Back EMF

Discussion in 'General Electronics Chat' started by EAI, Aug 28, 2010.

  1. EAI

    EAI Thread Starter New Member

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    Hello

    When connecting a power transistor to a ignition coil how is it possible to stop back emf.

    Placing a diode across the emitter and collector or use a zener diode across the emitter and collector.


    This circuit is very popular but usually there is so many problems with it as due to back emf destroying the lm555 or the transistor or both.



    [​IMG]


    Thanks EAI
  2. marshallf3

    marshallf3 Well-Known Member

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    The rectifier would go across the coil, cathode to the +12V, anode to the transistor's collector. I'd also use a rectifier with a decently fast switching time, the 3055 is a workhorse but it's old and probably a bit more vulnerable than most.

    A schottky diode with decent ratings would be a good choice.

    [Post has been edited to describe proper placement, I made a mistake in the original]
    Last edited: Aug 29, 2010
  3. EngIntoHW

    EngIntoHW Member

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    How come the rectifier should be connected from Collector (Anode) to Emitter (Cathode)?

    Didnt you mean a zenner diode?
  4. Ghar

    Ghar Active Member

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    Can you justify this?

    The entire point of a diode is give the inductor current a safe path to dissipate in. Your proposal doesn't do that.

    Check out this comparison simulation...

    BackEMF_schem.png

    BackEMF_plots.png

    As you can see that diode across the transistor doesn't do anything... the correct position is across the inductor.

    If you want to clamp across the transistor you use a zener.
  5. marshallf3

    marshallf3 Well-Known Member

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    My mistake, all my original post would do is feed it back into the power supply.
  6. EngIntoHW

    EngIntoHW Member

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    Hi Ghar,
    You did a wonderful job here, I'm glad to be part of this thread and learn from forum members here.

    I'd like to ask you a few questions please.

    How come you were applying 5V on the base?
    Wouldn't it damage the BE junction of the BJT?

    If you were to use a zener diode across the CE junction (Anode to Emitter, Cathode to Collector), would you need its Zener Voltage to be larger than V1 (25V)? (So the Zener wouldn't conduct when the BJT is in cutoff, right?).

    What Zener Voltage would you want in such application?

    And, could you give up on the Rectifier (across the relay) if you use a Zener diode across the CE junction?

    Thank you very much :)
  7. marshallf3

    marshallf3 Well-Known Member

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    You will absolutely have to have the rectifier across the ignition coil as that's where the back EMF is coming from and it's best to quench an offending signal at its source.

    I was just half asleep with another half dozen problems in my head when I spit that first post out, guess I should go edit it as not to accidentally mislead someone.

    What you're doing is shunting out the back EMF generated when the transistor is switched back off. When the current going through an inductance is abruptly interrupted it generates a back EMF (voltage in reverse) that can easily damage other things in the circuit. The principle of using a rectifier to catch it is even used in circuits that contain absolutely no active components at all. (such as relay banks)

    Herre's a little reading on the subject:
    http://www.ehow.com/how_5593373_calculate-back-emf.html
  8. rjenkins

    rjenkins AAC Fanatic!

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    Putting a diode directly across an inductor results in a slow turn-off as the current only decays due to circuit resistance.

    In circuits that need a fast switch-off, you should add a resistor in series with the flywheel diode. The resistor value depends on the inductor current and the voltage the switching device can safely withstand, above the supply voltage.

    eg. if the inductor is taking 2A on a 12V supply and the transistor is rated at 60V, a 22 Ohm resistor is about right; it will drop 44V @ 2A, adding to the 12V supply so the peak collector voltage will be around 56V (plus diode drop).

    The resistor is dissipating a peak power of 44(V) x 2(A) = 88W for a short time at each switch-off, so a power resistor of some form is essential.

    For a fast turn-off, the higher the resistor the better - so also use a transistor with a high Vce rating.
  9. Ghar

    Ghar Active Member

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    I just wanted to switch the inductor current in the simulation, this wouldn't have been a real circuit.

    You're right, the zener voltage would need to be higher than the supply or else it would always conduct. The value you pick depends on the transistor, you want it to be quite a bit lower than the maximum rating to protect it.

    I agree with marshallf though, in that it's probably better to keep the diode across the coil. Since the back EMF is really the inductor current, that zener would dissipate Vz*IL, while the diode would dissipate Vf*IL. Vz we already agree would be larger than 25V, while Vf is about a volt. However, that zener would reduce the current much faster than the diode, for the reasons that rjenkins mentioned. The inductor current will decay according to the simple equation:
    \frac{dI}{dt} = \frac{V}{L} = \frac{IR + V_f}{L}

    If you ignore the resistance you can see that a 25V zener would make the current decay 25 times faster than a 1V diode.

    Which is best depends on the application; the required decay time, the power levels, the level of protection needed...
    This kind of problem is almost an entire field which goes by the name of 'contact protection'.

    Also just as a general note I find that by far the best way to understand back EMF is to realize that it is the inductor current that's the problem. The inductor current will want to remain at the same value and will raise the voltage if impeded. By providing an easy path it can't complain and you will not have a voltage spike beyond those caused by the new current path's impedance.

    Edit:
    On another note, I do not follow that eHow article...
    They say calculate di/dt = V/L, but don't specify what V they're talking about.
    The next line says to calculate maximum V by multiplying di/dt*L.
    ...but you already needed V to calculate di/dt, since that's the exact same equation?
    Last edited: Aug 29, 2010
  10. The Electrician

    The Electrician Senior Member

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    If you "stop" the back EMF, you won't get very good performance.

    Your post describes this as an ignition coil driver. You haven't said what the application is, but assuming that it's an ignition coil for a gasoline engine, such coils have two windings, a primary and a secondary.

    They are a kind of transformer, and the voltages across primary and secondary are approximately in the same ratio as the turns ratio, typically something like 1:100. So, if you want to get 30,000 volts out of the secondary, you must have 300 volts across the primary.

    In other words, you must allow the back EMF to rise to several hundred volts. If you clamp it to a single diode drop by putting a diode across the primary directly, you won't get a spark out of the secondary.

    If you clamp the primary "back EMF" to around 60 volts, to protect a 2N3055, you will get very poor performance.

    What you must do is to use a transistor that can withstand several hundred volts when it's turned off.

    There are special transistors made just for this purpose; see, for example:

    http://www.onsemi.com/pub_link/Collateral/NGD15N41CL-D.PDF

    http://www.newark.com/on-semiconductor/mgp15n40clg/igbt-discrete-transistor/dp/26K4435

    These transistors have built-in protection and don't need another zener diode or other protective device, and they're easy to drive since they have a MOSFET-like input impedance.
  11. EngIntoHW

    EngIntoHW Member

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    Thank you so much guys,
    It's been great reading your answers! :)
  12. EAI

    EAI Thread Starter New Member

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    Hi

    Yes the ignition coil is for a gasoline engine, also testing sports ignition coils.

    I will take your advice and purchase the special transistors that can take this punishment.

    Just curious, is it possible to use a capacitor across the primary of the igtion coil or the emitor and collector to stop back emf just like that was used on the older model cars across the distributor to stop sparking or arcing across the points.

    Thanks EAI




  13. marshallf3

    marshallf3 Well-Known Member

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  14. EAI

    EAI Thread Starter New Member

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  15. Jaguarjoe

    Jaguarjoe Active Member

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    Here is how GM did it on millions of their cars:

    http://www.jaycar.com.au/images_uploaded/MC3334R0.PDF

    The MJ10012 is rated at 400 volts, the zener is 350V. There is no diode across the coil because it would slow it down too much. When my 12cyl Jag is cruising at 6000RPM, the ignition system is running at 600Hz.
    How is a 555 going to spark a gas engine? Unless it runs at one very fixed speed, how will the spark rate change with speed?
    Note that the MC3334 is obsolete and unobtainable. So is the MC79076 which is almost the same thing. However, you can buy a "4 pin HEI module" for about $15 which has that circuit in it.
  16. The Electrician

    The Electrician Senior Member

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    A capacitor is a good idea, but it won't "stop" the back EMF; it will just resonate with the coil and make a half-cycle resonant voltage wave. The ignition transistor has a certain amount of capacitance itself, but adding some more (around .1 to maybe .22 or .33 uF. You could even try a standard automotive capacitor) might help. It depends of the particular coil, and you might find it helpful to experiment with different values.

    The idea is for a half-wave resonant voltage to occur, with the peak just reaching about 350 to 400 volts.
  17. orbiter

    orbiter Member

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    Thanks for the input here guys. I just built this circuit and knew I'd built it correctly, however I was confused when the thing kept blowing 555 timer chips.

    Going to try again now with a <25ns diode in place as suggested earlier in the thread.

    Thanks again

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