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LED Questions (just two: )

Discussion in 'Electronic Basics' started by [email protected], Apr 15, 2007.

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  1. Guest

    Ok - since one person did offer here as a good place to ask some of my
    ignorance to knowledge questions concering my LED Brake Light fantasy:

    Ok, so I have learned that you vary voltage - preferable with pulse
    width? - to vary the lunines output of an LED
    AND that voltage drops from one LED to the next if wired in series
    by say about 2 volts...
    1. Then would not the LEDs in a series be dimmer the further on the
    series they are? (ie The 1st one brighter than the next with the last
    being the dimest)
    I know this must be wrong so
    2. What am I missing?

    (If you are so helpful to answer - just a link to info on the web
    would be enough. Typing is a pain!! : )
  2. Jan Nielsen

    Jan Nielsen Guest

    The leds will have the same brightness and share the voltage (about)
    evenly untill the voltage eventually drops too low to keep the leds
    turned on.

  3. Well you can vary the pulse width to vary the brightness, but it is the
    average current (not voltage) that is being varied. Generally a resistor
    (or some other current limiting device) is placed in series with the LEDs to
    limit the current flow to some predetermined maximum. If you hook LEDs
    straight to a voltage source without something to limit current, the LED
    will flow massive amounts of current and then they will likely burn out
    More of less yes depending upon the color.
    The part mandating that the current flowing thru devices arranged in series
    is the same in each device. Therefore each LED should be the same
    brightness if they are the same type.
  4. Guest

    So - it is the Current not the voltage that determines the
    Voltage can change and not affect the brightness just so long as it
    remaines with an operating range.
    Am I closer?
    Thanks for the replies
  5. LEDs should be driven from a current-limited supply,, not directly
    from a fixed voltage. The most common way to do this is to connect a
    resistor in series with the LED (or several LEDs), and connect the
    combination across a fixed voltage supply.

    The LEDs can be dimmed by pulsing the voltage.
    The voltage drop across an operating LED (red, yellow or green) is
    about 2 volts (red a bit less, green a bit more)
    The brightness of a LED depends on the current flowing through it, and
    the current is the same at all points in a series circuit, so if you
    have several identical LEDs in series, they will all get the same
    current, and will all be the same brightness.
    Peter Bennett, VE7CEI
    peterbb4 (at)
    new newsgroup users info :
    GPS and NMEA info:
    Vancouver Power Squadron:
  6. Guest

    Well - look Iike I sent my responce to the Author - Sorry!

    So, brightness would be determined by the Current.
    Voltage just need to remaine within the operating range of the LEDs.

    Am I closer?

    Thank you all for your help
  7. The voltage (all of it, whatever it is) will be "dropped" across your
    circuit. Some will be dropped by the LED (about 2V for each LED in series,
    just as you said), and the rest needs to be dropped by a resistor. Using
    Ohms Law, you can determine how much resistance you will need given the
    voltage being dropped and the desired amount of current flow for the LEDs
    (resistor and LEDs will flow same amount of current since they are in

    As an example, lets say we have a 5V voltage source and the LED we wish to
    illuminate has a forward voltage drop (Vf) of 2V. That means the resistor
    we will use will need to drop (5-2) 3V total. If we wish to pass 20mA thru
    the LEDs (this would be determined by looking at the datasheet for the LEDs
    you have, but 20mA is usually a safe amount of current), we would use Ohms
    Law to find the value of resistance that will drop 3V when 20mA is flowing
    thru it. Arranging Ohms Law such that R is on one side of the equal sign,
    we come up with R=E/I. So R (the resistance in Ohms) will be equal to E
    (voltage being dropped in Volts) divided by I (current in Amps).
    Substituting our numbers, we come up with R = 3 / .020 or R= 150Ohms.

    So using a 5V source, we can use a 150Ohm resistor to drop the 3V required
    to allow 20mA to flow thru both the resistor and the LED. The LED will drop
    the other 2V.
  8. LEDs work quite differently than incandescent lamps - they do not obey
    Ohm's Law (and I'm sure some pedant will jump on me for that

    The voltage across an operating LED is determined primarily by the
    chemistry (and therefore colour) of the LED, and not by the current
    through the LED, so an LED _must_ _not_ be operated from a fixed
    voltage, but rather from a higher-than-necessary voltage, with some
    means (usually a resistor) to limit the current to an acceptable

    If you want to operate a bunch of red LEDs (typical voltage 1.8 volts)
    from a 12 volt power supply, you might connect 5 LEDs in series, to
    give a 9 volt drop across the LEDs, then put a 150 ohm resistor in
    series to limit the current to 20 mA (9 volts across the LEDs leaves 3
    volts for the resistor - 20 mA in 150 ohms gives a 3 volt drop.)

    When reading LED datasheets, note that the voltage stated is typical,
    for some stated current, and the datasheet will usually state a
    maximum current, beyond which the LED may be damaged. LEDs will work
    fine well below that specified maximum current. Unless I _really_
    need maximum light from the LED, I'll usually plan on operating it at
    10 mA or less - that's well below the maximum rating for any LED I've
    looked at, and generally gives enough light, so I don't have to study
    the datasheet for the particular LED I have to calculate the
    appropriate resistor value.

    Peter Bennett, VE7CEI
    peterbb4 (at)
    new newsgroup users info :
    GPS and NMEA info:
    Vancouver Power Squadron:
  9. ehsjr

    ehsjr Guest

    No. The supply voltage can be much higher than the operating
    range of the LED, provided you keep the *current* within
    the operating range of the LED. Look at this circuit:

    +12V ---[470R]---+
    Gnd ------------+

    If the LED Vf (forward voltage) spec is 2 volts, that means
    that there will be ~2 volts across the LED when it is operated
    in its current range. The 470 ohm resistor will limit the
    current to ~ 21 mA. If the supply was 9 volts, you would
    change the resistor to 330 ohms. If the supply was 100
    volts, you would change the resistor to 4700 ohms. In each
    case, the resistor limits the curret through the LED to about
    21 mA.

    Focus on controlling the current through the LED - the voltage
    across it will automatically be correct, as long as you limit
    the current properly.

  10. Wim Lewis

    Wim Lewis Guest

    Yes. For a LED, the brightness is directly proportional to
    the current. (Other kinds of lamps behave differently.)
    Closer, but not quite there. The current depends on the voltage across
    the LED.

    The other thing I think you're missing is that "absolute" voltage
    isn't important, only voltage differences between one place and
    another are important. It's kind of like altitude --- climbing
    a flight of stairs at street level feels exactly the same as climbing
    a flight of stairs at the top of a tall building, as long as you
    don't look out the window. What matters is how tall or steep those
    stairs are. In other words, the difference in altitude between
    the top and the bottom, not the absolute altitude.

    People also talk about "voltage drop", which is just a way of talking
    about the difference in voltage on each side of a component. If one
    terminal is at 5v and the other is at 4v, then you can say that the
    voltage (or potential) has dropped by 1v as you move from one terminal
    to another.

    LEDs have a funny current/voltage relationship. If you put a small
    amount of voltage across them, (almost) no current will flow. If you put a
    bit more voltage, still no current will flow. If you keep increasing
    the voltage, then at some point --- around 2 volts, for a green LED ---
    the current will suddenly start to flow, and the LED will light up.
    Increase the voltage any more, and a huge amount of current will flow,
    and the LED will burn up, pfft!

    So you can't just attach the LED directly to a voltage source, because
    you're unlikely to get exactly the voltage that will light up the LED
    without destroying it. Plus, the voltage (the "forward voltage drop"
    of the diode) is different at different temperatures, for different
    LEDs, and so on. You need some way to keep the right amount of current
    flowing even if you don't know exactly what voltage to apply.

    The simplest way to do this is with a resistor. Unlike a diode, a resistor
    has a very simple relationship between current and voltage; they're
    directly proportional. In other words they follow Ohm's law.

    So, imagine this circuit:

    [ Battery + ]-----/\/\/\/\/\----->|------[ Battery - ]

    (that's a resistor and a diode in the middle there, with a battery
    connected to the ends). What do we know about this circuit? Assuming
    that it's operating correctly?

    - The voltage between [battery +] and [battery -] will be the
    battery's voltage. Let's assume we have a 12-volt battery.

    - The current through the LED will be 10 mA, because that's how
    much current we want. (The right amount depends on the LED ---
    bigger LEDs can handle more current.)

    - The voltage across the diode is unknown, but it's roughly 2 volts,
    depending on the kind of LED.

    - The voltage across the resistor is (current times resistance).

    So, we can subtract out the "roughly 2 volts" of voltage across the
    diode, and that tells us that the remaining voltage across the resistor
    will have to be "roughly 10 volts". It's a series circuit, so the
    current through all parts is the same: we've decided we want it
    to be 10 mA. Ohm's Law tells us how big the resistor must be: 10V/10mA
    equals 1 kOhm. Now we know enough to build the circuit and get
    10 mA flowing through the LED.

    Why does this work? Because now, if the diode's voltage varies by 0.1
    volt, or if the battery's voltage varies by 0.1 volt, the current through
    the diode doesn't change much. It stays around 10mA (maybe 9.9 mA, maybe
    10.1 mA, but that's close enough). Most of the variation is being taken
    up by the resistor, which has a very smooth variation of current
    with voltage, unlike the LED's very sudden reaction.

    Unfortunately, most of the power is being taken up by the
    resistor, too. Only about 1/6 of the energy is going to the LED,
    and the rest is going to the resistor (and turning into heat). For a
    brake light, this isn't much of a problem, but in other cases, it
    can be. In those cases, you can use more complicated techniques,
    like pulse-width modulation (where you keep the *average* current
    correct by adjusting the duration of pulses of current).
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