Electronic circuit analysis and design - Chapter 9

Tài liệu tham khảo Electronic circuit analysis and design | Electronic Circuit Analysis and Design 2nd edition Solutions Manual Exercise Solutions Chapter 9 pq Ad pi 4 i Ad 10J r2 -îoo kn . Acl a A 10 vi V vp V vt TTTÔ 0 025 mA 25 Zt 1U Kal ia ss îi 25 mA A a - io kft A A 20 kfl A 15 20 kft A 300 kft a A 2 A min 10 A max 19 so Aj b i1 max 4 t . . so f 1A fc 5 V Ad 103 -1------------ b. fi 5. i o 10 vc - Pa - pi Ad _ 5 - V 103 --------- C. fi pi f0 103 - f0 -2 V L V2 3 po 3 fo - A i P - P Vfl PJ - Pi Ad 3 - fi pi V c Maximum current specification is violated. We have We can write A A 10 kfl Want Agl -50 Set A A 50 left Acl - -50 -5 1 -5 A 4 _ A 1 __ L Ad A j A A 25 left Lee x A _12 -f 1 sT7ôr 1 l _ r_ O002 Ï-Aïï 5 x 103 A - left 12 x 5 x 103 J x - x 10 3 x . A x - 12 0313 25 kft A kft Electronic Circuit Analysis and Design 2nd edition Solutions Manual f Æh Ra Ra W - Z-vjj Ki - 3 Kj - îï mo Gï î2oo î7s i o - 1000 400 100 up 1500 iV mV Uo .4od 4 2 - V1 .4ûd t J - k l Q Vo -------VJ -1 or V V - - .4 o f- ûd V J . Vq Vj ij 3 and 12 5 Kj Itî We need 7 14. - . -W Jti Kl Ka Kî Set Rr 280 k l 280 Then Ri 40 kfi 7 280 A - 20 kil 14 280 R TT 80 kfl .4. N 1 v vn Vn J A fi. fi K. H fi I MO. R 3 1 3 1 Then fi -MQ 333K2 3 A . jX C A 11 -g I 5 vj Ri J Ri so that 4 Ri For p0 10 V t f 2 V 2 Then ii -5 50 jlA. fi 40 kQ I IOvh For vu 0 5vn Then Æa 160 kü we find Then V J Vû vn Vgfv î Vp 10v l 51 2 ta 2LZ . Rt 10-2 160 50 mA Test Your Understanding Chapter 9 Exercise Solutions So vo ij A3 Vi a x 10 3 A 4- A3 - kQ Let Aj kfl Then we want Rs _ Rf _ . 4 A2 Ai 5 Can choose Ai 10 kf and Rf 14 kil Rs A so t is iff 100 mA va tjAp We want - 10 - 100 x 10 e Ap Rr 100 kQ We may note that A3 3 Ap 20 2 and 2 Ri A 10 so that Ri _ Rf As Ai Then -y - -3 îi ÂT rrïdï ii 2 mA ÎlZl 2 x 10 3 200 V Vr t 777TH .