Chapter 5. Exercise Solutions. Microelectronics: Circuit Analysis and Design, 4 th edition Chapter 5 EX5.1 = 1 I. = βi EX EX5.3 = = I V EX5.

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1 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions xercise Solutions X5. ( β ).0 β 4. β hapter 5 β 40. α β X5. O 00 O n 3 β 0 or O 40.5 X β ( 50)( 3. 0)μ μ ( )( 3.). 85 ( 0.453)(.85) mw P X5.4 ( 80)( 3. )μ β ( )( ) μ X5.5 (a) (b) ( 0)( )μ β ( 0.44)( 5) μ 50 ( sat) μ 0.

2 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) When Q n is biased in saturation, we have 00 So, for 5., O 0. X5.7 ( β ) So μ 640 β μ ( 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 k Ω β 0 Q β.5 ( 3). k Ω 0.39 ( 0.5) X5.9 5 on ( ) (a) (b) (c) (d)

3 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions X5.0 on ( ) or 3.3 kω ( )( ).0 α or 8 μ ( 0.99)( ) 5 or 4.0 X5. γ ( sat) (a) 0 Ω k Ω 0.3 (b) Ω 0.08 X5. (a) 0, ll currents are zero and O 5. (b) 5, 0; 0, 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 ( 0.7) b b Then υ O 6.5υ 9.55 When υo υ 9.55 υ.438 Q-point υ Q

4 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions 5 0. υoq 0..6 Now Q 4.6μ 80 β Q t Q-point υ 5 ( )( )μ OQ ( ) k Ω X5.4 Q Q k Ω Q 0.80 μ β 50 Q M Ω 0.80 Q X5.5 (a) k Ω on β ( ) ( 3.3) (b) Q ( ) ( ) Q so (c) Q ( β ) 4.8 ( 5)( 0.5) ( 50 )( ) Q β Q β 5 Q β Q Then Q ( 0.396) ( )( ) ( )( ) Q 4.8 Q Q 4.8 μ ( 76)( 0.5) ( 75 )( ) ( 76 )( ) ( 0.335)( 4) ( 0.377)( 0.5) μ

5 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions X5.6 ( ) ( ) Q or which yields Q.08, 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 ) ( )( 3.0 ) 0.7 ( 5 )( )( 0. ) or We obtain 3 kω and then 3.93 kω X5.7 O β Q β Q ( 0.5) ( ) Q ( 0.5)( 0) ( )( ) Now 0.5 Q Q 3.33μ β β k Ω lso Then ( )( ) ( )( )( ) ( 0) 5 ( )( 0) 5 ( 30.)( 0) 5 Q Q 5 ( )( 30.) 0.7 ( )( ) ( 30.)( 0) or 67 k Ω and k Ω

6 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions X , O O O O k Ω 0. β 6 Q O ( 0.) 0. β 60 Q ( 0.) 0. 6 β 60 0 ( 3.3) k Ω X kω 50 ( 0) μ ( ) ( )( ).,.3 ( )( ) Now μ kω.0.5 ( ) kω.88 X5.0 We find Then kω 3 ( )( ) kω 0.05 lso ( )( ) Δ so 80 kω 0.05

7 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions and kω Then 6 kω 0.5 Test Your Understanding Solutions TYU5. β 60 (a) α β 6 50 α α 0.98 (b) β α β TYU β β 4 α 0.99 β 5 TYU5.3 α ( 0.995)(.0). 9 α β 6.6 α μ β 7.6 TYU5.4 (a) r ( 5 )( 0.8) 80 o 80 (b) (i) ro. 5 M Ω (ii) ro. 5 k Ω 8

8 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions TYU5.5 O t, O O (a) 75, 75 Then, at 0 0 ( ). 75 O O (b) 50, 50 0 ( ).06 m t 0, 50 TYU5.6 O O n β 3 O 00 ( 30) 39 so TYU < 0, (a) O 5 and P (c) 3.6, 0.64 transistor is driven into saturation, so ( sat) and < 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 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

9 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions TYU ( 3.3) μ β β 03 α β 04 TYU β 85 ( ) β μ β 86 0 ( 0.53)( 4) ( )( 8) 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 TYU ( β ) ( )( 0.5) k Ω 8.5

10 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions TYU5.3. (a)., β 9 on ( ) (.)( ) 0.7 ( 0.039)( 50). 56 (b) β 90 (.). 87 β (.)( ) 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 k Ω Q (b) Q 0.4 μ β β Now Q 5 ( ) β 80 Q 5 ( )( ) β 60 Q 5 (.667)( ). 67 So.67 Q 3.33

11 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions TYU Q β 75, βq ( 75)( ) Or β 50, ( 50)( ) Or Smallest Q Largest β 50, 4.96 kω β 75,.48 kω Q.5, 4.4 kω a nominal and , Q 5 ( 0.403)( 4.4) 3.33 Now for 0.806, Q 5 ( 0.806)( 4.4) So, for 4.4 kω, Q 3.33 TYU5.7 (a) k Ω (b) 30 () on Q μ β Q ( ) ( ) 5 ( 5)( ) β ; ( β ) Q Q Q Q Q ( )( 4) ( 0.508)( ) (c) 5 k Ω ; Q 4. μ Q 5 ( 9)( ) β Q ; Q ( ) Q ( 0.378)( 4) ( 0.38)( ) 3. Q β TYU β k Ω (a) ( )( ) ( )( )( ) β Q Q β Q β μ ( 0.4)

12 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions Q Q ( ) ( )( ) Now Q on k Ω ( ) ( ) Q ( 5.)( 5) ( )( 5.) 0.7 ( 0.407)( ) yields 66 k Ω ; (b).43 ; 5. k Ω Q ( β ) 5. ( 9)( ) Q k Ω 4.75μ β ; ( β ) Q Q Q Q ( 0.376)( 4.74) ( 0.380)( ) TYU5.9 ( ) 0 ( ) 5 0 β 0 Q Q 8.33μ β Q (). 008 β 0 0. β k Ω ( )( ) ( )( )( ) 05 Q Q (.008)( 0.5) 0.7 ( )( 6.05) So ( 0) 5 ( )( 0) 5 ( 6.05)( 0) k Ω k Ω TYU5.0 β β 0 (a) ( 0.5) or Q S e T ln T 4 S Q Q β ( )( ) 55 ( 0.06) ln μ

13 Microelectronics: ircuit nalysis and Design, 4 th edition hapter 5 y D.. Neamen xercise Solutions ( 0.479)( 4). 508 ( ) Q 60 6 (b) ( 0.5) Q Q Q μ β (.06) ln ( )( ) ( 0.459)( 4). 56 ( 0.90)

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( )( ) ( ) ( )( ) ( )( ) β = Chapter 5 Exercise Problems EX α So 49 β 199 EX EX EX5.4 EX5.5. (a)

( )( ) ( ) ( )( ) ( )( ) β = Chapter 5 Exercise Problems EX α So 49 β 199 EX EX EX5.4 EX5.5. (a) hapter 5 xercise Problems X5. α β α 0.980 For α 0.980, β 49 0.980 0.995 For α 0.995, β 99 0.995 So 49 β 99 X5. O 00 O or n 3 O 40.5 β 0 X5.3 6.5 μ A 00 β ( 0)( 6.5 μa) 8 ma 5 ( 8)( 4 ) or.88 P on + 0.0065