Question

please answer 1-3 thanks

1. Consider an n-channel MOSFET with tax = 6 nm, 1. = 425 cm /V., V,=0.8 V, and W/L = 15. What is the v of k,? (4 points) Identify the region of operation and find the drain current in the following four cases. (4 po each) (Total 20 points) Use &= 3.9 for Silicon. a. Vos = 2.5 V and Vps = 1.5 V b. Vos = 2.0 V and vps = 2.5 V c. Vos = 2.0 V and Vps = 1 V d. Vos = Vps = 2.0 V 2. Determine the voltages at all the three nodes (6 points), Vov (4 points) and the currents through all five branch (6 points). Assume saturation region operation and verify by your results. (4 points) (Total 20 points). Rai = 6.8 MS2, R2 = 3.9 M2, Rp = Rs - 4.7 k 2. The device parameters are V,= 1.0 V,k,(W/L) = 1 mA/V2, and neglect channel length modulation. V = +10 V Roz & 3. A transistor for which Is = 10A and B = 200 is conducting a collector current of 2 mA. Find VBE points). Also find Ise and Iss for this transistor (6 points each) (Total 18 points).

Answer 1

Que lo 4) given data tox = 6 hm.. is to un = 425 em2/v.s Vt = 0.80 W2 15: 00 Sun Vers = 25 V on Nos = 105U.. Nos 2. Vers - VT 7 Saturation region Vers LNT 7 cut off region Vos c Vw-va triode region __ (vers) in Breite wat Ups = 2.5-0.8 . Vos = Versa V = 17 vitin L ! c.21.0 4 x 10 kna? kn = Un Cox w Coxa - Cox's 3.9 x Eo Cox tox = 3.9x1:04 X10 in 2.3.45 810-4 floma Cox = 3.45 X 10" - 6 X 1619 = .575 X 10-20 = 5.75 m Hence it. Vus - Vay Nos Lit is en triode region ID = KN [(vers - Vt.) VAS - Vos 2356 {[l:})x (105) - (15) kn 200 L = o em Cox w .425 x 10:4 x 5.75X103215 3.666 X10-3 A. = 3.666X16? [ 1255.-1.125] = 3:66 X 1023 [1.425.]. 3 12 ) 3.66 M A E s.224 mA ma 103

(6) Nors I Love Vos = 2.5 Viens - Vip 2: 2-0.8. 3 . In = 3.66X169 [(1.2)X1 - 42) = 13.66 X10"? [12-05] 3 3.6681023.p.709 ID = . 2.566 im A . Nps 2.1.2 Hence in saturation region for saturation region 10 2 kN (vers-V+) ? = 3.666 X16 3 11.22 (d) Vas = Vos = av Ans - Vers-Vr = 2-0,8 = 1.2v - Vos (= 2) = (Vusa Vr) (3:1-2) Hence it is in saturation region T = 31639. MAI ID = KN (Vus - Vt) (C) VGS = 2 N. Vois FLV Vys-Vigan 2-0.8 Iliev - = 3.666X1003 [ rez] 2 ID = 2.639 manat Nos (iv) L (Nus-up) = 1.2 Hence it is in triode region for this | 20 2 kw (Wars Vt) Vas (Vas)

Date L ., Vbo slov Voys 5 VA- VSS = 3.45 - 470o los Ro24.7kr i Rur Desa 3.645. - VGS - 4200 Gisme Nos = Von - Ios the w a = KN (vos - Vi) ² iterhindar dari { Rs = 4.7km 3. 645 - Vgs = J X10-3 ( Vers - ) 4700 2350 2 News Ver = NooX RU2 ( Rent Rur 3.645 -VGS = 2.35, L Verst- QVGS J 3.645-Vorsi in 2035. Vos +2.35 – 4.7 VGS virs (235) - 3.780's -1.295 0 vus - 1.574 Vers uss ao - 10 x 3.9 x16 10.7 xrob 07 Vus. Coroot) = = 3,645 v +1,574 # 141,574)2-4*1*es 2x1 Vs = Ini x Rs . 21.574 & 2.4774- 2 las in saturation region IDs = kN (Vas von 11$74 0.69 2

Ves (choos) = 0.789 .345 Vos = Von a Noo-los : (Rs + R p.) 310-1535x1037 4.7+4.7) 2103 0,442 v News, = 1.132 V, Vis, 2 - I 10-9.099 is V Vos = 4.971v_boy far saturation region Vus Hence Vars = 1.132 NON 2 4.971v tota mation from equation o via Ins a 3.645 - Vas - 4700u NDS 2 Voa Vs 4,971 = Vp - 2.513 Fó Loa 7,484 vol = - 3.645 - 16132 v 4700 20.535 mA Nos Vasavg. for saturation La Oregion a los X Rs 4.971 1:139-1 2. 0:53 5 x 10" X 4.7 x 10 14.971_2:139 |_715 25 I operates in saturation 4 Hence MOSFET s viejien

Dat. No - No Rur Oul. 3. Is a lot ß = 200 10-1132 618x106 le = 2 x103 VBE 2.868 we know that 6.8 810-6 1,304 el A . VE Is e VI Ic = . - VBE 2x10-3 1076 e 26x103 o- VG ___3,9 106 2x103 VRE796610-3 a el 4006 = -0.99 HALUS NBE/26x10 3 ... 2x1013 1 IG a Tert. Den = I t I Ve id In cal of in 2013 a.. NBEN 266103 = 11,3044 - 0,29 1.013 MA 0.69 31 4 13 12.3025 = UBE 26410-3 G = 1.013 MA veya 1.132 U Vs = 9:513 v Vna 7.484 VL VAE = 796.294 X 1003 21.304 et VoE 1796 v Von=4.991 v I 12 = -0.29 MA

Ise - 9 dc = f lo le = la tla LB = 2 810 31 29.01mA 200 a 10mA = 0.01 mA LE = Les Giorg NBE/NT e fes (Ise) 2.0L .796 e 26x10 2.0)x107x Ise IS E = loosima VBENT Io a Iso e __0101 01013 2 Is 3 X 2 Iso = syA