HW6 Solutions

HW6 Solutions - Due Fri EE323 Homework 6 Solution Au/2004 5.1(20 points Refer to Table 5.1 on page 379 and note that > 0 EBJ forward-biased VBE =

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Due 11/19/2004, Fri 1 EE323 Homework 6--- Solution, Au/2004 5.1 (20 points) Refer to Table 5.1 on page 379, and note that: 0: EBJ forward-biased EBJ reverse-biased BE B E VV V > =− CBJ forward-biased CBJ reverse-biased CB C B V < So the operation mode is identified as follows: case EBJ CBJ Mode 1 0.7 BE V = , forward 0 CB V = , reverse Active 2 0.8 BE V = , forward 0.7 CB V , forward Saturation 3 0.7 BE V = , forward 0.7 CB V = , reverse Active 4 0.7 BE V = , forward 0.6 CB V , forward Saturation 5 0 BE V = , reverse 0.7 CB V , forward Reverse active 6 0.7 BE V = , forward 2.0 CB V = , reverse Active 7 0 BE V = , reverse 5.0 CB V = , reverse Cutoff 5.7 (20 points) Equations you may need to fill out the table are: / ,, , 1 B ET ECB C E B CS III I I I II e β αβ α =+ = = = = + Transistor a b c d e BE V (mV) 690 690 580 780 820 C I (mA) 1.0 1.0 0.130 10.10 73.95 B I ( μ A) 50 70 7 120 1050 E I (mA) 1.05 1.07 0.137 10.22 75 0.9524 0.9346 0.9489 0.9883 0.986 20 14.2857 18.571 84.1667 70.4286 S I ( 15 10 A) 1.0315 1.0315 10.922 0.2847 0.4208
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Due 11/19/2004, Fri 2 5.20 (20 points) (a) (5 points) 1 10.7 0.7 1 10 VV I mA K == (b) (5 points) 4 ( 10) 6 2.5 2.4 2.4 C V I mA KK −−− = ΩΩ Since β is very large, 1 1 α =≅ + such that 1/ 2.5 EC C I II m
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This note was uploaded on 02/21/2011 for the course ECE 205 taught by Professor Hemami during the Spring '08 term at Ohio State.

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HW6 Solutions - Due Fri EE323 Homework 6 Solution Au/2004 5.1(20 points Refer to Table 5.1 on page 379 and note that > 0 EBJ forward-biased VBE =

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