Question

1. Determine fio, fic and fus with ra = 100 ks Identify the actual cut off frequency. 2. draw the low-frequency response of the network using Multisim. 3. Discuss your results. 20 V Cw Cw 4 pF Cod = 8 pF = 6 pF C = 12 pF Com = 3pF 3.9 k92 220 kΩ 6.8 uF HE loss = 10 mA Vp = -6 V 1.5 k92 WE 1μF + 5.6 k 2 68 k22 21 2.2 k92 10uF +

Answer 1

Given J-FET Common sources ampuiffex De equivalent 400 3.9k1 quo a "ia 20x 68 220+68 Naa = Uo720 S Raa nhung 2.2k1 IDs tha Input kul equation vaca & Raca Vas - Toshg=0 vas Vac - Ioghs) as = uota - ID (2200) Ios Ioss [1- Vaso Up

loss - 10 A Up==bV Tos • Ipx; [+ น- 2200 ps 6 on 125 3+ 6,48644 - 36.41 164] (148864 -36.148) 100 โos (1-13แ) 2แแชร) x 155 15434326 - 16ใps 9 13uuuuuu16 + 3.192184 = 119.215 ใกง 13นนน!-41% - 140-2259s + 30921t4 -0 on solving above equation. 105 1.18u7A (68) 39 54u8 7 8 - for Ios= Tol8UMA Nasa Van IDs Rg )

Vas 4.22. [+lux 2.2] =-11.osur volts Wash > lvpl a no current flows So lps & T. 18 UMA -> for Vas 0.42-13.30u94) for Log 3:3048 ma 4.72-7.97056 Nas = 2.55056 volts Hasta luel current can flow. Vas 1-2-55056 V so IDs 3.30u8mA 9ma Trans conductance: avas slipss li- was ? duas Im = 2I oss vas Vp

2 loss ใน) - 10 lup) VI oss 9105516 > 40 x 1px16,330uf xไอ้ 6 Im 1.4LG2 14

y low frequency ac- equivalent of cs amplifier Smugs Roig a a D C3 + w Vgg B S gro w 2 R = 5.6 kr Resige 105HA Rg = 2.9 kn Vdg = looka RO = 3.9kn 2 = 104f C₂ =608uF Galuf Low cut-off frequency warot ci [ca & la care soc] Req, Wodit G is Req= Raathsig w Risig Raa

99bball 5 H3 920x64 xlb (90445yxth Rea - 51. นน 40 Rega Rsigt Rate = 105KA + 51.Juu Ku Rea - 55,uuu KJ 91x53.uuu!! x 16 ? Rege 3 10 25153.uuy วน 2 ร.9149 kg Wodot C [c, and ca] are shorted ** 3 (ในปี

dallio 100 x103 || 3-94003 ما 3 100x3.9 X10 (100+3,9) x103 100X 3,9 x10 103,9 Vallho = 3.753 kr Req Vallhot R 3. 753 %10%+ 506x103 Req = 9.353 kn fuel 9X+9= 353 x 203X 6081106 3 10 2.5024 Hz XX 9.353x6.8 1945 frc = 9.5094 HZ

Song lys + y = Iq = I x n t go this Worot cą [ca and Cg are short ciscited (voltage source is removed) 9 mugs Psig + 86 L ม 1, 1x & Rolla الم → Kel at a Ny + I, - Ix = Iva Box Rg Kel et b It 9 mgs- I = I x - x + 9 m Vgs ked at a

Į - In-V Rg Applying Kul to outpult loop Roll R = R' X+ Ig & dg + I (Rolla] ac + t VE * 55(:ܪܗܪ ܀ RS from input KOL Ny = - Vgg substitute in above equation NX- [Ix 3 - 9mx] 4s + (Ex - with a 14[+ + Hals + Som 925 + RU *[Last Ri] Rs Rs.

AR r Fe 1 + Pas[gm]+ Coast R') RS Rs. ( MR) १२. [1+2a59.m] + (As+R) Ix V Ix Rs 12511+ 4533 ci + + २०१R, -t5-91153 २१०० - १.२१४५03 Rear 2200 (11 toox |.१.62] + (100 + १.१५१) 2200 १२०० u.lu931 Req, ११.5055[[११.४५) 103 - ५१1०8166 KA

fis 97 Reg Ca 90x497.8166x10x106 105 2T XU97.8166 If 37.20166 HZ be actuall cut-off frequency will the one which is atleast u to 6 times greater 6 times greater than the frequencies other fra= 29119 H3 frez 2.50 2u H3 the: 37.20166 43 es appaoximately equal actual cutoff frequency fis: 37.901bb 3 to

frequency response 1A)= Amar Ife: 37.2016 Hz -i fus Amax = -9m [full Rolle Roll = 2.298 ku 100 k1|19.998241 -2.246 ki Amax = - 1,99 62 x 4² x 2.246 Xhos Amay = - 4.30u / Amaxl= 4.30u 4 304 1 - l'Al indb: 20 log 4.304 = 19.677 db

-> 75 tis 1 7² . :: it f => fc the fe 20 Wit fel=1 f for higher frequencies. is facta fo √1tf ا<< the » Las fal fet lAld In indby 12 677 2b k 37.2016 Hz f > acts as high pass filter at low frequency gain goes on decreases as far