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using a transistor to ignite an e-match (1amp) — Parallax Forums

using a transistor to ignite an e-match (1amp)

evergreenevergreen Posts: 43
edited 2006-05-08 04:14 in General Discussion
Hello, I've done some searching on this forum, but it appears no one has used a transistor to ignite an e-match (electrical match). Here's how I think they work (please let me know if I'm wrong).

A transistor has 3 connection points: base, collector, emitter

In order for me to ignite an ematch I need to do this:
-connect the 'base' to an I/O pin on the Stamp
-connect the 'collector' to a seperate battery (the positive end)
-connect the 'emitter' to one side of the e-match lead
-connect the second e-match lead to the negative side of the battery

Putting the I/O pin closes the circuit, and connects the 'collector' and 'emitter' internally.

Do I need resistors somewhere in there?
How do I know if a transistor can handle the 1 amp of current?

Thanks

Comments

  • Jon WilliamsJon Williams Posts: 6,491
    edited 2006-04-15 22:54
    The TIP120 Darlington should work fine -- refer to the attached circuit.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Jon Williams
    Applications Engineer, Parallax
    e-match control.jpg
    477 x 374 - 14K
  • evergreenevergreen Posts: 43
    edited 2006-04-16 15:42
    thanks Jon
  • evergreenevergreen Posts: 43
    edited 2006-04-23 20:57
    Does the ground for the TIP120 have to go to the negative terminal of the external 12v battery? Or can it go to the Stamp's ground?
    Thanks
  • Jon WilliamsJon Williams Posts: 6,491
    edited 2006-04-23 21:11
    The negative terminal of the 12v battery must be connected to the Stamp's Vss or the circuit won't work.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Jon Williams
    Applications Engineer, Parallax
  • Russ MillerRuss Miller Posts: 25
    edited 2006-04-24 03:59
    At the rocket factory we used a ballast resistor in the firing circuit to limit the current in case the match or squib shorted after firing. In our case the 1 Amp was an "all fire" rating but the resistance of the match itself was perhaps 1-3 Ohms, so with our 28Volt system we would have seen up to ~20A without the resistor. It may be worth looking at your specific matches and deciding if your system will survive the maximum current.
  • evergreenevergreen Posts: 43
    edited 2006-05-07 17:51
    Okay, pardon me for the n00b question, but... Where do the pins go? I currently have it set so that:
    Base goes to ground
    Collector goes to a pin (with a resistor inbetween)
    Emitter goes to one end of the e-match

    Negative terminal of battery goes to circuit's base
    Positive terminal goes to the other end of the e-match

    Is that wrong? Everytime the pin goes HIGH (which should close the circuit), the voltage drop down to around 0.15 volts, even with a fresh new battery (or two batteries in parallel, or two batteries in series).

    Please help

    Thank you
    Matt
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2006-05-07 18:15
    If you are referring to the Drawing e-match control.jpg: the Emitter goes to GND, the Base goes to the 470ohm, and the Collector is connected to one end of the incendiary device.

    *** Post Edit ***
    Looking at your first Post, you have it described (basically, but not wholly so)·the way that it should be, but in that preceding this you have it all jumbled.· What gives?

    One more thing -- the tab/heat-sink is the Collector, same as the middle pin [noparse][[/noparse]so, watch it!]

    Post Edited (PJ Allen) : 5/7/2006 6:32:02 PM GMT
    bang.jpg
    358 x 262 - 16K
  • evergreenevergreen Posts: 43
    edited 2006-05-08 04:14
    Whoa, I'm stupid. No wonder why it wouldn't work! But 'base' and 'ground' sound so smiliar! Thanks a lot Allen!
    Matt
    PS: I wonder why I didn't do what I said in the first post...
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