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

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1 hapter 5 xercise Problems X5. α β α For α 0.980, β For α 0.995, β So 49 β 99 X5. O 00 O or n 3 O 40.5 β 0 X μ A 00 β ( 0)( 6.5 μa) 8 ma 5 ( 8)( 4 ) or.88 P on mw X or 4.6 μ A 35 β ( 80)( 4.65) ma which yields 8.3 kω X5.5 (a) on 5.9 μ A A 0.59 ma + ma or 7.64 β ( μ ) ( β ) (b) 6.4 μ A 0 Transistor is biased in saturation mode, so ( sat) 0. ( sat) 0 0. or.45 ma ma X5.6 For 0 <, Q n is cutoff, O 9 ( 00)( )( 4) When Q n is biased in saturation, we have So, for 5., O 0. X5.7

2 8 ( β ) 30 ( 76)(.) on or 60. μ A β ( μ ) ( β ) A 4.5 ma + ma or 4.70 X5.8 For 4 and Q.5 ma, kω.5 Q + β 0 (.5 ).55 ma β 00 also on 0 Then 6.4 kω.55 X5.9 3 Q kω 75 Q ( 0.5) ma Q Q kω X on + 80 (a) ma ma 80 (b) ma 6.5 ma 80 (c) ma 0.64 ma 80 (d) ma 5.94 ma X5.

3 on 4.0 or 3.3 kω α ( 0.99) 0.99 ma or 8 μ A ( 0.99) 5 or 4.0 X5. + γ + ( sat) or 0 Ω 5 ma 5 v 4.3 kω P mw () X5.3 (a) For 0, All currents are zero and O 5. (b) For 5, 0; 0, 4.53 ma ma 0.6 O 0. (c) For 5, 4.53 ma; 8 ma, / 4 ma, O 0. X5.4 vo i βi Δ vo βδi +Δv i Δv Δ i ΔvO β Then Δv Let β 00, 5 kω, 00 kω Δ v ( 00)( 5 o ) So 5 Δv 00 vo Q pt Q 5 Then Q ma, Q so. Also, β ma Want Q-point to be X5.5 Q Q

4 .5 5 Q Q.5 or 0 kω 0.5 Q 0.5 Q ma β 0 Then.06 MΩ X5.6 (a) kω.5 ( 5) or.0 (b) Q + + β or Q μ A Q βq ( 50)( μa) or Q.4 ma Q ( + β ) Q Q.4 ma Q 5 Q Q 5 (.4) (.4)( 0.) or 3.3 Q (c) For β 75 Q 7.6 μ A.8 + ( 76)( 0.) Q βq ( 75)( 7.6 μa) or Q.3 ma Q + β Q μa or Q.34 ma Q 5 (.3) (.34)( 0.) or 3.4 Q X5.7 Q Q ( + ) or.5 5 Q ( + 0.) which yields Q.08 ma, Q.08 Q ma β β or 3.0 kω Now so ( 3.0)( 5) + We can write + + ( + β ) on Q Q

5 or + + We obtain 3 kω and then 3.93 kω X5.8 β 50, Q Q Q 0 5 ma Q ma β 50 ( 5) + ( + β ) ( β ) on Set 0. + We have Then Q (.)( 5)( 0.) which yields ( 0) kω so kω Now We obtain 6.7 kω and 3.69 kω X5.9 + Q ( 5) 5 Q so ma, and Q Q Q ma β 0 ( 0.)( β ) ( 0.)( )( 0.5) + or 6.05 kω We can write We have Q Q 0 5 and if we let Q Q ma, then we have () which yields 6.9 kω and 48.6 kω X5.0 Q + ( 0.5) ma β 40 or 6.38 kω 0 on

6 For 3, then.3 O β 40 O Q ( 0.5) ma + β 4 O + O.3.07 kω O X kω 50 ( 0) μ A ma,.3 ma Now ma 8.8 μ A.88 ma ma kω kω.88 X5. We find 40 kω Then kω 0.05 Also Δ so 80 kω 0.05 and kω Then 6 kω 0.5 Q Test Your Understanding xercises TYU5. β α + β 75 For β 75, α

7 5 For β 5, α TYU5. + β 80 so + β 8.5 then β Now β 80.5 α β 8.5 α ma TYU5.3 α β 99 α Now.5 μ A + β 00 and α ma TYU5.4 A 50 ro For 0. ma ro.5 M Ω For.0 ma ro 50 kω For 0 ma r 5 kω o TYU5.5 O + A At, ma (a) For A 75, O + O ma 75 Then, at 0 0 ( ) +. ma 75 (b) For A 50, O + O ma 50 0 At 0, ( ) +.06 ma 50 TYU5.6 O O so O n β TYU5.7 (a) For 0. < 0, O 5 and P 0 (b) For 3.6, transistor is driven into saturation, so ( sat) ma and 0.9 ma

8 0.9 Note that.4 < β which shows that the transistor is indeed driven into saturation. Now, 4.53 P + ( sat) mw TYU5.8 For 0 O 9.77 Then 9.77 ma and 0.95 ma 0.44 β 50 Now + ( 0.95)( 0.64) + or 0.85 Also P mw TYU ma 4 ( 0) And or α Now α and β α Also μ A and TYU ma μa 5 ( 50)(.8 μa).4 ma + 0 (.65)( 8) 0 (.397)( 4) ( 5.44) 6.4 TYU or + + ( + β ) Then + ( + β ) 0+ ( 76) or 5. μ A Also ( 75)( 5. μ A).3 ma and ( 76)( 5. μ A).5 ma Now TYU on + We have

9 on.. so 0 or 08. ma Then ( 0)( 0. 8) 8. ma 5. And kω 38 Ω 8. 8 TYU5.3 + on +.. ma ma 5 β 50 and (.).6 ma + β 5 Then (.) + + ( 0.043)( 50) or Now TYU5.4 (a) For v 0, i i 0, vo, P 0 v (b) For v, i 47. ma 0.4 ( sat) 0. i.38 A 5 vo 0. and P i + i ( sat) W TYU5.5.5 (a) For Q.5, Q.5 ma Q.5 Q Q.5 μ A β 00 Then 344 kω 0.05 (b) Q is independent of β. For Q, ma β β For Q 4, 0.5 ma 0.5 β β So 40 β 60 TYU5.6 Q ma 800

10 For β 75, β ( 75)( ) Q Q Or Q ma For β 50, Q ( 50)( ) Or Q ma Largest Q Smallest Q For β 50, 4.96 kω For β 75,.48 kω For a nominal Q ma and Q.5, 4.4 kω Now for Q ma, Q 5 ( 0.403)( 4.4) 3.33 For Q ma, Q 5 ( 0.806)( 4.4).66 So, for 4.4 kω, Q TYU5.7 (a) β 00 Q Q () 0.99 ma + β 0 Q Q 9.90 μ A + β 0 ( )( 50) Q or Q T ln ( 0.06) ln 4 S 3 0 or Then ( 0.99)( 5) 5.05 Then Q 5.05 (.3) 6.8 (b) Q Q ma, Q ma + β 5 ( 0.096)( 50) 0.98 β 50 Q Q () 0.98 ma + β ( 0.06) ln Q TYU5.8 Q Q + β and

11 Q ( 0) Q( 0.65) Q( 0.65) + β 0 ( 0.99) + β Q 5 ( 0.99) Q( 4) Q 5 Q Q( 0.65) + Q( 3.97) Q Q Q Q Then Q which yields 0 Q Q ma