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Does this Schematic look right? (Firing Electronic Matches)

Discussion in 'General Electronics Discussion' started by funman1, Aug 15, 2010.

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  1. funman1

    funman1

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    Aug 15, 2010
    ok so I have a microcontroller sending low voltage signal
    That is connected to an N channel Mosfet (IRF9910PBFCT-ND)
    That's what should drive power to the EMatch setting it off.

    But I also wanted a Continuity LED (Green) so when the ematch is connected I get a green light.

    Then when the unit is actively firing I wanted the Red LED to be on so I have another Mosfet P channel driven off the first to basically just reverse the power signal into a positive to drive the Red LED.

    Then I'm not sure about the diode on the right, I think I needed something to prevent backflow when not activley firing the FETs.

    Does this look right or am I way out in left field?

    [​IMG]
     
  2. Resqueline

    Resqueline

    2,848
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    Jul 31, 2009
    The circuit seems to do the job intended (although it's hard to read a diagram that's upside-down of ordinary conventions).
    The diode is backwards but I don't think it's needed anyway.
    The green led will be exposed to 10V in reverse when firing though. They usually survive that even if it's over ordinary spec's but you may want to do someting about it.
     
  3. funman1

    funman1

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    Aug 15, 2010
    How is the Green getting 10V backwards?
    Both the anode and cathode should be both pushed to ground durring firing?
    So it's not a bad thing to back feed Fets with reverse voltage on the gates?
     
  4. NickS

    NickS

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    Apr 6, 2010
    I too do not see why the diode "D" is used. The net shared by the anode of "D" the PFET gate and the NFET drain is blocked off from ever seeing a positive DC voltage, so I don't see the state of the PFET ever changing.

    **Edit** Sorry. I need to read more carefully.
    So the diode on the right does have to come out or your ematch will never fire. Right now the only time you draw current through the ematch is when the green diode is on which you said is when you do not want it to ignite.

    If you have a microcontroller to drive the gate of the N MOSFET I would use it to drive the LEDs as well. This would eliminate the PFET and would also make the circuit function without running current through the ematch when it is not supposed to ignite. I understand that it is a very small current but if I have fireworks/model rocket attached to the igniter then I want 0A current running though the igniter until its go time.
     
    Last edited: Aug 16, 2010
  5. funman1

    funman1

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    Aug 15, 2010
    OK So I have updated the picture (May need to refresh to see the change)

    OK so the green LED is for continuity, I have to run some very minimal current through the match to make sure it's good. I have this setup on my mechanical firing system, the LED does not draw anywhere near what it takes to ignite an ematch.

    So that's the green,
    then the red is for active firing, and I can not draw this from the microcontroller directly, because I have this schematic x10 for each channel on the box.
    And if I fire all at once, the LEDS current draw would be too much for the microcontroller, so I have to switch them seperately so as not to overload the microcontroller.

    But since the Neg is common on the red/green LED I can't use that to control when the LEDs come on, I'm using Neg control for the firing of the match. but then needed a second Gate to control the pos of the red LED

    Did that make any sense? :)
     
    Last edited: Aug 16, 2010
  6. NickS

    NickS

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    Apr 6, 2010
    Ok many channels makes a great argument for not driving it by uC. So now lets talk subtleties.

    (1) Do both sides of the Dual LED draw the same current? If not then you may consider tying the joined cathode to ground and bringing the current limiting resistor to each sided anode. Then you can control the current through each side independently. I find that the red is almost always brighter for the same current thus I prefer different resistors to equalize brightness.

    (2) I like the NFET you chose but have you read the datasheet to make sure that driving the gate with a logic level signal will give you enough current throughput(see attached pic). I am guessing that if the ematch is 2ohms and wants a 12 supply that it is counting on seeing about 6A to ignite. For that driving the Vgs at 3V is borderline and may cause the device to overheat. It may not be necessary but I would drive a cheap FET/transistor with logic and use that to drive the second with 12Vgs(see attached). Like I said you picked a good FET so you may not need to do this but I would.

    (3) What are you using for the PFET? P type MOSFETS usually require a negative voltage to turn on. ie 0Vgs is not enough. You could either replace the PFET with a buffer that comes off the uC line or you could look into JFET's which can conduct at 0Vgs but beware they can also have a high Rds on.
     
  7. NickS

    NickS

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    Apr 6, 2010
    Sorry I forgot to attach the pics.
     

    Attached Files:

  8. funman1

    funman1

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    Aug 15, 2010
    More or less, It really won't matter. They both look good with that resistor common'ed.
    BUT since I should never really "see" the red ones except for testing hahah, I really don't care. I'm not going to be anywhere near that box when it's actually firing, hance the wireless operation.

    They actually only need 1 amp to ignite, and a AAA battery does just fine with igniting them. And the pulse is for 1/2 second. I doubt I will be able to over heat them in 500ms once every month or so :)
    Also, the fets I chose for that were listed at TTL Logic Gates, So I assume it will be enough from the Controller?

    Here is more info on the ematches
    http://www.electricmatch.com/product_us.html

    Hmm, well then I may just go with my tried and true radio shack special PNP Transistors, and drive the Gate with the 0VG from the main mosfet to get positive switching for the LED?


    Thanks so much for the feedback...
    I know enough to get myself in trouble, but far from enough to ever become an expert :).
    This is by far my most complex unit built (Entire project; not just this circuit)
     
  9. NickS

    NickS

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    Apr 6, 2010
    Ok.

    Indeed it is a logic gate device but most of the performance plots are given for a Vgs of 10V. 500ms at 1A is plenty of time to kill a lesser MOSFET(they can be impressive fast blow fuses). That said, I looked over the datasheet again and I think yours will be just fine with a Vgs of 3.3V and Ids of 1A.

    Sure, I am partial to the 2N3906 but whatever you find at the shack will probably work.

    No problem, I like your project.
    -Nick-
     
  10. funman1

    funman1

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    Aug 15, 2010
    Yikes now you have me worried about those Fet.... Sometimes matches dead short after firing, is there any way to protect the Fets then?
    Also how many matches do you think can be fired on a single channel of the Fet?


    OK I have updated the Schematic above then would that work as designed?


    Here are some pics of the project so far,

    [​IMG]

    [​IMG]

    [​IMG]

    I have a video of it in action if you are interested, I have 90% or so of the software written for the module, It's working awesome on breadboard even wirelessly (With LEDs showing firing rather than fets just yet)
     
  11. NickS

    NickS

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    Apr 6, 2010
    Dead short, yikes. The way I would protect against that is by having a current sense resistor inline and a comparator hanging off it that has the ability to throttle the NFET gate low if the current goes to high. You could feed the output of the comparator and the uC into an AND gate as long as you configure the comparator to switch low when current spikes. Then of course you would drive the FET with the output of the AND gate.

    I will throw a schematic together later.

    How many matches were you planning on firing off one FET? And are you using both FETS from a single package at the same time for different channels?
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Along those same lines, I'd be looking at foldback current limiting
     
  13. Resqueline

    Resqueline

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    Jul 31, 2009
    In the first diagram; you can't replace the original p-fet with a pnp just like that, it'll need a base resistor at least.
     
  14. NickS

    NickS

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    Apr 6, 2010
    Steve- He hasn't said so but what if his power supply is nothing more than just a SLA battery? Also with multiple loads all drawing heavy current do you think it would be difficult for the power supply to distinguish between one load over currenting and 10 loads running normally(call it grand finale).

    Resqueline- Good catch, I missed that.
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I was thinking that you'd put foldback current limiting on each "channel"

    It could be as simple as a polyswitch.
     
  16. NickS

    NickS

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    Apr 6, 2010
    Sorry for the delay, things were crazy for a while. Here is what I had in mind for MOSFET protection.

    Now let me explain
    R8 represents the EMATCH
    Q2 R4 Q1 LED's and R2 should look familiar.
    S1 I used to simulate the uC drive signal

    R5 is the current sense resistor I added. This needs to be a 5W rated part. Under normal operation it will only dissipate ~50mW(1A^2*0.05 ohms) but at overload condition it will very briefly spike to 5W(10A^2 * 0.05ohm). The spike in current to 10A will produce a trip voltage of ~240mV across R5 this trips the state of the comparator U1A whose reference is set @ 238mV with R6 and R7.

    Now the next part may seem like extra bagage so let me explain. Driving directly to the gate from the comparator would briefly turn off the FET until the current bleeds off but then it would potentially come back on in the .5s window your micro is asserting high. The result would be a sawtooth of high current through the FET for as long as .5s which is no help at all. Thus we feed the comparator into a flip flop which holds the comparator state change for the duration of the drive cycle. This means if an over-current event occurs the FET drive is disabled until the flip flop is reset by a new rising edge from the micro via U3.

    This may be overkill but it will keep the FET's safe.
     

    Attached Files:

  17. funman1

    funman1

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    Aug 15, 2010
    Holy somkes... Hahah ok.....
    This will take a bit for me to digest and comprehend...
    Man now I really don't know if I have enough room on the PCB with all this extra added stuff.... Wow...

    Thanks a ton..
    Let me mull this over and get a grasp of it...
    Thanks again!! :)
     
  18. funman1

    funman1

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    Aug 15, 2010
    OK so after doing some more research it looks like many firing systems made commercially use this
    irfz44n

    This looks much better and robust, I think this might be able to handle a possible short better?
    What are your thoughts?

    Here is a video of where I'm at with this project right now.


    Thanks..
    ~Steve~
     
  19. NickS

    NickS

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    Apr 6, 2010
    It all depends(you really did have a nice FET already). What the IRFZ44N brings to the table is higher current handeling...BUT did you notice what the minimum Vgs threshold is? With a drive voltage of 3.3V you will not even turn this thing on(if you do you will smoke it because it will be in the linear range). You need a minimum of 4.5Vgs to guarantee 10A of drain current. To reach the specified 49A you would need a Vgs of 10v(per data sheet). So if you use this guy you will definitely want to use a FET gate driver between the micro and your FET that uses the 12V rail for Vcc.

    That aside I see that the approach you are taking is the brute force method. Instead of stopping a dead short you are going the route of helping to facilitate a good path for a dead short. I am sure the reason you could get away with this is because you are using very long runs of wire between the box and your payload. This added impedance probably is keeping the current in a safe range. Other wise 12V across 17mOhms(FET resistance) is over 700A which is impossible of course because everything in its path would vaporize. You add 1 ohm of cable and that current reduces to under 12A(quite manageable). If you go the brute force route just be sure you never test the thing with short wires.
     
  20. NickS

    NickS

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    Apr 6, 2010
    Just watched the vid. It looks like your project is coming along great. Keep up the good work. BTW what program did you use for building you GUI.
     
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