7.3 BIPOLAR JUNCTION TRANSISTORS 365
10
Load line
8
iC, mA
iB=μ100 A
80 Aμ
60 Aμ
40 Aμ
20 Aμ
iB= 0
vCE,V
(b)
6
4
2
0
0246810121416
Figure E7.3.1Continued
Corresponding to 65-μA interpolated static curve and load line [see Fig. E7.3.1(b)], we get
vCE 2 =1 V andiC 2 = 5 .5 mA. Hence,
Av=
vCE
vS
=
1 − 9. 4
2 − 1
=− 8. 4
and
Ai=
iC
iB
=
( 5. 5 − 1. 3 ) 10 −^3
( 65 − 15 ) 10 −^6
=
4. 2
50
× 103 = 84
EXAMPLE 7.3.2
Given that a BJT hasβ=60, an operating point defined byICQ= 2 .5 mA, and an Early voltage
VA=50 V. Find the small-signal equivalent circuit parametersgm,ro,andrπ.
Solution
gm=
ICQ
VT
=
2. 5 × 10 −^3
25. 681 × 10 −^3
= 97. 35 × 10 −^3 S
ro∼=
VA
ICQ
=
50
2. 5 × 10 −^3
=20 k
rπ∼=
β
gm
=
60
97. 35 × 10 −^3
= 616