Question

1. Hallar la respuesta en frecuencia en bajas frecuencias del siguiente circuito Indique claramente la localización de los polos y ceros del sistema y la ganancia a frecuencias medias 33k B = 120 Ig = 30044 Sin 2220 2.90

Answer 1

A A common-emitter amplition is given below: + SV T & R=3341 & RC = 4.7kn IMF vo SRL=5.6kd Re= 92 x1 * $3.46 lei Fo Given B=120, IE = 30. MA At low frequency, all the capacitors C, Cauf CE Comes into effect. That capacitors leads to lower cut-off frequencies in the freauency response of the amplifier To find the frequency response at low frequency obtain a b-model of above circult. And use lap lace transform for capacitor is the 6 Hore, hie internal resistance of c.E amplifier. = $ ve = B VI = 120x 26x1623 - I Kn TE 3 6 x1556 Vy= Theremal to tuge

calculated by The relationship between IL A RID is Conent divider ruke, IL = - BIR X Peta Ret Ret the Hore The oupud Voltage, 2 Vo= ILX R2 from eq ♡ voa - B Ibx R2 RC Ret Ret tice - - PL Rc & . R2 + RC It ß Ib ' 3C₂C Retre) Put Ti = C₂ (Retrw Vo= - Re Re Retrc - x B Ib No=-(Rell R) s Ib x 3T 1 + $T, At input side, The voltage v' is calculated as v'= Ibx hie+ (Rell I SCE) (Ib tB ID)

v'= thiet Rexto ICE (4B) 26 x (+ 1 RE+ dece* V = (hiet RE (1+B) ) 1 + $CERE Hors | hiet RE (1+B) 17 1 + $CE REI Tz = CERE u= From the input side, vi Riires ♡ Apply Voltage divider rule, we get u va vix Rill R2 Roll R2 + Rit to v'- RillR2 x vi (Rillre+Ri) Tit Sci (R, 11R₂ + R: 1 v=/Rill vi RillR+RiT 1 Put Tz = Ci(Rilke +R-) ST va (Rilla 1 * 353_) x Vi (Rill Retri) T Q

Put value of eq ☺ in ③ we get Rillra (Bill Ratri/ s T3 x vis hier! / - xvi - I biet 1+ST3 RE (1+B) 1+ ST2 Ib= ROLIR (RillR2 + Ri / sto Its T3 N X Vi het RE (+B) 1 + $T, Put the value of Ib in ex. ♡ we get Vo = - (RellR2) FXBT you alle. Rret 11 +3T) e festa ( Rollra ) (ST3 RollRetri / (+ STi) X Lt Vi hiet RE (1+B) 7 +3 Tz) Put the value at resistenics af in abure eg., we get Vos -921184 ST of Tz +13) ST Tict Re Ci41) ****, *(1+ata) v; (1 +3T2) AS RECITA) o hic tot Vo= - 22 1184 (its T2) 3-9x103 (1 +1201 s TiTo [ +3 Ti) (1+8 oz)

- 0.47x TiTzx (l+ST2) 8² Cits Ti) (1+5 T3) Hore, To= Ca ( Rc + R2) = IME( 4.7K+ 5.6k) Ti = 10-3x1013 wie = = 97.11 Taz CERE = 204F X 3.9K Tz = 0.078 I w2= 12-8 sec rad /sec Z Tza Cil RillR2 + Ri) Elu (331122 + 5) x103 T3 = 0.6182 sec ws = 1/2 2 55 ras lecce eroes of the trans for function, 3=0,0, -1/16 s = ,9 - 1908 Poles of the trans for function, $- - --97-13-55

-97-1 -55 - 12-8 l y all Calculate mid frequency hain.. . To calculate hair at mid-frequentes, Capacitors are shorted coreuiled I because of their less reactance. So, Xc; = XC2 = XCENTO The h-modd at mid frequency i → . Ib Voed &chic Rilla § } Rc 8 R2 Vo ABIL by-pass capaciter The effect of Re is 360 because of Shosted. (ic XCE =o) Now at output tide, The output voltage, Vocap Ib (RCH Re] VOS-B(Rellez) Ib

At input vide, Voc = Ibx hic G Apply voltage dinder blo vi & Voe, - o + o Vi r ilirali-hie Ube = Vix R Rall hie . Rit Rill Rallthie From eq. G um hie - So Ibx hie= Vi (R | Rall hic) Rit( Rillkallthic) Iba vix (Rillkall hie) Rit (Rill Rallhie) Since, RI/R2 = 13.2k1 Ib= Vix chie - ble (Ritchie) thie = lka Rill R₂ > hic, Iba Vi So R II Rall hic Rit hic & Put value of 0 in eg. ①, westet Vos - P(Rell R2.) x Vi - B (4.2k f5.6k) Rithie 5k+ Ik Mis haunt the one - 512 hie