Question

Homework 1 Problem 4: Assume transistor parameter are Vtp = -0.4 V. 25,(W/L)2 = 15, (W/L)2 = 5, and 1 = 0. Hp Cox = 20 A/V2, (W/L). = V3V a) Determine lo, IreF, and Vsi, VsG3. b) What is the largest value of R such that My remains biased in the saturation region?

Answer 1

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Given data: Vip = -0.40; In Cox = 204A/V2; (W) = 25 ; (W)2 =15. (W)3=5 and VA 500. Given Ciruit is as follows: V-3v 2 from the figure ; it is clear that V+ = US9 = \s2 = 3V And Van = VG2 = "D=153 VG3 0 = U23 CGmund) TREF Sly.

Ime = Im3 = TREF ents of transistor Me and Transistor M3 ; hle gel Un cox ( W), [Was - 1 Mol] = Lun cox (]; [vas – Vip]* 9k (g) [3-va, Joujº: 2 x8 x[1-6 - 04/? 5[3.-14,-0-17- {15-04] Squaring on Both Sides; We get- with vs [22.6- Vy] = 152-04. (:'V4, =V83] Scanned with

YGL = 1.9V VGH = VG2 = Von = 153 = 1.9V. Tref = Imz - Mon cox [u]; [U53 – 04]?. =_x29x 5x [1.9-04]+=> 115-5x106A. Therefore ; Jeep = 115-SMA 186), - \s, - \ => 3-1.9 ->1.10. = (66) Vissiz = - VGz => .9-0 => 1.90 .

2 Io = Pa Ime - locox [w], [use, -14vel] - 20M XIS [1.1-04]+=> 73 SMA Yrosiz : VAS) 2 - Vip = (1.9-39+0.4 => Vo2 -3--0-1 VO2 => 2.3V. Value of R: Where Io = 73.5MA and Voz = 2.3V R = 2,35 = 31.29612 12.5

@Therefore ; Jo = 73.5MA Tref = 115.5MA- VSG = 111 VSG3 = 1.90. O The largest Value 04 R such that Me remains biased in Saturation Region : IR = 31.29KR Scanned with II

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