555/Transistor Help

Discussion in 'The Projects Forum' started by gregbernard, Jun 11, 2011.

  1. gregbernard

    Thread Starter New Member

    May 21, 2011
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    0
    Hi folks,

    Relatively new here and this is my first post so please correct me if I've done anything wrong.

    Looking for a bit of a sanity check here since its been oh so long since I've done any serious design. I am looking for advice/critique of a simple 555/transistor circuit I'm working on for a laser optics application. In a nutshell, the requirements are if the user wants to open the shutter to enable the laser on the optics table, there is an approximate 10s delay that sounds an alarm, and then the shutter opens. So, I used a 556 to create the 10s delay, and the other half of it to invert the output since the Bogen tone generator is active low. See http://www.bogen.com/products/pdfs/specialelectronicspdfs/TG4Cm.pdf.

    Once the 10s is over, I use a 555 to generate a 0.25s pulse to open the shutter. Pretty easy so far but the shutter is a 24V device not 12V, so I put a 2N3904 in there to drive it. That's where I'm a bit shaky. I'm not sure if my very simple setup is correct. It works, but is it correct or just 'good enough'?

    The shutter is a Lasermet LS-10, 24V. http://www.lasermet.com/laser-beam-shutters.php Since they sell a complete laser safety system I have not been able to get a lot of good technical information, but I have a question into them if the shutter should be a inductive load or not or even how much current or power it needs.

    From their literature is the following:

    DB-9 pinout
    1 +12V DC power to shutter
    2 0V
    3 Remote open input
    4 Not Used (Internally connected to pin 1)
    5 ‘Open’ status output
    6 ‘Closed’ status output
    7 - 9 Optional electrical Interlock option - see below.

    5 and 6 give the status of the shutter,

    If using remote control with a switched 12V supply connect the switched +12V supply + to pins 1 and 3 and – to pin 2. The shutter will open immediately the supply is turned on.

    If using remote pushbuttons to control the shutter use a normally-closed button in series with the supply for the ‘Close’ function, and a normally-open pushbutton between pins 1 and 3 for the ‘Open’ pushbutton. Both switches need only switch for approximately 0.25 s to close or open the shutter.

    Note that the literature says 12V but there is a note saying it still applies to the 24V version.

    I'm attaching my schematic, and have two questions. First, and most importantly, is the 2N3904 part to the shutter correct? (BTW, I'm not sure if the resistor is a 10K but I'm not there so I can't double check. I will though.) That's what I really need help with.

    Secondly, and more of an incidental/curiousity, can I replace the second half of the 556, which is inverting the first half, with a 2N3906 to invert the signal? That way I only need the one chip. I know, two transistors and one chip vs. one transistor and two chips so there is no real savings. Like I said, more of a curiousity.

    If I need to supply any more information or am unclear about anything, please let me know and I'll clear it up ASAP.

    Thanks all for your help.
     
  2. iONic

    AAC Fanatic!

    Nov 16, 2007
    1,420
    68
    Welcome to AllAboutCircuits Greg,

    I am not the smartest tack in this shack, but will attempt to give you my take on the circuit.

    I've rarely, if ever seen a case where the NPN's emitter was not tied to ground, but I may be wrong. And if you need a +24 pulse to the "Shutter Trigger" then I would suggest a PNP (3906) in place of the 3904 as in the example below.

    [​IMG]
     
  3. gregbernard

    Thread Starter New Member

    May 21, 2011
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    0
    IONic,

    Thanks for the reply. I think the missing ground off the transmitter is an 'oops' on my part in the schematic. I know its connected on the bread board in the lab. Brain fart on my part.

    Whats the advantage or difference of the 3906 compared to the 3904? I have both of them by the boatloads so can use either one.

    Thanks for your help.
     
  4. Pencil

    Active Member

    Dec 8, 2009
    271
    38
    IONIC,

    Isn't using a PNP off of a 555 monostable configuration going to
    give him +24V before triggering and "ground" for .24 sec after triggering?

    Remember, the 555 will be low before triggering. PNP will be on, shutter trigger will be high.
    Upon triggering the 555 will go high, PNP will turn off, shutter trigger will be low for .24 sec.

    You will need to invert, or drive shutter directly from 555.
    EDIT: Scratch that. He cannot drive directly from 555, he needs 24V.

    Also let's see some resistors on the bases of these transistors before
    we blow something up.

    I think Bill Marsdens blog has some examples of transistor drivers for use
    with the venerable 555.
     
    Last edited: Jun 12, 2011
  5. iONic

    AAC Fanatic!

    Nov 16, 2007
    1,420
    68
    The OP says the circuit works as he drew it, although drawn without the NPN 3904 emitter to ground. with such a circuit there is always 0V to the shutter trigger!

    Sticking with the NPN 3904, grounding the emitter and taking the output from between the 10K resistor and the collector of the transistor should be the right way.
     
    Last edited: Jun 12, 2011
  6. Potato Pudding

    Well-Known Member

    Jun 11, 2010
    684
    92
    This is how I would connect the 555 to the shutter and 24V supply. The main fix is switching a higher voltage than the signal from the 555. You were using a follower that would have only allowed 12V to the shutter.

    [​IMG]
     
  7. gregbernard

    Thread Starter New Member

    May 21, 2011
    4
    0
    iONic,

    I really did fudge up the transistor part of the schematic. I had a 100K to ground, and the signal line coming off the collector as you said. Sorry for the confusion - my fault. Too much time on the 555 drawing...

    Potato Pudding,
    You are right - I just measured it. 12V to the shutter! Yikes! It works but if I trigger it often enough I find that every so often the shutter does not 'click' open so much as it is half open. Thanks for pointing that out - that would definitely come back to haunt me later down the line!

    I'm in the lab today but have a bunch of optics to put on the table so I'll have to push off your solution until tomorrow. It makes a great deal of sense to me now that I see it :)

    Thanks everybody for your help. I'll report back in a day or two to on the results.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    1,728
    If the shutter is opened via a solenoid, then there needs to be reverse-EMF diodes (or at least clamping diodes) on the inputs, or the transistors may be destroyed.

    I would also use a base return resistor on Q2 to ensure that it gets turned off. 2.2k to +24v from the base would be OK.
     
  9. Potato Pudding

    Well-Known Member

    Jun 11, 2010
    684
    92
    True enough Sarge. Here are the recommended revisions.

    [​IMG]
     
  10. gregbernard

    Thread Starter New Member

    May 21, 2011
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    0
    Interesting. I penciled in the corrections that Sarge suggested before I read Potato Pudding's reply. All the components are the same except Potato placed the diode across the Trigger and ground. I placed a reverse biased diode between the Trigger and 24V and a 0.01u cap from Trigger to ground.

    After looking up the diode Potato used I know I need to change mine to a Schottky, but am curious if I was way out of the park (again)? Would my modifications have worked? Or did I put the diode in the wrong place to protect the transistor?

    Schematic attached.

    BTW, thanks all for your help. Most appreciated :)!
     
  11. Potato Pudding

    Well-Known Member

    Jun 11, 2010
    684
    92
    It is precautionary and not everyone even bothers with precautions.

    It doesn't take much of a Capacitor to smooth out the spike from anything but larger coils with the caution that it creates a tank circuit that might ring or oscillate.

    The Diode as you have it connected is not going to help much.

    You normally need to connect across the coil with polarity to block the drive voltage.

    The way you have it might actually help damp the ringing from the LC tank circuit, but it would allow feed back of spikes to the 24 volt power supply which could (but not too likely) damage the power supply. It would actually work as a regenerative brake as long as the 24 volt supply had enough capacitive filtering.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Either way will work.
    If the load is a coil (we're not certain whether it is or not at the moment, but the diode won't hurt anything if it is not; if it IS, the absence of the diode will destroy the transistor) then connecting a reverse-biased diode to either ground and trigger or trigger and 24v will effect the same remedy. When the current is switched off, the current through the coil keeps moving for awhile, which causes the polarity across the coil to swap. If a reverse-EMF diode is not present, the voltage can reach a very high peak value, which will (sooner or later, probably sooner) destroy the transistor. The reverse-EMF diode provides an alternate current path, so the reverse voltage normally doesn't exceed around 1v.

    A 1N400x series diode will work just fine for this project. If the frequency of shutter trip were fairly high (above 400Hz), then you would need to consider a Schottky or other fast recovery diode, but for this slowly cycling project you would never know the difference.
     
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