Question

VDD VDD 10V R3 12k R1 >600k 1uF 2N7000* 1uF SRL 10k + Vs ( 0.1 Vpk 1kHz R2 > 450k R4 50 03 10uF Fig. 1. CS Amplifier: schematic For the circuit of the figure, determine the upper 3dB frequency fu analytically using the method of open-circuit constants (e.g., text-book, pp. 733-736, Sec. 9.4.3-4.). Use the following parameters of the transistor: kn = 20 u A/V?, WIL = 250, V1= 2 V, Cgs = 20 pF, Cgd = 10 pF). Neglect the channel length modulation effect.

I have a CS amplifier and I have to find the upper 3dB frequency using the method of open-circuit constants with given values.

Answer 1

Cinci't Diaguem Vop Kor 2011 Alu? 12 RB 33 Cgs = pop cgd = lopt Tinct nc Analysis Copautoizs are open, Use thorenin vas= Vg - is =up (450) = 6.2.8V I = Umlox u (Yos-Y)² Io = unlox W (VG-Us - Vt) ² 2 we know Vos q us Io = 20x250 (428-105K - 2v)? In = 5000 (2.28 - 50001n)? Ip: fope ( . 1 + 25x11 - 2 28 poIP) 10 - 25950 + 125x10 I, - 114x10 ID

125*10*1 - 114x70I, +252 50 =0 114x10 +1 a 114x100 If Using x - 62 Jozsac Io = 114x109 12996101? 12975X10/2 2x 125x10? ja 211 = 114x10 2508109 109.42 250 250xlo 10 (1444.503 11850416-3 (m) so ID = 0.47ma (m) 0.43 mA we know am 2km (Uus-en) Us= I RI - 047x5 2x Rout, arel - 2.35 conside Vers= 2.13 as s u USIRS =64345 2+52.15 vos-193 Im = 24 20x108 0.13 Vus=4.28-2.35 Im - 5.2410 Alv Vass=2013 AC Analysis Ground voltages, shorst capacitance = 1.93

midt fraquency analyn's Capacitors are shounted millor copautesouro u potrdod en cado a as there is is girocoded Ren} Tas ir IT test cgs Open cismit constants Basic ideo is to replace capacitance of interest with to get the desired resistance VIS Applying kuL v. - It Rth=0 v = I, Rth & Rena R. 11R2 R, R2 R, +R2 1 T 600x10x 45001 1050xips T = Rth lgs Tath = 257.14k & 257.14%10 x 20x10" 5142.8710} 1 1 - 109 599 32296.78 2 м 2731445142.8710 5 9 3.09X1O X10 - 3.094100 161 = 30.9KHZ

Now replace cad with as an Is Jest voltage a IX SI Define Vgs in terms of & FIT Vgs = vg-vs As us=0 gs=vg Va=IR - Vg- I Roch KCL at Disain (highlighted Det) It gets = 1x)-(17 hom small signal Analysis & ve- VoIP - I kth & ID = IP th & Ip - 2 substituting 6 in 2 = Ilth & Ro (It & Games) 4= I Rith + I Rp + am I Rth Rp. I= Rent Rot 9m Rth Rol I 2= Rent Rotam Rebro R = RullR3 - RLXR3 10x 12x10 22 x 103 5.45 KN = AD RetR3

2 - Rent Rotam Rth Ro وی را به پا x253 ا . و25 = F + 5 5 + 5 . 2 x = 282.8 + 72 87. 3 = 262 35 4 32 27. 37 م 72 102 . 37 22 2 : 2 Cad ?اهر 3۶۸ox ,وهه ور و دوم 29 - 3 . 4 / 5 / 2 و در و2. هجمہ : 1 5 . 03 های -۳ کد ms 02. دو2 ملعو2 = 2 + 3 12 : 38 3 . 8و 5 n ولا -6 ا + )3IT ا ( ر2 3143383 . 87 5( : /e9 ۹ 237.06 = 2 . 6 و 15 / 5 : 2 . 37 م - ماه FH2 22 . 3

3dB frequency = 20 log IBw/ =2010g 120300/ = 20*4.30 3 dbtro 86 30B Frequency=86
I had given a detail explanation of finding 3db frequency using open time constant method by providing the small signal analysis and proof of how it need to be done along with calculations.

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