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BJT in Saturation Region – Example 1

I n t h e C E T r a n s i s t o r c i r c u i t s h o w n e a r l i e r V B B = 5 V , R B B = 1 0 7 . 5 k Ω , R C C = 1 0 k Ω , V C C = 1 0 V . F i n d I B , I C , V C E , β a n d t h e t r a n power dissipation using the characteristics as shown below s i s t o r

Here even though IB is still 40 μA; from the output characteristics, IC can be found to be only about 1mA and VCE 0.2V(VBC 0.5V or base collector junction is forward biased (how?))

iC

100 μA

10 mA

80 μA 60 μA

β = IC / IB = 1mA/40 μA = 25< 100

40 μA 20 μA

0

15

20V vCE

BJT in Saturation Region – Example 2

In the CE Transistor circuit shown earlier VBB= 5V, RBB= 43 k Ω , R C C = 1 k Ω , V C C = 1 0 V . F i n d I B , I C , V C E , β a n d t h e t r a n s i s t power dissipation using the characteristics as shown below o r

Here IB is 100 μA from the input characteristics; IC can be found to be only about 9.5 mA from the output characteristics and VCE 0.5V(VBC 0.2V or base collector junction is forward biased (how?))

β = IC / IB = 9.5 mA/100 μA = 95 < 100 Transistor power dissipation = VCEIC 4.7 mW

Note: In this case the BJT is not in very hard saturation

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