Given: Voltage-divider biased, npn BJT with 1. VCC = 11V 2. R1 = 40kohms 3. R2 = 7kohms 4. RC = 3kohms 5. RE = 1kohms 6. BDC = 126 ** Solve for: 1. VB = Blank 1 V; 2. VE = Blank 2 V; 3. IE = Blank 3 mA;
Given: Voltage-divider biased, npn BJT with 1. VCC = 11V 2. R1 = 40kohms 3. R2 = 7kohms 4. RC = 3kohms 5. RE = 1kohms 6. BDC = 126 ** Solve for: 1. VB = Blank 1 V; 2. VE = Blank 2 V; 3. IE = Blank 3 mA;
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Question
Answer. 1,2,3
![USE APPROXIMATE ANALYSIS
USE 2 decimal places, no commas.
Given:
Voltage-divider biased, npn BJT with
1. VCC = 11V
2. R1 = 40kohms
3. R2 = 7kohms
4. RC = 3kohms
5. RE = 1kohms
6. BDC = 126
**
Solve for:
1. VB = Blank 1 V;
2. VE = Blank 2 V;
3. IE = Blank 3 mA;
4. VCE = Blank 4 V;
5. VC = Blank 5 V;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77d0c56c-6985-4220-9314-742fe3066780%2F4764ffaf-575c-4fda-849f-247b94efd872%2Fmmd5y6j_processed.png&w=3840&q=75)
Transcribed Image Text:USE APPROXIMATE ANALYSIS
USE 2 decimal places, no commas.
Given:
Voltage-divider biased, npn BJT with
1. VCC = 11V
2. R1 = 40kohms
3. R2 = 7kohms
4. RC = 3kohms
5. RE = 1kohms
6. BDC = 126
**
Solve for:
1. VB = Blank 1 V;
2. VE = Blank 2 V;
3. IE = Blank 3 mA;
4. VCE = Blank 4 V;
5. VC = Blank 5 V;
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