tailieunhanh - Microelectronic Circuit Design Third Edition - Part III Solutions to Exercises

CHAPTER 10 Page 509 Vo = 2Po RL = 2(20W )(16Ω) = V Io = AP = Av = Vo = = Vi Vo Vi I A = = A I i = = = Ai = o = = RL 16Ω RS + Rin 10kΩ + 20kΩ I i ( A) Po = = PS () Checking : Ap = = Page 510 (i ) AvdB = 20 log(5060) = dB AidB = 20 log = 140 dB APdB = 10 log. | Microelectronic Circuit Design Third Edition - Part III Solutions to Exercises CHAPTER 10 Page 509 Vo yl2PoRL h 20W 16Q V A Vo 253_ x103 V o L V v A V I V 25 .31 A I V SV A ụ 58 o B l 16Q i ii RS Rm 10kQ 20kQ It pA p A _ t I. P 0 005V 0 167pA 4 79x10 Checking Ap o Ap Page 510 i AvdB 20log 5060 dB AidB 20log 140 dB APdB 10log 107 dB ii AvdB 20log 4x104 dB AidB 20log 169 dB APdB 10log 130 dB Page 511 i The constant slope region spanning a maximum input range is between vj and the bias voltage Vj should be centered in this range Vj A V V. vi - V and vi - V. For vj the slope is 0. Av 0. ii vO VO vo For vi 0 vd V VO 14V and Av 40. Vo AV 40 4V vO volts VO 14 V 1 Page 519 g11 20kQ 76 50kQ 0262 ụS g21 76 50kQ _ 1 . 1 .75 . oo 50kQ 20kQ 20kQ . Rin MQ A g21 g11 11 g12 g22 20kQ 20kQ Rout J 262 Q g22 Page 520 i P - A A- A - I --1--r 50k0 80 V 2 100W 80 - 40 V 40 5 ko A -00 80J -A 46 800 ri 40V 2 ỵ Oo J P IR 2 - W A - í40VY 50 Mp - X w í 80 í zzz 40 k 5k0 J l A 5k0 5k0 J 80 80 80J A -160 000 P - ỏ - 8 f A1 -100 W A - i4 5k0 5k0 - X 107 2 2 80 80 A Page 521 4. 4- 300s s 5000 s 100 Zeros at s 0 and s Poles at s -5000 and s -100. Page 523 4. w- 2n x 106 s 5000n -400 . . 5000n cnlTT_ --- Amid - -400 fH - 2-50 kHz A s 2n 1 5000n BW - fH - fL - kHz - 0 - kHz GBW - 400 - MHz 2 Page 524 A J 5 l- 50 I52 -4 J 52 - 2 2 4 52 - 20log - dB A J5 - à 52 - 4 tan -2 5 52 - 2 - 0 - - A .J 1 -50 112 - 4 11 - 12 - 2 2 4 12 20log - dB M j1 - 4- -4 tan 1 -2 1 12 - 2 - 180 - - 243 -117 Page 524 zz Av jrn 20 1 - a j 20 12 M2 ỵA j 420 - tan 1 1 - 0 - A . 1 1 20 12 1 - .