# Fundamentals of Microelectronics

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7.1 V GS = V DD = 1 . 8 V V DS > V GS V T H (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 T H ) 2 (1 kΩ) > V GS V T H W L < 2 . 04

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7.3 V GS = V DD I D (100 Ω) V DS = V DD I D (1 kΩ + 100 Ω) > V GS V T H (in order for M 1 to operate in saturation) V DD I D (1 kΩ + 100 Ω) > V DD I D (100 Ω) V T H I D (1 kΩ + 100 Ω) < I D (100 Ω) + V T H I D (1 kΩ) < V T H 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 T H = 2 I D,max V DD I D,max (100 Ω) V T H = 0 . 588 mS

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7.5 I D 1 = 0 . 5 mA V GS = V T H + radicalBigg 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
7.6 I D = 1 mA g m = 2 I D V GS V T H = 1 100 V GS = 0 . 6 V V GS = V DD I D R D R D = 1 . 2 kΩ

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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 T H ) 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 T H V G V DD V DS 1 kΩ + 200 Ω R S = V DS + V T H V DS = 0 . 6 V V GS = 1 V I D = 1 2 μ n C ox W L ( V GS V T H ) 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

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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 T H ) 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 parenleftbigg 1 2 μ n C ox W L ( V GS V T H ) 2 I R P parenrightbigg (2 kΩ) + 50 mV V DD I R P (30 kΩ) = V DD parenleftbigg 1 2 μ n C ox W L ( V DD I R P (30 kΩ) V T H ) 2 I R P parenrightbigg (2 kΩ) + 50 mV I R P = 1 . 380 μ A R P = 50 mV I R P = 36 . 222 kΩ

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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 T H ) 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 T H ) 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.

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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 T H | ) (1 + λV X ) = 372 Ω

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7.17 (a) Assume M 1 is operating in saturation.
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