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Am I right?

Discussion in 'Electronic Design' started by Michael, Feb 15, 2007.

  1. Michael

    Michael Guest

    I have some 1 watt LED's that I want to flash on and off.

    I have a 317 current regulator supply set at 300 ma for 3 of them in series.

    Now, I know there are alot of professional designers on this group and was
    wondering if I couldn't get an opinion on my approach to this, as it makes
    sense to me.

    I know many decisions are made based on parts cost but either way I do this,
    the cost is about the same....the cost aspect is not what I'm concerned
    with.....just if I'm making "the right choice".

    I initially thought I'd use a 555 timer to have a constant flash rate on an
    optoisoltor LED and control on/off with the LED ground.....then use the opto
    output and a sink transistor to control the DC path.

    Cons: what if anything were to change with the transistor and present a
    change in the DC series circuit?....could I get an "on resistance" that
    would be low enough and STAY in a specific window? With 300 ma flowing
    through, some "potential" for disaster would be there.

    So....

    Why not this....which makes more sense to me....

    Control the AC to the transformer with the same setup....flash a triac
    driver opto LED with a 555, control on/off with the opto LED ground, use a
    triac to switch on and off the transformer primary?

    That's what I have breadboarded now...works fine.

    Don't you think it would be the proper way to go as far as a design
    decision?......it lets the cc power supply be stable and avoids risk?

    The only risk on doing it the AC way is an AC line surge but I'll use a
    400 - 600v triac good for 1 amp or so.

    Thanks.
     
  2. Eeyore

    Eeyore Guest

    Can you provide a link to data on them ?

    Graham
     
  3. Dave Pollum

    Dave Pollum Guest


    Maybe I missed something, but why did you need the 555 output to be
    optically coupled to the LED driver transistor? Why not just connect
    the ouput of the 555 to a transistor that can handle 300 mA? A 2N4401
    (NPN) will handle 600mA. Or find an appropiate FET to do the job. It
    seems to me that controlling the AC side of the power supply makes
    things more complicated than needed.
    HTH
    -Dave Pollum
     
  4. Michael

    Michael Guest

    I want the optocoupler because the control pair (ground and LED cathode)
    will be in an "industrial" environment.

    Rather safe than sorry, I suppose.

    I just thought having the transistor in the DC path would potentially be
    more risky.

    One example....
    not all 2n4401's are alike? ...what about that and the influence on the
    "constant current" requirements of the supply...or any possible changes in
    the characteristics of the transistor over time?

    If I dealt with the AC side, that would be eliminated because as long as the
    transformer secondary stayed within a certain window the 317 wouldn't care?
     
  5. Use a PIC.

    --
    Many thanks,

    Don Lancaster voice phone: (928)428-4073
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
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  6. Dave Pollum

    Dave Pollum Guest

    OK, so I understand why you want the optoisolator. So the output of
    the 555 drives the opto's LED, which turns on the opto's transistor.
    That transistor then controls a driver transistor (i.e. 2N4401) that
    handles the 300 mA current of the LEDs. According to a graph in the
    Fairchild 2N4401 data sheet (http://www.fairchildsemi.com/ds/2N/
    2N4401.pdf), when it's saturated its Vce is 0.25 volts. That graph
    shows that the current gain is only 10, so the base current would need
    to be 30 mA. These values differ from what is stated in the tables,
    so some experimanting would be required to determine what you need.
    And as far as I know, all 2N4401's are the same. I suggested a 2N4401
    because I knew it could handle the current, it's a common part, and
    it's cheap. There may be other transistors that are more suitable for
    your project.
    HTH
    -Dave Pollum
     
  7. Dave Pollum

    Dave Pollum Guest

    Don,
    It seems like the cure for everything except the common cold is either
    a 555 timer or a PIC ;)
    -Dave Pollum
     
  8. Michael

    Michael Guest

    SORRY you guys, I didn't think this through before posting.

    I liked the idea of controlling the AC side and keeping the series path in
    the power LEDS clear but I would need the power supply ON all the time for
    the 555 etc.

    Wasn't thinking.

    I'll breadboard a transistor to open/close the DC path....an FET would be
    better, for lower on resistance?

    Also, an FET would have less current drive requirement, right?

    I'm going to try and switch 300ma....that's quite a bit of
    current.......maybe they make an opto with a "second side" that will do the
    trick.
     
  9. Rich Grise

    Rich Grise Guest

    That's not an ungodly amount of current to switch.

    Are you running the 317 as a current regulator? That way you won't need
    dropping resistors (assuming 300 mA is the right forward current for the
    LED).

    Then, a low-side switch can be anything you want that can handle the
    volts and mA. ; you might want a bleeder resistor directly to ground
    from the 317 output - when you open it, it will go as close to the
    supply V as it can, trying to push 300 mA through an open. =:-O

    There might be a more sophisticated circuit, like constant current
    with a voltage limit, but you'd have to look that up.

    Good Luck!
    Rich
     
  10. The cure for a 555 timer is a PIC.
     
  11. rebel

    rebel Guest

    Is there a compelling reason why you don't use the opto output to pull down the
    reference pin of the 317?
     
  12. Gibbo

    Gibbo Guest

    Sarcastic f*c^er :)
     
  13. John Fields

    John Fields Guest

     
  14. Tom Bruhns

    Tom Bruhns Guest

    Why make things difficult by adding another pass element, when you
    already have a perfectly good one in an LM317?

    To easily control the current, since the LM317 is there anyway, why
    not just use it? Then the control current to switch things on and off
    is quite low. All you have to do is run the output of the LM317
    through a resistor R1 to the string of LEDs. R1 = 1.25V/LED_current,
    or about 4 ohms for 300mA. Then put, say, a 1k resistor between the
    junction of R1 and the top LED cathode, and the LM317 control pin.
    Now if you pull down the LED control pin, the LM317 just delivers a
    bit over a milliamp to the pull-down device. If the voltages are
    appropriate, you can just use a 555 timer, or you could use some other
    arrangement of transistors or whatever. You can even drive the
    control point on the LM317 with an opto isolator output directly.
    Surely you don't have to worry about the LED current with three in
    series across about 1.25 volts, in the "off" state. You do have to
    account for the LM317's 50uA nom., 100uA max control pin current,
    which will drop through the 1k resistor when the LEDs are on. It
    would make a max. difference of less than 10%, assuming that 1k
    resistor, if you just ignore it.

    Cheers,
    Tom
     

  15. The cure for a PIC is a 555.


    --
    Service to my country? Been there, Done that, and I've got my DD214 to
    prove it.
    Member of DAV #85.

    Michael A. Terrell
    Central Florida
     
  16. Funny how you can get away with that. ;-)
     
  17. John Fields

    John Fields Guest

    Programmed to look like a 555?

    OK, that works for me. ;)
     
  18. Michael

    Michael Guest

    Yeah, I'm going to look at that approach later and breadboard something.

    I've never really tried to fiddle with the control pin (adj?) pin on the 317
    but it has occured to me in the past that something could be done with it.

    Even in it's current regulated setup where the output is taken from the
    input pin, what you're saying is I can pull the adj pin low and effectively
    turn off the voltage, right?

    Make sure there is some resistance there (1k like you say).....you wouldn't
    want it see a straight shot to ground?

    Just take the opto output (npn out).....connect the emitter to ground and
    the collector through 1 k to the adj pin.

    Sounds like the way to go.

    I have the opto trigerring a 4401 right now and all is fine....put +5 DC on
    the opto emitter and using a 560 ohm on the collector to drive its
    base....switching the 300 ma and not even warm.

    Oh, and I love Picbasic but I'll take the 23 cent 555 for this !

    I'm using 2 optos at once, one for the dc switch and the other npn out to
    reset the 555 on each control closure.
     
  19. Michael

    Michael Guest

    I meant to say output from adj pin...not input.

    output...power resistor....to adjust

    and adjust = current reg output
     
  20. Tom Bruhns

    Tom Bruhns Guest


    Um, the 1k resistor I suggested goes between the top of the LED string
    and the LM317 adj pin. You yank down directly on the adj pin. The
    LM317 output will follow 1.25V above that. So if you manage to fully
    ground the adj pin, the output voltage will be 1.25V. 1.25V across a
    series string of 3 LEDs will cause a current of perhaps nanoamps to
    flow in the LEDs, depending on what kind they are. But it will be at
    most a tiny current, even it they are pretty hot.

    When the adj pin is allowed to "float", it will be held by the 1k
    resistor to very nearly the voltage at the top of the LED string. The
    LM317 output pin will be 1.25V above that. So to get 300mA current in
    the LEDs, you put a resistor between the output pin and the top of the
    LED string: 1.25V/300mA is about 4 ohms.

    When you pull the adj pin low, the thing which pulls it low must
    handle the current out of the adj pin of the LM317, at most 100uA
    (from the data sheet), plus the current in the 1k resistor. Since the
    LED current is very low now, with the adj pin held low, there will be
    essentially no drop in the 4-ohm resistor, so 1.25V drops across the
    1k resistor, causing 1.25mA current. So the net current in the "low-
    puller" (whatever it may be) is 1.35mA at most. You do want to make
    sure you only pull DOWN on the adj pin to turn things off. Do NOT
    connect it directly to the output (pin 3) of a 555, that can pull it
    UP. That would try to turn on the LEDs harder than you intend.
    Instead, put a diode in (anode to the adj pin) so you can only pull
    down if you have any doubts about your circuit. If the LEDs you're
    driving have even a moderately high voltage drop, you can safely use
    just about any silicon signal diode. You don't need a diode if you're
    driving the adj pin low using an open-collector or open-drain
    transistor.

    You shouldn't need any capacitors on the output end of the LM317 for
    all this to work.

    Cheers,
    Tom
     
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