Electronics-I Lecture 25

Electronics-I Lecture 25 - Example5.10 =100,vBE=0.7V...

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Lecture 25 Electronics I 1 β  = 100, v BE  = 0.7 V V BB =15(50)/(100+50) = 5  V   R BB = (50)(100)/(50+100)  = 33.33 K V BB  = I B R BB  + V BE  +   I R E I B  = I E /( β +1) I E  = (V BB  – V BE )/[R +   R BB /( β +1)] = 1.29 mA I B  = I E /( β +1) = 0.0128 mA Example 5.10
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Lecture 25 Electronics I 2 V E  = I R =1.29x3 = 3.87 V V B  = V BE  +   I R E V B  = 0.7+1.29x3 = 4.57 V I C  = I E α  = 0.99x1.29     = 1.28 mA V C  = 15 – I C R C V C  = 15 –1.28 x 5 = 8.6 V V is higher in potential than  V by 4.03 V, Reverse  biased so Active mode Example 5.10
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Lecture 25 Electronics I 3 Example 5.11 V B1  = 4.57 V,     I E1  =1.29 mA I B1  = 0.0128  μ A, I C1  =1.28 mA Assuming negligible current  I B2   flows V C1  =15 – I c1 R C1 V C1  =15 – 1.28x5k= 8.6 V V E2  = V C1  + V EB(Q2)   V E2  = 8.6+0.7 = 9.3 V I E2  =(15 – 9.3)/2 = 2.85 mA I C2  =  α  I E2 = 0.99x2.85=2.82 mA V C2  = I C2  R C2  = 2.82x2.7= 7.62 V V C2  is lower than V B2  by 0.98 V,  reverse biased so Active mode  
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