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1) how did the soution for the collector current become VCC/(Re+Rc)? please explain 2.)instead of using the boxed solution, would it be alright to use KVL instead to solve for the collector current? If yes, please show the equation for the KVL.

1) how did the soution for the collector current become VCC/(Re+Rc)? please explain 2.)instead of using the boxed solution, would it be alright to use KVL instead to solve for the collector current? If yes, please show the equation for the KVL.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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1) how did the soution for the collector current become VCC/(Re+Rc)? please explain

2.)instead of using the boxed solution, would it be alright to use KVL instead to solve for the collector current? If yes, please show the equation for the KVL.

R₂ 0 R₁ O + Vcc Rc IEC RE 3.26 In the circuit of Fig. 3-8(a), RE = 300 S2, Rc = 500 2, Vcc = 15 V, B = 100, and the Si transistor has ß-independent bias. Size R₁ and R₂ for maximum symmetrical swing if VCEsat ~ 0. For maximum symmetrical swing, the quiescent collector current is Ico 1 Vcc 2 RE+RC 15 2(300+500) -= 9.375 mA Standard practice is to use a factor of 10 as the margin of inequality for ß independence in (3.8). Then, RB BRE 10 and, from (3.7), and (100)(300) 10 V BBV BEQ+Ico(1.1 RE) = 0.7+(9.375 x 10-3)(330) = 3.794 V Equations (3.5) may now be solved simultaneously to obtain R₁ 3 x 10³ 1-3.794/15 15 3.794 R₂ = RB RB 1- V BB/VCC Vcc VBB -= 3kg = 3 × 10³ : 4.02 ΚΩ · 11.86 ΚΩ
Transcribed Image Text:R₂ 0 R₁ O + Vcc Rc IEC RE 3.26 In the circuit of Fig. 3-8(a), RE = 300 S2, Rc = 500 2, Vcc = 15 V, B = 100, and the Si transistor has ß-independent bias. Size R₁ and R₂ for maximum symmetrical swing if VCEsat ~ 0. For maximum symmetrical swing, the quiescent collector current is Ico 1 Vcc 2 RE+RC 15 2(300+500) -= 9.375 mA Standard practice is to use a factor of 10 as the margin of inequality for ß independence in (3.8). Then, RB BRE 10 and, from (3.7), and (100)(300) 10 V BBV BEQ+Ico(1.1 RE) = 0.7+(9.375 x 10-3)(330) = 3.794 V Equations (3.5) may now be solved simultaneously to obtain R₁ 3 x 10³ 1-3.794/15 15 3.794 R₂ = RB RB 1- V BB/VCC Vcc VBB -= 3kg = 3 × 10³ : 4.02 ΚΩ · 11.86 ΚΩ
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