EXAMPLE 23 The emitter-bias configuration of Fig. 61 has the following specifications: %3D Solution: Ic = = 4 mA VR Vcc- Vc %3D
EXAMPLE 23 The emitter-bias configuration of Fig. 61 has the following specifications: %3D Solution: Ic = = 4 mA VR Vcc- Vc %3D
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Transcribed Image Text:• E:19
%07 l. LI
*Feedback bias: Change Rg to Rc in above equation
28 V
"Co- Hut. IC = 8 mA, Ve = 18 V, and B = 110. Determine Ro Rg, and R.
Solution:
R.
Ve = 18 V
Ico = Hc = 4 mA
VR Vcc- Vc
Rc =
Ice
28 V - 18 V
Ice
RE
2.5 kn
4 mA
FIG, 61
Example 23.
Vcc
Rc+ RE
28 V
= 3.5 kl
and
Rc + Rg =
8 mA
RE = 3.5 kn - Rc
- 3.5 kn - 2.5 kfl
= 1kn
Ice 4 mA
= 36.36 µA
110
Vcc- VBE
R + (B + 1)RE
Vce - VBE
and
R + (B + 1)RE =
Vec- VRE
R =
- (B + 1)RE
with
28 V - 0.7 V
- (111)(I kn)
36.36 HA
27.3 V
1I1 kn
36.36 HA
= 639.8 kn
For standard values,
Rc = 2.4 kn
Rg = 1 kn
Ry = 620 kn
Equations
Ic = Blp
VBE = 0.7 V,
Ig = (B + 1)/ Ic
Fixed bias:
Vcc-VBE
RB
Ic = BlB
Emitter stabilized:
Vcc- VBE
Rp + (B + DRE
R, = (B + 1)RE
Voltage-divider bias:
ETh - VBE
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