Question

1.In the circuit below, if VCC = 5V, ?? = 4?Ω, ?1 = 60?Ω, ?2 = 40?Ω, ?? = 3?Ω, ?? = 4?Ω, ? = 75, ?? = 80?, ??? (??) = 0.7?

a. What kind of amp is the circuit? Why is that? What are its features?
b. Find the VECQ and ICQ values. Draw the load line.
c. Draw the small signal curve circuit of the circuit. Find all the intermediate steps of your analysis and find the voltage gain, input and output resistors.

Į REŽ R

Answer 1

(a) common collector amplifier used between amplifier stages for impedance matching. It has high current gain and low Voltage gain.

DC Analysis: • open circuit all capacitors and draw the ciocult A p5V 60k Zakr -0.7 +YI cokr HD Thevenin equivalent between A, e points RTA = 60k 11 çokn = 60x40 . 60+40 - too R2=2462) 2400 kr Golr x 5v 60Ел ook rt coke = 3V Now redrawo circuit agaios

e ca akn yte = (+1) IG -3V of VCEO 2012 m Apply KUE COEB lopp. 418 + 3KR XIO (p+1) + 0.7 +26+R(IB) - 43 V = 0 IB = 2.30 3kx76 + 2CkX! 2.9.13 LA in fcq= 95% 9-13 UA s tea= 0.685 MA-75a = 0.693 mA VEDO = SV - 3* *(B+1) 18 - SV - 36 x 7-6 X9-13 UA PEcQ = 2.98V]

Transconductance 980 = TE vp = Theron al voltage : = gomv at room temperature gos. 3 0.685mA 26mu som = 26-68 ms. ] load line e vec = VECO * IERE (Veea, ted) ECC

(c) small segonal Analysies.. T5 - = 2.81 km 26-66 ms Reb v At out put sede Yo = VA 800 =lis. 2 KR 0.693 mA = rollrellau) xf@+1).18) - (115.29 kn) || airll Gir) (-76**s) Vo = -1280336k+8 - 0 at inpet side using voltage division rule Voz (RJI R21114) XNS (Rith Albo) +RS

2.52 kr Vo = — xus 2052kr +482 V = 0.387 Vs - 0.383 VS 208kr substitute ② 00387 xks Vo = -128.3 76KXS - 2.81kr Voltage galos A = 0 b = -17068 Rib = do directly Input resistance = 2.81K Dupat resistance oshort all independent voltage sources thers eB=0 5 opeos dependent current source (BtDip=0 . = 115,27 k liar Ro = roll Re) ~2.8k2