Fet current

Discussion in 'General Electronics Chat' started by mentaaal, Apr 1, 2008.

  1. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Hey guys, we are currently doing FETS again in class and there is something i dont understand. IN the schematic i have attached, the question asks to design this circuit for Id to be 1mA. I dont understand how this is possible as the internal resistance of the FET is 100kOhm. for 1mA to flow through the FET, the way i see it (which is completely wrong) is that the voltage across it will be (1mA * 100000) which is ridiculous and not the 10 volt my lecturer got.

    My lecturer kep saying something about the resistances being in parallel and i completely understand this from the output impedance point of view (looking in from the vout) but without this circuit connected to anything, how does this work?

    Help , i am lost!
     
  2. veritas

    Active Member

    Feb 7, 2008
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    attachment?
     
  3. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Crap Sorry!
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    Didn't you look at the datasheet for that very old FET?

    Its drain to source current is from 5mA to 12mA when its gate voltage is the same as its source voltage.
    It conducts less when its gate voltage is negative and is almost completely cutoff when its gate voltage is -1.0V to -8.0V.

    So simply feed a negative voltage to its gate until its current is 1mA.
     
  5. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    sorry audioguru, we were not given a spec for a jfet. I just drew this schematic quickly on orcad and that must of been the first jfet i could find.

    Its more the circuit theory that i am struggling with.
     
  6. Audioguru

    New Member

    Dec 20, 2007
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    Look at the datasheet for any N-channel junction depletion-mode FET. It conducts its max current when its gate voltage is equal to its source voltage.
    It conducts almost nothing when its gate voltage is negative with respect to the source voltage.
     
  7. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Yes i am aware of that, but i guess what i mean is what is the meaning of rd? Is this the internal resistance of the FET when it is in cutoff? I am sorry if these are really stupid questions but we are being taught this module almost entirely with formulae, as such, trying to understand this all is quite a challenge!
     
  8. Audioguru

    New Member

    Dec 20, 2007
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    The current of a FET is used in calculations, not its resistance. Its drain voltage is 10V when 1mA flows in its 10k drain resistor that is connected to 20V.

    I don't know what is Rd. Ask your teacher.
     
  9. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    my lecturer says its the internal resistance of the FET. He says this can have a value of 100kOhm???
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    The voltage at the gate of the FET adjusts how much current it passes. Then simply use Ohm's Law to calculate its resistance.

    The old JF1033B FET can have a max current of 12mA with 10V Vds when its gate voltage is 0V. Then its resistance is 10V/12mA= 833 ohms.

    Its cutoff current is 10uA with 10V Vds when its gate voltage is -1V to -8V. Then its resistance is 10V/10uA= 1M ohm.

    You can adjust its gate voltage so that its resistance is 100k ohms if you want.
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    The 100kohms may be the small-signal dynamic impedance when the FET is biased into the active region. It means that, if you change Vds by 1V, Id will only change by 10uA.
     
  12. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Thanks for all the help you guys, and Ron i think you are spot on the money there! So, you mean that changing Vds wont really do an awful lot but obsiously changing the gate voltage will change the current through the fet by the transconductance right?
     
  13. Dave

    Retired Moderator

    Nov 17, 2003
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    Id is effectively independent of Vds when the FET is saturated. If you change Vgs such that Vgs minus the FET threshold voltage is greater than Vds then the FET falls into the triode region where Id is much more susceptible to changes in Vds (almost like resistor for small values of Vds).

    Dave
     
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Here is a simulation of a 2N4416. It is not necessarily the same as the JFET in your problem, but notice the slope of the constant gate voltage curves. The slope, dV(ds)/DI(d) is the dynamic drain-source resistance. I am including the .asc file, in case anyone wants to play with the sim.
     
  15. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Thanks guys, that really clears this problem up for me! This is my point entirely! My lecturer never really told us what rd is! So i was under the impression that that was just some constant internal resistance. Thats why i was always wandering how could the FET get current anywhere near even a milliamp! And when i asked the teacher this he just waffled on something about being in parallel!

    So let me just state what you guys have told me and please correct me if i am wrong:

    Rd doesnt really come into play when the FET is in the triode region
    This resistance is only important when the FET is limiting current through the FET
    We work out the currrent going through the FET by V/R of the load resistor

    Thanks for all your help on this, i thought i was going mad!
     
  16. Audioguru

    New Member

    Dec 20, 2007
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    Wait a minute.
    The FET operates at only 0.5mA so its dynamic resistance is many megohms, not just 100k.

    Ron, your ASC file has junction transistors, not a FET.
     
  17. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Hmmm... Not sure how that happened. I have replaced it with the correct file (in the original post). I hope.:rolleyes:

    Thanks, AG.:)
     
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