16574
CHAPTER 16
16.1
4.02
k
Ω
1
+
0.15
(
29
1

0.03
(
29
≤
R
≤
4.02
k
Ω
1
+
0.15
(
29
1
+
0.03
(
29

4.48k
Ω ≤
R
≤
4.76k
Ω
16.2
I
C
1
=
I
S
1
exp
V
BE
1
V
T

I
C
2
=
I
S
2
exp
V
BE
2
V
T

I
C
2
I
C
1
=
I
S
2
I
S
1
exp
V
BE
2

V
BE
1
V
T

∆
V
BE
=
V
BE
2

V
BE
1
∆
I
S
=
I
S
1

I
S
2

I
S
=
I
S
1
+
I
S
2
2

I
S
1
=
I
S
1
+
∆
I
S
2
I
S

I
S
2
=
I
S
1

∆
I
S
2
I
S
a
( 29
I
C
2
=
I
C
1
:
∆
V
BE
=
V
T
ln
I
C
2
I
C
1
I
S
1
I
S
2
=
0.025ln 1
( 29
I
S
1
+
∆
I
S
2
I
S
I
S
1

∆
I
S
2
I
S
=
0.025ln
1.05
(
29
0.95
(
29
=
2.50
mV
b
( 29
∆
V
BE
=
0.025ln
1.10
(
29
0.90
(
29
=
5.02
mV
b
( 29
I
S
1
I
S
2
=
1
+
∆
I
S
I
S
1

∆
I
S
I
S
=
exp
V
BE
2

V
BE
1
V
T
=
exp
0.001
0.025
=
1.04
→
∆
I
S
I
S
=
0.02
or
2%
16.3
a
( 29
For
v
1
= 0 =
v
2
,
V
BE
1
=
V
BE
2
and the collector currents are the same.
So,
V
OS
= 0.
Only the base currents will be mismatched.
b
( 29
I
C
1
=
I
S
1

0.025
(
29
exp
V
BE
V
T
1+
V
CE
V
A

I
C
2
=
I
S
1
+
0.025
(
29
exp
V
BE
V
T
1+
V
CE
V
A
∆
I
C
=
I
C
2

I
C
1
= 0.05
I
S
exp
V
BE
V
T
1+
V
CE
V
A

I
C
=
I
C
1
+
I
C
2
2
=
I
S
exp
V
BE
V
T
1+
V
CE
V
A
V
OS
=
∆
I
C
g
m
=
V
T
∆
I
C
I
C
=
0.025
V
0.05
(
29
=
1.25
mV
c
( 29
I
C
1
=
I
S
exp
V
BE
V
T
1+
V
CE
V
A
1
+
0.025
(
29

I
C
2
=
I
S
exp
V
BE
V
T
1+
V
CE
V
A
1

0.025
(
29
∆
I
C
=
I
C
2

I
C
1
2245
I
S
exp
V
BE
V
T
11.025
V
CE
V
A

1
+
0.975
V
CE
V
A
=
I
S
exp
V
BE
V
T
0.05
V
CE
V
A
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document