Connect with us

LEDs in SERIES

Discussion in 'Electronic Basics' started by chris, Jun 3, 2004.

Scroll to continue with content
  1. chris

    chris Guest

    I'm going to be wiring some LEDs in series. they are 1.2v 100ma IR
    LEDs.
    Let's say I use 4. 4x1.2v = 4.8 volt source. would the total amp draw
    still only be 100ma? Is this better than wiring them in parallel as
    far as current goes?
     
  2. Wayne

    Wayne Guest

    It depends on your source voltage and current capability.
    For instance, for your series circuit, if you have a 4.8V 1A
    supply, you will fry your leds unless you limit the current
    to a safe level ( 100ma ).
     
  3. Wong

    Wong Guest

    Think that you will need a current-limit-resistor. For example, IR LED
    in series base on your assumption:

    4.8 - 1.2 - 1.2 = 2.4 = Voltage drop on resistor.
    Since 100 mA is required, therefore 2.4/100m gives you 24, so you
    need 24 ohm resistor. In practical, 24 ohm is not common so you may
    choose any other values around that.
     
  4. Amnon

    Amnon Guest

    Hi
    Since you want to use four LEDs , each with 1.2V drop, this means a total of
    4x1.2=4.8V drop.
    Thus your power supply must be a bit larger than 4.8V to allow the addition
    of a current limiter serial resistor as well,in serial to these 4 LEDs.
    For power supply of Xvolt (where X > 4.8) , the serial resistor value will
    have to had a value of R=(X - 4.8)/0.1
    to guarantee a limit of 0.1amp (100mA).

    Working in parallel means that X can be smaller now (X>1.2v) but it should
    supply 4x100mA=400mA, instead of only 100mA, and you will need four
    resistors instead of only one. However, if one LED blows, the other will
    still lit, where in serial connection all will go off.
    Amnon
     
  5. dB

    dB Guest


    No, it isn't a 4.8V source, it's 4.8V across the diodes. The source
    voltage will have to be higher and you will have to connect a resistor
    in series to linmit the current to your required value.

    The same value of current flows out of a series circuit as flows into
    it. It will "only" be 100mA if the value of the series resistor has
    the appropriate value.


    If you run the diodes at 100mA and use a series arrangement than the
    total current will be 100mA. If you use the parallel arrangement the
    total current will be 400mA. It is up to you to decide which is
    "better".



    You must get used to thinking of a l.e.d. as a current operated
    device, not a voltage operated one.
    A l.e.d. requires a current within a certain range to function
    correctly. When a current within this range flows through it a
    voltage will be developed across it. The value of this voltage will
    change very little if the current varies throughout the specified
    range.

    If you try to operate a l.e.d. from a voltage source you will be
    asking for trouble and you shouldn't be surprised if you are rewarded
    with it. Your l.e.d. will probably have a very bright life - but a
    very short one.

    This series of articles may help you:

    http://p199.ezboard.com/fbasicelectronicsfrm5.showMessage?topicID=16.topic
     
  6. chris

    chris Guest

    I'm going to be wiring some LEDs in series. they are 1.2v 100ma IR
    "Current operated device" - That gets me every time!
    I am having trouble getting past the idea the LEDs are not light bulbs
    and don't have their own resistance. I'm always thinking voltage and
    not current.
    I think that since I'm limited by the amount of remaining current I
    have available from the transformer, I'll do the series option, using
    a resistor to set the current.

    BTW, nice series of articles in that link - very helpful.
     
  7. A 1A supply only means the amount of current you can pull from it
    without overheating the transformer. He would have to limit the
    current using a resistor if he only had a 100mA supply.

    The issue is that when the LEDs get hotter, they can pass LOTS more
    current. However, resistors don't have this flaw, so using, say, a 6V
    supply, with a 12 ohm resistor, would work better.

    Another issue is that when manufacturers say 100mA, they probably mean
    maximum. The OP might want to limit the current even further to
    increase LED life.

    Regards
     
  8. C M Baker

    C M Baker Guest

    I asked the same thing a couple of months back and most people
    responded with the formula (Supply Voltage minus forward voltage) =
    voltage drop.
    Voltage drop divided by max forward current = value of limiting
    resistor.
    This works fine for 1 LED, but as soon as you put two or more in
    series, that formula does not work. I tried several ways to work out
    what the correct value of the series current limiting resistor should
    be and ended up by using a bread board and milliammeter and starting
    with a high resistor and reducing the value until the meter showed the
    correct forward current for the LEDs used. Even with this method,the
    voltage applied to the series is affected by the internal resistance
    of the meter, so opt for a series current limiting resistor that is of
    the nearest standard value, higher rather than lower, this will extend
    the life of your LEDs. I have successfully built several mains failure
    emergency backup lights using LEDs in series running off a 12volt
    battery. Hope this information helps.
     
  9. C M Baker

    C M Baker Guest

    See the series of articles 6th April 2004
     
  10. This calculation should work for several LEDs in series, just as well
    as it works for single LEDs - but you should use the "typical" forward
    voltage drop, not the maximum.

    Note that the formula with several LEDs in series is Supply voltage
    minus (number of LEDs times typical voltage drop). You should then
    use "voltage drop divided by desired current = resistance" (_not_
    maximum current!!).

    Also, it is rarely necessary to run LEDs at their maximum rated
    current - I find most LEDs give quite adequate light with under 10 mA,
    although their rated maximum current may be 20 - 30 mA. (and if I plan
    on using well under the maximum current, I don't need to bother
    checking the specs every time.)



    --
    Peter Bennett, VE7CEI
    peterbb4 (at) interchange.ubc.ca
    new newsgroup users info : http://vancouver-webpages.com/nnq
    GPS and NMEA info: http://vancouver-webpages.com/peter
    Vancouver Power Squadron: http://vancouver.powersquadron.ca
     
  11. C M Baker wrote:

    It does. You just add the voltage drops of all the LEDs together.

    Say, you want to operate a red (1.8 V), green (2.4 V) and blue (3.5 V)
    LED in series. The total voltage drop across them would be (1.8 + 2.4 +
    3.5) V = 7.7 V (the LED act as resistors in series, the total voltage
    dropped is the sum of the individual drop voltages, the current flowing
    is identical through all of them).

    If you use a 9 V supply, you have to drop 1.3 V across the current
    limiting resistor. If you want 20 mA to flow through the diodes, you'd
    use 1.3 V / 0.02 A = 65 Ohm, or rather the next higher standard value.
    The power dropped over the resistor would be 1.3 V * 0.02 A = 26 mW,
    thus a standard 1/8 watt resistor would do just fine.

    As always, there is a little trick to save oneself the calculation: If
    you use a J-FET instead of a resistor to limit the current, the current
    depends only on the amplification of the FET (up to the maximum voltage
    of the FET, that is), not on the supply voltage. A BF245A will supply
    about 10 mA, a BF245B 15 mA and a BF245C about 18 mA:

    o +5..30 V
    |
    |
    | |S
    |----
    G | BF245
    ------->|----
    | | |D
    | |
    -------------
    |
    __|__
    <-- \ | /
    <-- \|/ LED
    -----
    |
    |
    |
    --- Gnd


    If you have a AC supply, you can use a capacitor as a blind resistor to
    limit the current through the LED. Since current and voltage are out of
    phase, little power is dropped across the capacitor, limiting heat
    developed. This is usefull as "power on" indicator in mains circuits:


    o 230 V, 50 Hz
    |
    |
    ----- 100 nF, 1000 V =, 630 V ~
    -----
    |
    |
    ---
    | |
    | | 10 Ohm, 1/4 W
    | |
    ---
    |
    |
    -----------
    __|__ | 1N4148
    <-- \ | / -----
    <-- \|/ /|\
    ----- / | \
    LED | --|--
    -----------
    |
    |
    o neutral

    Note that in this circuit, all components have contact to mains, thus
    appropriate care is required. The capacitor needs to be a type certified
    for use across mains. For countries using other mains voltages or
    frequencies the value of the capacitor needs recalculation.

    Shouldn't we put this in a FAQ somewhere, I recall answering this
    question a couple of times.
     
  12. C M Baker

    C M Baker Guest

    As I sit here at my computer, I have 6 LEDs, specification of which
    is: Forward Voltage 1.7v, Max If 100ma, running from a 13.7v supply,
    in series. According to calculations, current limiting resistor should
    be, (13.7 - (6 x 1.7)) = 3.5v using V=IR, 3.5/.1 = 35 ohms. To limit
    the current to approx 100ma, I have had to put a 100 ohm & 120 ohm in
    parallel,(to limit heat buildup in the resistors) giving a calculated
    resistance of 54.5 ohms. At switch on, this gave a current of 96.5ma
    which has gradually risen to 100.5ma during the last hour.
    My 'research' and my friendly man at the local MAPLINS store say that
    as LEDs are not straight forward resistive components like light
    bulbs, they don't follow Ohms Law directly, and all my testing on a
    bread board also indicates the same. There appears to be two 'camps'
    concerning series LEDs, one saying yes they do follow Ohms Law and one
    saying no they don't. Someone out there must surely have the
    definitive answer or formula. I am not trying to be rude to those
    emminent people who say they do follow Ohms Law, but why do they do it
    for you and not for me?
     
  13. Rich Grise

    Rich Grise Guest

    It depends what you mean by "follow ohm's law." For any given voltage,
    at some current, there's an equation R = E * I, so, _at that particular
    voltage and current_, the component is acting like it has a resistance
    of R ohms. But that value changes for different voltages and currents
    in an LED. It's said to be non-linear. So, the camp that says, "It
    doesn't follow Ohm's law" just mean that R in that equation is not
    constant, i.e. it doesn't behave like a plain "ohmic" resistor.

    The reason your current increased is that Vf decreases as they heat
    up. This also changes the effective resistance, even at a constant
    current.

    Hope This Helps!
    Rich
     
  14. John Fields

    John Fields Guest

    ---
    As Rich pointed out, your LEDs have a non-linear voltage VS current
    characteristic, as well as a negative temperature coefficient of
    resistance.

    Unfortunately, the friendly man at your local MAPLINS store misled you
    with the part about the light bulb being a straightforward resistive
    component. It isn't, and because of the positive tempco of tungsten,
    a plain old 120V 100W lamp will look like about 14 ohms when it's cold
    and rise to about 144 ohms when it's hot.
     
  15. LEDs (and other diodes) do not follow Ohm's Law - that is, the voltage
    across the LED does not vary directly with current like a resistor.
    The voltage across a forward-biased LED is fairly constant, and is
    determined by the chemical composition of the LED (which also
    determines its colour). The voltage will increase slightly with
    increased current.

    The "advertised" forward voltage of your LEDs is probably a typical
    value, and may vary slightly between production lots.
     
  16. Ohmic resistance has a voltage drop proportional to the current
    through the resistance, and that factor of proportionality (volts per
    ampere) is called ohms. Nothing is precisely ohmic (perfectly fixed
    proportion between voltage and current) regardless of the current.
    Some materials are pretty good over a temperature range that is
    useful, and they make resistors out of these materials. Any PN
    junction (silicon, germanium or exotic materials like used in LEDs)
    have a completely different relation between the voltage drop and the
    current. Current through these devices is more accurately described
    as an exponential function of voltage drop (or conversely, the voltage
    drop is related to the logarithm of current) but this effect is also
    dependent on temperature (as you found as your LEDs warmed up from the
    I*V power they consumed minus the energy they emitted).

    Here is a data sheet for an LED that is perhaps similar to the ones
    you are using:
    http://www.vishay.com/docs/81009/81009.pdf

    Figure 4 describes the typical voltage current relationship at room
    temperature. Note that one of the scales is logarithmic and one is
    linear. If a simple logarithmic relation between current and voltage
    drop existed, this graph would be a straight line on this kind of
    graph, and it is pretty close for very small currents up to about
    rated current of 10^2 milliamps. For higher currents, the series
    resistive drops of the wire bonds and the material on each side of the
    junction starts to show itself, and the graph heads off toward a more
    resistive line.

    Figure 5 shows how the voltage drop associated with one point on the
    line from figure 4 (the 10 milliamp point) varies with temperature.
    You can assume that this variation is reflected all along the current
    line (with small error).
     
  17. Everybody has the answer (all correct), but not the formula. :) Here
    it is:

    If
    --- = exp(Vf/(m*Vt)) - 1
    Is

    If is the forward current, and Vf is the forward voltage.

    'Is' is the reverse leakage current, and depends on the device, and
    also the temperature. Vt is a combination of terms (it equals kT/q,
    where k is Boltzmann's constant, q is the fundamental electric charge,
    and T is the temperature in degrees K) which turns out to equal 25.3mV
    at 20C. m is a 'fudge factor' which is between 1 and 2.

    'Is' is usually quite small, and also depends on temperature. Its not
    easy to figure it out for a discrete device, which makes this
    relationship somewhat unusable except for doing things like predicting
    that increasing the voltage by 60mV will increase the current 10x,
    since in the ratio, 'Is' is divided out.

    This relation is called the Ebers-Moll equation.

    Regards,
    Bob Monsen
     
  18. C M Baker

    C M Baker Guest

    Many thanks to all those who have replied, especially Bob Monsen,
    whose answer was most enlightening.
    Since Monday, we have had three power cuts, two lasting over 4 hours,
    during which time the emergency light system functioned well. I am now
    onto the "Mk lll" version, replacing the 12v halogen lamps with 5mm
    8000mcd white LEDs, 3 of which appear to produce as much light as one
    5w quartz, but with much lower power consumption and at comparable
    cost, one LED being £2-99 one 5w quartz being £4-99.
    I certainly know where to come to get sensible and helpful advice.
    THANKS. Colin.
     
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

-