Multiple Camera Firing

Discussion in 'The Projects Forum' started by JDR04, Sep 11, 2013.

  1. JDR04

    Thread Starter Active Member

    May 5, 2011
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    Hi Folks, please refer to my attached schematic.

    I currently have a break beam circuit that works nicely with my Canon 30D camera. The camera is connected via SFH618-2 optocoupler which is part of the break beam circuit. That part of it all works well.

    What I would like to do is connect another two 30D cameras without modifying the breakbeam circuit at all. In total, 3 cameras connected via their remote triggering cables.

    So, the attached schematic is how I think I should do it but would appreciate any comments, suggestions and of course any criticisms.

    Thanks -JDR04
     
  2. wayneh

    Expert

    Sep 9, 2010
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    When that middle SFH618-2 transistor turns on, it shorts the power supply to ground. Probably not what you meant.

    You want Q2,3 & 4 to all be on all the time except when Q1 is on, activating the transistor?
     
  3. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Hi wayneh, thanks for your observations. I'm not sure I understand this properly.

    I want Q2,Q3,Q4 to all come on at the same time as Q1. Q1 is the output of the beambreak circuit. So when it triggers I would like all the other Q's to trigger as well.

    You mention how Q3 will short to ground when it is on. Will this not be the same for Q2 and Q4 when they are on?

    So, how do I get around this problem of them shorting to ground? Any suggestions are welcomed.

    Thanks again-JDR04
     
  4. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    You need to connect the cathodes of all the leds to Q1 collector,so that they all fire at the same time.
     
  5. Sensacell

    Well-Known Member

    Jun 19, 2012
    1,129
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    Put the 3 LED's in series, with one current limiting resistor, connect them to the collector of the input isolator, ground the emitter.

    Or just connect the 3 LED's to the output of your break beam circuit and save the cost of the unnecessary 4th opto-coupler.
     
  6. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Thanks Sensacell, just want to know if Q2,3 and Q4 are all in series, wont there be a bigger time differenece between Q2 and Q4 being triggered?
     
  7. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Thanks dodgydave, I've attached what I think you mean.Is my schematic correct??
     
  8. wayneh

    Expert

    Sep 9, 2010
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    That looks much better. Have you checked that the "main" transistor can handle the combined current of the 3 LEDs?
     
  9. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Thanks wayneh. Appreciate your time.

    I've redone the circuit (attached) with what you pointed out in mind. If I'm reading the datasheet correctly (attached) the collector of Q1 can only handle 50mA.

    So, 166 ohm resistors R1, R2 and R3 give each LED 50mA. 9V -0.7v/0.050mA = 166 ohms.

    Collectively Q1 is getting 150mA from the parallel circuit of Q2, Q3 and Q4. Way, way too much!

    My theory is another 166 ohm resistor (R4) will reduce the collective amperage down to 50mA.

    Not sure if my reasoning is correct so any help is appreciated. Thanks again. JDR04
     
  10. wayneh

    Expert

    Sep 9, 2010
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    R4 would not be needed if you simply double R1,2 & 3. If you want to keep R4, I'd put it on the high side, between +9V and the LEDs. I'm not sure if this matters to the opto's transistor, but it would matter if that were a normal NPN acting as a switch. Placing the resistor below the emitter raises the voltage required at the base to turn on the transistor, since it needs to be at emitter + ~0.7V.

    Are you sure the total current is now OK? It's good that you've reduced the individual LED currents below their peak ratings. It's never good to run on the ragged edge.
     
  11. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Hello wayneh, thanks for the help so far.

    If I'm reading the datasheet correctly, Q1 can handle a maximum of 50mA. So, Q2 ,Q3 and Q4 need to collectively produce about 15mA each so as not to destroy Q1.

    I calculate that a resistor of 560 ohms is required for Q2, Q3 and Q4. A 560 ohm resistor on each one produces 15 mA x 3 = 45mA in TOTAL.

    Could you tell me if my calculations are correct please;

    9V - 0.7V = 8.3V.
    8.3V/0.015 =553. Prefered ressitor value 560 ohms.

    So, I have two questions for you or anybody else;

    Have I read the datasheet correctly? (attached in previous post)
    Are my calculations correct?

    Thanks for your timeand help - JDR04
     
  12. wayneh

    Expert

    Sep 9, 2010
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    Assuming R4 is gone, there is a 1.1V typical voltage drop across each emitter and another 0.25V across the detector (both per the data sheet), so ~1.3V between +9V and ground. So 7.7V/0.015A=513Ω. Your 560Ω should be fine. To be honest, even a 1kΩ would probably be fine. You might consider it, to save power.

    Speaking of that, how sure are you of the 9V stability when under load? If it's a 9V battery, it'll likely be drawn down quite a bit after some time.
     
  13. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
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    Thanks for that, had'nt thought of the battery issue. Now that you mention it, I would imagine the 9V battery will not last very long.

    I think the best thing for me to do now is construct the circuit and test it to see what the results are and I'll post them for you to see.

    Thanks so much for your time and help.Its appreciated. JDR04
     
  14. wayneh

    Expert

    Sep 9, 2010
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    No substitute for data! If it was me, I think I'd experiment with how low a current I could use to the LEDs and still get reliable switching. This would dramatically reduce battery power consumption. I think there is also a possibility to put two, maybe even all 3 of the LEDs in series.
     
  15. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Hi wayneh, sorry for the delayed reply.

    Connecting in series was mentioned before. My concern is if the optocouplers are in series does that not mean the 3rd one will infact trigger later than the first one?

    I'm also concerned that if Q2 fails then Q,3 and Q4 would fail to trigger as well.I though using them in parallel would mean they all trigger exactly the same time or is the time delay really small? I wonder if I could measure it on a scope?? Any suggestions.

    Your suggestion of using a low current as possible is great and I'll experiment with that. Thanks for the tips.

    Thanks again
     
    Last edited: Sep 19, 2013
  16. wayneh

    Expert

    Sep 9, 2010
    12,103
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    Nope, current flowing is identical throughout any series circuit loop, and that happens at the speed of light. Lightbulbs in series all turn on at the same instant.

    Actually, a series arrangement might give a miniscule improvement in synchronization for this reason. If there is any delay in current development in one device relative to another, in series they all to wait for the slowest device. No current can flow through one unless it also flows through all. In parallel they just fire up at their own rate. I have no idea if this makes any difference, I'm just pointing out the potential.
     
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  17. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
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    Thanks for that. I tried measuring it on the scope and the difference was so miniscUle I couldnt really determine its value properly.

    Anyway thanks for all your help and time. I'm going to move the project into the real world and build it all up proper.

    Thanks to everyone on the forum especially wayneh. CheerS
     
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