Question

F3 The below circuit is part of a two-stage amplifier. Ic, is 0.1 mA. a) Determine Ica and calculate the overall gain Volvi b) Use the method of zero-value time constants to find the dominant pole frequeney in K Ha. Capac tors: Ccs=3pF V T 100A Cu = 0.3pF Z2k CA= 4.2 pF ink sk - 3n

For part a, solve for Ic2 instead of Ic3

Answer 1

93 berary Ru per il olmA 3 Ink IT 2K. Pez VCI 5k Unz C01 プ - 7k les P = 0.1mA. Ve= - Pux Rc, = 6- CO.lx11) = 4.3V V = 162 4.3 U Une = lebo 0.7V Vez = 0.7 + Ub2 = 0.7 + 4.3 : 5V 6 - Ve 6-5 2x10² 0.5 mA ak factorget, Mannrasm Assme that ß = 100 Then Pcz :) Pec :600)xos mA. B+ Per ;0.495 MA

ani B. Vi 14 ) 100x 25. 3 2512 rar 100X25 5.05 kr Ich 0.495 Small signal model (short all de sources to ground ). 5K ovo. Ru VIL. s osta Ilin 25k Imun Siak. Rez I gm. Vh2 712 w W 77 RE 2kr I for first stage E amplifier A, Uin. vni Vinx 25 Vi= gm. Va Rei 0.83 llis 25+5 V -0.83 gm, Ru Vin gm, a 0.1 =) Ia UT 250 iso 25 A, a 3. 17 x 10 -0.83 x ? 250 :-56.44

For second stage CE am amplit. vi Jin Vo=-gma Vaz .Rc2 Nña Eva 6 s.orkal - Bin Igm vill in (Bilincs RE₂ =214 Vo= - Bin. Rea Vi= Pb.rhet (B+D PG. RE. v=Pn î raz + (8 +D RE) Limo Pina YA 2 + (8 +DREL Az V. - B. Req Thz + (8 +D RE2 - 100 X 710 - 3.38 5.05x10 + 10) * 28109 alwall gown Ava 4 x A₂ --56.44 x 3.38 Vin Av 190.7672 (h) For frequeny aanlys Ru=17k. 3 2kRE R& ov Vs (1 7K : Rez Ces

(2. wro JMV) cos miller capacitnice, it can Iranited 4 hish Frequency small anyml model, th PS SRL ch Teal 3pF (y = 0.3pf, LTE 4.2 pF Cu and Ca me me both ilp and lp side. ZRE Ces to (4, Culin Ch [THAI Cy, out= (y = 0.3pf Ailə gm, Ru - 3 ㅗ x 17 410 250 ly, in 0.3 [1+68) PF 20.7PF (win = Cay (1+(AI) (mont = Cay= 0.3PF (A₂l = 3.38 Lule Pn = 0.3 x4,38 PF = 1.314 PF

S 3 Y6% lo Cuild Cyout IR hel 0 ff, (due to its and Curen) ZRE, t fui - 4.1666 але, с R, = rallRs = 25k 11 sk (,= (t + Canin 4.2 pF + 20. 7 PF 24.9 PF fH, = 27 4.17 x 10 x 26.9810? = 1.532 MHz fH₂ (due to Cuy, out, (cs and Cupain ) (CA open cint), 1 the 20 2₂ C2 VE2 (2= Cui ont + C + (Main 0.3 PF + 3PF + 1.314 PF 4.614 PF RG C2 unt 36B+1) RED tu R2 - Reill Craz + (p+1) REZ) = 176 || (5.05k + 202k) 15.71 K ftipa 2.195 Mit 25x15.31718x 4.6148162

fitz I due to a fine eneris NIL Rei CK (3 = (t = 4.2PF 20 R₂13 Ros (RestRu) | me (2k +17+) || 5-ost. 3.989 kr 3 (8+1) RE: f43 . 28x3.89x18 x 4.2x102 9.74 M H₂ fina fita = due to Mout and les (4 = (4 out tles 21 RyC4 R4 = R(2 0.3PF + 3 PF - 3:3 PF 7 ke fite 25 x 9 x 13 x 3.301012 = 6.89 Mit cominent frequeny a Mindft, , the {fu, ifte, fun, fue} domimt frequens, fH. fr. = 1.532 MHz