Solutions _ Tutorial 6_Transistors.pdf

Solutions _ Tutorial 6_Transistors.pdf - Manora Caldera...

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Unformatted text preview: Manora Caldera Electrical Systems Tutorial 6 – Solutions 1. (a) Given that IB = 2.8A and IE = 325A, using 1 Using Hence, 115 , 115 2.8 0.322 89, Same as (a) and using the values given, 0.9889, 2. (a) Given that IC = 726A and IE = 732A, using 0.9918. Using (b) Hence, . , As , 0.9914 , As (b) 116. . 1 1.78 , 0.9918 121 . 6 , Same as (a) and using the values given, 0.9800, 49, 59.22 3. For the input loop (B-E loop) applying KVL gives Given that 4 4.7 0.7 , 10 1 4 52.98 This gives Hence, 4.7 10 4.7 10 52.98 10 5.23 1 Manora Caldera 4. For the input loop (B-E loop), applying KVL gives 4 20 . 0 and hence 165 Applying KVL to the output loop (C-E loop), 6 100 6 Therefore, 100 , As, 5. 100 50 165 10 5.175 and 5.175 50 165 10 8.25 For the input loop (B-E loop), applying KVL gives 5 10 5 10 0.2 0.7 0.2 1 . and hence 0 142.38 . 0.2 0 0.2 0.2 1 101 142.38 10 2.876 For the output loop (C-E loop), applying KVL gives 12 0.5 This gives 6. 0.5 12 0.5 100 2.876 142.38 10 2.876 2.005 For the input loop (B-E loop), applying KVL gives 1 120 . and hence 0 2.5 For the output loop (C-E loop), applying KVL gives 20 10 and hence 20 10 2.5 18 As 10 hence, 1.8 2 Manora Caldera 7. For the output loop (C-E loop), applying KVL gives 9 5 0.4 Using voltage divider rule, 2 6 2 3 0.75 For the input loop (B-E loop), applying KVL gives 0.4 Therefore, 0.75 0.7 50 and as 0.4 , . 0.125 0.6219 Hence Then, 9 8. 5 0.4 As the current through as Note that 9 5 200 0.6219 10 0.4 0.125 8.33 , it can be said that is same is the current that goes to the collector. Applying KVL to the B-E loop, Applying KVL to the C-E loop, and this gives, Substituting this to the KVL equation obtained from B-E loop gives, 12 Solving for gives, 1 5 100 0.7 10 1 100 100 12.296 10 5 4 12.296 12 1 5.791 3 ...
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