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# Fundamentals of Microelectronics

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4.4 According to Equation (4.8), we have I C = A E qD n n 2 i N B W B parenleftBig e V BE /V T 1 parenrightBig 1 W B We can see that if W B increases by a factor of two, then I C decreases by a factor of two .

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4.11 V BE = 1 . 5 V I E (1 kΩ) 1 . 5 V I C (1 kΩ) (assuming β 1) = V T ln parenleftbigg I C I S parenrightbigg I C = 775 μ A V X I C (1 kΩ) = 775 mV

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4.12 Since we have only integer multiples of a unit transistor, we need to find the largest number that divides both I 1 and I 2 evenly (i.e., we need to find the largest x such that I 1 /x and I 2 /x are integers). This will ensure that we use the fewest transistors possible. In this case, it’s easy to see that we should pick x = 0 . 5 mA, meaning each transistor should have 0 . 5 mA flowing through it. Therefore, I 1 should be made up of 1 mA / 0 . 5 mA = 2 parallel transistors, and I 2 should be made up of 1 . 5 mA / 0 . 5 mA = 3 parallel transistors. This is shown in the following circuit diagram. V B + I 1 I 2 Now we have to pick V B so that I C = 0 . 5 mA for each transistor. V B = V T ln parenleftbigg I C I S parenrightbigg = (26 mV) ln parenleftbigg 5 × 10 4 A 3 × 10 16 A parenrightbigg = 732 mV

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4.15 V B V BE R 1 = I B = I C β I C = β R 1 [ V B V T ln( I C /I S )] I C = 786 μ A

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4.17 First, note that V BE 1 = V BE 2 = V BE . V B = ( I B 1 + I B 2 ) R 1 + V BE = R 1 β ( I X + I Y ) + V T ln( I X /I S 1 ) I S 2 = 5 3 I S 1 I Y = 5 3 I X V B = 8 R 1 3 β I X + V T ln( I X /I S 1 ) I X = 509 μ A I Y = 848 μ A

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4.21 (a) V BE = 0 . 8 V I C = I S e V BE /V T = 18 . 5 mA V CE = V CC I C R C = 1 . 58 V Q 1 is operating in forward active. Its small-signal parameters are g m = I C /V T = 710 mS r π = β/g m = 141 Ω r o = The small-signal model is shown below.

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ch04sol - 4.4 According to Equation(4.8 we have IC = AE qDn...

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