Question

Q2. Find an estimate for the lower and upper cut-off frequencies for the amplifier circuit shown in Figure Q2. You may use B - 100. g. 50 ms and r: -2 k22 in your calculations. The device capacitances are: Che = 25 pF, C = 10 pF and Cee = 1pF. For the low-frequency: where the are the time constants for the capacitors. PA 1 For the high-frequency: Στην (25 marks) 18% 0.47F 0.474F 221 100K olt 5 Figure Q2 Answer:

Answer 1

22/8 I 7.9kn 0:47AF mett GG RB Xbox/l18k) C2 22 118 I TH 0.4FM IK 100k Rin * Nozó CEL . • Low Frequency pole. when one capacitance is considered other is asteemedl apentiruited'. Resistance seen by Reqa (Roll Rin) Rin = (1+13) [rett*] (9.9k/) 103 %) rattr] Rega = 9 kr [rat 101 I Rin (1+) [ 148 103 kr 9x103 * 0.47*166 @tta Rege Ccq Its = 4.245 msec Resistance seen by CC2 + 100k Rega= { 0 + re 7 // ex} Re (0.114114) kit look | Rega O. 105ks tlook 100.14k Rega CC2 = 100.41 */03 760.47 *10-6 ta Tg - 47.049 msec Wizdo = tega - Epitoon have a look red sec WL 3dB 256.82 rad/sec

High frequency modelo рі y funga Vi are= NA HB = 2 = 19.8.2 101 a fetare re acelook Che Ebe Сce = 25PF = LOPE tk LPF 1 ㅗ = Vo Vi i tre 0.98 V/ [LC] miller approximation for Coe va we use VL Vo Vo Che 1 2 6 (1+r) e (th) Vo Vi Av= New circuit is I Blo Viº So I can he (HAN) P re = 19.8.2. Vo luke I ccelt Gellt ceg 1 Rin dk ceqz * ceqi ceqi Cbc tebellt av) = 10 + 25 (1+0.98) = 59.5 PF (RB // Rein) = 69.911103) K (9.9/1103) = 96.2. Reqs TEALA TAL Ceqi Req, 9x108x595*10th 535.5*109 sec 8pt = 53505 msec

Ceqa Av Cee ttbe [141 ={It has 194 +25PF 57.5PF Reqa RB Litß oc] l1 ex} ll look (00/05/1100) Rear = 1052 50410 nsee 105*5105*1672 Tp2 = Rega Ceq lol WH BAB TP. + EP2 535:5450410 TWHRAB 10848 #106 sad /see