ANALOG ELECTRONICS 1 DR NORLAILI MOHD NOH

Transcription

1 24 ANALOG LTRONIS TUTORIAL DR NORLAILI MOHD NOH

2 . 0 8kΩ Gen, Y β β 00 T F 26, (a)deterne the dc ltages at the 3 X ternals f the JT (,, ). 0kΩ Z (b) Deterne g,r π and r? (c) Deterne the ltage gan ( IA Y / s )f ths aplfer crcut f a sgnal surce wth a ltage s anda2kω nternal resstr s a) dc ltage at, and. cnnected t X and an 8 kω lad resstr I α I β F I 0.99A s cnnected t Y. Z s cnnected t β F grund. I 0.99A 9.9μA (d) If the r s neglected, calculate the β 00 percentage f errr n yur calculatn f the ltage gan. I A I 099A I 0 I R A(8k) 2.08 I 0-0k 9.9μA(0k) A 2

3 b) Gen: β β 00 T A F alculated: I α I βf I 0.99A β F I 0.99 A g T 26 β 00 g A 00 I rπ Ω r 000.0Ω 3

4 2kΩ Y s 0kΩ g rπ rο 8kΩ 8kΩ X y 8kΩ 0 Y 0kΩ Z c) Y a g r //8k //8k s s (r π //0k)s g (r//4k) 2kr π //0k s I 0.99 A g T 26 β 00 g A 00 I 0.99 rπ Ω r 000.0Ω Ω IA 4

5 d) If r s neglected, A 0.038(0.5097)4 k Hence, the percentage f errr % 3.96% 396% 2kΩ Y s 0kΩ g rπ 8kΩ rο 8kΩ y 5

6 2. Deterne the sgnal ltage at the. The sgnal surce ges a ltage f 0 rs and has a resstance f 300Ω. 22kΩ 2 3 ut 22kΩ dc equalent crcut : 2 I n 6.8kΩ 560Ω 2 R L 6.8kΩ I 560Ω ac equalent crcut : 300Ω Gen β F 50; β 60; 0.7 and 26 T n 6.8kΩ 22kΩ g be rπ be rο R L 6

7 2 22kΩ 3 ut s 22kΩ 6.8kΩ 2 560Ω 2 R L Methd I R R IN() I I Assue ( β ) I R <<, IN() I I R R where I ( β )I R 2 Assue β F >> R,R βf R 6.8kΩ R R IN() I 560Ω R Snce β F 50, RIN() IN() 50(560) 84kΩ F 7

8 A cn rule-f-thub s that f tw resstrs are n parallel and ne s at least ten tes the ther, the ttal resstance s taken t be apprxately equal t the saller alue. 6.8k//R IN() (2) 6 6.8k//R 22k IN() (The exact alue s 6.8k//84k k Ω) IN 6.8k () 84kΩ, (2) 68k 6.8k 22k Snce R Ω βf α βf I A T 26A β 60 g I I I I A g rπ Ω A 22kΩ R 2 68kΩ 6.8kΩ R 2 I 560Ω R R IN() 8

9 6.8k (2) 6.8k 22k I I A g 0.456A r Ω π 6.8k//22k // rπ be 6.8k//22k // rπ 300 R 6.8k 22k.0989k T n Hence, R T Ω Ω be be n rs 300 kω n 6.8kΩ 22kΩ rπ g be be rο R L 9

10 Methd 2 TH TH 6.8k (2) 6.8k 22k R 22k//6.8k Ω I ( ) I (560) 0 I ( ( )560) βf I A I A R I αi 3.565A g I 0.37A rπ ( ) T β.670 kω g rs be n s 22kΩ 6.8kΩ 560Ω R TH ut 2 R L I TH 560Ω 0

11 nclusn Usng R (nce t s knwn that R ) R R IN() 0 R 2 wll result n a be that s ery clse t the ne btaned fr Methd 2 n

12 3. Select a nu alue fr the etter-bypass capactr f the aplfer ust perate er a frequency range fr 2kHz t 0kHz Slutn The alue f the bypass capactr ust be large enugh s that ts reactance er the frequency saple f the aplfer s ery sall (deally 0Ω) cpared t R. A gd rule f thub s that X f the bypass capactr shuld be at least 0 tes saller than R at the nu frequency fr whch the aplfer ust perate. 2 Fr ths crcut, 0X R X R Ω 0 0 X 2 π f 2πfX 2 π(2k)56.42μ F 22kΩ 6.8kΩ Ths s the nu alue fr the bypass capactr fr ths crcut. A larger alue can be used, althugh cst and physcal sze usually pse ltatns Ω 2 R 2

13 At the axu frequency f 0kHz, 2 π (0k).42 μ X.2Ω At the axu frequency f 0kHz, X.2Ω and ths alue s R /0 where R 560 Ω If the axu frequency s used nstead t deterne, then μF 2 π (0k)56 At the nu frequency f 2kHz, X π (2k)0.2842μ Ω 0 X 2800 and ths s > R. Hence, n rder t calculate the nu alue f the bypass capactr, the X ust be 0 tes saller than R at the nu frequency. 3

14 4. alculate the ltage gan wth and wthut the bypass capactr. 2 a) Wth capactr 22kΩ 3 n 300Ω 68kΩ 6.8kΩ 560Ω 2 ac equalent crcut : 300Ω n 22kΩ 6.8kΩ g be rπ be rο 4

15 300Ω n 22kΩ 6.8kΩ g be rπ be rο Open crcut ltage gan : a ( ) gbe r //k n n Assung r >>k, We knw be a g be( k) n 22k//6.8k//rπ 22k//6.8k//r 300 π n r π Ω be n Fr Questn 2 (ethd ), and hence g( k ) 0.754n a n Fr Questn 2 (ethd ), g 0.456A a

16 If there s n kΩ 3 n 300Ω 6.8kΩ 560Ω ac equalent crcut 300Ω n 22kΩ 6.8kΩ g be rππ be 560Ω rο 6

17 b) Wthut bypass capactr : 300 Ω b g be rπ be n 68kΩ 6.8kΩ 22kΩ 560Ω R' e rο If the effect f r s neglected, the pen crcut ltage gan : a n g be ( k) be e ( g ) be b be 560 be be ( g ) 560 r π ( g )560 r be π ( 0.456) be be ( ) k

18 300Ω n 22kΩ b g be rπ be 68kΩ 6.8kΩ R' 560Ω e ( 22k//6.8k//R' ) n 22k//6.8k//R' be( rπ R' ) kΩ b be 494.7n n n n Hence, f there s n 2, the pen crcut ltage gan s ery uch reduced. 8

19 5. nnect a 5kΩ lad t the aplfer shwn belw. Fnd the erall gan. ac equalent crcut 2 300Ω n a R be T 300Ω 22kΩ g be n R be Τ 3 6.8kΩ 560Ω g k//5k n n RT R 300 n T be( ) 6.8k//22k//r π 5kΩ 2 5kΩ be 0.754n a 0.456( )(0.754) 3333)( Fr Questn 4, when there s n lad, a Thus, pen-crcut ltage gan s the axu gan that can be acheed by an aplfer. Fr Questn 2, g 0.456A, rπ , RT Ω Ω 9

20 nclusn nnectng a lad t the utput wll reduce the gan. When there s n lad (.e. R L ), the gan f the aplfer s at ts axu. At R L,the gan s sad t be the pen crcut t ltage gan. 20

21 n 6. Fnd R, a, a, and a p. Gen β F β 75 and assue that the capactances reactances are neglgble at the frequency f peratn. T 26 and 0.7. Neglect r. rs 0kΩ 0kΩ 0 2 kω 0 kω n 5kΩ R ac equalent crcut R b b g rπ be be β b 0k//k Ω R RIN() R β 75kΩ F (k) 0R IN() 2 0k (0) 20k I 4.3 k 4.3A I 75 I (4.3) A R α b (76) r π (26) Ω kΩ R n br π ( b β b) k//6.0642k R b b kΩ b R r ( ) b π β 2

22 0 n rs 0kΩ 0kΩ kω 2 0 kω a n ( b β b ) br π ( b β b) ( β) r π( β) R ac equalent crcut n R b 5kΩ β g rπ g be rπ be β b 0k//k Ω 75 (26) I 22

23 a n b g rπ be be β b 5kΩ 0k//k Ω R ( b β) br π ( β)b b 5k ( b β) br π ( β) b( ) b 5k ( ) 5k

24 a p n b rπ g be be β b 5kΩ 0k//k R Ω0909Ω ( b β) br π ( β) b( ) b 5k ( ) 0909) 5k a a a Fr a cnfguratn, p a a r π r a Ω ( b β ) n R 76 r π ( β ) k

25 Usng Methd 2 t deterne I : 0 R Usng KL at the nput lp : 0kΩ p R 2 0kΩ 0k 20k (0) 5 0k//0k 5kΩ a a a I (5k) 0.7 I (k) 0 I (5k) I ( β)(k) 4.3 I 4.3 8k A I βi 4.65A rπ Rb ( ) kΩ R 5k//6.0924k kΩ 76( ) ( ) 76(4.8495k) ( ) a a

26 7. Gen : β F 250, T 26 and 0.7. Fnd the nput resstance, shrt crcut current gan and pwer gan fr the aplfer shwn belw. ac equalent crcut 00μF 56kΩ 2 μf 2kΩ 0 2.2kΩ 3 μf 0kΩ rπ 56 k Ω //2 k Ω be n g be 2.2kΩ 0kΩ n The 56 kω//2 kω s shrted t grund and can be neglected n the sall sgnal del. The analyss beces dffcult because the current surce s between the utput and the nput nde. Hence, use the T-del. 26

27 T-del e k k n R IN() re be Re e β R α F 250kΩ 250k >0 2k 2 k 0 2 k56 k I.0647A I.0604A ( ) r ge e 2.2kΩ R L 0kΩ e g g I T g r e r π I.0604A g g A 250 e a Ntce that the 56 kω//2 kω are shrted t grund and can be neglected n the sall sgnal del. 0 n g e(2.2k//0k) e g 2.2k//0k2k//0k ( ) a g r Ω R k// Ω R r e ( 2.2k) 56kΩ 2 μ F 00μF n 2kΩ 2.2kΩ 3 μf 0kΩ 27

28 e n k re ge e 2.2kΩ R L 0kΩ a s ge a 0 s e g k r e be KL at nde : e k e e k r e k r e e kΩ 2 μf 00 μ F n Shrt-crcut current gan f a aplfer. s a current buffer. 2kΩ Open-crcut ltage gan, Hence, fr a aplfer, pwer gan ltage gan. a a a 89.76(0.974) p s a g (2.2k) kΩ 3 μf 28 0kΩ

29 μnx W 8. Gen k A 2, GSQ 6.4, 2 L I DQ 2.75 and r O 50kΩ. Deterne g, R wth and wthut r O (cpare the result), and a wth and wthut r O (cpare the result). 2 2kΩ 0MΩ μf Slutn W gμnx ( GS t ) 2k ( GSt ) L μnx W 2 2 I D ( GS t ) k ( GS t ) 2 L 2 μf g ID k 2k g S 29

30 2 2kΩ G 0MΩ D 0MΩ μf μf gs ggs 50kΩ 2kΩ R gs KL at nde D : ggs 50k//2k (0M) gs g gs (0M) 50k//2k gs R S 0M g 50k//2k 50k//2k 0M 50k//2k g 50k//2k R 2.425M Ω 30

31 G 0MΩ D gs ggs 2kΩ R S Wthut r : 0M 2k 500 g k R MΩ Obseratn : Snce r >0R D, R wth r wthut r wll be qute clse. 3

32 a gs G 0MΩ D g 50k//2k (0M) gs gs 50k//2k 0M ( g (0M) ) 50k//2k.6248(0M) M g 0M ggs 50kΩ 2kΩ S a Wthut r, [ g (0M) ] M 500 2k k The tw results are clse. Hence, the effect f r s nt ery sgnfcant t the calculatn f a. 32

33 9. Deterne the utput ltage f 0.8 and r 40kΩ. Gen k0.4x0-3 A/ 2 and t 3. The dc ltage at surce s.2. Assue that bdy effect s neglgble MΩ 3.3kΩ 3 0MΩ.2kΩ 2 ac equalent crcut G D g gs gs 0MΩ 40MΩ S 40kΩ 3.3kΩ 33

34 G D ggs gs 0MΩ 40MΩ 40kΩ 3.3kΩ dc-analyss a g gs(40k//3.3k) gs ag g? ( ) W ( ) L μnx W L 3( ) g μ S n x GS t k g GS MΩ 0MΩ.2kΩ 3.3kΩ MΩ 0MΩ 3.3kΩ.2kΩ? GS G 0M ( ) 0M40M 30 6 S.2 GS g (.8).44S a.44 ( ) a (0.8)

35 0 (a)????? D del R sg sg D O GS G S sg R n R D DD SS 2 R G 0 ID 0.5A D DDIDRD 0 0.5(5k)2.5 5 D I R L Gen: DD SS 0 I 0.5A RD 5kΩ.5 t μnxw A A L 75 μ W ( ) 2 L μ W ( O ) 2 L 2 O 2 (0.5)2 I n x 2 D GS t n x 2 ID R D DD SS S O I GS t GS GS G S

36 (b) g? r O? R sg sg DD R D 2 I R n SS I μnx W 2 D ( GS t ) 2 L W g μnx ( GSt ) L R L g ( ) I 2 D GS t 2ID 2(0.5) A g r GSt A 50k I Ω D 36

37 R? R? a? (c) O R sg sg R D DD 2 R L R sg gs G ggs sg S r sg R R D L sg I R g R b bs n SS S sg r D T-del R sg sg R g D sg gsg R // D R L G 37

38 R sg S sg r D sg gsg R // D R L g sg R Gen : G R sg KL at nde S, sg g sg r KL at nde D: RL 5kΩ R g sg 50Ω r 50k Ω g A R 5kΩ D Snce χ s nt gen, assue the bdy effect s neglgble. R? a sg r R D//RL g r r R D//RL r R D//RL g r RR r r R D//RL g r g r ( ) 38

39 r r R D//RL R g ( r) g r r D L R r R //R ( gr) r R D//R L g r g r 50k R 50k 7500 ( 5) 50k k 50k 2 R D L R Ω R//R 7.5kΩ r 50kΩ g A 39

40 R sg S r D sg g sg gsg R D R L R' t O t 0 sg KL at nde S : g R sg t sg sg sg r sg sg g Rsg r r t g r Rsg r t G R sg R ' sg g S R r sg G D t gsg R ' t 40

41 S r D t R sg sg g sg gsg t KL at nde D : t sg r t sg g r r t g t r r g r Rsg r g t sg t t G R ' t t g t r r g Rsg r t 50k 50k 50 50k 6 t t t R ' k5528k Ω O O t R 5k// k 3.696kΩ 4

42 a sg R sg sg R R sg g sg D// L r sg S r r D g sg gsg R // D R L R G R D //RL g // r r g R D //R L r R D//R L r k k R R sg D L sg sg sg.05 R sg R Rsg sg a sg sg sg sg sg sg sg sg a a