Current conduction of a MOSFET in saturation region

Discussion in 'Analog & Mixed-Signal Design' started by Babun Pal, Jul 28, 2016.

  1. Babun Pal

    Thread Starter New Member

    Jul 13, 2016
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    Hi,

    Could anyone please explain me how would the current conduction take place if the channel of a MOSFET is pinched off?

    Thanks in advance.
     
  2. AlbertHall

    Well-Known Member

    Jun 4, 2014
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  3. dl324

    Distinguished Member

    Mar 30, 2015
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    Leakage current.
     
  4. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    Nope - this is normal forward conduction. See the explanation in post #2.
     
  5. dl324

    Distinguished Member

    Mar 30, 2015
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    Nope, it's leakage current. WIth zero gate bias, all MOSFETs will have leakage current. This is especially true for short L and thin gate oxide devices.
    upload_2016-8-6_19-7-45.png
     
    Last edited: Aug 6, 2016
  6. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    From: http://www.circuitsgallery.com/2012/08/working-of-MOSFET-java-applet.html
    Here is the picture of a MOSFET with the channel pinched off (ringed in red line). It is conducting 6mA of drain current.
    [​IMG]

    • Additional increase of VDS causes the pinch off point to shift towards the source, dropping the effective length of channel. This effect is called channel length modulation.
    [​IMG]
     
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  7. dl324

    Distinguished Member

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    It seems that I confused JFET terminology with MOSFET...:eek:
     
    Last edited: Aug 7, 2016
  8. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    OP is specifically talking about MOSFET conduction while the channel is pinched off - the situation depicted in the diagrams I posted.
    Channel pinch off is NOT what happens when the gate voltage is below the threshold.
    From the link I posted in post #2:
    "What happens between the pinch-off point and the Drain:
    The Gate-to-Substrate voltage in this region is not enough for a formation of the inversion layer, therefore this region is only depleted (as opposed to inverted). While depletion region lacks mobile carriers, there is no restriction on current flow through it: if a carrier enters the depletion region from one side, and there is an electric field across the region - this carrier will be dragged by the field. In addition, carriers which enter this depletion region have initial speed.

    All the above is true as long as the carriers in question will not recombine in the depletion region. In n-type MOSFET the depletion region lacks p-type carriers, but the current consist of n-type carriers - this means that the probability for recombination of these carriers is very low (and may be neglected for any practical purpose).

    Conclusion: charge carriers which enter this depletion region will be accelerated by the field across this region and will eventually reach the drain. It is usually the case that the resistivity of this region may be completely neglected (the physical reason for this is quite complex ... )."
     
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