Fundamentals of Microelectronics

Info iconThis preview shows pages 1–19. Sign up to view the full content.

View Full Document Right Arrow Icon
7.1 V GS = V DD = 1 . 8 V V DS > V GS V TH (in order for M 1 to operate in saturation) V DS = V DD I D (1 kΩ) = V DD 1 2 μ n C ox W L ( V GS V TH ) 2 (1 kΩ) > V GS V TH W L < 2 . 04
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
7.3 V GS = V DD I D (100 Ω) V DS = V DD I D (1 kΩ + 100 Ω) > V GS V TH (in order for M 1 to operate in saturation) V DD I D (1 kΩ + 100 Ω) > V DD I D (100 Ω) V TH I D (1 kΩ + 100 Ω) < I D (100 Ω) + V TH I D (1 kΩ) < V TH I D < 400 μ A Since g m increases with I D , we should pick the maximum I D to determine the maximum transconduc- tance that M 1 can provide. I D,max = 400 μ A g m,max = 2 I D,max V GS V TH = 2 I D,max V DD I D,max (100 Ω) V TH = 0 . 588 mS
Background image of page 3

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

View Full DocumentRight Arrow Icon
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
7.5 I D 1 = 0 . 5 mA V GS = V TH + r 2 I D 1 μ n C ox W L = 0 . 612 V V GS = 1 10 I D 1 R 2 R 2 = 12 . 243 kΩ V GS = V DD 1 10 I D 1 R 1 11 10 I D 1 R S R 1 = 21 . 557 kΩ
Background image of page 6
7.6 I D = 1 mA g m = 2 I D V GS V TH = 1 100 V GS = 0 . 6 V V GS = V DD I D R D R D = 1 . 2 kΩ
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
7.8 First, let’s analyze the circuit excluding R P . V G = 20 kΩ 10 kΩ + 20 kΩ V DD = 1 . 2 V V GS = V G I D R S = V DS = V DD I D (1 kΩ + 200 Ω) I D = 600 μ A V GS = 1 . 08 V W L = 2 I D μ n C ox ( V GS V TH ) 2 = 12 . 9758 13 Now, let’s analyze the circuit with R P . M 1 10 kΩ V DD 20 kΩ 1 kΩ I D + I R P R P I R P 200 Ω R S V G = 1 . 2 V I D + I R P = V DD V DS 1 kΩ + 200 Ω V GS = V G ( I D + I R P ) R S = V DS + V TH V G V DD V DS 1 kΩ + 200 Ω R S = V DS + V TH V DS = 0 . 6 V V GS = 1 V I D = 1 2 μ n C ox W L ( V GS V TH ) 2 = 467 μ A I D + I R P = I D + V DS R P = V DD V DS 1 kΩ + 200 Ω R P = 1 . 126 kΩ
Background image of page 9

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

View Full DocumentRight Arrow Icon
7.9 First, let’s analyze the circuit excluding R P . V GS = V DD = 1 . 8 V V DS = V DD I D (2 kΩ) = V GS 100 mV V DD 1 2 μ n C ox W L ( V GS V TH ) 2 (2 kΩ) = V GS 100 mV W L = 0 . 255 Now, let’s analyze the circuit with R P . M 1 R P I R P 2 kΩ 30 kΩ V DD V GS = V DD I R P (30 kΩ) I R P = V GS V DS R P = 50 mV R P V GS = V DD ( I D I R P ) (2 kΩ) + 50 mV V DD I R P (30 kΩ) = V DD p 1 2 μ n C ox W L ( V GS V TH ) 2 I R P P (2 kΩ) + 50 mV V DD I R P (30 kΩ) = V DD p 1 2 μ n C ox W L ( V DD I R P (30 kΩ) V TH ) 2 I R P P (2 kΩ) + 50 mV I R P = 1 . 380 μ A R P = 50 mV I R P = 36 . 222 kΩ
Background image of page 10
Background image of page 11

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

View Full DocumentRight Arrow Icon
Background image of page 12
7.12 Since we’re not given V DS for the transistors, let’s assume λ = 0 for large-signal calculations. Let’s also assume the transistors operate in saturation, since they’re being used as current sources. I X = 1 2 μ n C ox W 1 L 1 ( V B 1 V TH ) 2 = 0 . 5 mA W 1 = 3 . 47 μ m I Y = 1 2 μ n C ox W 2 L 2 ( V B 2 V TH ) 2 = 0 . 5 mA W 2 = 1 . 95 μ m R out 1 = r o 1 = 1 λI X = 20 kΩ R out 2 = r o 2 = 1 λI Y = 20 kΩ Since I X = I Y and λ is the same for each current source, the output resistances of the current sources are the same.
Background image of page 13

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

View Full DocumentRight Arrow Icon
7.13 Looking into the source of M 1 we see a resistance of 1 g m . Including λ in our analysis, we have 1 g m = 1 μ p C ox W L ( V X V B 1 − | V TH | ) (1 + λV X ) = 372 Ω
Background image of page 14
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
Background image of page 17

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

View Full DocumentRight Arrow Icon
7.17 (a) Assume
Background image of page 18
Image of page 19
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 / 85

ch07sol - 7.1 VGS = VDD = 1.8 V VDS &gt; VGS VT H (in order...

This preview shows document pages 1 - 19. Sign up to view the full document.

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