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D-MOSFET construction and working

Short answer

A D-MOSFET has an n-channel diffused between n+ source and drain on a p-substrate, with insulated gate. At VGS=0, ID=IDSS. Negative VGS depletes channel, reducing ID; positive VGS enhances it, increasing ID. Saturation drain current: ID=IDSS(1−VGS/VP)².

Describe the construction and working principle of DMOSFET with the help of drain characteristics curve, transfer characteristics curve and mathematical expression.

The depletion-type MOSFET (D-MOSFET) has a channel already built in during manufacture, so it conducts at VGS = 0 and can be operated in both depletion and enhancement modes.

Construction (n-channel)

  • A p-type substrate with two heavily doped n⁺ regions (source and drain).
  • A thin, lightly doped n-type channel is diffused between source and drain.
  • A thin SiO₂ layer insulates the metal gate from the channel.
      S      G      D
      |   +--||--+  |
   +--+---+-------+--+---+
   | n+  n-channel   n+  |
   |    (built in)       |
   |   p-type substrate  |
   +----------+----------+
              B

Working principle

  • VGS = 0: with VDS applied, electrons flow from source to drain through the built-in channel; ID = IDSS.
  • VGS < 0 (depletion mode): the negative gate repels electrons from the channel, so the channel narrows (depletion), and ID falls below IDSS. At VGS = VP (pinch-off, negative), the channel is closed and ID ≈ 0.
  • VGS > 0 (enhancement mode): the positive gate attracts more electrons, widening the channel; ID rises above IDSS.
  • For small VDS the device acts as a voltage-controlled resistor (ohmic region); when VDS ≥ VGS - VP, the channel pinches off near the drain and ID becomes nearly constant (saturation/pinch-off region).

Characteristics

Drain characteristics:

 ID
  |         ________ VGS = +1 V (enhancement)
  |       _/________ VGS = 0 (IDSS)
  |      / ________  VGS = -1 V
  |     / /________  VGS = -2 V
  |    / /
  +---+--+----------- VDS

Transfer characteristic: a parabola crossing the ID axis at IDSS (VGS = 0), reaching ID = 0 at VGS = VP, and continuing to positive VGS.

 ID
  |                 /
  |              _.'
  |   IDSS ---.-'
  |        .-'
  |  _.--'
  +-+---------+--------- VGS
   VP         0

Mathematical expression

In the saturation region, for both modes (VP ≤ VGS):

ID = IDSS (1 - VGS/VP)²        ;  VDS ≥ VGS - VP

For VGS > 0 the same equation holds (ID > IDSS). Transconductance: gm = (2 IDSS/|VP|)(1 - VGS/VP).

✓ Answer checked by Claude Opus 5.5 (high effort)

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