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Example 18-6 (Semiconductor charge carrier analysis) A silicon wafer (Group IVA) is doped with 2 x10º boron (Group IIIA) atoms/cm³, 1.5 x 101 phosphorus (Group VA) atoms/cm. Calculate the electron and hole concentrations (carriers per cm.) Given: n, = 1.5 x 101/m³ . %3D

Example 18-6 (Semiconductor charge carrier analysis) A silicon wafer (Group IVA) is doped with 2 x10º boron (Group IIIA) atoms/cm³, 1.5 x 101 phosphorus (Group VA) atoms/cm. Calculate the electron and hole concentrations (carriers per cm.) Given: n, = 1.5 x 101/m³ . %3D

Introductory Circuit Analysis (13th Edition)
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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Please help me understand where (1.5•10^10)^2 over 0.5•10^16 comes from to get 4.5 •10^4. The notes don’t make it quite clear so please help me out with this. thank you.
Example 18-6 (Semiconductor charge carrier analysis) A silicon wafer (Group IVA) is doped with 2 x101 boron (Group IIIA) atoms/cm³, 1.5 x 1016 phosphorus (Group VA) atoms/cm. Calculate the electron and hole concentrations (carriers per cm.) Given: n, = 1.5 x 101º/m³ . EX./8-6 Notice that the silicon is doped with both acceptor and donor atoms. Acceptor atoms 16 SILICON V Donor atoms N= 2 x10' B ATOMMS ni = 1.5 X/0 CARRIERS 1.5 x106 P ATOMS Think of acceptor & donor atoms canceling each other: Na – Na = 0.5 x 101/cm It looks like 0.5 x 1010 acceptor atoms/cm³ are added to the silicon. NET REESULT : Revised sketch: 16 GET Na = 0.5x10 B ATOMS CC P-7YPE SEMI CO NDUCTOR P>>n Na = 0 NEUTRALITY: Na + + P → pN, =|0.5x10 16 HOLES NEGLECT MAJORITY ;CARRIER CONC. MASS A CTION LAW: np = n; Notice: p >> n; checks out! Solve for n: (1,5×10°/cc)² in = ni MINORITY 4.5 x10 e- CARRIER 0.5x10'6/cc c CONC.
Transcribed Image Text:Example 18-6 (Semiconductor charge carrier analysis) A silicon wafer (Group IVA) is doped with 2 x101 boron (Group IIIA) atoms/cm³, 1.5 x 1016 phosphorus (Group VA) atoms/cm. Calculate the electron and hole concentrations (carriers per cm.) Given: n, = 1.5 x 101º/m³ . EX./8-6 Notice that the silicon is doped with both acceptor and donor atoms. Acceptor atoms 16 SILICON V Donor atoms N= 2 x10' B ATOMMS ni = 1.5 X/0 CARRIERS 1.5 x106 P ATOMS Think of acceptor & donor atoms canceling each other: Na – Na = 0.5 x 101/cm It looks like 0.5 x 1010 acceptor atoms/cm³ are added to the silicon. NET REESULT : Revised sketch: 16 GET Na = 0.5x10 B ATOMS CC P-7YPE SEMI CO NDUCTOR P>>n Na = 0 NEUTRALITY: Na + + P → pN, =|0.5x10 16 HOLES NEGLECT MAJORITY ;CARRIER CONC. MASS A CTION LAW: np = n; Notice: p >> n; checks out! Solve for n: (1,5×10°/cc)² in = ni MINORITY 4.5 x10 e- CARRIER 0.5x10'6/cc c CONC.
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