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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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