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Am I right?

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

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  1. ehsjr

    ehsjr Guest

    It can't work the way I understand your description - I'm
    obviously not getting it. A schematic would help. Here's
    a possibility for yanking down - feel free to modify
    it to show what you have in mind. The 1K resistor from
    the top of the LED string to the adj pin escapes me, so
    it's not in the diagram, but it should work as drawn.

    Ed
    I -> ~300 mA
    -----
    Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
    ----- | the top of the LEDs.
    Adj [4R]
    | |
    +------------+ <-- total LED vF when the NPN
    | |a is off, Vce when the NPN
    | [LED] is conducting.
    /c |
    YDC >---| [LED]
    \e |
    | [LED]
    | |
    Gnd ------------+------------+

    YDC = "Yank Down Control"
     
  2. Tom Bruhns

    Tom Bruhns Guest

    Hi Ed,

    OK, yours will work, BUT you have to sink 300mA to do it! If you put
    the resistor in, then you only need to sink 1.25V/1k + the adj pin
    current, max 100uA. With the 1k in there, the LED string sees 1.25V
    plus Vce(sat), say 1.35V, but so what? Even red LEDs have negligible
    current when hot at 0.5 volts each. Much less power hungry to add the
    1k in there. The 1k is nominal, of course; it could be 220 ohms, or
    2.2k ohms, though with 2.2k, I'd start worrying about the voltage drop
    across the 2.2k when the LEDs light; that increases the drop across
    the 4 ohm resistor by about 0.1 volts. 1k seemed a reasonable
    compromise to get to low current in the "yank down control
    transistor." Revised version, as I had posted, below. Thanks for
    providing the framework to edit!

    Cheers,
    Tom
     
  3. Tom Bruhns

    Tom Bruhns Guest

    Hi Ed,

    OK, yours will work, BUT you have to sink 300mA to do it! If you put
    the resistor in, then you only need to sink 1.25V/1k + the adj pin
    current, max 100uA. With the 1k in there, the LED string sees 1.25V
    plus Vce(sat), say 1.35V, but so what? Even red LEDs have negligible
    current when hot at 0.5 volts each. Much less power hungry to add the
    1k in there. The 1k is nominal, of course; it could be 220 ohms, or
    2.2k ohms, though with 2.2k, I'd start worrying about the voltage drop
    across the 2.2k when the LEDs light; that increases the drop across
    the 4 ohm resistor by about 0.1 volts. 1k seemed a reasonable
    compromise to get to low current in the "yank down control
    transistor." Revised version, as I had posted, below. Thanks for
    providing the framework to edit!

    Cheers,
    Tom
     
  4. Tom Bruhns

    Tom Bruhns Guest

    Ooops. I failed to edit the top line in that diagram I posted
    earlier. The current is about 300mA when the LEDs light, but only
    about 1.35mA when they are off. The LED current is essentially zero,
    the current in the 1k resistor is 1.25V/1k, and the adj pin current is
    100uA max.

    Cheers,
    Tom
     
  5. jasen

    jasen Guest

    triacs don't like transformers.
    This seems easier to me.

    +V -----[LM317]--[4R2]-+--->|-->|-->|--- 0V
    | | | LED X 3
    ...... | | (300ma)
    . 555. | |
    .output---|<-+--[120R]---'
    ...... 1n914
    | etc
    |
    --+-- 0V

    if the 555's discharge pin is unused it can be used instead of the output
    pin and the diode can be omitted.
     
  6. jasen

    jasen Guest

    +---[120R]---+ <--- add resistor here
    the resistor reduces the current through the transistor to ~11mA when the
    transistor is on, and when the transistor is off has essentially no effect.



    Bye.
    Jasen
     
  7. ehsjr

    ehsjr Guest

    Thanks, Tom. Neat. Picture, words, 1000 !

    Yes the circuit I drew always pulls 300 mA. Just
    couldn't do the words/picture translation to get
    what you were saying.
    Yes! Much better, and not just from power consumption,
    but heat production is lower with your circuit.

    Ed
     
  8. ehsjr

    ehsjr Guest

    Thanks!

    Ed
     
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