Question

The transistor parameters in the circuit of Figure 4 are Beta = 65, VBE (ON) = 0.7V and VA = 50V. Determine a) Zi, b) Zo and c) Av using the Pi model. Be sure to draw the equivalent circuit in AC.

+5V 10k VO 8 330 w Zo 220K ୮ zi 100k 16k -5V

Answer 1


BJT AC- easi valut of BJT Cpi-model) dic с B IB f DVC + VCE Gr gearbe {ao V be B + OVE E VBE VIE Here Base B Base CA collector B - Base c Collector E → Emitter E Ic VI Emitter CUTE 25x1c3u) gn= lan IE = Ipt Ic В gm do Ic= = Biel ß → current gain VA To IE= Ie + P IB E = (B+ 1) IB В.

Given ciawit diagram Given that +5V Beta-CB)= 65 VBE CON) = 0.7V JOK2 ovo 3302 it VA = 50v 220k 12 8 PLOKR 16_2 71111 -SV first of all we have to do De- Analy sis : to find te for Ac-circuit : Se, for De- thalyis all capacitor will works as open circuit: Then the given ven Circuit will be follows: 09 +5v lokalite IB TOCK 3 16k_2IE TINT -51 By applying kVh in loop Lai 100k (Ig) + VBECoN) + 16kCle) 55-0 look(IP) + Vee CON) + 16k (3+1)(7e) #5=0 (: Je = (B+1) 76)

look TB +VBECON) +16% (B+1) IB = 5 100k 18+ 16K (B+] IB = 5 - VBECON) > Jo ( 100k+ 16KCB+0) lok(B) = 5-Voz Cow) IB = 5-0.7 5– VBE CON) look+ 1ők (3+1) M 100k+ 16K (65+1) 4.3 IB 4.3 100K+1056K 100K + 16k(66) 4.3 1156K IB = 4.3 1156X163 - 3.719x10-6A IB = 3.79X 106A So, we know that Ic=B IB Ic = (65)(3.71x109) * Ic=0-2417x103A Ac- Analysis : for AC-analysis we need Al-equivalent circuit diagrama In AC- aralyis all the capacitors will works as short-circuit De-voHoges AC- Bruivalent ciswit for given cirwit will be as follows: and all will work as Grounds Then the .

Ac- Carnivaleut circuit : (pi-model) B. 3302 w ovo f 20 122062 Vg Art Invbe & 3 look Voe lok-23 Zin E TIL Here gm Im = Ic - 9.669 X10 x10²S 0.2417x168 25x103 11 UT gm = 9.668 X10 38 ..Ar I w 65 9.668x10-3 6723.21-2 gn * Alt = 6723-21-12 50 206868.0.12 5o = VA Ic 0.2417X10-3 90= 206868 m need to apply at To find zi we all counces : some test voltage '2;' place by Let the voltage Supply Value bo vx and current that source Esiroving a by ila then [2;= VX ix

So, the circuit diogram will be as follows: ix Vy B ovo. of V Rc 1002 galbe San ao Whe 220K2 10K 12 7711 By applying kan at Vx node: Vx TÖOK 2 Vx + ix - an → Vx Vx [look + in in V ehte + 100K -in [ Tort) (look Vx ano XTOOK ix (6729-21) x 100x103 6723.9+(100X103) ATC + 100K -

VA 672.321 X106 6299.66 » 630022 106723.21 Vx la = 6300-12 Zio Vx ia 63002 2;= 63002 To find zo we need to abnove all source independendent test a sources from our apply ac earivalant circuit at voltage source at zo place. Let that voltage sunce value be Vy Then curent from that sounce be ty 20= Vy ily D) Grcuit diagram for finding 20 : ly 330-2 look Voe 1gabe {cdo loka Vy TH Here we can say TK that be =ov, since both sil terminals well conneced to ground

By applying Kch at y node: Vy Ny + galbe = y So JOKI iy Vy So + Vy IOK + In(0) → Vy do M Vy lok + jy Vy [to + yox] = iy + ܠ Vy to lick) Oktoo h! hi e Joxlok = 206868 X 10X103 COX103)+(200868 ly yokto - ལ Vy iy 2068.68X106 216868 = 9538.8952 E Vy ly 9538.9922 20= Vy = 9538.89-2 ly > 20 = 95 38.94-12

Ava Vo Vs For finding Av we need to use our Al-equivalent circuit diagram; Vx B 3302 ovo f E on 100k-2 Voe? VE gulbe & Q0 (bk 22012 TI TI 771 from above circuit diagram: x = Vbe By applying Kch at Vx noder VX – Vs + Vx + = 330 100K-2 + + VX la Vs 330 330 JOOK Vx 1 II. i 330 + + VS 330 100K eu 330 =16[330) Va [the parent tama) = v[ [(3:03 x16+9+ (0.01x16)+(0.1467xo+) = vs [3.3 X10-3 VX [ 3.1884x103) - Vs [3:03 x108] Vx

Vx =Vg E 3.03 Xi -3 3.1887x10-3 =Vs (0.9502) by applying kch at Vo node: Vo + vo 220K lok + You + gabe ao vs [loekt nok + Bot] gombe o Toxic Oefe]=- 6.668x10°3) (0.9502) > v[6.1803) + (9.1865386X16-3) Vo + 10XL03 + 920x103 +(0.004545x103)+ (0.0048 34x10 Vo [0.109399x10-3] ca 9.1865336x103V5 vo Vs 9.1865-338 x108 0.109379 X1003 -83-988 Vo Vs -83.988 Av= No Vs = -83.988 Av= -83-988