Question

Capacitor values can be assumed to be large enough in the small-signal analysis frequency band. (Hint: Consider separating the bias and small-signal signals.) 1. Find the bias current flowing into the transistor (λ=0) 2. Find the small-signal voltage gain Vout/Vin value. (λ=0.1V^(-1))

Vpp=20V T VBias=3V Rp=10k Rg=90k 2 Vout VIN - HE Rsig=10812 B=1mA/V2 Vth=1V Rs=50012

Answer 1

Voorov VBISH=3V P-10 kr &&B=90k Vinth VB. Bimal? VH - ۱۷ Vic Rs=suon D a=0 Assume Transistor is in saturation region. 2 1 = (B) (Vas-VH) B=1MAN, Vus + VB-vs, VB =VBlas YB = 3V -> Ves = 3-vs 207 BV loun Visias: Vp RB 9015 vs ver Suor

so, Ip = (m) (3-V5-0 į (im) (2-vs) Ź Cim) (4+1-4V5) clearly, I = Is so = Is - Vs-o 500 from and Vs (im) (44 V7-405) llo? Tooo 500 Vetodesou) (V) Vs = Ý (4+vsus) 15_LV3+4-V)4 = 0 US-8Vstu=0 solving this -> Vs= 7464, 0.5358 But Vas > V "TH so, Vs =0.5358

So, Tof Vs 500 ID 0-535898 500 Ip = 0.0010716 II = 16 of 16 MA (2) To rol- 20 (0.1) (1.0916m) ro= 9331.84 gm = 2 IP Vers VGH = (2) (1.0716m) (3-0.5358-1 = 1.4638m So, small Signal model is Vei lp Vo t Rsig lokh genilgs 240 Vin Jus

so clearly, No e fom Vgs) (roll Ro) Vgs = van Vo --50 (roll) Vigis Vo. goo Cool120) Vin Vo Vin :-(1.4637m) (9331.84 X 10k) 933.sy took Vo = -7.0655 Vio gain = -7.0655