FET replaced with Bipolar...Help?

Discussion in 'General Electronics Chat' started by messi_tnt, Jul 31, 2011.

  1. messi_tnt

    Thread Starter New Member

    Jul 22, 2011
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    Hi there,

    I need to ask if it is a common practice when replacing power FET with Bipolar transistor to add a (1-2 k) resistor to the base of the Bipolar one??
    If yes, Why??


    Regards
     
  2. tom66

    Senior Member

    May 9, 2009
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    The base of a transistor is like a diode, it will conduct current, so a resistor needs to be added to limit this current and set the current gain of it. A fet conducts almost no current on the gate so the resistor is not necessary.
     
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  3. messi_tnt

    Thread Starter New Member

    Jul 22, 2011
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    thank you for your quick answer.. one more thing, does the value of the resistor depend on the application (as you said : it sets the current gain)
    or there is a common value to be used ?

    Regards
     
    Last edited: Jul 31, 2011
  4. tom66

    Senior Member

    May 9, 2009
    2,613
    214
    Yes, you need to calculate it according to the desired collector current.

    Collector current (Ic) = Hfe * Base current (Ib)

    If you have a 5V logic input, the B-E drops 0.65V then the voltage across the resistor will be 4.35V. If your transistor has a gain of 100 (look up exact figure in the datasheet for your transistor), and you want (up to) 1A collector current, your resistor needs to pass 10mA across 4.3V.

    Ohms law tells you V = IR so rearrange to get R = V/I and plug in your values. 4.35 / 0.01 = 435 ohms. The closest lower E12 value is 390 ohms, but if you use E24 resistors you can use 430 ohms. You should go lower to allow for a margin of error.

    Be aware that the power dissipation in this 390 ohm resistor is given by V^2 / R or I^2 * R giving 47.41mW which is acceptable for a 1/4W or 1/8W through hole resistor and a 1/10W 0603 SMD resistor. But at larger collector currents a more powerful base resistor may be necessary.
     
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  5. messi_tnt

    Thread Starter New Member

    Jul 22, 2011
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    This explanation is great.. I think that this is all what I need to know regarding this issue
    Thank you again



    regards
     
  6. Adjuster

    Well-Known Member

    Dec 26, 2010
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    300
    If the base resistor value has to be made very small to get the necessary drive, you need to be sure that the gate (or whatever) that is driving it can actually supply the current.

    If there is not enough current available to drive an ordinary bipolar transistor properly, bearing in mind that the saturated gain level may be as low as 10 in some cases, one alternative is to use a Darlington type. These have much higher gain, but bear in mind that they are slower, and have a bigger saturated "on" voltage.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    When you are using a transistor as a saturated switch (like driving a relay, turning on an LED, etc) you need to use 10 for the gain of the transistor, not the hFE specification.

    If you attempt to use the hFE specification instead of 10, you will very likely wind up with a high Vce (voltage from the collector to the emitter) and power dissipation in the transistor.

    Also, Vbe (voltage from the base to the emitter) goes up as Ib (base current) increases; at very low base current you can use 0.63v, but better to start off with 0.7v. Also, you want to use a transistor with at least twice the Ic rating that you need; otherwise the saturation voltage will be quite high.
     
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  8. messi_tnt

    Thread Starter New Member

    Jul 22, 2011
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    Great and useful advice as always from SgtWookie, and thank you all for your helpful posts.

    Regards
     
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