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 (BE 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 (BE loop), applying KVL gives
4 20 . 0 and hence 165 Applying KVL to the output loop (CE 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 (BE 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 (CE 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 (BE loop), applying KVL gives
1 120 . and hence 0
2.5 For the output loop (CE 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 (CE loop), applying KVL gives
9 5 0.4 Using voltage divider rule,
2
6 2 3 0.75 For the input loop (BE 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 BE loop, Applying KVL to the CE loop,
and this gives,
Substituting this to the KVL equation obtained from BE 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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 Transistor, Resistor, Harshad number, Noncototient, loop, be loop

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