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luxeon led dimmer circuit

Discussion in 'Electronic Basics' started by andy, Oct 2, 2005.

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  1. andy

    andy Guest

    I've built the following circuit on breadboard as a dimmer circuit for 3
    luxeon III Star ultrabright LEDs:

    any comments welcome. It does work at the moment, but when the battery is
    low, the maximum current drops from 0.9 A to 0.7 A, and the op-amp swings
    near to the top rail. The one thing I'm thinking of changing before I
    solder it up is to change the 1 ohm sense resistor for a lower value
    (maybe 0.5 ohm) so that the transistor has a bit more headroom to work
    with, which should cure this I think.

    The LEDs are /very/ bright by the way - 3 of them are enough to light a
    small room well enough to read by. The light is a bit unkind on the eyes,
    but not too bad. 3 at full current are about equivalent to an 11 Watt
    energy saving fluorescent light bulb, which puts them in roughly the same
    bracket as far as efficiency goes.
  2. Does the BD139 get very hot? I agree that you can use a lower value
    current sense resistor. But you also have to scale the reference
    voltage down by half. I think you don't need the first opamp to
    supply the divider, since it is a constant load, so it can be factored
    into the zener bias design. Also, the divider can be very much higher
    value resistors, since the opamp draws almost no current from it. For
    instance, a 10k pot in series with a 100k resistor (to use with a .5
    ohm sense resistor). You could use the opamp to regulate the current
    to the zener, instead, to make that voltage almost completely
    independent of the battery.

    The idea is to have the opamp output drive the zener through a current
    limiting resistor, while the zener is also tied to non inverting
    input. The inverting input is connected to a voltage divider (could
    also be your output adjust divider) between the opamp output and
    ground, which sets the voltage gain of the opamp to something a little
    more than 1. Lets say you use a divider that creates 5.6 volts when
    the opamp produces 8.4. (say, a 10k resistor to the opamp output and a
    20k to ground) To bias the zener with about 5 mA, the resistor from
    output to zener would have to be about 2.8/.005=560 ohms. you may
    have to add a 10k pull up resistor to the opamp output to make sure it
    starts positive when the power is applied. You tap off the lower half
    volt across the lower part of the 20k resistor to produce the 0 to .5
    volt needed for the current reference.
  3. andy

    andy Guest

    Yes, but not as bad as the other power components, so probably not
    desperately in need of heat sinking. (The LEDs get hot enough to scald
    when run at full current, so I'll probably heat sink them with some
    aluminium sheet. They already have mini heat sinks built in though.)
    Yes, I was assuming that.
    Thanks for the idea - I'll give it a try.

    cheers, andy.
  4. Bob Monsen

    Bob Monsen Guest

    Another way to control brightness is to use what is called a PWM circuit.
    Basically, you give your pass device a pulse every once in a while, and
    control the percentage of time it is on. Since you aren't just burning up
    energy with a resistor, it is often cooler and more efficient.

    A simple circuit for this consists of a cmos 555, an N-channel JFET, a
    cap, and a pot:

    | |
    | ||--'
    .--------. | || J1=BF245B
    .------+Vss Vcc+----' .->||--.
    | | CN555 | | |
    | | | | |
    | | tr & th----------o------o
    | | | R=470 |
    | .---+OUT DISC+---o-\/\/\/\--------o
    | | | | | ^ |
    | | '--------' '----' |
    | | |
    | | |
    | '-Gate of logic level NMOS ----- C=1uF
    | -----
    | |

    Call the current through J1 'I'.

    The period is going to be the sum of the charging time

    Tc = (1/3 * Vcc) * C / I

    and the discharge time is

    Td = R*C*ln((2Vcc - 3RI)/(Vcc - 3RI))

    The duty cycle, which is what you are interested in, will obviously be


    since OUT is high during Tc.

    The odd thing is that if you simplify this, the duty cycle doesn't depend
    on the size of the capacitor; it only depends on vcc, Id(J1), and R. Thus,
    you can pick a cap that is small enough so you don't see a flash, but not
    too so the pass transistor requires too much dynamic current. The duty
    cycle defined by this monster:

    D = ---------------------------------------------
    Vcc + 3 R I ln((2 Vcc - 3 R I)/(Vcc - 3 R I))

    So, when R is 0, D is 1, and when R is Vcc/3I, D is 0

    That means that when R=Vcc/3I, the thing will simply stop, with output
    low. The reason is that the discharge pin won't be able to pull the
    trigger pin lower than Vcc/3. When R=0, it won't take any time to
    discharge the timing node (well, almost no time) so D <- 1.

    The N-JFET will vary as to how much current it will source in this
    configuration. If it is sourcing too much, put a small resistor between
    the drain and the point where the gate attaches; this will lower the
    current. However, you need to make sure that your new resistor times the
    current isn't bigger than Vcc/3. If it is, the output will get stuck
    trying to pull the timing cap up to 2/3 Vcc, which means output will be
    high and your pass transistor will be on all the time.

    If you can't find a JFET, you can use a couple of PNP transistors as a
    reasonable current source, like this:

    | |
    | [Rxx]
    | |
    e |
    c |
    | e
    | c
    | |
    [Ryy] '----- current out

    Rxx sets the current to near I = 0.615/Rxx. Ryy should provide 1/10 of the
    current through Rxx, so

    Ryy = 10*(Vcc - 1.3)/I
  5. redls1bird

    redls1bird Guest

    i am also working with some leds (lumileds superflux). My project i
    to control a cicuit of leds that will be a parking light/turn signal
    The light while illuminated as parking light mode is set at on
    brightness, and gets almost twice as bright when turn signal is on.
    The only diagram i can findf for this particular circuit i
    but being a beginner, im not sure how to construct this complicate
    of a circuit. i know i should use pc board (or so ive been told)
    but it is still pretty intimidating. Also, anyone know where i ca
    get a 15v zener diode?
  6. ehsjr

    ehsjr Guest

    For the 15 v zener
    Cat #1N4744 from
    4 for $1.00

    The circuit at the url does not give you the values for 3
    of the resistors, nor the part to use for the darlington
    PNP transistor.

    Here's a different way to do it:

    ------- 1N4001
    Tail -----in| LM317 |out---+---->|----+
    ------- | |
    Adj [240R] |
    | | |
    +----------+ |
    | |
    [POT] 5K |
    | |
    Gnd |
    ------- |
    Turn -----in| LM317 |out---+---->|----+--- To led array
    ------- |
    Adj [240]
    | |
    [POT] 5K

    Here's the LED array:

    +----+----+----+---- Vcc
    | | | |
    V V V V
    --- --- --- ---
    | | | |
    V V V V
    --- --- --- ---
    | | | |
    V V V V
    --- --- --- ---
    | | | |
    V V V V
    --- --- --- ---
    | | | |
    | | | |

    All resistors are 1/2 watt. You can adjust the
    pots for the desired brightness, and then replace
    them with the next closest standard value resistor.

    The 51 ohm value was determined assuming Vf of 2.5 volts
    and max current of 70 mA for the LEDS - if your LEDS
    are different, a new value needs to be computed.

  7. redls1bird

    redls1bird Guest

    i dont want to sound like a total [email protected]$$, but i cant understand tha
    kind of diagram!!! lol i really appreciate the help and the circui
    appears to be much less complicated, but where can i get some help s
    i can better understand that diagram
  8. andy

    andy Guest

    Not sure which circuit you're talking about, but the basics are:

    - lines mean an electrical connection between one part and another. A
    circle means a connection between three lines (you never show four lines
    connecting, so a crossing without a dot means the wires don't touch)

    - square boxes are resistors. resistors with 3 connectors and
    an arrow on the middle one are variable resistors (pots)

    - two lines with a space between are capacitors

    - a circle with an arrow in it is a diode. If there's a lightning arrow
    off it, it's a light emitting diode.

    - a circle with a bar in it an 3 lines coming off is a transistor (the
    line with the arrow is the emitter, the line into the middle of the bar
    is the base, and the other line is the collector)

    - triangles are 'operational amplifiers' - high gain amplifiers which you
    'program' with external components to make them into reliable amplifiers
    or control elements. These usually come in Dual In Line IC packages, often
    with more than one on the same chip.

    - a box with several lines coming off is a more complex chip, which could
    do any number of things, so you have to look up the chip number.

    The idea of these diagrams is that they show just the 'logic' of the
    connections between the components. Given a circuit diagram, you should be
    able to wire up the real parts to make a working circuit. Or else you can
    use these diagrams to look at the circuit before you build it and see how
    it's going to work, or show other people, etc.

    any decent electronics book or website should tell you enough to make
    sense of diagrams like this, at least enough to build the circuit and
    understand it at a simple level.
  9. ehsjr

    ehsjr Guest

    Is it the ascii art that you have trouble viewing?
    You need to set your newgroup reader to read the messages
    in a fixed width font such as courier.
    When you see something like this: -[240R]- it is a
    240 ohm resistor. [51R] is a 51 ohm resistor
    --->|--- is a diode - so is V
    It was used as an LED (light emitting Diode)

    [POT] is a potientiometer
    and the LM317 is a three terminal voltage regulator
    whose pins are labeled in, out and Adj

    Does that help?
  10. redls1bird

    redls1bird Guest

    very much so. for some reason i just couldnt wrap my brain aroun
    that. its definitely less complicated than the original circuit
    found. ill give it a shot
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