Question

help me please

Question: 20 V Cwi Cw 220 k12 4 pF C - 8 pF 6 pF Ces = 12 pF Ca - 3pF 3.9 k2 6.8 uF HE loss = 10 mA Vp = -6 V 1.5 k92 WE 1uF + 5.6 ks2 68 k12 w 2.2 ks 10uF Determine fio, fic and fus with ra = 100 k2 Identify the actual cut off frequency and draw the low-frequency response of the network using Multisim. Discuss your results.

Answer 1

vaca e Rua Vas - Ios Ry=0 ج Given J-FET common sources amplifier Dc- equivalent 900 wo 3.9k1 quo a 20% 68' Vaat 220+68 S Raa Maca - = 4otau 2.2k A los tha Input KUL equation vas Vaca-Ilog Rs) vas = usta - Ing(2200) Ios: Loss [1- last

loss = 10mA Up --bv 10x103 +4072-22001ps 6 100 IDs = [ite 1+0878667 - 3662667 Ias] Q [1.78667 -3 66.667 Pos] 100Ips 웃 (1078664) Â [36626671)- 9x l-78 667x 366-667003 100 IDS 2 134464_uu Io + 3.299189 - 1910.995 785 1 Buuuuuu Iž - 141.925 Thos +3.19918920 on solving above equation. Ios = 7.18UMA (08) 383048 mA - for Ios = 7o 18UMA Nasa Vace IDs Ry

vas: 4-12-[7. 1$ux 9.9) = -lloonud volts Wish > lvpl a no current flows So Ips & T. 18 UmA - for Vas= lof (3.3048x2.9) - for IDs 3.3048 mA. = 4.72-7.97056 Nas=~2,55056 volts Mast elupl - current can flow. Vas -2.55056 V So Ios - 3.30u8 mA Trans conductance 9m “ใps ovas o lips li-Vante? duas Im = 2 loss hol - Vas Np

(6) 2 loss นา ID lup) VI oss 9 10s519 (6) - x byา 32548 x16 6 1.3163 %

→ low frequency ac- equivalent of cs amprifier gmgs Rsig a a Cz + Vgg win Rua Ro w Rs la Yds = 100kn R = 5.6 kr Rsige 15th Rg = 2.9 kn RO = 3.9kr C = 104f C₂ =608MF C, aluf - how cut-off frequency warot ci [caq & la are soc] - o Req Wooit į is Req= Rantasig) Risig

bg Кл 720kalle 990x68x104 [220+68]x103 Reca 51.quu kr Req= Rsigt Race = le 5 kd + 51.9uuka Req = 53.4HUKA fra ax 2KX53.uuuxlox 106 2X Regal 1023 21x53.uny fa= 2.9779 43 -> Wooot Cg [c, and ca] are shooted ra RO no

100 x103 || 3.9/10 Falloo 100X3,9x109 (100 +3,9) x103 100x3.9 KIDS 103,9 Vallho = 30753 kr Reqi Vill Rot R 3.753 xlo? +506x103 Req = 9.353 kr fue QR Rea Coz 9X+9.353020² 68x106 100 2.5024 8Xx9.3534 618 frc = 9.5094 HZ

Werot cą [a and Cong ase Shoxt circuited is removed) (voltage source 9 mugs t IL & Rolla IX aaa دلم uu RS Kel at a J VY x +I- Ix toolt Rs Kel let b It Imgs- I = Ix-x + 9 m Vjj ked at c t I myys Im Ygs + I = I = I x x R8

Į - dy- atles Applying kul to output loop X+ Iq Ods + Į (Rolla- Ve + Imigs] &s + (1x y RS Rs - from input KOL y = =Vgg substitute in above equation Vy {Ix- - 9mY] ais I. - Vx ist Rg Rs. Yo [it pe sites topik ] - 7 (2+ 3

Pas+R 1 bu Ix I + Pas[9.J+ bastre) Rg Rg Cras the २. [1+24:59.7] + (55+R) Rs १२.[1+ 459.3 (As+R) V Ix + PORN २१०० Rol|R. =t-al|sas) - १.२१४५03 Req= 2200 (14 toox [.१/62+ ) (100+ १.११) ११०० १२०० 4.10934 Req, : ११.5055[११.32) - 1 103 Req : ५१1०8166 KA

fis 97 Req Req ca 2X X4 91.8166x 10x10 6 105 24 au 21.8166 37. 20166 HZ be y actuall cut-off frequency will the one which 6 times greater than the is atleast u to other frequencies fue- 2.50 2u H3 fraz 2.977943 fg = 37.201663 → actual cut off frequency as appoosimately fig= 37.20166 43 equali to

→ frequency response IALE Amar fue = 37.2016 HZ 1.1 - 3 ful Amax = -9m [rall Roll Re Roll = 2.998 ku 100 kr 119.998 Kell = 2.246 kl Amax = - 19968 *18² x 2.246x10 Amax = -4.30u / | Amaxl= 4.30u IA 40304 pl=j3zoruny tal indba 20 log u.304 = 12.677 db

-> 11- stell Jit Singh . lit the fe no ... it f>> fc Elal f for higher frequencies. if fefe the >> 11t fc as fd fc f | Alah lal indon 12 677 db 37.9016 H3 low frequency > acts as high pass filter at gain goes on decreases as far