MOSFET in place of Bipolar

Discussion in 'General Electronics Chat' started by lokeycmos, Jun 18, 2010.

  1. lokeycmos

    Thread Starter Active Member

    Apr 3, 2009
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    i built a ignition coil driver based on 555 and 2n3055. [​IMG]
    id like to swap out for a mosfet because of lower internal resistance, but unsure how to calculate the output resistor. i tried a irfp250 but got no output at all. do i have to put a pull down resistor across the gate and source? and could you tell me what the pull down resistor does? (just saw that on another site)TY
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    1,728
    You have the coil polarity reversed.

    With the printing on the face of the IRFP250 so that you can read it, they are
    Gate Drain Source

    The tab is also connected to the drain.

    Use a 56 Ohm resistor rated for about 1W. You could use a somewhat higher value resistor if you wanted to. A pair of 110 Ohm resistors in parallel would get you close.

    It's a good idea to use a 10k resistor from the gate to the source just to make certain that if the gate drive resistor fails, the gate doesn't float.

    If you exceed the Vdss rating of the MOSFET, you'll probably kill it after a few cycles.
     
  3. lokeycmos

    Thread Starter Active Member

    Apr 3, 2009
    432
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    thanks for the info! (i know about the polarity, i pasted it from a site) how did you know to use 56ohm and 1 watt? is there a formula? TY:)
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    I figured you will be powering it from a 12.7v source.

    The 555 loses 1.3v from V+ to pin 3 due to the Darlington follower.

    It has an output source/sink capability of 200mA.

    So, (12.7v-1.3v)/200mA = 57 Ohms.

    I just ballparked the power. Were it a constant load, you'd actually need a 5W resistor.
    P=EI, so (12.7-1.3) * 200mA = 2.28; double it for reliability and you wind up with 4.56W. 5W is the closest you can get.

    However, if the MOSFET fails, you frequently get a short from the gate to the drain. In this case, you really want the resistor to burn up to act like a cheap fuse.
     
  5. lokeycmos

    Thread Starter Active Member

    Apr 3, 2009
    432
    7
    thank you much! very helpful!
     
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