I need the answer quickly • E:19%07 l. LI*Feedback bias: Change Rg to Rc in above equation28 V"Co- Hut. IC = 8 mA, Ve = 18 V, and B = 110. Determine Ro Rg, and R.Solution:R.Ve = 18 VIco = Hc = 4 mAVR Vcc- VcRc =Ice28 V - 18 VIceRE2.5 kn4 mAFIG, 61Example 23.VccRc+ RE28 V= 3.5 klandRc + Rg =8 mARE = 3.5 kn - Rc- 3.5 kn - 2.5 kfl= 1knIce 4 mA= 36.36 µA110Vcc- VBER + (B + 1)REVce - VBEandR + (B + 1)RE =Vec- VRER =- (B + 1)REwith28 V - 0.7 V- (111)(I kn)36.36 HA27.3 V1I1 kn36.36 HA= 639.8 knFor standard values,Rc = 2.4 knRg = 1 knRy = 620 knEquationsIc = BlpVBE = 0.7 V,Ig = (B + 1)/ IcFixed bias:Vcc-VBERBIc = BlBEmitter stabilized:Vcc- VBERp + (B + DRER, = (B + 1)REVoltage-divider bias:ETh - VBE

The solution can be achieved as follows.

Answered: EXAMPLE 23 The emitter-bias…

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

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

I need the answer quickly

• 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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