Question

The amplifier shown in Figure below has Rsig = R1 =1 k12, Rc= 1 k-2, RB = 47 k22, B = 100, 1=0.54 mA, Cu = 0.8 pF, and f1 =600 MHz (a) Find gm and r7. (b) Neglecting ro, find the midband voltage gain from base to collector (neglect the effect of RB). (c) Use the gain obtained in (c) to find the component of Rin that arises as a result of RB. Hence find Rin (d) What is the midband gain? (e) Find Cin (f) Find fi +1.5 V Rc² cc

Answer 1

page-1 solution. DC anouysis for DC capactor will act as open docuit l.sv ß=100 Re kr uzkr IB nom t RB - - ic - 1.5+ Rc (FBtIc) +RBIBT 0,7= 0 | (Ic= $16) ] 1000 (I 18 +100113 ) + 4700u I p = 0.8 14800 of B=0.8 IR = 0.8 وهل5 IB= 0.8 148000 = 5.405 M A Ic = BIB = 0.5 405 mA Parta- I'm = V = 0.54053 25 X 10 -3 = 0.021625 let Vr = 25mv Sen ß - 100 0.02162 - 4625-346 - 4.625kr

page-3 For remaining parts mid band voltage gain for AC midband capacitor will act as short circut w 47K Rs RB RESIK RB can be booken into two parts by miller's theorem ISS min WWW V ĮRB — Тр ~RB Refik & RL IK A Au= Vice =-9mRo Vbe A=-9m Ro=0.02162 x 494-7 = -10.6954 R = RB – 47K Ro= RellRull RB Ro= Illik1476 Ro=0.5x47- = d.1947k 47.5 11.6954 = 4.0186k 8MP AI QUAD CAMERA not on realme 5 Pro

parte = Page 3 Rs=1k * UIK RPKR & R IK Vsign 4.01866 Riss Rinput = (RB=4.01864 se= 4.625k RB 1182 4.0186 X 4.625 4.0186+4.625k 2.1502k ) Rinput = Ris = Ris - Rst RB 118 lkt 2.15024 = 3.1502K Part b - Voltage gain. be Vs Av = von = Voe. Vbe = 69mRo) Rí Rstri Ro= RB || Re/R= uzkllikllik = 0.4947k Rs=1k Ri= 2.1502k Im= 0.02162 Av = 60.02162X0.4947 X1000) x 2.1502 3.1502 Av – 7.300 laul = 7.3

page-4 part d - midband gain. Av= (-ImRo) Ri - 10.02162x 494.7) × 2.1502 RsARs 142.1502 AV = -7.3. Avl = 7.3 T partF- FH = ? fт – Coo MHz Fr= BFH = hreft B= hre=100 600MHZ = loofH FH = 6ooPHz— 6м HZ P code For = 1 2282 Catca] Se= 4, 625k Fr= 6MHZ Catch I 22 x 6x100x4.625 X103 - 174.27X109 Catcu = 5.7382 P.F caF 4.9382 P.F