Question

5. Determine the voltages at all the three nodes, Voy and the currents through all five branches, Rai - 6 MS2, RG2 - 4 MS2, Rp Rs - 5 k92. The device parameters are V, - 1.0 V,k,(W/L) - 1 mA/V?, and neglect channel length modulation. VOD +10 V Rai §

Answer 1

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VPD = lov VID Ral & 6MM & Ro=5kn + vah VDS X vas {4M {Rs=5kn z Re Vp = 1 Volt, kn=1mA/v2 Nmos nmos state (region) ON LINEAR SATURATION OFF Vas>,UT | Vastur | Vas>,UT | Udsk Vas-VT Vos), Vas- Vt iso es en aces. Van is kan vas. where, kn = Ź Ph Concu

Let Assume, In ssume, Is = drain current os = Drain to source veltage Vų = Gate voltage [sa source current ID = Is (when nmos is oN) VT = 1 velt. Apply voltage divider et va Vy= VDDX RG2 - 10x4x106 RG1+ RG2 68106+4X106 Ug = to xa Xtec - evolt loxtob VGTVT means nmos is on Assume nmos is in saturation region Is = kn { vas-Vr]? kn = 1mAlva

VDD = lov VID Ro=5k2 4V VDS ④ Vas YIS 3 Rs=sku RS= SKA Is = kn (vas. Ut) o Apply KVL in vas loop VG-Vas- Is Rs = 0 (Is=Io) Vas= VG-ID RS - from equation 2, place vag value in equation 2 Ip = kn (va - IDRs-V+)² Is = kn (VG - Up - ID Rs) ² (A - B) ² = A²+ B² – 2 AB

Is = kn [(Na-V)² + 1 8 23 - 2 (VG-14) Io Rs] Re-arranging : Vu=40, Talv, kn=1.5mA/v2 1o?- (75 + ) mA 10-6 ImaIo I ²_ 1.2 I (mA) - 1.2 (mA²) =0 Quadratic equation : Io= -bt5b²uac 2a Io + 1.2 + 51.44+7,2 I= 41.2 +2193 - 1.2 + 2.93 2 UVU U I = 2.065 mA, Io=-0.865 mA

check which is correct Is ID= 2.06 mA Vas= Va - IDRS = 4- 2x5X103 Vas = 4-10 = -6 No way if vas (-6V) then Nous will not on Io = – 0.865mA Vas= VG-IDRs = 4 + 86565 Vas= 4+4.32 = 8.32 u mean Ipa - 0.865 mA) Is= – 0.865mA) V4 = UV Vo = VDD-IXRO 10-4-32= 5.68v Vo=S.GOV

Let Assume, me, Ig= drain current In- os = Drain to source veltage u = Gate voltage sa source current ID = Is (when nmos is on) Vy = 1 velt. Apply voltage divider et va V = VDOX RG2 - 10x4X106 RGI+ RG2 6x106+48106 = to X4 X 106 = 4volt lexus VGTUT means nmos is on. Assume nmos is in saturation region Is = kn [ vhs-up]² kn = 1 mA/v2