Download Example 3.2 Consider a chrome-silicon metal

Example 3.2
Solution
Consider a chrome-silicon metal-semiconductor junction with Nd
= 1017 cm-3 . Calculate the depletion layer width, the electric field
in the silicon at the metal-semiconductor interface, the potential
across the semiconductor and the capacitance per unit area for an
applied voltage of -5 V.
The depletion layer width equals:
2ε s (φ i − Va )
xd =
qN d
2 ×11.9 × 8.85 ×10 − 14 × ( 0.3 + 5)
=
= 0.26 µm
1.6 × 10 −19 × 1017
where the built-in potential was already calculated in Example
3.1. The electric field in the semiconductor at the interface is:
qN d x d
E ( x = 0) =
εs
=
1.6 × 10 −19 × 1017 × 2.6 × 10 − 5
11.9 × 8.85 × 10
−14
= 4.0 ×10 5 V/cm
The potential equals:
φ ( x = xd ) =
qN d x d2
= φ i − V a = 5 .3 V
2ε s
And the capacitance per unit area is obtained from:
ε
11.9 × 8.85 ×10 − 14
Cj = s =
= 40 nF/cm 2
−
5
xd
2.6 × 10