9 BJT DC Biasing - Solution

# 9 BJT DC Biasing - Solution - Tutorial BJT DC Biasing(1 Go...

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 15

9 BJT DC Biasing - Solution - Tutorial BJT DC Biasing(1 Go...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online