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 + 8.88.47 mw X5.4 +.8 or 4.6 μ A 35 β ( 80)( 4.65) 0.369 ma 5 0.369 which yields 8.3 kω X5.5 (a) on 5.9 μ A 0 00 5.9 A 0.59 ma + ma 0 0.59 4 or 7.64 β ( μ ) ( β ) 0.597 6.5 (b) 6.4 μ A 0 Transistor is biased in saturation mode, so ( sat) 0. ( sat) 0 0. or.45 ma 4 +.45 + 0.064.48 ma X5.6 For 0 <, Q n is cutoff, O 9 ( 00)( )( 4) When Q n is biased in saturation, we have 0. 9 5. 00 So, for 5., O 0. X5.7
8 ( β ) 30 ( 76)(.) on or 60. μ A + + + β ( μ ) ( β ) 4.58 75 60.3 A 4.5 ma + ma 4.57 0.4 4.577. or 4.70 X5.8 For 4 and Q.5 ma, + 0 4 4 kω.5 Q + β 0 (.5 ).55 ma β 00 also on 0 Then 6.4 kω.55 X5.9 3 Q 0.5 9. kω 75 Q ( 0.5) 0.467 ma 76 + + 3 0 Q Q 3+ 6.89 kω 0.467 X5.0 5 + on + 80 (a) 5 + + 0.9859 ma 4 0.96 ma 80 (b) 6.3 +.75 ma 6.5 ma 80 (c) 6.3 +.6657 ma 0.64 ma 80 (d) 6.3 +.97365 ma 5.94 ma X5.
on 4.0 or 3.3 kω α ( 0.99) 0.99 ma.0 0.99 or 8 μ A ( 0.99) 5 or 4.0 X5. + γ + ( sat).5+ 0. 5 or 0 Ω 5 ma 5 v 4.3 kω P + + 5 0. 3.7 mw () X5.3 (a) For 0, All currents are zero and O 5. (b) For 5, 0; 0, 4.53 ma 0.95 0. 8 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.5 5 00 Q 5 Then Q 0.005 ma, Q 0.005 00 so. Also, β 00 0.005 0.5 ma Want Q-point to be X5.5 Q Q
.5 5 Q Q.5 or 0 kω 0.5 Q 0.5 Q 0.00083 ma β 0 Then.06 MΩ 0.00083 X5.6 (a) 9.5.8 kω.5 ( 5) + 9+.5 or.0 (b) Q + + β.8 + 5 0. or Q 9.375 μ A Q βq ( 50)( 9.375 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 76 7.6 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 0.039 ma β 50 0. + β 0. 5 0. or 3.0 kω Now so ( 3.0)( 5) + We can write + + ( + β ) on Q Q
3.0 5 0.039 3.0 5 0.039 0. or + + We obtain 3 kω and then 3.93 kω X5.8 β 50, Q Q Q 0 5 ma Q 5 0.0333 ma β 50 ( 5) + ( + β ) ( β ) on Set 0. + We have 0 5 + + Then 0.0333 Q (.)( 5)( 0.) which yields 0.806 + ( 0) kω so 0.806 3.0 kω Now 0. 5 0. 3.0 + We obtain 6.7 kω and 3.69 kω X5.9 + Q ( 5) 5 Q + 4.5 + 0.5 so ma, and Q Q Q 0.00833 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, 5 0.5 + + 0.00833 6.05 + 6.05 0 5 then we have () which yields 6.9 kω and 48.6 kω X5.0 Q + ( 0.5) + 0.65 ma β 40 or 6.38 kω 0 on 0 5 0.65
For 3, then.3 O β 40 O Q ( 0.5) 0.439 ma + β 4 O + O.3.07 kω 0.439 O X5. 50 00 33.3 kω 50 ( 0) 5.67 50 + 00.67 5. μ A 33.3 + 0. ma,.3 ma 5.3 5.74 3.5 0.5 Now 0.5+...90 ma 8.8 μ A.88 ma. 0.088.0 ma 0.5 4.08 kω.0.5..5.9.9 5.97 kω.88 X5. We find 40 kω + + 0.05 Then 0.5 +.7 3.7 3 34 kω 0.05 Also 0.5 + 4 + 5.7 Δ 5.7.7 4 4 so 80 kω 0.05 and 40 80 34 6 kω + 4+ 4 9 + 9 Then 6 kω 0.5 Q Test Your Understanding xercises TYU5. β α + β 75 For β 75, α 0.9868 76
5 For β 5, α 0.99 6 TYU5. + β 80 so + β 8.5 then β 80.3 0.00960 Now β 80.5 α 0.9877 + β 8.5 α 0.9877 8 70 ma TYU5.3 α 0.990 β 99 α 0.990.5 Now.5 μ A + β 00 and α 0.990.5.3 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 0.9868 ma 75 Then, at 0 0 ( 0.9868) +. ma 75 (b) For A 50, O + O 0.9934 ma 50 0 At 0, ( 0.9934) +.06 ma 50 TYU5.6 O O so O n β 3 00 30 39 TYU5.7 (a) For 0. < 0, O 5 and P 0 (b) For 3.6, transistor is driven into saturation, so 3.6 + ( sat) 0. 4.53 ma and 0.9 ma 0.64 0.440
0.9 Note that.4 < β which shows that the transistor is indeed driven into saturation. Now, 4.53 P + ( sat) 4.53 + 0.9 0. 5.35 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 + 0.95 + 9.77 6.98 mw TYU5.9 + 0 6.34 0.95 ma 4 ( 0) And or 0.930 0 0.95 α 0.9839 Now α 0.9839 and β 6 0.930 α 0.9839 Also 0.930 0.95 5 μ A 6.34 7.04 and TYU5.0 0.6 ma 8.65.8 μa 5 ( 50)(.8 μa).4 ma + 0 (.65)( 8) 0 (.397)( 4) 0 5.44 ( 5.44) 6.4 TYU5. + + 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 + 8 +.3.5.5 6.03 TYU5. 5. 5. + on + We have
5 0 7 5 on.. so 0 or 08. ma Then ( 0)( 0. 8) 8. ma 5. And 0. 38 kω 38 Ω 8. 8 TYU5.3 + on +.. ma 0.043 ma 5 β 50 and (.).6 ma + β 5 Then (.) + + ( 0.043)( 50) or 5.06..8 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) 0.047 +.38 0. 0.7 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 60 0.05 4 For Q 4, 0.5 ma 0.5 β β 40 0.05 So 40 β 60 TYU5.6 Q 0.005375 ma 800
For β 75, β ( 75)( 0.005375) Q Q Or Q 0.403 ma For β 50, Q ( 50)( 0.005375) Or Q 0.806 ma Largest Q Smallest Q For β 50, 4.96 kω 0.806 4 For β 75,.48 kω 0.403.5 For a nominal Q 0.604 ma and Q.5, 4.4 kω 0.604 Now for Q 0.403 ma, Q 5 ( 0.403)( 4.4) 3.33 For Q 0.806 ma, Q 5 ( 0.806)( 4.4).66 So, for 4.4 kω,.66 3.33 Q TYU5.7 (a) β 00 Q Q () 0.99 ma + β 0 Q Q 9.90 μ A + β 0 ( 0.0099)( 50) Q or 0.495 3 Q 0.99 0 T ln ( 0.06) ln 4 S 3 0 or 0.630 Then 0.495 0.630.3 0 ( 0.99)( 5) 5.05 Then Q 5.05 (.3) 6.8 (b) Q Q ma, Q 0.096 ma + β 5 ( 0.096)( 50) 0.98 β 50 Q Q () 0.98 ma + β 5 3 0.98 0 ( 0.06) ln 0.69 4 3 0 0.98 0.69.6 0 0.98 5 5. 5..6 6.7 Q TYU5.8 Q Q + β and
Q ( 0) Q( 0.65) Q( 0.65) + β 0 ( 0.99) + β Q 5 ( 0.99) Q( 4) 5 3.97Q 5 Q Q( 0.65) + Q( 3.97) 5 3.805 + 5.7 Q Q Q Q Then 3 5.7 3.805 Q which yields 0 Q Q ma