Question

Q. 4. (a) Draw a small signal equivalent circuit by replacing the BJT with its hybrid-mt model for the circuit shown below. Also derive an expression for Zin when la is neglected. (10) + Vx

Answer 1

84) cirunt small signal BJT - small 1 Vio MIT om vi Given Gruit is Here Base and collector are short arcuited In Vx Amize Input Impedence Zx = x Ix. By Nadal analysis Ist vx ATI + Im Vx=0 Ve [ + qm ] = #x Zix Yx AT it gm ro CS Scanned with CamScanner

- DC quivalent You tlov UBE :0.7 V =z6mv. R, SHOKA Hok 4 le Ro 3skn 18 Reallen bok 123 3 bok lov lov 6V (upk/1605/ 24Kn 10x x 60 9 w x 686 Leto notho 6+3723 hu 0-7 56 X 6-8.7 s fahlst 151x5K) 23 Given ckt. Given f 150 c-open HK {5k osk 5-3 23 4.6.8 MAY R8+ (1+8)RĘ (244151x50K IC: 82B5) 150X6.84A 1.02MA & IB tac DC: 102 MA, u 6.8un +1.02mA 26 = 1:0268mA Tarcmsconductance Im :) - h2r 39:23m8s abmy. 150 CS Scanned with 93 m ) 3-89kn CamScanner upxbo. $ 245 pro Apply KUL RB :) (uoll601) bw: 2318H1) +0.7 +143)28N5K 5.3 77: Plam

Ac equivalent circuit Small Signal ego poder RS BS + va 4 Dana + RL, [VS. + ws. RE SALVO pijp simpedance Re: Pu/1821] 876.+®. 20 = 11+3) 2b vin: 2 79 + (1+B) [rellPL)22. = 10 + (14p)./ 85(182) : 3.82k+ (151) (Elli Die 25 Rib ving 4 3.89415 +151x Rib. 129-6Kn i's RP = PillR211 Ribe :-) Moll 6016)] 129.6le tak Ri, 20-25kn : 20.25TH 2oput impedomier pergu-25'll 2) M VS: 0.035 Vs. Bit RS 20.25 3 20.25+ UK No: 28114B) ( REJI RL) -) 2B(151) (s/l«) >; dB(151) () Vin: 88(197) + (148){RE 180) Mo. - IB1125.8;) h 181 3.824 15144 151%()) -) 28 (1996) 1958 28 129-628 y you 125,8 vo 125% 1296 129-6? vin X01835 VS VO CS Scanned woltage gain av 50.84 CamScanner VS

Ac equivalent clauit 151 small clann output Resistonte Ro: 11 RE/IRL THB Effect of R1, R2, RS not consided) 3.8211 SKUILIS If considered 25:3|156)!!!4. portat e RapR21183)) 11 8511 84 CŠ Scanned with1+B 24 SSI RO CamScanner