not understanding this inductor circuit (with pictures!)

Discussion in 'General Electronics Chat' started by jamjes, May 21, 2010.

  1. jamjes

    Thread Starter Member

    May 10, 2010
    35
    0
    As time goes on I seem to be here more and more often.. :D

    I've built a modified circuit I found online which I can use for home made inductors, for experience, research, just for the heck of it, whatever:

    [​IMG]

    I'm not quite sure if the 391Ω resistor is correct, since I just trim it to get about 40 kHz, but it shouldn't matter that much. The 4R7Ω is a big 50W one.

    The symbol for the MOSFET isn't right. Its an avalanche rated 900v 5A sucker. (Datasheet attached) This could be where I'm going wrong if I've got the wrong one?



    So basically I rig up the scope and get this output:

    [​IMG]


    Which just confuses the heck out of me.

    Shouldn't it be showing some sort of deflection on the up, as well as the down??


    In addition, if I connect to TP3, I get no deflection of any sort. What's going on here?
    (EDIT: I get a +10v deflection, sorry)

    Thanks -- I'm probably being a bit dim.
     
    Last edited: May 21, 2010
  2. ifixit

    Distinguished Member

    Nov 20, 2008
    639
    108
    Tell us about the inductor. i.e. winding resistance, core type, inductance?

    It seems to deflect up 150V. Look harder for the down part.

    Regards,
    Ifixit
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Well, you don't have a reverse-EMF or "flywheel" or recirculation diode across the inductor. When the MOSFET cuts off, there is no place for the current to go, except via your 50w 4.7 Ohm resistor and the MOSFET's body diode.

    Basically what you want to do is turn ON the MOSFET for a short amount of time, and then let the current through the inductor die out. You might need to have the MOSFET turned on for only 1% to 10% of the time.

    Measure the voltage across your 4R7 diode. If the voltage starts rising quickly, you know your inductor is becoming saturated. The voltage increase across that resistor should be a constant slope; if it suddenly gets steep, that's saturation.

    Check out Ronald Dekkers' "Flyback for Dummies" page:
    http://www.dos4ever.com/flyback/flyback.html
    It's a really great resource.
     
  4. jamjes

    Thread Starter Member

    May 10, 2010
    35
    0
    I have been reading that page SgtWookie, but I just can't seem to tie up the theory with reality.

    ifixit:
    The voltage at TP3 is actually changing slighty, about 100mV over the cycle.


    my inductor is 232 turns of enameled copper 34 SWG (0.236mm) on a ferrite core. Coil resistance is 11.9Ω using a multimeter.

    <br />
A_L = 6700 \rm{nH}<br />

    So the calculated inductance is...

     L = N^2 * A_L

     L = 232^2 * 6700 * 10^{-9}

     L = 361 \rm{mH}


    After placing a diode over +12v and TP2 I get more interesting results with a known inductor (120uH). Curves and all. That bit certainly worked :)

    So -- is my inductor too.. inductive??
     
    Last edited: May 22, 2010
  5. ifixit

    Distinguished Member

    Nov 20, 2008
    639
    108
    Hi,

    Not much energy will be stored in a inductor of that size with only a charge time of 12.5uS ( half of 1/40000). Try 4Hz. You should see a big difference. One time constant is approximetely 32mS.

    Regards,
    Ifixit
     
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