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Q:Triac Switch

Discussion in 'Electronic Basics' started by Chris, Sep 15, 2005.

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  1. Chris

    Chris Guest

  2. Andrew Holme

    Andrew Holme Guest

    No.

    1. The +5V rail is shorted to ground through the base-emitter junctions of
    Q1 and T3
    2. T2, R2 and T5 are redundant

    Why is the Q output AC-coupled? Is it to pulse the opto-triacs to save
    power?
     
  3. Andrew Holme

    Andrew Holme Guest

    You can just connect the opto-LEDs, through current-limiting resistors,
    directly to the Q output.
     
  4. Jasen Betts

    Jasen Betts Guest

    it's an unusual looking circuit (there's a short circuit through the two
    first output transistors) what's the intended purpose of C3

    you can simpply connect the optocouplers (and resistors) above and below
    the output of an NE555 to have them firing alternately.

    you have the 555 configured to be on for upto about 4 seconds (adjustable)
    and then off for about 15 miliseconds (or approx one mains cycle) - is
    that what you want?

    if you want an even mark-space (on and off for the same length of time)
    connect the 250K pot between pins 3 and 2 and disconnect pin 7 and remove
    R10

    something like this

    0V +--------------+--+5V
    | | . . . . . | . . . . .
    +------(---.1 . . 8.--+-[220R]---.1 6.
    | 22u | . ~ . .MOC3022.
    +--||--(-+-.2 7. +--------.2 5.
    | | . NE555 . | . .
    +------(-(-.3 6.--+ | .3 4.
    | | | . . | | . . . . .
    +---+ +-(-.4 5. | |
    | | | . . . . . | | . . . . .
    | v | | +--------.1 6.
    +-[250K]-+------------+ | .MOC3022.
    | | +--.2 5.
    +-----------------------+ | . .
    | .3 4.
    0V--[220R]--+ . . . . .

    The stuff on the other side of the optocoupler seems needlessly complex
    but I guess that depends on what the load is you intend to drive. doing
    that part the way you plan won't cause any problems.


    It appears that you copied the a "solid state relay" schematic from
    somewhere (and got atleast one copy wrong)

    You don't need the whole solid state relay input stage if the input
    has a predictable voltage range. a simple resistor is sufficient...
    the 3022 has a 10mV input
    sensitivity and Fairchild suggests feediing it about 15mA in case it
    degrades a little, the internal LED has a voltage drop of about 1.5V
    so 15ma at 3.5V comes to 233 ohms so 220 is close enough.

    Bye.
    Jasen
     
  5. Bob Monsen

    Bob Monsen Guest

    You can probably get by with fewer transistors and lower operating
    current.

    The MOC3022 needs at least 10mA to trigger, so you work backwards
    from there. Its input forward voltage is less than 1.5V, and a saturated
    2N4401 is something like 0.3V. Thus, you will have 3.2V, and you want to
    have 10mA through it, so you need at most 320 ohms of resistance. Make it
    300 ohms. Passing 10mA through a signal NPN will probably require 500uA
    into the base, so use an 8.2k resistor from Q to the base. This will
    reliably turn on a MOC3022 when Q is high, and turn it off when Q is low.

    Now, the other side of the coin is turning one on when Q is low. To do
    that, you use a PNP transistor in an inverted configuration, emitter to
    5V, collector to the MOC3022, and the other MOC input terminal to a 300
    ohm resistor which then goes to ground. Again, connect the base to Q using
    an 8.2k resistor, and your MOC3022 will turn on when Q is low, and turn
    off when Q is high.

    When you turn off the 555, both branches will go off. If you want it to
    stay on after the 555 goes away, use what is called a latching relay.
    There are two coils, and they latch the relay one way or the other.

    Good luck.
     
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