366 SEMICONDUCTOR DEVICES
EXAMPLE 7.3.3
Considering the circuit shown in Figure E7.3.3(a), find the state of operation and operating point
if the BJT hasβ=80 and other typical values of a silicon BJT at room temperature.
(a)
(b)
E
B
C
+
+
+
−
−
−
RB = 60 kΩ
RE = 4 kΩ
vBE
iB
VCC = 20 V
iE
iC
vCE
E
B
C
+
+
+− −
−
60 kΩ
4 kΩ
vBE = 0.7 V
iB
20 V
iC =
80 iB
iE =
81 iB
vCE
Figure E7.3.3
Solution
Let us check the state of operation through some preliminary calculations. Application of KVL
yields
vBE=vCE−RBiB=VCC−REiE−RBiB
If we assume the saturated state, thenvCE=VsatandiB>0, so that
vBE=Vsat−RBiB< 0. 2
which is in violation of the saturation condition:vBE=Vγ= 0 .7V.
If we assume the cutoff state, theniB=0 andiE=iC=ICEO, so that
vBE=VCC−REICEO∼=20 V
which is in violation of the cutoff condition:vBE<Vγ.
Having thus eliminated saturation and cutoff, the active-state model is substituted, as shown
in Figure E7.3.3(b).
The outer loop equation gives
20 − 60 iB− 0. 7 − 4 × 81 iB= 0
whereiBis the base current in mA. Solving,