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LEDs in Series with LM334 Current Source

Discussion in 'Electronic Basics' started by Mike H, Apr 30, 2004.

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  1. Mike H

    Mike H Guest

    I'm working on a circut design to replace the LED's that came in a
    parking-light assembly on an automobile. The manufacturers design used
    some HP designed circuit that simply used current limiting resistors to
    drive the leds. For whatever reason something shorted out and things
    became warm enough to take the isulation off the power connection at the
    circut board, melting the board itself.

    Time to build a new circut with new white LED's.

    Considering the voltage variability of a automobile alternator I've
    decided to build into the new circut a current source rather than a using
    a current limiting resistor.

    Digging around it appears the LM334 is a good solution. I have 20 LED's
    to light in this array.

    12 volts with an estimated 3.2 volt forward voltage on the LED's allows me
    3 leds per series.
    3 leds per series means 7 series assemblies.

    As I'm completely new to this, the question that remains for me is what is
    the value of the resistor I choose to set the current output of the LM334?

    The LM334 has an upper limit of 10ma, but people in the model trains world
    have been able to run it at 12ma. That is the goal for me as well, get an
    output from the LM334 of 12ma.

    Any comments or suggestions are welcomed.

    Here is a link to the circut.
    http://www.frontiernet.net/~miketoni/circut.jpg

    The ideas for this circuit came from the circut on this website:
    http://www.pollensoftware.com/railroad/circuit.html

    Thanks.
     
  2. CFoley1064

    CFoley1064 Guest

    Subject: LEDs in Series with LM334 Current Source
    Hi, Mike. As long as the car's voltage regulator is working properly, you
    shouldn't see that much variability in your power supply. You'll see around
    12VDC with the car motor off, and 13.6 to 14.3V with the motor on.

    An LM334 is good for 10 mA, and possibly 12 mA, but I think you're going to
    need more than 12 mA to get the type of brightness you want with the white
    LEDs.

    If you really want to use a current source for the LEDs, you might want to try
    something like this (view in fixed font or M$ Notepad):

    Multiple Current Sources
    VCC
    o----------o-----------o-----------.
    | | | |
    | | | |
    V .-. .-. .-.
    - | | | | | |
    | | | | | | |
    | '-' '-' '-'
    | | | |
    V | | |
    - |< |< |<
    | .---| .---| .--|
    o-. | |\ | |\ | |\
    | | | | | | | |
    | | | | | | | | And so on...
    | '--o-----------o------------o--------->
    | | | |
    | | | |
    | | | |
    .-. V ~ V ~ V ~
    | | - ~ - ~ - ~
    | | | | | ~
    '-' V ~ V ~ V ~
    | - ~ - ~ - ~
    | | | |
    | V ~ V ~ V ~
    | - ~ - ~ - ~
    | | | |
    | | | |
    o----------o-----------o-----------o
    |
    ===
    GND

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    The diodes can be 1N4001s, the resistor in series with the diodes can be 1K,
    and the current sense resistors are chosen as 0.7V/I(LED). For 20 mA, that
    would be 0.7V/.02A = 33 ohms. The transistors can be any TO-92 PNP with V(ceo)
    of 60V or greater, your choice.

    Good luck
    Chris
     
  3. Mike H

    Mike H Guest

    (CFoley1064) wrote in
    In general, the concern is what happens when the voltage regulator dies.
    The existing circuit, that failed, relies on resistors to limit current
    flow. In my case something catastrophic happened that damaged the PCB.
    But more commonly, the LED's in the original circuit simply die to to
    excess current.


    I haven't received any LED's yet to verify brightness. My goal was to run
    them below their rated current as what I have read thus far leads me to
    believe that their rated current is generally a maximum and to be avoided.

    This application uses the LED's with a reflector and a lenses that
    disburses the light in a wider range than a LED normally would. It is
    this reflection of light that is visible in the marker lens. Since there
    is a right and a left, and I only need to replace a right one. The goal
    would be to adjust the current so that the right side would match the left
    side.

    The stock circuit (on the left side) was developed in 1993, so I'm
    guessing the stock design is not all that "bright" in comparison to the
    LED's available today. But I may be wrong.

    Regarding the circuit you proposed.

    If've done some further research and am wondering if all of the components
    are necessary. Here is a second circuit diagram I drew where the LED's
    are placed into an array, and then the current into that Array is
    controled by a single device.

    http://www.frontiernet.net/~miketoni/circut2.jpg

    Would that work?

    Could that be translated to what you proposed?


    I.E. reducing the number of components in the circuit.

    And can you clarify the calculations for me in your diagram. Where did
    the value for .7 come from for the LED?

    Thank you for your assistance.
     
  4. John G

    John G Guest

    Unless I missed something your new circuit requires 70
    milliamps thru the LM334 and its maximum allowed is only
    10ma.
    Surely there are larger constant current regulators that
    would work here but I have not got time to research this
    morning.
    The LM334 was most likely used in the model train because he
    needed to get the ONE led and regulator in a small space.
    You should not have this restriction.
     
  5. CFoley1064

    CFoley1064 Guest

    Subject: Re: LEDs in Series with LM334 Current Source
    Hi again, Mike. The second circuit shows several strings of three LEDs in
    parallel, driven by a current source. That won't work too well, because minor
    changes in the Vf of the LEDs will mean that the current will get hogged up by
    the strings of LEDs that have the lowest combined forward voltage. That means
    major differences in brightness between LED strings. If you match forward
    voltages, it might work pretty well, but as a paractical circuit, no. (I guess
    the business with having two LEDs in one string is a typo -- that will limit
    forward voltage to 6.4V, and only those 2 LEDs will light. Not to mention that
    the LM334 has no chance of pumping enough current for all those LED strings.).

    Let's take a look at the original "HP" circuit for best and worst cases (view
    in fixed font or M$ Notepad):

    +12V from regulator ~ ~ ~ GND
    ___ ~ ~ ~
    o---|___|--->|---->|---->|------o

    Let's assume you've got a series resistor of 200 ohms +/- 5%, an LED forward
    voltage between 3.0V and 3.4V with a 3.2V nominal, and a regulator voltage that
    varies between 11.7V (low battery, motor off) and 14.5V (high regulator, motor
    on).

    Nominal LED current: (13.8V - (3 * 3.2V)) / 200 ohms = 21 mA
    High LED Current: (14.5V - (3 * 3.0V)) / 190 ohms = 28 mA
    Low LED Current: (11.7V - (3 * 3.4V)) / 210 ohms = 7.14 mA

    As you can see, high case to low case is almost a 4:1 ratio. Assuming the
    motor is on, you can assume a minimum of 13.5V, which would give a low case of

    Low LED Current: (13.5V - (3 * 3.4V))/210 ohms = 15.7 mA

    Which gives you less than 2:1 when the motor is on. If you can get better
    control over the LED Vf (after all, HP has been making LEDs since the 70s --
    they know what they're doing), that means a better ratio.

    Now let's assume your voltage regulator has gone south. In that case, your car
    voltage may go as high as 16V or so. When that happens, you're killing your
    battery, and bad things would happen to incandescent lights too because of the
    higher power. In that event your nominal current would be:

    Nominal LED current = (16V - (3 * 3.2v)) / 200 ohms = 32 mA

    That is 50% over nominal, which isn't a problem if you're at 20 mA. If you're
    at or over 40 mA, it might be fatal, depending on the LED.

    That gets us back to the current source. The idea is a good one -- keep the
    same current going through the LEDs despite variations in voltage. Here's a
    segment of the circuit:

    12V
    .-o-.
    | |
    | |
    D1 V .-.
    - | |33 Ohm
    | | |
    | '-'
    D2 V |
    - | Ve
    | |<
    Vb o-|
    | |\
    .-. |Vc
    1K| | |
    | | V ~
    '-' - ~
    | |
    | V ~
    | - ~
    | |
    | V ~
    | - ~
    | |
    '-o-'
    GND

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    You're probably familiar with the assumption that a forward-biased diode will
    have about 0.7V across it. Two diodes in series, then, will have about 1.4V
    across them. So, Vb is 1.4V less than the 12V. By the same token, the forward
    biased PNP transistor will act so there is 0.7V from the emitter to the base.
    That will mean that the transistor will tend to act so that Ve is 0.7V less
    than the 12V. Your job is to choose the emitter resistor so that 0.7V across
    it will give you the current you need. That would be (0.7V)/0.02A = 35 ohms,
    or 33 ohms to the nearest standard value.

    Of course this is a first cut approximation. Collector current is emitter
    current plus base current (which should be less than 1/100th of emitter current
    with a stiff base voltage source). More importantly, Vf of the diode and Vbe
    of the PNP transistor is somewhat dependent on junction temperature. But under
    normal circumstances in an automotive environment, Vbe will only drift by about
    0.2V, and more importantly, the Vf of the diode will tend to track Vbe if
    they're the same temp. Without going through the math, I can tell you that the
    current source will give far better results than the series resistor alone,
    like the example above. Also, a jellybean SOT-23 PNP transistor and a 1/8W SMT
    resistor will cost a lot less than an LM334. The voltage source of the two
    diodes and the 1/4W resistor is used by all the diode strings.

    The main issue here is the voltage burden of the current source. However, even
    if you assume 3.4V per diode, the current source can drive a 10.2V burden as
    long as the voltage source is above 11.7V or so. Below that, the transistor
    will begin to saturate, and current will go down. That should be acceptable --
    people are used to having lights dim a little when the car is starting. The
    main reason automotive manufacturers didn't use current sources is cost and
    reduced reliability due to circuit complexity. Also, they specify LED Vf, and
    generally will not accept delivery of LEDs with excessive Vf (which can be
    controlled, to an extent, in manufacture).

    Good luck
    Chris
     
  6. Mike H

    Mike H Guest

    (CFoley1064) wrote in

    ....
    Thanks for the feedback and information. This should give me plenty of
    information to work with.

    Thanks
     
  7. One other thing about LEDs in parallel is that they will hog more current as
    they get hot, because their TC is negative. Thus, as they get hotter, their
    Vf gets smaller, causing even more current to get put through the hot one,
    leading to even more current, heat, etc, etc.

    To overcome this problem, I'm a fan of the multistage current mirror. You
    can regulate the current for all strings quite easily by using your current
    source like this:

    VCC
    +
    |
    +----+-------+---------+--------+------+---------------+
    | | | | | |
    | | | | | |
    |< |< |< |< |< |<
    +-----------------------------------------------------|
    | |\ |\ |\ |\ |\ |\
    | | | | | | ... |
    | | | | | | |
    +-----+ V V V V V
    | - - - - -
    .-----. | | | | |
    | | | | | | | |
    | | | V V V V V
    | \|/ | - - - - -
    '-----' | | | | |
    +----+-------+---------+--------+------+---------------+
    |
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


    The box with the arrow is your LM334 current source. If you use it like
    this, all the strings in the system will have the same current sourced by
    the current source, minus a tiny bit for the biasing.

    Its easy to see why this is. Notice that the voltage from emitter to base is
    equal on all of the transistors. That means that, according to transistor
    theory, the current through them will be equal if they have the same temp
    (and Vce, which is a far less important issue.) If you mount them all on the
    same chunk of steel, they will stay pretty much the same temperature, so the
    current will be pretty much the same.

    The nice thing about this scheme is that its independent of Vcc. If Vcc
    fluctuates, it still regulates the current properly, since the current sink
    isn't dependent on Vcc.

    You can also use a resistor as your current source, but you lose the Vcc
    independence if you do that. That can apparently be an issue in an
    automotive environment, because the voltage can spike quite a bit.

    Regards,
    Bob Monsen
     
  8. Mike H

    Mike H Guest


    Lots of information here and I greatly appreciate it. It gives me more to
    think about than I can currently grasp.

    I believe that Chris' comment regarding the current limitations of the
    LM334 are valid. I did some initial search for another current source
    that is as simple as the LM334 but with higher capacity and didn't have
    any luck. I'm not sure there would be any advantage with finding one that
    is more complex as Chris showed me a design that works with simple
    transistors.

    In the mean time I've located a LED that has a lower Forward Voltage of
    2.2v at 20ma. Thus I should be able to support 5 LEDs in each series.

    Question 1 - Is this not a good idea to grow series?
    Question 2 - Here is how his circuit looks extended out, keeping the 1k
    resistors and 33k resistors. Am I extending this properly?

    o---------------------------------------
    +12v | | | | | | | |
    V .-. V .-. V .-. V .-.
    - | | - | | - | | - | |
    | | | | | | | | | | | |
    | '-' | '-' | '-' | '-'
    V | V | V | V |
    - | - | - | - |
    | |< | |< | |< | |<
    -| -| -| -|
    | |\ | |\ | |\ | |\
    .-. | .-. | .-. | .-. |
    | | | | | | | | | | | |
    | | V | | V | | V | | V
    '-' -\\ '-' -\\ '-' -\\ '-' -\\
    | | | | | | | |
    | V | V | V | V
    | -\\ | -\\ | -\\ | -\\
    | | | | | | | |
    | V | V | V | V
    | -\\ | -\\ | -\\ | -\\
    | | | | | | | |
    | V | V | V | V
    | -\\ | -\\ | -\\ | -\\
    | | | | | | | |
    | V | V | V | V
    | -\\ | -\\ | -\\ | -\\
    | | | | | | | |
    o---------------------------------------
    -12v
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


    Question 3: Are your (Bob) comments regarding the current source related
    to the above design in that the current flow through each parallel section
    above may vary as each one is being managed within each grouping of 5?


    Oh... and re-reading the information from Chris, I wonder if my circuit
    above should have looked like this: Rather than that^
    o---------------------------------------
    +12v | | | | |
    V .-. .-. .-. .-.
    - | | | | | | | |
    | | | | | | | | |
    | '-' '-' '-' '-'
    V | | | |
    - | | | |
    | |< |< |< |<
    -|---------|---------|---------|
    | |\ |\ |\ |\
    .-. | | | |
    | | | | | |
    | | V V V V
    '-' -\\ -\\ -\\ -
    | | | | |
    | V V V V
    | -\\ -\\ -\\ -\\
    | | | | |
    | V V V V
    | -\\ -\\ -\\ -\\
    | | | | |
    | V V V V
    | -\\ -\\ -\\ -\\
    | | | | |
    | V V V V
    | -\\ -\\ -\\ -\\
    | | | | |
    o---------------------------------------
    -12v
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
     
  9. No, the one above is not a current mirror. The idea behind a current mirror
    is to use the current through one transistor to control the current through
    the rest of the transistors. The thing above is simply 4 separate circuits
    which don't affect each other (assuming the 12V doesn't sag, which it won't
    in a car.)

    The circuit below is a form of current mirror. However, I think you'll get
    better results with the one I posted, which also uses fewer parts, and has a
    larger voltage compliance (meaning you can possibly use more LEDs per
    string.) The one below also has problems with voltage spikes, which my
    posted circuit does not have.
     
  10. Mike H

    Mike H Guest

    ....

    O.k. Sounds like I'm learning some but not quite enough. Taking your
    circuit design. Assuming I need 20ma of current through the system. That
    eliminates the LM334, Correct? If not, can you demonstrate how it could
    be retained?

    So if the LM334 is removed, then what goes in it's place to limit the
    current?

    The circuit below simply puts the number of LEDS in question in place and
    adds a blank space for the current control.

    The LED's in question have a 2.2v Forward Voltage when run at 20ma.

    With this design are we still assuming "any" TO-92 PNP transistor with a
    V(ceo) of 60V or greater?

    I truly apprecate all the assistance and explinations.


    VCC
    +
    |
    +----+-------+---------+--------+------+
    | | | | |
    | | | | |
    |< |< |< |< |<
    +-------------------------------------|
    | |\ |\ |\ |\ |\
    | | | | | |
    | | | | | |
    +-----+ V V V V
    | - \\ - \\ - \\ \\
    | | | | |
    | | | | |
    .---------. V V V V
    | | - \\ - \\ - \\ - \\
    | | | | | |
    | | V V V V
    | | - \\ - \\ - \\ - \\
    | | | | | |
    | | V V V V
    | | - \\ - \\ - \\ - \\
    | | | | | |
    '----+----' V V V V
    | - \\ - \\ - \\ - \\
    | | | | |
    |---+--------+---------+--------+------+
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
     

  11. Here is a current source that can replace your LM334, made from a few
    discretes.

    VCC
    +
    |
    |
    | current in
    | here = 20mA
    | |
    .-. |
    1k ohms | | |
    | | |
    '-' |
    | |
    | |
    | |
    | |/
    +-------| NPN (like 2N3904)
    | |>
    | |
    1N914 V |
    - .-.
    | | | 33 ohms
    | | |
    V '-'
    1N914 - |
    | |
    | |
    +---------+
    |
    |
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    You only need one of these to control all of the strings. (it goes in
    the box above.) The way it works is that the NPN transistor has a Vbe
    of 0.7V. The two diodes have a combined Vf of 1.4V. Thus, there is
    0.7V across the 33 ohm resistor, which means 0.7/33 = 21.21mA will go
    through the NPN transistor.

    You can also use a common voltage regulator, like an LM317. Its
    simple:

    .------------.
    I in In| |Out
    ---------| LM317 |----+
    = 20mA | | |
    | | .-.
    | | | |
    '------------' | | 62 ohms
    |Adj '-'
    | |
    | |
    +----------+
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    The LM317 is designed to keep the voltage at Out equal to Adj + 1.25.
    It'll put out whatever current it needs to to do this. Since our 68
    ohm resistor is between Out and GND, then we have

    Iin = 1.25/62 = 20mA.

    For this circuit, you need to ensure that the Iin is greater than
    about 4V. That isn't really an issue with your circuit, though.

    You can get LM317s in TO-92 cases, just like a small transistor.

    Note that your voltage compliance (Vc) is going to be less than 12V
    (probably more like 11V) so if the forward voltage of your LEDs is Vf
    at 20mA, then each string will only support Vc/Vf LEDs. If you put in
    more than that, the strings won't light up.

    Regards
    Bob Monsen
     
  12. Mike H

    Mike H Guest

    [email protected]_s54:
    ....
    ....


    Thanks again for the assistance. Busy stamping out software bugs in two
    different projects started by others and left to me to fix.

    I've placed your example into mine and just want to verify the
    connections. Does the below look correct?

    You also mention a common voltage regulator. If I were to use such a
    device do I then get away from my goal of ensuring a specific current to
    the leds regardless of voltage?

    Thanks



    VCC +
    |
    +--------+-------+---------+--------+------+
    | | | | | |
    | | | | | |
    | |< |< |< |< |<
    | +-------------------------------------|
    | | |\ |\ |\ |\ |\
    | | | | | | |
    | | | | | | |
    .-. +-----+ V V V V
    | | | - \\ - \\ - \\ \\
    | | | | | | |
    '-' | | | | |
    | |/ V V V V
    +------| - \\ - \\ - \\ - \\
    | |> | | | |
    | | V V V V
    V | - \\ - \\ - \\ - \\
    - .-. | | | |
    | | | V V V V
    | | | - \\ - \\ - \\ - \\
    V '-' | | | |
    - | V V V V
    | | - \\ - \\ - \\ - \\
    | | | | | |
    +--------|-------+---------+--------+------+
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
     
  13. I hate it when that happens...
    This looks fine, except there are too many LEDs, I think, unless you
    are using the standard 1.7V Vf leds, which are not very bright. I
    thought you were using those high-brighness white LEDs, which are more
    like 3.5V Vf. Just make sure the forward voltage across all the LEDs
    is less than about 11V, and you should be fine (with Vcc >= 12V.) You
    can add more strings if you need to. I'd say to keep below about 10
    strings with this circuit.

    Note that the value of the resistor on the bottom left (parallel to
    the two series diodes) controls the current through each of the
    strings. Less resistance means more current. The relation is current
    times the resistance = the voltage drop across a diode, or

    R = 0.7/I

    where I is the current you'll get through each parallel string.

    The voltage regulator circuit I described was really just a constant
    current sink; its another way to do what the left bottom side of the
    circuit does, except using a single IC + resistor instead of two
    resistors, a transistor, and two diodes. It also wastes less current,
    cause it doesn't need the bias provided by the resistor + diode pair.

    Regards,
    Bob Monsen
     
  14. Mike H

    Mike H Guest

    Currently I'm looking at a Super-Red LED that has a typical forward
    voltage of 2.2v, and a MAX of 2.6v. I thought I wanted white until I
    thought about the fact the lens in the assembly is already red. Why waste
    energy that will be filtered?

    2.6v * 5v = 13v. Woops. I guess I'll be adding another string and
    decreasing the count per string.


    Regarding the LM317 source you described. Could you describe for me how
    it works as a current source as it appears to be a regulator of voltage.

    Is the theory that by using such a device, voltage is fixed, thus a
    resistor can be again used for current limitation. Thus I would no longer
    need to concern myself with the fear of possible voltage issues in an
    Automotive electrical system


    Would the below be how I would integrate the LM317 source then? (I didn't
    change the LED count per string yet assume a reduction to 4)



    VCC +
    |
    +--------+-------+---------+--------+------+
    | | | | |
    | | | | |
    |< |< |< |< |<
    +-------------------------------------|
    | |\ |\ |\ |\ |\
    | | | | | |
    | | | | | |
    +-----+ V V V V
    | - \\ - \\ - \\ \\
    | | | | |
    | | | | |
    +---------+ V V V V
    | ____ - \\ - \\ - \\ - \\
    | | | | | | |
    +--|LM317-+ V V V V
    |____| | - \\ - \\ - \\ - \\
    | .-. | | | |
    | | | V V V V
    | | | - \\ - \\ - \\ - \\
    | '-' | | | |
    | | V V V V
    +----+ - \\ - \\ - \\ - \\
    | | | | |
    +-------+---------+--------+------+
    |
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
     
  15. No. Consider a device which, given a range of input voltages, can
    produce a given voltage. Now, connect that fixed voltage to ground
    through a resistor R. By ohms law, the current through the resistor
    will be I = V/R. Since V is fixed at about 1.25V for an LM317 with a
    grounded "adj" pin, and since the amount of current through the 'adj'
    pin is, essentially, zero, then the total current sinked (sunk?) by
    this device is V/R. Voila, its a current source! The LM317 can take
    voltage differentials up to 30V, so given any voltage from 30V down to
    about 3V, the LM317, hooked up as shown through a resistor, will sink
    the same amount of current.
    Yes, thats it. the resistor R is going to be 1.25/I, where I is the
    current you want through each string. So, for example, if you want to
    pass 20mA, you use 1.25/0.020 = 62.5 ohms for R. Anything near that
    will be fine.
     
  16. Robert C Monsen wrote:

    [snip]

    The above schem assumes that all five PNP transistors have the same Vbe
    drop, which would be a valid assumption if they were all on the same
    chip in the same package. But they're usually not, and they are not
    thermally coupled. So there could be a wide variation in the current of
    each string of LEDs.
     
  17. How much do you think? What percentage?

    Regards,
    Bob Monsen
     
  18. Rich Grise

    Rich Grise Guest

    Off the top of my head, a 2N2907 has a beta from 100-300, so I guess
    300%.

    But that's a very quick off-the-cuff WAG - but I've never really understood
    the theory behind a current mirror, which is what this looks like.
    Personally,
    I would have done just "ordinary" current sources, like this:

    +V
    |
    ----o----
    | |
    [D] |
    | [R]
    [D] |
    | / E
    ------B|
    | \ C
    [R] |
    | [LED]
    | |
    ----o----
    |
    -V

    But the other circuit is probably more efficient.

    Cheers!
    Rich
     
  19. Rich:

    Current mirrors just depend on the fact that, for equal transistors,
    if the Vbe is equal, the current is equal. Its as simple as that. So,
    setting a group of transistors with the same Vbe means they'll allow
    the same current through them. Use the first one (the one with the
    base-collector connection) to set the current, and everybody else is
    following along. Its a neat trick.

    I'm guessing that for the OPs application, the diference in current
    between transistors from the same batch will be less than about %5. If
    its more, and he cares, he can use a small resistor between the
    positive rail and the emitter, since he can ensure that resistors are
    %5 much more easily. However, that decreases the voltage range the
    source will generate, meaning less LEDs he can support. Since the
    worst that can happen is that one of the LED strings will be slightly
    brighter than the others, I'd say that its not an issue.

    Regards,
    Bob Monsen
     
  20. Mike H

    Mike H Guest


    Ok. To review.

    The goal here is to use a current source. Initially the thought was to
    use the LM334 as a current source to drive each string of LED's. Then the
    idea became a mirrored Current source. But the LM334 current limits put
    it outside the realm of driving LED's that need 20mA to reach their
    brightness.

    A constant current source is prefered due to the variations that can occur
    within the automobile charging system, and the delicate nature of LED's

    Information from CFoley1064 and Robert Monsen has been very helpful. So
    now it's time to look at the preliminary design and look for issues.

    Below is diagram using a LM317 voltage regulator in combination with a
    resistor to ensure a fixed current source regardless of changes in
    automobile voltage output.

    Here are the planned devices to use:

    12 Volts (as high as 14 v is possible, as low as 11)
    1-EA LM317 (http://www.national.com/pf/LM/LM317.html)
    1-EA 62ohm Resistor .1 watt 1% tolerance
    6-EA 2N2905A PNP Transistors (Has Vceo of 60)
    20-EA RL5-R3545 Super-Red LED Foward Voltage of
    2.2 typical, 2.6 max (http://www.superbrightleds.com/leds/r2_specs.htm)

    When wired as shown below, the goal is to achieve a current level of 20mA
    across all of the LED's in a fairly even manner to achieve near equal
    brightness across all of them. The theory being that changing the
    resistor value can impact the amount of current flowing as it will impact
    the setpoint of the voltage regulator.

    Questions:
    1) Are the devices listed above appropriate for the goals?
    2) What is a good source for the above components. (LED's I know where to
    go)
    3) Any other comments or changes to the circut below recommended?

    Thanks



    VCC +12
    |
    +--------+-------+---------+--------+------+-------+
    | | | | | |
    | | | | | |
    |< |< |< |< |< |<
    +-+-------+---------+--------+------+-------+-|
    | |\ |\ |\ |\ |\ |\
    | | | | | | |
    | | | | | | |
    +-----+ V V V V V
    | - \\ - \\ - \\ \\ - \\
    +----------+ | | | | |
    | ____ | | | | |
    | | | V V V V V
    +--|LM317---+ - \\ - \\ - \\ - \\ -\\
    |____| | | | | | |
    | .-. V V V V V
    | | | - \\ - \\ - \\ - \\ -\\
    | | | | | | | |
    | '-' V V V V V
    | | - \\ - \\ - \\ - \\ -\\
    +------+ | | | | |
    | | | | | |
    +---------+------+---------+--------+------+-------+
    |
    |
    ===
    GND -12
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
     
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