Question

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3. For each of the circuits shown in Fig. P5.57, find the labeled node voltages. The NMOS transistors have Vtn = 0.9V, ka 6,-) = 1.5m A/V2.(20 points) +5V +5V +2.5 V ZIKA UL 21 ξΙΚΩ -2.5 V Willie (a) Fig. P5.57

Answer 1

Given Vth = 0.90 +250 ASV Kilw) = 1.5malur No, = 50 VG = 2.50 Va = ov Vso - I, (IK) = -2.5 Vse -2.5+ ID CIK) both transistors are in IV 2.50 Assuming Saturation Io = Io = K (W) (VGs Veran) Vas, ? o- Vse 1.5m (2.5-Ig(K) -0.9) = 2.5 -14) 1.6 - 16k) (1.33K) Is = 2.56 + 1b (100) – (3.2k)I, I (106) (4.53K) Is +2.56=0 Iss 66 LA con 3.868mA But for Io = 3.868mA, Vs = -2.5+ (3.86) - 1.36 > VG Vs > G > Violates saturation condition so, Io = 0.661 mA V = v = -2.5+(0.661) = -1.839 By Current equation for a 0.661 m = knw [ vas, - Van] Vas, = 2.5-V, 0.66% = 1.5m (2.5-V, -0.9)

0.881 = (1.6-1) 1.6- V F 0.930 = 2.538 V we which W₂ = -1.839 v) can observe that for both transistor Vegsth <NS is required for saturation. atsu VG = Us, = 5-I (1K) AM Øiş e Q I Since o o are diode Connected, they will be in Saturation. & lkn Io = kw) (vas-Venn 2 Ip = (0, -V₂ -0.9 -0 1.5m (UGS = VG-vs for a 2ID = (V₂ - V₃ -0.9 - ②. lom Vz - I, (k) -☺ v = S-Io(k) - 0 6 (5 - Iplik) -0.9-V2) from Of ( No - Io(IK) -0.9) ₂ - Istik)-09 5- Dotik) - DT-V₂ = 2V₂ = 5 V₂ = 2.5V

From ero ZID 2 1m = t 5 -I (IK) – 2.5-0.92 (1.6* (IK))" = (4.3k) ID Ip = 0.661 mA (or ) 3.868 mA But if I = 3.868 , VDS, I VDs, will become negative =) violates saturation ID = 0.66lma, V = 5-ID (IK) 2 5-0.661 = 4.339v N = I lik) = 0.661 V = 2.5V