounded strange but I was confused
    by the markings on the battery.
    They were "LRG AM3 1.5V".
    I guess LRG just means large and AM3 means short curcuit current.
    I was further thrown off by trying to read the current with my
    multimeter from terminal to terminal.
    I guess that I was createing a short curcuit which was trying to empty
    the battery.

    (BTW, my multimeter only reads up to 250mA so do you think reading a
    3A short curcuit for no more than 2 seconds would have damaged the
    MM?)

    The exact specs for the LEDs were 2.8V-Min. 3.2V Typ. 3.8V-Max..
    So if intend to run the LED at 2.9V then I should still be able to run
    1 LED off 2 1.5V cells right?
    ..1 divided by .02 = 5 Ohm reistor?
    So 2 AA batteries and a 5 Ohm resistor will run 1 of these LEDs for
    150 hours?
    Or I could wire several LEDs in parallel so long as a 5 Ohm resistor
    prefixed every led?

    Do LEDs suffer any damage from being underpowerd?
    Since we are talking about batteries then the constant drain would be
    a problem.
    I have a 3 white LED headlamp and I have run the batteries in it all
    the way down several times.
    It uses 2 flat 3V lithium batteries.
    It still seems to work ok.

    BTW, radio shack sells their white LEDs for 6$ a peice.
    If you are used to payin that much then you should check out the link
    I posted earlier.
     
  8. Bill Bowden

    Bill Bowden Guest

    I have a little dual white LED flashlight that uses 3 AA batteries.
    The leds are in parallel, connected directly to the battery,
    with *no* resistors. These LEDs apparently have a voltage range
    of 3 to 4.7 volts or so.

    -Bill
     
  9. I doubt that the designation LRG AM3 has anything to do with size or
    amperage.
    Here is a summary of some Duracell specs:
    http://www.web-ee.com/Component Pages/battery_info.htm
    Note that the MN1500 AA cell is rated for 2850 milliampere hours and
    its maximum rated load is 43 ohms for a load current of 28
    milliamperes. Now, .028*43=1.2 volts, so this implies that a typical
    internal resistance must be dropping the rest of 1.5 volts. So the
    internal resistance is about .3volts/.028 amperes or 10 ohms. So the
    short circuit current should be about 1.5 volts / 10 ohms = .15
    amperes. I am quite sure that you will get more than this with a
    fresh cell, but this is typical for the mid life point.
    Here is another site that summarizes battery characteristics:
    http://www.zbattery.com/zbattery/batteryinfo.html

    Note the resistance of the alkaline AA cells.
    Yes. Current meters have very low resistance.
    If it reads more than zero then you haven't blown the fuse in series
    with the current meter function.
    Yes, the required voltage varies with temperature and lot to lot.
    The two cells with their 20 ohms of total internal resistance will
    push some current through the LED, but the exact amount is hard to
    predict, except that it will be low.
    The resistor is in addition ot the 20 ohms of internal resistance, and
    at this low voltage the current will be much lower than 20
    milliamperes, so the battery life will be very long.
    Yes. But if the LEDs vary all over the specified voltage range, you
    can expect the one with the lowest drop to be drawing much more
    current than the one with the highest drop.

    This is why I recommended that you use 3 cells and a much higher
    resistance in series with each. It makes the current matching much
    better and more independent of LED variations and sag in battery
    voltage.
    Just the opposite. Their light output is roughly proportional to
    their current, but their lifetime goes up dramatically as the current
    goes down.
     
  10. John Popelish wrote:
    (snip)
    Correction. I mistook the ohms listed as the cell internal resistance
    (which seemed high to me). Then I read the note above the list that
    says the listed resistance is the load used for the test data (I
    assume this means the ampere hour test data that took the batteries to
    the cut off voltage).
     
  11. BobGardner

    BobGardner Guest

    I tried to read up on electrical stuff but there is a lot I dont
    =======================
    Most of it is derived from Ohms Law.... E=IR (voltage=current x ohms). The gist
    of it is, voltages in series add, but the current in a series circuit is the
    same in all series elements of that circuit... makes sense if you think about
    it. The trend is to use special led driving voltage regulators that drive
    strings of several (6 or 7) leds in a string, 20ma each string. The regulator
    keeps the right voltage and current even as the battery voltage starts
    dropping. You could use a 12v battery and drive 3 leds in series with a
    resistor... 12V from batt - 9v across leds leave 3v/20ma for the resistor to
    soak up... 150 ohms. You could drive as many of these 3 diode and a 150 ohm
    resistor strings as you can afford.
     
  12. peterken

    peterken Guest

    could use a voltage regulator and a resistor...
    but why not use just a current source ? They are easy to come by...
    gets you the same result (stable current through led's while dropping supply
    voltage) but cheaper circuit

    one more thing.... led voltage is depending on color type, from say 1.2V
    upto say 2V
    best define setpoints of leds using current i.s.o. voltages, suince
    calculations with led voltage might get you faulty resistor values




    =======================
    Most of it is derived from Ohms Law.... E=IR (voltage=current x ohms). The
    gist
    of it is, voltages in series add, but the current in a series circuit is the
    same in all series elements of that circuit... makes sense if you think
    about
    it. The trend is to use special led driving voltage regulators that drive
    strings of several (6 or 7) leds in a string, 20ma each string. The
    regulator
    keeps the right voltage and current even as the battery voltage starts
    dropping. You could use a 12v battery and drive 3 leds in series with a
    resistor... 12V from batt - 9v across leds leave 3v/20ma for the resistor to
    soak up... 150 ohms. You could drive as many of these 3 diode and a 150 ohm
    resistor strings as you can afford.
     
  13. Terry

    Terry Guest

    The OP does not appear understand the relationship between volts and amps
    (basic Ohm's Law).
    So there's little chance will understand the niceties of powering LEDs,
    variations to specification and current limiting.
     
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