Electric shutter circuit question - ready for timer?

Discussion in 'Darkroom Developing and Printing' started by John, Jun 15, 2008.

  1. John

    John Guest

    I have an electric shutter. The circuit works the shutter when testing
    with a simple switch, but because I know nothing about electrics, it
    puzzles me.

    Here is a drawing of it: http://www.digoliardi.net/electrics2.gif

    I am dangerously ignorant of electrics so I'm asking: How does it work
    that one AC pole goes to one pole of the orange thing (capacitor?) and
    then also go around it? It's there a short there?

    The electro magnet works just fine. Nothing gets hot even with a long
    'on' time (for focusing). Do you think it is safe to hook up to the
    timer (F-Stop timer by Darkroom Automation)?

    Thank you,

    PS: The shutter is on the back of the lens board because if it were on
    the front it would make changing the F-stop almost impossible. I've also
    made an extension of the F-stop ring to permit adjustment from below the
    lens. (Shutter is from a long-rollfilm camera and it fits in a well
    fitting enclosure around the rear of a Red Dot Artar.) Can provide
    pictures or a build-article if you like.)
    John, Jun 15, 2008
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  2. John

    John Guest

    Thanks for the help, Nicholas. Clearly I'd better leave this to a
    professional to build. I don't want to make an expensive mistake after
    wasting my time screwing up the build.

    I'll stick to analog photography, nuts, bolts, machines.

    John, Jun 15, 2008
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  3. No, the capacitor is charge from the AC line with a resistor and a
    rectifier diode.

    You have some problems here that will cause trouble down the line:
    1) I am assuming that "TR R210" is a rectifier - the number comes up with
    zilch when googled. It should be rated 700V (~2x Vp-p) or greater. The
    standard popcorn part would be a 1N4007.

    2) The capacitor needs to be rated for a minimum of 200VDC continuous - it
    will charge to 170V in this circuit.

    3) You should have a fuse in the circuit. It is hard to say what size
    without knowing the solenoid rating. If the Lectrohm is rated for 5W then a
    good guess is the circuit needs about a 1/8 amp with the shutter closed. A
    1/4 amp slow-blow fuse should be adequate. If you don't have a fuse, and
    there is a short in the shutter solenoid or capacitor, then there will be 33
    watts dissipated in the Lectrohm. This is a very unfortunate number if the
    Lectrohm is rated for 5W - the Lectrohm will get yellow hot but not fail and
    you have yourself a fire starter. You should test the fuse, making sure it
    will blow, by shorting the solenoid and closing the switch: make sure the
    fuse blows before smoke starts to rise from the Lectrohm. The fuse gets a
    bit of a workout in this circuit and may fail from fatigue every few years.
    In lieu of a fuse you can use a 50W Lectrohm.

    I am assuming the circuit works as follows, though I have no idea of the
    shutter solenoid's ratings, so some of it is guestimation:

    The capacitor is normally charged to 170V and it sits there at this voltage
    until the shutter is switched on. No current is drawn from the 120VAC once
    the capacitor charges.

    When the switch closes the capacitor provides a shot of 170V power to close
    the shutter solenoid.

    Solenoids need a lower voltage to hold closed than they need to activate. By
    reducing the holding voltage the solenoid coil can be made smaller - the
    solenoid doesn't have to be sized to take full voltage all the time.
    Continuous full voltage will overheat the coil in a solenoid designed to be
    used in this fashion.

    When the shutter is closed, and the capacitor has partially discharged, the
    circuit provides a much lower current to hold the solenoid closed. The 200
    ohm resistor drops some of the voltage going to the solenoid and capacitor.

    When the switch is opened the shutter solenoid opens and the capacitor is
    charged back to 170V for the next shutter closing. The recharging takes less
    than half a second.

    Connecting the circuit to a timer:

    First, a problem: a timer is wired to supply 120V when exposing. You can't
    wire the timer's socket across the switch without major pyrotechnics.

    The "standard" way to connect this circuit to an enlarging timer would be to
    leave the switch closed and plug the 120V leads into the timer. This may not
    work very well.

    If the timer controls the 120V then the capacitor doesn't get fully charged
    and the shutter solenoid doesn't get a high voltage kick to slam it closed.
    Without the higher voltage the solenoid may not work or work sluggishly and

    You need to wire a relay where the switch is located - the coil of the relay
    goes to the timer and the relay contacts are wired across the switch. You
    need a SPST relay with a 120VAC 60Hz coil. You can get one at Radio Shack.
    Nicholas O. Lindan, Jun 15, 2008
  4. John

    jch Guest

    The circuit diagram looks correct. It is a half wave rectifier with an
    80 microFarad storage capacitor. The capacitor is always kept charged
    through the 200 Ohm resistor by the silicon diode (TR210). When the
    switch to the solenoid is open, the DC voltage across the capacitor will
    be about 169V (square root of 2 times 120V). When the switch to the
    solenoid is closed, DC current will flow through the magnetising coil,
    thereby opening the shutter. The actual voltage seen by the solenoid
    coil will be lower than 169V because of a small drop across the 200 Ohm
    resistor. The drop depends on how much current the coil requires. The
    capacitor could be of a higher voltage rating to be safer. If the
    solenoid "hums" due to the half wave nature of the circuit, you could
    increase the size of the capacitor to, say, 150 microFarad, or use a
    full wave rectifier. You should put a master power switch on the hot
    side of the the AC line (black wire in USA/Canada). It would be no
    problem plugging this device into a timer since the current draw will be
    in the order of 250 mA (my guess).
    jch, Jun 15, 2008
  5. I used to design electronic circuits, much more complicated than this one.
    N.B.: that was a long time ago.

    IMO, the AC line slowly charges the capacitor and when the switch is
    pressed, the energy stored in the capacitor is what fires the shutter (like
    electronic flash circuits). The current through the resistor does not
    directly fire the shutter. In fact, with the circuit as shown, the resistor
    should have high enough resistance so the shutter solenoid will release
    after tripping the shutter so it can try again.
    Jean-David Beyer, Jun 16, 2008
  6. John

    jch Guest

    Indeed, you need that initial higher voltage to "kick" the solenoid,
    then hold it with a lower voltage. The above solution using an
    interposed relay is the correct way to do the job. Not complex, just a
    bit more wiring. I also concur about the need for a fuse, and the
    higher voltage rating of the capacitor. I would use a full wave
    rectifier. For this service, they are about 3/8 inch in diameter, and
    1/4 inch high, having four leads; live, neutral, plus and minus. It
    would be good to know the DC resistance of the solenoid coil. With that
    information we can check the sizing of the resistor in terms of Ohms and
    wattage rating.
    jch, Jun 16, 2008
  7. Quick question: what is the provenance of the circuit?
    Is it the one that was in the long-roll camera that supplied
    the shutter?
    Nicholas O. Lindan, Jun 16, 2008
  8. John

    John Guest

    Yes. The circuit minus the parts for 'Instant' shutter with flash.
    John, Jun 16, 2008
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