9 BJT DC Biasing - Solution

# 9 BJT DC Biasing - Solution - Tutorial - BJT - DC Biasing...

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Tutorial - BJT - DC Biasing (1) Go over - Examples in Chapter 4 Sections 4.11, 4.12 (2) Chapter 4 problems - Numerical problems - 1, 2, 4, 5, 6, 9, 10, 12, 16, 22, 28, 31, 41, 45 Answers (1) 32.55 μ A, 2.93mA, 8.09V, 8.09V, 0.7V, 0V (2) 3.2mA, 1.875k Ω , 282.5k Ω , 6V (4) 5.93mA (5) (a) 7mA, 21V (b) 812k Ω (c) 10.75V (d) 136 (e) 0.992 (f) 7mA (h) 36.55mW (i) 71.92mW (j) 35.37mW (6) 29.18 μ A, 2.92mA, 8.61V, 13V, 5.12V, 4.39V (9) 5.13mA (10) 2.33k Ω , 133.33, 616.67k Ω , 40mW, 37.28mW (12) 21.42 μ A, 1.71mA, 8.17V, 9.33V, 1.16V, 1.86V (16) 2.43mA, 7.55V, 20.25 μ A, 2.43V, 3.13V (22) 15.88 μ A, 1.91mA, 9.12V (28) 6.2 μ A, 0.744mA, 7.91V, 7.07V (31) 3.32mA, 4.02V, 4.72V (45) 17.5 μ A, -13.53V

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Tutorial - BJT - DC Biasing - Solutions 1. (a) 16 V 0.7 V 15.3 V 470 k 470 k Q CC BE B B V V I R - - = = = Ω Ω = 32.55 μ A (b) Q Q C B I I β = = (90)(32.55 A) = 2.93 mA (c) Q Q CE CC C C V V I R = - = 16 V - (2.93 mA)(2.7 k Ω ) = 8.09 V (d) V C = Q CE V = 8.09 V (e) V B = V BE = 0.7 V (f) V E = 0 V 4. sat C I = 16 V 2.7 k CC C V R = Ω = 5.93 mA
2. (a) I C = β I B = 80(40 μ A) = 3.2 mA (b) R C = 12 V 6 V 6 V 3.2 mA 3.2 mA C R CC C C C V V V I I - - = = = = 1.875 k Ω (c) R B = 12 V 0.7 V 11.3 V 40 A 40 A B R B V I - = = = 282.5 k Ω (d) V CE = V C = 6 V

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## This note was uploaded on 01/11/2011 for the course EE 2106 taught by Professor Ytchow during the Spring '07 term at City University of Hong Kong.

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9 BJT DC Biasing - Solution - Tutorial - BJT - DC Biasing...

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