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Induction Coils

  1. Mar 4, 2007 #1
    Hi all. I recently read an article by a MIT prof. about RLC etc ckts in which he has given a spark plug example. I think the analysis given by him is faulty. The link to the document is here:

    http://ocw.mit.edu/NR/rdonlyres/3EBFD5FD-DCF5-422A-A7E3-E1312C74153D/0/trns1_rc_lc_rlc1.pdf

    I have also attached the same pdf with the post.

    On the 4th page he has given a single inductor automobile spark plug circuit (fig 5 of document) example.

    According to his calculations the voltage induced across the inductor is 24K. Which is huge and I think his way of calculating is wrong.

    Actually, if we analyze it correctly. The time constant of the circuit is L/R = 0.002 sec (2 ms) Hence the amount of current that can be developed in the inductor within 1us is calculated to be around 0.00119A and hence the max voltage across the inductor can be calculated as L(di/dt) ie:
    0.01(0.00119/1u) = 11.97 Volts. This seems to be more realistic and much more accurate.

    This is why in actual sparking circuits we use a secondary induction coil to multiply the voltage to about 20K Volts.

    Please comment on whether my analysis is correct.

    One last question - Once the switch is opened the inductor circuit is broken it can be alalysed as an R-L ckt with infinite resistance and hence zero time constant. Is this why theoritically we have infinite voltage developed across the inductor?
     

    Attached Files:

    Last edited: Mar 4, 2007
  2. jcsd
  3. Mar 5, 2007 #2

    berkeman

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    Staff: Mentor

    I don't like his analysis, but his answer is more realistic than your 12V answer. Think about what a spark plug is -- it's a controlled spark source, and the arcs that are sustained across the output are around 10kV to 20kV in order to be able to jump the gap.

    The problem I have with his analysis is that he doesn't explain what determines the output time constant of the kickback waveform. When the switch is opened, what you are left with is the energy in the inductor, and the parallel capacitance across the inductor. The output voltage reaches a peak and comes back down (and would ring the other way if there were no loss), and that kick voltage waveform is basically the positive half of a sine wave, with the period determined by the parallel LC components. The capacitance comes from the physical wires and capacitance of the plug itself. Of course, once the voltage gets high enough to start the arc, the voltage plunges quickly because the impedance across the arc is very low. The arc is sustained as long as the energy is dumping out of the inductor.

    Quiz question -- to match the arc time that he indicates, what would the parallel capacitance have to be?
     
  4. Mar 5, 2007 #3
    Induced voltage across the inductor is

    [tex]v = L \frac{\Delta i}{\Delta t} [/tex]


    By collapsing the current in the shortest possible time, as t approaches 0, the induced voltage approaches infinity.
     
  5. Mar 5, 2007 #4
    I didn't bother reading through any of it. I think the whole thing is pointless because there is no secondary winding which puts the high voltage spike across the switch contacts (points) which would never allow the voltage to reach 24KV. They would arc as soon as they have opened a fraction of an inch.
    -
    However, the spike generated on the primary side of an ignition coil on a gas engine can easily reach a hundred or more volts when the points open. How high it gets depends on how the secondary coil is loaded, how well arcing is controlled at the breaker points, the type of steel used in the core of the coil, those are just a start.
     
  6. Mar 6, 2007 #5
    Thanks all for the answers.
    The problem I have with the pdf analysis is:
    1) The way he arrives at 24K volts seems faulty. Here again there are two things that I find wrong.
    a) According to the formula V = L delta(i)/delta(t).
    He puts delta(i) = 2.4 A and delta(t) = 10u secs
    First things is within 10u secs the current will not drop by 2.4 A.
    You need to have the capacitance value to calculate this.
    b) How did he get the 10u sec value for the inductor current dissipation.

    2) The 12 volt analysis that I have done is using the L-R values only (this is also wrong) and hence I am getting a 12 V spike. Practically the inductor will be in series with a capacitor forming an LCR and there will be another secondary coil which will give the voltage boosting. The Primary LCR is only for the sustained oscillations so that the arc remains for a longer time.

    Another point by averagesupernova is very correct about the sparking of the switch. This is another reason (electrical isolation) that the secondary coil is used. The main reason being the voltage multiplication.

    Please comment on this analysis. Thanks
     
  7. Mar 6, 2007 #6

    marcusl

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    Gold Member

    All mechanical distributors have a capacitor across the points to help limit arcing when they open. Even so, the big voltage across them causes arcing, which eventually eats up the contacts. That's why the points need regular replacement.
     
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