Spark Plug+Coil+Controller = ARRRRGH

Discussion in 'The Projects Forum' started by Cloud9, Apr 16, 2013.

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  1. Cloud9

    Thread Starter Active Member

    May 11, 2009
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    Hi All,

    I am so frustrated at this point since I have been at this problem for what seems like ages now.

    I have a controller circuit which activates a FET board (contains 2 12N10L and 2 MBR40250G diodes. These diode/fet pairs activate 1) gas valves 2) ignition coil attached to spark plug.

    Initially, I got it all worked out using a particular coil. Things work great but alas, the coils are no longer available. The specs were primary res 3.5ohm, secondary 14kohm. Not sure about the ratio.

    Now I have moved to a Blaster SS 8207 coil. It's high performance. Primary res 0.355ohm, secondary 4.4Kohms and 70:1 ratio.

    The first thing I notice is that I have to limit the input current. No problem there. Even when I dry fire (no gas applied), things seem to work ok but once gas is applied, at some point, my FET controlling the coil fails (STUCK ON) and this in turn sticks my gas on..... at which point I run screaming.

    I understand why my spark FET failure causes a failure on the gas, even though I REALLY dont like it. My question is WHY does my spark FET fail.

    I have attempted to raise my input resistance (via ballast resistor) all the way up to 8 ohms. I see a marked decrease in spark strength although it still fires well.... until it LOCKS.

    I am kind of stuck looking at the secondary resistance. My new coil is so much lower, do I need to increase it? I use really short plug wires.

    My theory is that for whatever reason, the kickback from the coil is overcoming the diode, and ravaging the FET. But why would it, the diode is rated for up to 250V blocking voltage and 40A.

    Additionally, I noted that with no ballast my initial coil that works great has a current of 3.8A vs the new blaster that doesnt work at 8A.

    On my FET board I have the diode in reverse polarity across the source and drain.

    Any help would more than greatly appreciated.

    Chris
     
  2. wayneh

    Expert

    Sep 9, 2010
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    Schematic?
     
  3. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Something's fishy here. This sounds like an automotive enhancement project to me......

    This thread will probably be closed, unless you can give a good explanation of what you're doing.
     
  4. bertus

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  5. Cloud9

    Thread Starter Active Member

    May 11, 2009
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    Sorry for not including the schematic of the FET board, it is attached.

    What sounds fishy?

    This has nothing to do with automotive. I have made a gun fire simulator that uses O2 + LP + SPARK to go boom.
    [​IMG]http://www.freeimagehosting.net/hmekx
    [​IMG]
     
    Last edited: Apr 16, 2013
  6. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    The condenser/capacitors on a points ignition are rated at 400V-450V, so your diodes are much under rated. Diodes are often used in circuits that are twice or more than the expected voltage. I'd look for diodes with a minimum 600V reverse voltage, more wouldn't hurt.
     
  7. DerStrom8

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    Feb 20, 2011
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    Good to know, that was my mistake. I should not have jumped to conclusions.
     
  8. #12

    Expert

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    The "fishy" part is that this website does not allow automotive discussions, a fact that new visitors are often unaware of because so many, "click here or you can't use the software" requirements have taught people that reading the text is a waste of time.
     
  9. Cloud9

    Thread Starter Active Member

    May 11, 2009
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    Thanks for the input so far. DerStrom, no worries, I dont think many folks are stupid enough to do projects like this haha.

    ShortBus, duly noted. Any indication as to why the would be such a drastic difference in operation (or lack thereof) between the coils?

    Also, I wouldnt expect more than 80ish V coming back from the primary. Or would I?

    Seems like I measured at one time and saw 40V spikes which was surprising enough.
     
  10. Cloud9

    Thread Starter Active Member

    May 11, 2009
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    I would like to clarify since it is not shown on the schematic.

    I have J2 jumpered across 1(TRG1) and 2(TRGx) so that trigger 1 also fires the FET 2.

    ----

    While I got such experienced folks pondering my project. I would love to solve the problem of the spark FET (always the culprit) failure causing my valve FET to stick on. Since the outputs are separate, the issue has to be the gate of the spark fet (after failure) leaking a voltage high enough to trigger the valve FET (both gates are tied). Could I solve this with diodes at the gates to keep voltage from coming FROM the gates? Is there something I am overlooking that could be an issue?

    Generally speaking once the spark FET is gone, and my valve FET is stuck on, replacing the spark FET solves both issues.
     
    Last edited: Apr 16, 2013
  11. #12

    Expert

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    Energy = 1/2 L(I squared)
    There is no part of that equation that says the voltage is limited. You have to dump that energy or the voltage (theoretically) approaches infinity.

    On another point, if your trigger signal does not go to an active ground connection, the 1Meg resistors will slow down the turn-off time. I would say to reduce them to the 10k area.

    The slow turn off might be the saving grace of this circuit in that the slow turn off time allows some of the energy to escape through the mosfet.
    Another useful equation is: energy = 1/2 C (V squared). A capacitor across the switch can absorb some of the energy and limit the peak voltage. The capacitor will need a resistor in parallel to dump that energy over a few seconds so you can fire the machine again.
     
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  12. WBahn

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    Mar 31, 2012
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    I'd recommend rethinking your controller so as to incorporate some arming and failsafe features. One of them might be a state machine that won't let the signal for the spark be generated while the valve is open and other state is that once the valve is closed and the system is armed, the valve cannot be commanded to open again until the system is reset following the completion of a cycle.
     
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  13. wayneh

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    So the spark MOSFET actually fails to "on"? Maybe the schottky isn't fast enough to protect the MOSFET, and/or the spike coming back from the coil is too much for the MOSFET and diode.

    I don't quite understand how the triggers are linked, as the schematic doesn't show that. I don't get how a failure of the spark MOSFET causes a problem with the valve MOSFET's trigger. Have you measured voltages? Is voltage from the coil reaching the gate of the coil's MOSFET after failure? That's bad!
     
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  14. shortbus

    AAC Fanatic!

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    when you have 12V on the primary and it makes 50KV on the secondary. When the secondary voltage collapses it also creates a voltage back into the primary. The "rebound" voltage won't be at the same ratio though. Because the HV current isn't enough to create as much magnetic force in the core.

    In your circuit, how close is J1 to the mosfet gates? Should be less than 1". That and the 1k ohm gate resistors are too high. Between 4R7 and 47R ohms is kind of a starting point for gate resistors. Almost never as high as 1k.

    Low side gate drivers would probably help to, unless your trigger source is a totem pole arrangement. You have to sink the voltage fast to shut the mosfet off fast. It's the off time that will make your spark strong.
     
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  15. Cloud9

    Thread Starter Active Member

    May 11, 2009
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    I will try to address all of these responses at once, but Im sure I will forget something.

    Wayneh - I posted an edit above about the trigger link. J2 is actually a jumper where I can select FET2 trigger as either TRG1 or TRG2. TRG1 is used for both FET inputs (jumpered 1 to 2).
    Unfortunately, the answer to "does the FET fail on?" is yes. When the spark FET dies it is on. I was thinking with the spark FET fails and allows a voltage to come OUT of it's gate, which in turn activates the valve FET. The output is not the cause (I dont think) regarding the valve FET being stuck on (until the spark FET is removed or replaced).
    Also to my disadvantage, is when the spark FET is stuck on (along with gas), the first instinct is to kill the power, alas, the field colapses in the coil and BOOOOM. Scares the bejeezus out of me. Tommorow I will get some actual voltages for you.

    WBahn - I had considered that. That is the original reason I made 2 seperate triggers. There are a few safeties built in. 1) I have an arm signal the operates a relay the feeds the TRG signal to my FET board. 2) The actual controller that generates the fire signals outputs to a non-retriggerable one-shot. Technically this means the board could never try to hold the FIRE signal on longer than a few ms. I learned this the hard way since every once in a blue moon, the spike from the spark plug could reset or lock up my board. That *seems* to be solved. I think your suggestion is great, my only concern is blowing my inputs on my control board that would be monitoring the dreaded spark FET gate.

    #12 - I had not considered any of that. At some point I plucked those resistor values from the air and they seemed to work. I know, I know, I deserve a good lashing for it. The actual TRIGGER signal from my control board comes via a ULN2008 darlington driver. I am not sure if this could be considered an 'active ground'. Could you comment? I knew that I needed a fast signal to get my spark proper, and I *ASSumed* that a high value of res for the gate bias would be beneficial to a quick activation since I needed less current to overcome the ground. I kinda thought maybe 1M was a little high since, when I touch the gate, I trigger, ouch. "capacitor across the switch..." would this also be in parallel with the diode across the drain/source, or are we talking gate/source?

    shortbus - The distance from J1 to the gates is probably less than 1" but... Do I have to consider the unwound length of the trace? Do I consider the length of the wire feeding J1? As mentioned earlier, I dont suppose I had a whole lot of theory behind the resistor values. I assumed the higher value could help protect any feedback from triggering the FET. Why is 1K too high? It's certainly a huge difference from the 4.7ohms you suggest. Could you give me a little insight into this?
    I cant remember why I chose high side drivers, I think it was becuase of the added complexity of driver circuits versus a logic input. I dont *think* speed is an issue here because it really sparks spectacularly and I saw the effects already of using a FET that was too slow... a very unreliable spark.

    Thank you all SO much for your insight so far!!!
     
  16. wayneh

    Expert

    Sep 9, 2010
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    A diode would put an end to that, at least.
     
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  17. Cloud9

    Thread Starter Active Member

    May 11, 2009
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    I agree (I hope) and it's at least a safer way to fail. Hopefully we can get to the bottom of the problem.
     
  18. #12

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    I used the word, "switch" because you are using the mosfet as a switch. The mosfet is the switch.

    Let's do a few basics. A mosfet has capacitance as a dominant feature of its gate. Evey switching action is about charging or discharging that capacitance. Suppose the gate has 1000pf. If you use a 1 meg resistor to drain the "go" current from that gate, it will take t=-RC LN Vo/dV
    t = - 1e6 x 1e-9 Ln (2/5) assuming the shut off voltage is 2 and the driving voltage is 5
    .9 milliseconds
    In most cases, fast switching is the rule, and this setup can not switch any faster than 1KHz. Even for a car engine, that is dismally slow. Your coil was not made to run at that speed.

    When you do get the switching speed up, the energy in the coil has to go somewhere when you open the switch. If there is no capacitance in the switching circuit, the voltage (theoretically) approaches infinity. Methods to deal with that include attaching a diode across the coil so the reactive energy goes back into the 12 volt power supply. Another way is to put a capacitor across the switch, like in an old car ignition system with "points". The size of the capacitor determines how high the voltage gets by acting like a storage place for current. I already gave you the math for that. You will have to measure the inductance and the current (energy). If you use a capacitor to dump the shut-off energy, you need a resistor in parallel with it to dump that current slowly to ground. Way slower than a microsecond, more like a tenth of a second to a whole second for what you're doing.

    Here's a circuit that does a direct conversion from 1/2 LI^2 to 1/2 CV^2
    Maybe it will show you the relationship.

    http://forum.allaboutcircuits.com/blog.php?b=478
     
  19. Cloud9

    Thread Starter Active Member

    May 11, 2009
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    #12 - I am not up on my theory as well as I should be but do understand I am using the mosfet as a switch - lol :p give me a little credit. My question was when you say "across" the switch, but I guess when I think about it, the 'switch' portion of the FET is the Source/Drain.

    Does placing the diode across the FET serve the same purpose as placing it across the coil. I have tried placing a diode across the coil, and the result was no spark. Most likely it's due to the diode I used.

    I also looked into RC snubbers and even tried a sample circuit with no positive results. I have 0 dobuts that did not have the proper values.

    You suggest I should measure the inductance and amperage. The coil inductance is 6.9mH. Do I have to consider the inductance of the FET and diode as well? Which amerpage do I need? The short circuit current is about 8A. As far as the spike current, I have been really scared to hook up my expensive (to me) scope rated at 2000V to this circuit. :eek:

    Thank you so much for your comments.
     
  20. #12

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    Using 1/2 LI^2 = 1/2 CV^2
    and full saturation current is 8 amps and the inductance of the coil is 6.9mh

    If you energize the coil until it saturates at 8 amps, then shut off the mosfet instantly, you will find these capacitors will be charged to these voltages:

    .1uf = 2100V
    .47uf = 969V
    1uf = 664V

    The capacitor will oscillate with the coil for a while, using the resistance of the coil to waste the energy as heat. Meanwhile, you should have a discharge path to ground (for the capacitor) of about 1% of the coil current when 12V is applied.
    12V/.08A = 150 ohms.
    Truth is, that will discharge the capacitor really quickly and you could get away with, say, 1500 ohms, considering you are not firing this thing a thousand times a second.
     
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