Fundamentals of Microelectronics

Info iconThis preview shows pages 1–22. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4.4 According to Equation (4.8), we have I C = A E qD n n 2 i N B W B p e V BE /V T 1 P 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 .
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
4.11 V BE = 1 . 5 V I E (1 kΩ) 1 . 5 V I C (1 kΩ) (assuming β 1) = V T ln p I C I S P I C = 775 μ A V X I C (1 kΩ) = 775 mV
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4.12 Since we have only integer multiples of a unit transistor, we need to Fnd the largest number that divides both I 1 and I 2 evenly (i.e., we need to Fnd 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 ±owing 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 p I C I S P = (26 mV) ln p 5 × 10 4 A 3 × 10 16 A P = 732 mV
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
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
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
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
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 18
Background image of page 19

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 20
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. B r π + v π E g m v π C (b) I B = 10 μ A I C = βI B = 1 mA V BE = V T ln( I C /I S ) = 724 mV V CE = V CC I C R C = 1 . 5 V Q 1 is operating in forward active. Its small-signal parameters are g m = I C /V T = 38 .
Background image of page 21

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 22
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/03/2009 for the course ECSE 2050 taught by Professor Monahella during the Spring '08 term at Rensselaer Polytechnic Institute.

Page1 / 62

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

This preview shows document pages 1 - 22. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online