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(2)

iB

iC

100 μA

100 μA

10 mA

80 μA

BJT in Saturation Region – Example 2

17

0

60 μA 40 μA 20 μA

0

5V vBE

Input Characteristics

20V vCE

Output Characteristics

BJT in Saturation Region – Example 3

In the CE Transistor circuit shown earlier VBB= 5V, V = 0.7V BE B , RBB= 107.5 kΩ, RCC = 1 kΩ, VCC = 10V, β = 400. Find I I C ,VCE,

and the transistor power dissipation using the characteristics as shown below

By Applying KVL to the base emitter circuit

IB =

VBB VBE R BB

= 40μA

Then I and V But V C CE CE

= β I B = 4 0 0 * 4 0 μ A = 1 6 0 0 0 μ A = V C C - R C C * I C = 1 0 - 0 . 0 1 6 * 1 0 0 0 = - 6 V ( ? ) cannot become negative (since current can flow only

from collector to emitter). Hence the transistor is in saturation

18

9

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