409
Bipolar Junction Transistor Compact Models
excess majority carriers = total excess minority carriers. Therefore, for an npn-BJT
with |V BC | > 0; V BE = 0, we can show
Q
qA p x N x dx
qA n x
n
N x
dx
R
R
a
x
x
R
i
a
x
x
E
C
E
C
=
−
=
−
∫
∫
( )
( )
( )
( )
2
(11.98)
Again, from Equation 11.47, we can show that the reverse base transit time
for BJTs is given by
Q
I
R
B R EC
= τ
(11.99)
Therefore, the normalized reverse injection charge is given by
q
Q
Q
I
Q
Q
I
q
V
v
r
R
B
B EC
B
BR
B
SS
b
BC
kT
=
=
=
−
0
0
0
1
τ
τ
R
exp
(11.100)
Equations 11.81, 11.90, 11.97, and 11.100 represent the components of the normalized base charge q e , q c , q f , and q r , respectively, in terms of measurable
device parameters. We will substitute these components of base charges in
Equation 11.78 to solve for q b in the following section.
Evaluation of q b : Substituting for q e , q c , q f , and q r from Equations 11.81, 11.90,
11.97, and 11.100, respectively, in Equation 11.78 we can show the expression
for total normalized charge as
q
V
V
V
V
Q
I
q
V
v
Q
b
BE
AR
BC
AF
f
B
SS
b
BE
kT
r
B
= +
+
+
−
+
1
1
0
0
τ
τ
exp
I I
q
V
v
V
V
V
V
q Q
SS
b
BC
kT
BE
AR
BC
AF
b
f
exp
−
= +
+
+
1
1
1 τ
B B
SS
BE
kT
r
B
SS
BC
kT
I
V
v
Q
I
V
v
0
0
1
1
exp
e xp
−
+
−
τ
= +
q
q
q b
1
2
(11.101)
where we defined
q
V
V
V
V
q
I
V v
I
V v
BE
AR
BC
AF
f ss
B E
kT
rSS
BC
kT
1
2
1
1
= +
+
=
(
)−
+
τ
τ
exp
e xp( (
)−
1
0
Q B
(11.102)
Bipolar Junction Transistor Compact Models
excess majority carriers = total excess minority carriers. Therefore, for an npn-BJT
with |V BC | > 0; V BE = 0, we can show
Q
qA p x N x dx
qA n x
n
N x
dx
R
R
a
x
x
R
i
a
x
x
E
C
E
C
=
−
=
−
∫
∫
( )
( )
( )
( )
2
(11.98)
Again, from Equation 11.47, we can show that the reverse base transit time
for BJTs is given by
Q
I
R
B R EC
= τ
(11.99)
Therefore, the normalized reverse injection charge is given by
q
Q
Q
I
Q
Q
I
q
V
v
r
R
B
B EC
B
BR
B
SS
b
BC
kT
=
=
=
−
0
0
0
1
τ
τ
R
exp
(11.100)
Equations 11.81, 11.90, 11.97, and 11.100 represent the components of the normalized base charge q e , q c , q f , and q r , respectively, in terms of measurable
device parameters. We will substitute these components of base charges in
Equation 11.78 to solve for q b in the following section.
Evaluation of q b : Substituting for q e , q c , q f , and q r from Equations 11.81, 11.90,
11.97, and 11.100, respectively, in Equation 11.78 we can show the expression
for total normalized charge as
q
V
V
V
V
Q
I
q
V
v
Q
b
BE
AR
BC
AF
f
B
SS
b
BE
kT
r
B
= +
+
+
−
+
1
1
0
0
τ
τ
exp
I I
q
V
v
V
V
V
V
q Q
SS
b
BC
kT
BE
AR
BC
AF
b
f
exp
−
= +
+
+
1
1
1 τ
B B
SS
BE
kT
r
B
SS
BC
kT
I
V
v
Q
I
V
v
0
0
1
1
exp
e xp
−
+
−
τ
= +
q
q
q b
1
2
(11.101)
where we defined
q
V
V
V
V
q
I
V v
I
V v
BE
AR
BC
AF
f ss
B E
kT
rSS
BC
kT
1
2
1
1
= +
+
=
(
)−
+
τ
τ
exp
e xp( (
)−
1
0
Q B
(11.102)