Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions xercise Solutions X5. ( β ).0 β 4. β 40. 0.0085 hapter 5 β 40. α 0.999 β 4..0 0.0085.95 X5. O 00 O n 3 β 0 or O 40.5 X5.3 0.7 430 β ( 50)( 3. 0)μ.3 0.453 3.0 μ 0. 453 3 ( )( 3.). 85 ( 0.453)(.85) 0. 838 mw P X5.4 ( 80)( 3. )μ 3.3 0.7. 400 β 5 0.8 ( )( ) 83 3.5 μ 3.3 0.8 5.5. X5.5 (a) (b) ( 0)( )μ 3.3 0.7 50 β 4 0. 44 ( 0.44)( 5). 3.3 3.3 0.7 0.7 μ 50 ( sat) 3.3 0. 0.6 5 4 μ 0.
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions X5.6 0 < 0.7, Qn is cutoff, O 9 ( 00)( 0.7)( 4) 0. 9 5. When Q n is biased in saturation, we have 00 So, for 5., O 0. X5.7 ( β ) So 3.3 0.7 3.6 μ 640 β μ 0. 49 ( 8)(.4) ( 80)( 3.6) β 8 ( 0.49) 0. 5 β 80 [ 3.3 ( 3.3) ] ( 0.49)( 0) ( 0.5)(.4) 3. 5 X5.8 Q 3 0.7 8.4 k Ω 0.5 0.7.. 5 β 0 Q β.5 ( 3). k Ω 0.39 ( 0.5) 0. 39 X5.9 5 on ( ) 80 5 0.7 0.9859 4 0.96 (a) 80 6.3.75 6.5 (b) 80 6.3.6657 0.64 (c) 80 6.3.97365 5.94 (d)
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions X5.0 on ( ) 4 0.7 or 3.3 kω ( )( ).0 α 0.99 0.99.0 0.99 or 8 μ ( 0.99)( ) 5 or 4.0 X5. γ ( sat) 5.5 0. (a) 0 Ω 5 5 0.30 50 50 5 0.7 4.3 k Ω 0.3 (b) 0.08 5 5 0.7 0 4 Ω 0.08 X5. (a) 0, ll currents are zero and O 5. (b) 5, 0; 0, 0.7 4.53 0.95 0. 8 0.6 O 0. (c) 5, 4.53 ; 8, / 4, O 0. X5.3 n active region, υ O mυ b m 6.5 t υ 0. 7, υo 5 5 6.5( 0.7) b b 9. 55 Then υ O 6.5υ 9.55 When υo 0. 6.5υ 9.55 υ.438 Q-point.438 0.7 υ Q 0.7.069
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions 5 0. υoq 0..6 Now.069 0.7 Q 4.6μ 80 β 0 4. 6 0. 5535 Q t Q-point υ 5 ( )( )μ OQ.6 5 0.5535 ( ) 34 4. k Ω X5.4 Q Q.8.4.7 k Ω 0. 0. Q 0.80 μ β 50 Q.8 0.7.65 M Ω 0.80 Q X5.5 (a) 85 35 4. 8 k Ω 35 35 85 on β ( ) ( 3.3) 0. 965 (b) Q ( ) ( ) Q so (c) Q 0.965 0.7 ( β ) 4.8 ( 5)( 0.5) ( 50 )( 0.0067) 0. 396 Q β Q β 5 Q β 50 3.3 0.396 4 0.395 0.5. Q Then Q ( 0.396) 0. 395 ( )( ) ( )( ) 53 0.965 0.7 Q 4.8 Q Q 4.8 μ ( 76)( 0.5) ( 75 )( 0.0048) 0. 335 ( 76 )( 0.0048) 0. 377 3.3 ( 0.335)( 4) ( 0.377)( 0.5). 89.67 μ
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions X5.6 ( ) ( ) Q or.5 5 0. which yields Q.08,.08 0.039 β 50 ( 0.)( β ) ( 0.)( 5)( 0.) or 3.0 kω Now so ( 3.0)( 5) ( ) ( ) Q on β Q We can write ( 3.0 )( 5 ) ( 0.039 )( 3.0 ) 0.7 ( 5 )( 0.039 )( 0. ) or We obtain 3 kω and then 3.93 kω X5.7 O β Q β Q 5 0 0.5 0 5 50 ( 0.5) 0. 5033 ( ) Q ( 0.5)( 0) ( 0.5033)( ) 3. 99 0 Now 0.5 Q Q 3.33μ β 50 0. β 0. 5 30. k Ω lso Then ( )( ) ( )( )( ) ( 0) 5 ( )( 0) 5 ( 30.)( 0) 5 Q Q 5 ( 0.00333)( 30.) 0.7 ( 0.5033)( ) 5 3. 93 ( 30.)( 0) 5 3. 93 or 67 k Ω and 67 30. 36. 9 k Ω
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions X5.8 0.7, 0.7.6 0. 9 O O O O 3.3 0.9 0 k Ω 0. β 6 Q O ( 0.) 0. β 60 Q ( 0.) 0. 6 β 60 0 ( 3.3) 3.3 0.7 0.6 0.6 k Ω X5.9 50 00 33.3 kω 50 ( 0) 5.67 50 00.67 0.7 5. μ 33.3 0 ( ) ( )( ).,.3 ( )( ) 5.3 5.74 3.5 0.5 Now 0.5 0.7...90 8.8 μ.88. 0.088.0 0.5 4.08 kω.0.5 ( )..5.9.9 5.97 kω.88 X5.0 We find 40 0.05 Then 0.5 0.7.7 kω 3 ( )( ).7 3 34 kω 0.05 lso 0.5 4 0.7 5.7 ( )( ) Δ 5.7.7 4 4 so 80 kω 0.05
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions and 40 80 34 6 kω 4 4 9 9 Then 6 kω 0.5 Test Your Understanding Solutions TYU5. β 60 (a) α 0. 9836 β 6 50 α 0.9934 5 α 0.98 (b) β 54. 6 α 0.98 0.995 β 3.3 0.995 TYU5. 0.60 0.005 0.65 0.60 β 4 0.005 β 4 α 0.99 β 5 TYU5.3 α ( 0.995)(.0). 9 α 0.995 β 6.6 α 0.995.0 0. μ β 7.6 TYU5.4 (a) r ( 5 )( 0.8) 80 o 80 (b) (i) ro. 5 M Ω 0.08 80 (ii) ro. 5 k Ω 8
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions TYU5.5 O t, O O 0.9868 (a) 75, 75 Then, at 0 0 ( 0.9868). 75 O O 0.9934 (b) 50, 50 0 ( 0.9934).06 m t 0, 50 TYU5.6 O O n β 3 O 00 ( 30) 39 so TYU5.7 0. < 0, (a) O 5 and P 0 3.6 0.7 4.53 (c) 3.6, 0.64 transistor is driven into saturation, so ( sat) 0. 0.9 0.440 and 0.9.4 < β 4.53 Note that which shows that the transistor is indeed driven into saturation. Now, P ( sat) ( 4.53)( 0.7) ( 0.9)( 0.) 5.35 mw TYU5.8 0 O 0.7 0.7 9.77 9.77 0.95 Then 0.44 β 50 and ( 0.95)( 0.64) 0.7 Now or 0.85 lso P ( 0.95)( 0.7) ( 9.77)( 0.7) 6.98 mw
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions TYU5.9 3.3 3.3.7 0.575 4 ( 3.3) 3.3 0.7 0.60 0 0.60 0.575.5μ 0.575 β 03 0.005 β 03 α 0.99038 β 04 TYU5.0 5 5 0.7 0.5375 8 β 85 ( 0.5375) 0. 53 β 86 0.5375 6.5μ β 86 0 ( 0.53)( 4) ( 0.5375)( 8) 3. 575 TYU5. ( ) ( ) on β or 0.7 ( β ) 0 ( 76)( ) Then or 5. μ ( 75)( 5. μ ).3 ( 76)( 5. μ ).5 lso and Now 8 (.3)(.5) (.5)( ) 6.03 TYU5. 5 5..8 5 0.7.8 0.5 0 ( β ) ( )( 0.5) 8. 5.8 0.54 k Ω 8.5
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions TYU5.3. (a)., 0. 039 β 9 on ( ) (.)( ) 0.7 ( 0.039)( 50). 56 (b) β 90 (.). 87 β 9 5 5 (.)( ) 3. 8 TYU5.4 (a) v 0, i i 0, vo, P 0 v 0.7 i 47. (b) v, 0.4 ( sat) 0. i.38 5 vo 0. and P i i ( sat) ( 0.047)( 0.7) (.38)( 0.) 0.7 W TYU5.5 5 Q 5.5 (a). 5.5 Q Q 0.4 μ β 0 5 5 0.7 43 k Ω Q 0.004 (b) Q 0.4 μ β 80 0. 8336 β 60. 667 Now Q 5 ( ) β 80 Q 5 ( 0.8336)( ) 3. 33 β 60 Q 5 (.667)( ). 67 So.67 Q 3.33
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions TYU5.6 0.7 Q 0.005375 800 β 75, βq ( 75)( 0.005375) 0.403 Or β 50, ( 50)( 0.005375 ) 0.806 Or Smallest Q Largest β 50, 4.96 kω 0.806 4 β 75,.48 kω 0.403.5 0.604 Q.5, 4.4 kω a nominal and 0.604 0.403, Q 5 ( 0.403)( 4.4) 3.33 Now for 0.806, Q 5 ( 0.806)( 4.4).66.66 So, for 4.4 kω, Q 3.33 TYU5.7 (a) 440 30 5 k Ω (b) 30 () 5. 76 440 30 on Q 3.364 μ β Q ( ).76 0.7 ( ) 5 ( 5)( ) β 0.5046 ; ( β ) 0. 508 Q Q Q Q Q ( 0.5046)( 4) ( 0.508)( ). 47 5 (c) 5 k Ω ;. 76.76 0.7 Q 4. μ Q 5 ( 9)( ) β Q 0.378 ; Q ( ) Q 0. 38 5 ( 0.378)( 4) ( 0.38)( ) 3. Q β TYU5.8 0. β 0. 5 5. k Ω (a) ( )( ) ( )( )( ) β 0.4 50 Q Q β Q β 5 50.667 μ ( 0.4) 0. 407
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions Q Q.7 5 0.4 0.407 ( ) ( )( ) 4. 74 Now Q on k Ω ( ) ( ) Q ( 5.)( 5) ( 0.00667)( 5.) 0.7 ( 0.407)( ) yields 66 k Ω ; 66 5. 9. 6 (b).43 ; 5. k Ω Q.43 0.7 ( β ) 5. ( 9)( ) Q k Ω 4.75μ β 0.376 ; ( β ) 0. 380 Q Q Q Q ( 0.376)( 4.74) ( 0.380)( ). 84 5 TYU5.9 ( ) 0 ( 4.5 0.5) 5 0 β 0 Q Q 8.33μ β Q (). 008 β 0 0. β 0. 0.5 6. k Ω ( )( ) ( )( )( ) 05 Q Q (.008)( 0.5) 0.7 ( 0.00833)( 6.05) 3. 746 5 So ( 0) 5 ( )( 0) 5 ( 6.05)( 0) 5 3.746 6. 9 k Ω 6.9 6.05 48. k Ω TYU5.0 β β 0 (a) ( 0.5) 0. 479 or Q S e T 0.479 0 3 0 3 ln T 4 S Q 0.479 Q β 0 0.00066 75 0. ( )( ) 55 ( 0.06) ln 0. 5937.066 μ 0.55 0.5937 0.7487
Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions ( 0.479)( 4). 508 ( 0.7487). 6.5.508 Q 60 6 (b) ( 0.5) 0. 459 Q Q 3 0.459 0 0 4 3 0 0.459 Q 4.098 μ β 60 0.004098 75 0. (.06) ln 0. 5935 ( )( ) 307 0.307 0.5935 0.90 ( 0.459)( 4). 56 ( 0.90). 4.5.56