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8 BJT Basic Principle - Solution

8 BJT Basic Principle - Solution - Tutorial BJT Basic...

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Tutorial - BJT Basic Principle - Solutions Fig. 3.10 Solution (a) r av = 0.9 V 0.7 V 8 mA 0 V I Δ - = Δ - = 25 Ω (b) Yes, since 25 Ω is often negligible compared to the other resistance levels of the network.

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Fig. 3.8 Solution (a) I C I E = 4.5 mA (b) I C I E = 4.5 mA (c) Negligible: change cannot be detected on this set of characteristics. (d) I C I E
Fig. 3.7 Fig. 3.8 Fig. 3.10

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Fig. 3.7 Fig. 3.8 Solution (a) Using Fig. 3.7 first, I E 7 mA Then Fig. 3.8 results in I C 7 mA (b) Using Fig. 3.8 first, I E 5 mA Then Fig. 3.7 results in V BE 0.78 V (c) Using Fig. 3.10(b) I E = 5 mA results in V BE 0.81 V (d) Using Fig. 3.10(c) I E = 5 mA results in V BE = 0.7 V (e) Yes, the difference in levels of V BE can be ignored for most applications if voltages of several volts are present in the network. Fig. 3.10
Solution (a) I C = α I E = (0.998)(4 mA) = 3.992 mA (b) I E = I C + I B I C = I E - I B = 2.8 mA - 0.02 mA = 2.78 mA α dc = 2.78 mA 2.8 mA C E I I = = 0.993 (c) I C = β I B = 0.98 1 1 0.98 B I α α = - - (40 μ A) = 1.96 mA I E = 1.96 mA 0.993 C I α = = 2 mA

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Fig. 3.12 Solution I i = V i / R i = 500 mV/20 Ω = 25 mA I L I i = 25 mA V L = I L R L = (25 mA)(1 k Ω ) = 25 V A v = 25 V 0.5 V o i V V = = 50
Fig. 3.14 (a) Fig. 3.14 (b) Solution (a) Fig. 3.14(b): I B 35 μ A Fig. 3.14(a): I C 3.6 mA (b) Fig. 3.14(a):

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