Question

Can I get help with part g through part r, please?

Given the following Common-Emitter Amplifier with Voltage-Divider Bias: The transistor is a Silicon device with B160 (Beta 160) and transistor r.= 30 kn: 20 V 34.7k LAL = Vou 35.10 31.1k LRE $3900 03 - Find the following quantities in parts a through where means Quiescent DC): a) Thevenin Voltage and Thevenin Resistance Re b) Base Current la Collector Current le d) Emitter Current le e) Transistor Collector-Emitter Quiescent Voltage Vera f) Collector to Ground Voltage Ve Collector Saturation Current la hi Collector-Emitter Cutoff Voltage Valand draw DC Load Line showing saturation, cutoff, and quiescent points Find the transistor emitter dynamic resistance, and then find the Note that he is the transistor ac resistance looking into the base of the transistor, that is 18+1) , where is the beta of the transistor Find the following AC quantities: Assume that the bypass or is CONNECTED as shown in the schematic above and find: impedance at base k] Input Impedance 2 1) Output Impedance m) Voltage Gain As m) Find the Output voltage Vos if Vis an ac sine wave with a peak voltage value of Vip o Drow the input and output voltage waveforms with peak voltages labeled 10 mV is REMOVED from the circuit drown above (without capacitor C3) and find: Assume that the bypass capacitor pl Voltage Gain Aw G] impedance at base Z input Impedance

Answer 1

voltage gain with Re resistor reduces than without feedback resistor(Re), but stability increases with Re.

Given P2160; 8 = 30k For DC analysis capacitances => Open the capacitors. offer s'impedences ---------- Vec-20V 4.74 glor R24 {3904 - Đ Thevinen vollage at Rg is o v th= Vect the Verx Rz 20x l' ~3.8V O RitR2 4.7+1.1 Rina Rill R2 2 4:7k|| 1.11 - 8900 equalent circuit at base of the transistor vec & glow Rath=890.2 Yic. [ +" the T3.80 IRT M Th +IB Vth 3900 Assunne transistor is operating in Active region and UBE=0.7V -3.8 + 890IB + VRE+ (3+1) ER390 = 0 IB = 3.8 -0.7 890+(16070) 390 - 48.68 UA o collector current I c = BIB 160GIB = 7.788 mA @ Emailler current IEQ = (B+1) IB ~ 7-837 mA Τ Τ 2ον I coke Apply KVL to outputp loop. -9Q + Rclcat Ve Eqt IEQ 390 = 0 Veeu = 90 - Rc Icq-IEQ390_985V to (6+1) IR = I EQ NEQ RE

☺ Vc = Vceat Re Te Q =91857 7.83X16*3905 12.912 V Collector Saturation current (I csat) itp. Siste ( Assume Veesatz0.80) [" olp side ( Assume VCES D23) histratuBest Rth -20+ Rc Ic AVC Esat tigre=0 c. Rc - I 1 VT & RE 890 + +61 396 – 3,670.81 LB Riht (Btore Blindat 47-1871A ilp side: -3.8+ Ryth Is + VBE Sat + JEREMO From © & eliminate IERE 0-2 -20 + RcIct VC Esat +3=8 - Rth IB &VBEST=0 -20 +ReIcsat + 0.2V +3.8–890x Icsak 0.8=0 Rutes 6+0-1+smX Po (910-890) I sat - 20-0.2-3.8+0.8 Icsat = 16.8 ~ 18.5mA 910-5.5625 TIesat = ls som sat=18.5mA Ic i I csat 7.786 MA 021 UCESCA 9.85v Vce Cute CE (1) rear 30k .6mv = 26m ~ re= i X8/6/4 - 3317 7.837. MA hie - Vt ZL 30 (B+1) Be ~535

then Assuming bypass Capacitor is 'Shorted. equivalent corcuit. ac SR sex ş. Be lezb ® Zi - - RiliR! CBtijde = (4.7k)||(1.1K)11(5358) = (890)||(535) ~ 3350 » 2b=input impedence at the base = reli+ß) S =5351 Zo = Rollro- R dira gon (91011 30k) . ~8832 e voltage gain (Ärz vout = current gainy out input impedence Vin = B x (Rel ro) B R z 160 x 883 11R241 (2+1) re 335 -9422 335 for a peak ile sinusoid of 10 mv olp voltage = Arxiomu = 4220mu = 4.22 0. o ipon (Wollfen) i P loma a time a lomal- /p 4.2 / - 4-22VF

Assuming by pass Capa cites is removed ac equivalent circuit becomes e n I + (p+1) re på 2008Rc... (Rilles Blik T RE Y re 26 Uib (BtiDreibt (B+)Reil ih 26 Zin. Zb (i/o impedence at base) = (B+l)ret (8+1) RE - 535+(161) (390) = 63-325 kr. o imput impedence (21) = ; - (R, URs)" (CB+1)6et(B+RE] = (890) 11 [63-3251] ~ (RIUR2) = 8902 D 12 Ava Current gain x output impedence input impedence. = B x ZocRc) = 160 x 910 E 890 164 890