Model Rocket Igniter from Logic Level - Output Stage Question

Discussion in 'The Projects Forum' started by Renbo, Apr 12, 2009.

  1. Renbo

    Thread Starter Member

    Apr 12, 2009
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    Greetings!

    I am working on a project to control, via logic circuits, the ignition of model rocket igniters. My goal is to isolate the logic circuits from the higher voltage/amperage ignition circuit, and to do it with solid state components.

    In order to work with a range of different igniters that have different voltage requirements, I would like the output stage to be designed with that flexibility in mind.

    I have a circuit diagram of an output stage idea. My first question is if it looks workable as shown, or if I am way off base. The other questions involve what methodology or formulas I might use in the selection of the power transistor, and the calculation of R1, R2 and R4 (if the circuit idea looks sound).

    [​IMG]

    Thanks in advance for any thoughts!

    Sincerely,

    Reynolds Kosloskey
     
    Last edited: Apr 12, 2009
  2. Wendy

    Moderator

    Mar 24, 2008
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    Last time a model rocket question came up I was suprised how many people popped out of the woodwork that liked that hobby. I count myself among them.

    The diagram looks good. The only question I would have is the 50V for the current source. When I was around 11 years old or so I used AC power on my ignitor (stupid, and I made sure my parents didn't know). The ignitor blew, much like a fuse, so fast that I didn't have reliable ignition. Estes ignitors are much different now, but the idea is to make the wire glow, not blow. You may want to dial that voltage down, and focus on current.

    *************

    On second thought, you need a safety interlock. I don't see it in your schematic.

    Eliminate R2, not needed. What is the power transistor you are planning on using, it needs to be as high a gain as you can get, since you will pull several amps through it. I'm guessing as many as 10A, so you will need around 200ma into the base, which is pushing it for the optocoupler. Why the optocoupler, may I ask?

    Another thing just occured, most times the alligator clips are live when they are thrown against the blast deflector. You may need to limit the current, or have some other plan for the contingency. Also, the simple versions have a continouity detetion.
     
    Last edited: Apr 13, 2009
  3. Audioguru

    New Member

    Dec 20, 2007
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    R2 in your circuit does nothing, is not wanted and is not needed. It should be replaced with a peice of wire.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    Made me think, shorting the BE of Q1 would make a good safety interlock. You also need to show where the 10' of wire is in the schematic.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    I'm not so keen on your circuit. If the optocoupler fails, R4 will pull the base of Q1 high, lighting the ignitor.

    It would be better to have the output of the optocoupler in series with R4, and omit R2.

    Go with lower voltage. Somewhere between 6v and 12v will work just fine for a standard ignitor.

    Don't forget that you must have a safety interlock switch (a removeable key switch is best, but a removeable phono plug/jack combo will work), a Safe/Armed indicator (audible and/or visual), and a fire button that won't stay closed without someone pushing on it.

    If you are planning on automating a launch sequence, you'll also need a big ABORT button that will stop the sequence, just in case a big gust of wind comes along, someone enters the launch area, or an aircraft enters the airspace near you.
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    Looking at it, I question the optocouplers use. I understand the base concept, keeping the wires short in the high current end for maximum current. Remember, we are taking a process that is more about current than voltage.

    If I had had the moxie I might have done something similar (with a relay) as a kid.

    Question for the OP: What kind of ignitors were you thinking of?
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    If you're looking to control more than one ignitor, you might consider using a ULN2003B or ULN2803B driver IC, which contain (respectively) seven or eight Darlington transistor pairs, with built-in limiting resistors for the inputs to the bases. These ICs can sink up to 500mA current per channel.


    If you need more current, there's also the ULN2067B, which has four Darlington pairs, and can sink 1.5A per channel. Newark carries these: http://www.newark.com/jsp/search/productdetail.jsp?SKU=89K1136&CMP=AFC-OP
     
  8. Renbo

    Thread Starter Member

    Apr 12, 2009
    15
    2
    Thanks for all the feedback!

    O.K., SgtWookie, I agree I had not thought about the failure mode of the coupler, and I wouldn't want to have the output driven live in that event. So I think I understood your ideas on the different approach and have modified the schematic, but also thought I'd need to add a pull-down resistor (R2) to the idea to make sure the power transistor remains switched off with no input from the ILQ5. I also omitted the resistor between the ILQ5 output and the base of Q1:

    [​IMG]


    As far as a safety interlock, my thought was that this unit would have multiple output stages, each independently controlled, and I wanted the interlock to arm/disarm the bank. I was thinking I'd achieve this with the interlock switch merely disconnecting the +0-50V supply from the load rail.

    As for the optocoupler, the choice to include that digs a little deeper into the purposes behind this unit. In addition to using this unit to power igniters, I thought I might also want to try it out during the Christmas season and have it drive from each output a string of twinkle lights in series (calculating the number of twinklers by dividing 120v by the number in the original series set to get each twinkler's voltage, then dividing the 50V by that number to get the number of twinklers per 50V output). I would want to be able to turn these strings on and off very rapidly in some cases, and I wanted to isolate all that potential switching noise from the logic side of things.

    Yet another idea for this unit would be to use it on the 4th of July to ignite some fountains off. I didn't mention this use originally because in some states even fountains are not legal to use, so I stress that *I AM NOT SUGGESTING THAT ANYONE USE THIS CIRCUIT FOR THAT PURPOSE*. But it does let you know another reason for the higher voltage, as I would likely try to set those fountains up with fuses that have a piece of thin nichrome wire wrapped around them, and would be experimenting with how much juice that would require.

    Thanks again for the input! Let me know if these circuit changes are in line with what you were thinking SgtWookie!

    Sincerely,

    Reynolds
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    I've been interest in rockets and similar things since Sputnik blasted into orbit...

    I'm going to hop up on my soapbox for a minute or two, so please bear with me. ;)

    When dealing with launch systems, K.I.S.S. rules. (KISS = Keep It Simple, Stupid!)

    And I'll tell you why, too.

    Back in the 1970's, there was a USMC Gunnery Sergeant who built a device that was supposed to save time prior to launching aircraft that were armed with missiles. It was mandatory to check the umbilical cables to the AIM-9 Sidewinder missiles with a multimeter to ensure that there was no stray voltages that might cause an unexpected missile launch.

    He had built his own "stray voltage tester" that connected between the missile umbilical and the missile itself. While he was in the process of testing for stray voltages with his tester, an electrician was in the back seat of the F-4J Phantom that the missile was mounted on, re-seating the circuit breakers by pulling out their "stems" and pushing them back in. Apparently, the Gunnery Sergeant had not planned for this combination of events, and the rocket motor of the missile ignited, sending the missile bouncing down the flightline and right through a hangar that was filled with aircraft and personnel. :eek: Thankfully, the missile did not have a warhead, and no aircraft nor personnel happened to be in the path of the errant missile.

    It's occurances like this that will really make you appreciate safety and simplicity.

    Please don't make your launcher any more complex than it absolutely has to be. Safety interlocks, warning lights and/or audible warnings are a must. Continuity indicators add a "cool factor" and can be very helpful to ensure you have a good connection through your ignitor(s).

    You really don't need an optocoupler/optoisolator for this project. It adds unneeded complexity. If you just have one or two ignitor circuits, you can do with TIP120 Darlingtons and an input resistor; they have the pull-down resistors built in. Same thing with the ULN2067B, except they already have the 2.7k input resistor.

    The fewer physical parts that your launcher is built from, the better. If you keep it simple, and have spare parts on hand, you may actually be able to rescue a possible "scratch" on launch day because something broke in the launcher.

    The more complex you make it, the less chance that you can fix it in time to save a "launch window".

    I've attached a schematic of how to wire a TIP120 Darlington with a safety interlock switch and an LED to indicate continuity through the ignitor.
     
  10. Wendy

    Moderator

    Mar 24, 2008
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    I'd add a 0.5Ω on the emitter, to limit a surge if the alligator clips short to the blast deflector. This would would limit the current to 25A, which should be plenty to fire an ignitor. A shorted transistor after such an accident would be no fun at all.

    A question for Wookie, why not a MOSFET?
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    I suggest that instead of a 0.5Ω resistor, that you use a 2A to 5A fuse between the battery + and the safety interlock switch. In the event of a dead short, the fuse will protect the circuit, and you won't lose much voltage across the fuse. A typical ignitor may only have a resistance around 2 Ohms; adding a resistor in the circuit may cause delayed or "hang-fire" ignition.

    While MOSFETs perform extremely well as switches once installed in a properly designed circuit, they do not handle exposure to static electricity well at all. One "zap" and they're toast. They may fail shorted; particularly if the gate gets shorted to the drain.

    In an application such as this, safety is the most important consideration, in which reliability is a major component. The leads to the ignitor may be dozens of feet long; and one of them would be connected to the drain of the MOSFET. Having an unconnected drain connected to a long wire that'll be flopping around in a bag/box/case/etc between uses is cause for concern.

    Even the smallest model rockets could cause injury if inadvertently ignited. Safety must be kept foremost in mind, particularly when dealing with pyrotechnics of any kind.

    An old quote from the National Rifle Association which certainly applies here, is: "Safety is no accident". A truly great double entendre.
     
  12. pmyshkin

    New Member

    Apr 24, 2009
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    SgtWookie-

    Thanks for the circuit, it was just what I was looking for. I have breadboarded up this circuit with the digital input to the NPN coming from a PIC microcontroller (and only a 6V supply--4 'AA' batteries). When I run a test of the ignition, it fails. The output from the PIC goes high momentarily and then immediately shuts off (I'm assuming, I don't have a scope, just a multimeter, so presumably the node goes low after current starts flowing through the NPN). If I short the NPN directly to the PICs power, the ignitor lights as expected, and if I remove the NPN and just test the output of the PIC, it goes high for a couple of seconds as expected. So, is this some issue with the NPN drawing too much current from the PIC? I'm thinking of adding an op-amp between the PIC and the NPN (LM741) now.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    Post your circuit.

    If you omitted the resistor from the PIC output pin to the base of the transistor, you're overloading the PIC's output.

    6v is too high to power a PIC with. Their max rating is 5.5v.

    You need a diode between the batteries and the PIC's supply, with a good-sized capacitor across the PIC's supply pins to power the PIC while the batteries are busy powering the ignitor. Otherwise, the PIC will reset.
     
  14. pmyshkin

    New Member

    Apr 24, 2009
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    I have attached the schematic--sorry for the messiness, I'm new to TinyCAD. As you can see I do have the 2.7K resistor to the base of the NPN and of course I'm powering the PIC through a diode.

    I don't currently have a cap between the powers, though I understand that would help. But I don't think that is currently my problem. I guess when I test again I'll measure the current flowing through the NPN to see where it's at. Otherwise I'm also considering just using a relay.

    Thanks again for your help and I really appreciate the comments.
     
  15. pmyshkin

    New Member

    Apr 24, 2009
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    I now see my problem. The PIC is shutting off the output pin when it senses the collector of the NPN going low, which I was thinking would happen once the fuse was blown, but now I realize it happens once current starts flowing (obviously, since collector is now in essence shorted to ground. I think I can solve it in PIC re-programming.
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    You still need caps across the PIC's Vdd & Vss pins. I'd use a 0.1uF and a 220uF or larger.

    Otherwise, it's quite likely that the ignitor load will pull the battery voltage low enough momentarily to cause the PIC to reboot.
     
    Last edited: May 16, 2009
  17. pmyshkin

    New Member

    Apr 24, 2009
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    I understand the potential need for the cap. Excuse my ignorance, but why the two caps?
     
  18. hgmjr

    Moderator

    Jan 28, 2005
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    The large valued cap is used to filter the low frequency ripple on the supply and the lower valued cap is used to filter the high frequency noise on the supply since the larger caps are not that efficient at filtering high frequencies.

    hgmjr
     
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