550
21 Diodes
E V
E F
E i
E C
semiconductor
e B
V>0
metal insulator
e S
neutral
region
depletion
region
inversion
region
V i
Q n
Q m
-eN A
x
(x)
-d
0
x
w
-d
0
w
Q m / i
x
w
-d
0
s
(x)
V
V i
(a)
(c)
(d)
(b)
Fig. 21.34 Ideal MIS diode at inversion: a band diagram, b charge distribution, c electric field and d potential
the space charge is due to ionized impurities (depletion) and F is dominated by the second term in
(21.85a), i.e. s ∝
√
s . For B ≤ s ≤ 2 B the diode is in the weak inversion regime and still
s ∝
√
s .
Eventually the dominating term of the second bracket in (21.85a) f 1 = n p 0 / p p 0
exp(β) = (n i /N A )
2 exp(ββ) becomes comparable and exponentially larger than the dominating term of the first bracket f 2 = ββ. Solving f 1 = f 2 for = γ γ B yields the equation
(N A /n i )
γ −2
= γ ln(N A /n i ) and a value γ larger than and close
19 to 2. Thus for s > >
inv
s ≈ 2 B
(using (21.78)),
inv
s ≈
2 kT
e
ln
N A
n i
,
(21.88)
strong inversion starts and the space charge is given as Q s ∝ − exp(ββ/2). For the case of strong
inversion the band diagram is shown in Fig. 21.34 together with the charge, field and potential.
The bias voltage V at the MIS diode (or the total voltage drop across the MIS diode) is
19 For N A /n i = 10 4 , 10 6 and 10 8 , we find γ = 2.33, 2.25 and 2.20, respectively.
21 Diodes
E V
E F
E i
E C
semiconductor
e B
V>0
metal insulator
e S
neutral
region
depletion
region
inversion
region
V i
Q n
Q m
-eN A
x
(x)
-d
0
x
w
-d
0
w
Q m / i
x
w
-d
0
s
(x)
V
V i
(a)
(c)
(d)
(b)
Fig. 21.34 Ideal MIS diode at inversion: a band diagram, b charge distribution, c electric field and d potential
the space charge is due to ionized impurities (depletion) and F is dominated by the second term in
(21.85a), i.e. s ∝
√
s . For B ≤ s ≤ 2 B the diode is in the weak inversion regime and still
s ∝
√
s .
Eventually the dominating term of the second bracket in (21.85a) f 1 = n p 0 / p p 0
exp(β) = (n i /N A )
2 exp(ββ) becomes comparable and exponentially larger than the dominating term of the first bracket f 2 = ββ. Solving f 1 = f 2 for = γ γ B yields the equation
(N A /n i )
γ −2
= γ ln(N A /n i ) and a value γ larger than and close
19 to 2. Thus for s > >
inv
s ≈ 2 B
(using (21.78)),
inv
s ≈
2 kT
e
ln
N A
n i
,
(21.88)
strong inversion starts and the space charge is given as Q s ∝ − exp(ββ/2). For the case of strong
inversion the band diagram is shown in Fig. 21.34 together with the charge, field and potential.
The bias voltage V at the MIS diode (or the total voltage drop across the MIS diode) is
19 For N A /n i = 10 4 , 10 6 and 10 8 , we find γ = 2.33, 2.25 and 2.20, respectively.