Electronic – MOSFET Depletion Region Widening


I do understand why when drain voltage (\$V_D\$) increases, the channel carriers decrease (the channel narrows), specially near the drain island, eventually reaching pinch-off phase.

What I don't get is why the depletion region is supposed to widen, as (\$V_D\$) increases.

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Best Answer

The schematic shown in question corresponds to an N-channel-MOSFET. The substrate (or body or bulk) of which will be a P-type semiconductor. So source and drain, which are N+ semiconductor regions will form PN junctions with the substrate. The depletion region width of a PN junction increases as reverse bias voltage is increased. Since the voltage applied at drain is positive, the junction becomes reverse biased and hence the width increases with \$V_D\$.

The source and substrate terminals are shorted together to the same potential, that is why depletion width at the source side remains the same while that at the drain side is increasing with drain voltage.