CHAPTER 9
9.1
Since
V
REF
= −
1.25
V
, and
v
I
= −
1.6
V
,
Q
1
is off and Q
2
is conducting.
v
C
1
=
0
V
and
v
C
2
= −
α
F
I
EE
R
C
≅ −
I
EE
R
C
= −
2
mA
(
)
350
Ω
(
)
= −
0.700
V
9.2
I
C
2
I
C
1
=
exp
∆
V
BE
V
T
⎛
⎝
⎜
⎞
⎠
⎟
⇒ ∆
V
BE
=
0.025ln
0.995
α
F
I
EE
0.005
α
F
I
EE
=
0.132
V
(a) v
I
=
V
REF
+ ∆
V
BE
= −
1.25
+
0.132
= −
1.12
V
v
I
=
V
REF
+ ∆
V
BE
= −
1.25
−
0.132
= −
1.38
V
(b) v
I
=
V
REF
+ ∆
V
BE
= −
2.00
+
0.132
= −
1.87
V
v
I
=
V
REF
+ ∆
V
BE
= −
2.00
−
0.132
= −
2.13
V
9.3
Since
V
REF
= −
2
V
, and
v
I
= −
1.6
V
,
Q
2
is off and Q
1
is conducting.
v
C
2
=
0
V
and
v
C
1
= −
α
F
I
EE
R
C
≅ −
I
EE
R
C
= −
2.5
mA
(
)
700
Ω
(
)
= −
1.75
V
Note that Q
1
is beginning to enter the saturation region of operation, but V
BC
= +0.15 V is not
really enough to turn on the collectorbase diode.
(See Problems 9.5 or 5.61.)
9.4
v
I
=
V
REF
+
0.3
V
⇒
Q
1
on; Q
2
off.
I
C
1
=
α
F
I
EE
≅
I
EE
=
0.3
mA

I
C
2
=
0
v
C
1
=
0
−
I
C
1
R
1
+
R
C
(
)
= −
0.3
mA
3.33
k
Ω+
2
k
Ω
(
)
= −
1.60
V
v
C
2
=
0
−
I
C
1
R
1
= −
0.3
mA
3.33
k
Ω
(
)
= −
0.999
V
91
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