CS计算机代考程序代写 v BE /VT Ex:6.1iC=ISe 􏱼 􏱽

v BE /VT Ex:6.1iC=ISe 􏱼 􏱽
iC2
I ISB=S=
􏱼I 􏱽
VBE =VT ln C =25ln
IS
=25×29.9336 = 748 mV
Ex: 6.6
Exercise 6–1
10−16
=10−18A 100
vBE2 −vBE1 =VT ln
􏱼 0.1 􏱽
􏱼1mA􏱽 10−16
VCC 􏰔 􏰀5 V RC
β
iC1 vBE2 =700+25ln 1
= 642 mV
vBE3 =700+25ln 1 = 758mV
Ex:6.2∴α= β β+1
50 <α< 150 50+1 150+1 0.980 < α < 0.993 Ex:6.3 IC =IE −IB = 1.460 mA − 0.01446 mA = 1.446 mA α= IC = 1.446 =0.99 IE 1.460 β = IC = 1.446 = 100 IB 0.01446 IC = ISevBE/VT I = IC = 1.446 S ev BE /VT e700/25 = 1.446 mA=10−15 A e28 Ex: 6.4 β = α and 1−α 􏱼10􏱽 10 􏱌A BC bIB E Forα=0.99, β= IB=IC =10=0.1mA 0.99 1−0.99 IC = 10 mA =99 vBE =690mV IC = 1 mA For active range VC ≥ VB, RCmax = VCC −0.690 IC = 5−0.69 1 = 4.31 k􏱹 Ex:6.7 IS =10−15 A AreaC =100× AreaE ISC = 100×IS = 10−13 A β 99 Forα=0.98, β= 0.98 Ex: 6.8 iC = ISevBE/VT − ISCevBC/VT foriC =0 =49 β49 SSC 1−0.98 IB=IC =10=0.2mA IevBE/VT =I evBC/VT ISC = evBE/VT IS evBC/VT I =10−16 A, β=100,I =1mA =e(vBE−vBC)/VT Ex: 6.5 Given: SC􏱼I􏱽 Wewrite ∴VCE =VBE −VBC =VT ln SC 􏱾􏱿 IS ISE =ISC/α=IS = 1+ 1 ForcollectorArea=100×Emitterarea β 􏱼100􏱽 =10−16 ×1.01=1.01×10−16 A VCE =25ln 1 =115mV Ex: 6.9 IC = ISevBE/VT − ISCevBC/VT IB = IS evBE/VT + ISCevBC/VT Ex: 6.12 Exercise 6–2 β βforced= C􏲀 <β ISevBE/VT +βISCevBC/VT = β ISe(vBE−vBC)/VT − ISC ISe(vBE−vBC)/VT +βISC eVCEsat/VT −I /I 􏰀1.5 V VBE I 􏲀􏲀􏲀 2 mA 􏰀 R C VC 􏰔 0.5 V IB sat = β ISevBE/VT − ISCevBC/VT =β SCS Q.E.D. I 􏰔(2)mA Ea 􏰒 􏰒1.5 V VE􏰔􏰒VBE RE eVCE sat /VT β = 100 + β ISC /IS e200/25 − 100 forced = 100 × 0.2219 ≈ 22.2 Ex: 6.10 e200/25 + 100 × 100 2 mA E IS /a B ISevEB/VT 􏰔aiE C RC = 1.5−VC = 1.5−0.5 IC 2 = 0.5 k􏱹 = 500 􏱹 SinceatIC =1mA,VBE =0.8V,thenat IC =2mA, 􏱾2􏱿 VBE =0.8+0.025ln 1 = 0.8 + 0.017 = 0.817 V VE =−VBE =−0.817V 2 mA 2 IE= α =0.99=2.02mA IE = VE −(−1.5) RE RE = −0.817+1.5 =0.338k􏱹 􏰒10 V Eα 2.02 I = IS evBE/VT 2 mA = 5110−14evBE/VT = 338 􏱹 Ex: 6.13 Thus, 50 VBE =25ln 103 ×51×1014 􏱼2 50 􏱽 = 650 mV 􏰀10 V 5 k􏰵 VC VE 􏰔 􏰒0.7 V 10 k􏰵 IC= β IE=50×2 β+1 51 I IB C = 1.96 mA IC 1.96 IB = β = 50 ⇒39.2μA Ex: 6.11 IC = IS ev BE /VT = 1.5 A ∴ VBE = VT ln 􏲁1.5/10−11􏲂 = 25 × 25.734 = 643 mV IE 􏰒10 V IE = VE −(−10) = −0.7+10 10 10 = 0.93 mA Assuming active-mode operation, IB= IE = 0.93 =0.0182mA β + 1 50 + 1 = 18.2 μA IC =IE −IB =0.93−0.0182=0.91mA VC =10−IC ×5 = 10 − 0.91 × 5 = 5.45 V Since VC > VB, the transistor is operating in the active mode, as assumed.
and
VC =−10+1.65×5=−1.75V
Since VC < VB, the transistor is indeed operating in the active mode. Exercise 6–3 Ex: 6.15 Ex: 6.14 VB VB = 1.0 V Thus, IB = VB 100 k􏱹 VE =+1.7V Thus, 􏰀10 V IE 100 k􏰵 IC = 0.01 mA 􏰒5 V IC 2 mA VE VC RC 􏰔 1 k􏰵 IB 5 k􏰵 VE VC 5 k􏰵 The transistor is operating at a constant emitter current. Thus, a change in temperature of +30◦C results in a change in VEB by △VEB =−2mV×30=−60mV Thus, △VE = −60 mV Since the collector current remains unchanged at αIE , the collector voltage does not change: △VC =0V Ex: 6.16 Refer to Fig. 6.19(a): iC = ISevBE/VT + vCE (1) ro Now using Eqs. (6.21) and (6.22), we can express ro as VA ro = ISevBE/VT SubstitutinginEq.(1),wehave IB I = 10−VE = 10−1.7 =1.66mA E 5k􏱹 5 and β+1= IE = 1.66 =166 IB 0.01 ⇒β=165 α= β = 165 =0.994 β+1 165+1 Assuming active-mode operation, IC =αIE =0.994×1.66=1.65mA 􏱾v􏱿 1+ CE 􏰒10 V iC =ISevBE/VT whichisEq.(6.18). VA Q.E.D. Ex:6.17 ro = VA = 100 IC IC AtIC =0.1mA, ro =1M􏱹 At IC = 1 mA, ro = 100 k􏱹 AtIC =10mA, ro =10k􏱹 Ex: 6.18 △IC = △VCE ro where ro = VA = 100 = 100 k􏱹 IC 1 △IC = 11−1 =0.1mA 100 Thus, IC becomes 1.1 mA. Ex:6.19 Ex: 6.20 For VBB = 0 V, IB = 0 and the transistor is cut off. Thus, IC =0 and VC =VCC =+10V Ex: 6.21 Refer to the circuit in Fig. 6.22 and let VBB = 1.7 V. The current IB can be found from IB = VBB −VB = 1.7−0.7 =0.1mA RB 10 Assuming operation in the active mode, IC =βIB =50×0.1=5mA Thus, VC =VCC −RCIC = 10 − 1 × 5 = 5 V which is greater than VB , verifying that the transistor is operating in the active mode, as assumed. (a) To obtain operation at the edge of saturation, RC must be increased to the value that results in VCE =0.3V: Exercise 6–4 VBB B VCC 􏰔 10 V RC 􏰔 10 k􏰵 􏰀 VCE 􏰒 I IC RB 􏰔 10 k􏰵 􏰀 VBE 􏰔 0.7 V 􏰒 (a) For operation in the active mode with VCE =5V, IC = VCC −VC = 10−5 =0.5mA IB = IC = 0.5 =0.01mA = 0.7 + 0.01 × 10 = 0.8 V (b) For operation at the edge of saturation, VCE =0.3V IC = VCC −VCE = 10−0.3 =0.97mA RC = VCC −0.3 RC10 IC β 50 VBB =VBE +IBRB = 10−0.3 =1.94k􏱹 5 (b) Further increasing RC results in the transistor operating in saturation. To obtain saturation-mode operation with VCE = 0.2 V and βforced = 10, we use IC = βforced × IB =10×0.1=1mA The value of RC required can be found from VCC −VCE RC= I C = 10−0.2 =9.8k􏱹 1 Ex: 6.22 Refer to the circuit in Fig. 6.23(a) with the base voltage raised from 4 V to VB. If at this valueofVB,thetransistorisattheedgeof saturation then, VC =VB −0.4V SinceIC ≃IE,wecanwrite 10−VC =VE =VB−0.7 RC RE RE RC 10 IC 0.97 IB = β = 50 =0.0194mA VBB =VB +IBRB = 0.7 + 0.0194 × 10 = 0.894 V (c) For operation deep in saturation with βforced = 10, we have VCE ≃ 0.2 V I = 10−0.2 =0.98mA C 10 IB = IC = 0.98 =0.098mA 10 βforced VBB =VB +IBRB = 0.7 + 0.098 × 10 = 1.68 V Thus, 10−(VB −0.4) Ex: 6.23 VB 􏰔 􏰀4 V IC =5IB = 10−(VB −0.5) (1) 4.7 IE = 6IB = VB − 0.7 (2) 3.3 Dividing Eq. (1) by Eq. (2), we have 5 = 10.5−VB × 3.3 6 VB − 0.7 4.7 ⇒VB =+5.18V Ex: 6.25 Refer to the circuit in Fig. 6.26(a). The largest value for RC while the BJT remains in the active mode corresponds to VC =+0.4V Since the emitter and collector currents remain unchanged, then from Fig. 6.26(b) we obtain IC =4.6mA Thus, RC = VC − (−10) IC = +0.4+10 =2.26k􏱹 4.6 4.7 = ⇒VB =+4.7V Exercise 6–5 VB −0.7 3.3 􏰀10 V 0.5 mA RC 0.5 mA RE VC 􏰔 􏰀6 V VE 􏰔 􏰀3.3 V To establish a reverse-bias voltage of 2 V across the CBJ, VC =+6V From the figure we see that RC = 10 − 6 = 8 k􏱹 0.5 and RE = 3.3 =6.6k􏱹 0.5 where we have assumed α ≃ 1. Ex: 6.24 Ex: 6.26 􏰀10 V 1 mA 1 mA RE RC VE 􏰔 􏰀0.7 V VC 􏰔 􏰒4 V 􏰀10 V IC 􏰔 5IB 4.7 k􏰵 VB 􏰒 0.5 VB 􏰒 0.7 3.3 k􏰵 􏰒10 V VB IB For a 4-V reverse-biased voltage across the CBJ, VC =−4V Refer to the figure. IC = 1 mA = VC − (−10) RC ⇒RC = −4+10 =6k􏱹 1 RE = 10 − VE IE Assuming α = 1, 10−0.7 RE = 1 =9.3k􏱹 6IB The figure shows the circuit with the base voltage at VB and the BJT operating in saturation with VCE =0.2Vandβforced =5. Ex: 6.27 Refer to the circuit in Fig. 6.27: IB = 5−0.7 =0.043mA 100 To ensure that the transistor remains in the active mode for β in the range 50 to 150, we need to select RC so that for the highest collector current possible, the BJT reaches the edge of saturation, that is, VCE = 0.3 V. Thus, VCE =0.3=10−RCICmax where ICmax = βmaxIB = 150 × 0.043 = 6.45 mA Thus, RC = 10−0.3 =1.5k􏱹 6.45 For the lowest β, IC = βminIB = 50 × 0.043 = 2.15 mA and the corresponding VCE is VCE = 10−RCIC = 10−1.5×2.15 = 6.775 V Thus, VCE will range from 0.3 V to 6.8 V. Ex: 6.28 Refer to the solution of Example 6.10. IE = VBB − VBE Ex: 6.30 Exercise 6–6 􏰀15 V 2.78 mA 2 k􏰵 􏰀9.44 V 2.75 mA IC3 /b 2.7 k􏰵 Q2 VC2 IC3 Q3 VE3 470 􏰵 (1) IC3 a From the figure we see that VE3=IC3×0.47 α VC2 = VE3 +0.7 = IC3 ×0.47+0.7 α A node equation at the collector of Q2 yields 2.75 = VC2 + IC3 2.7 β Substituting for VC2 from Eq. (1), we obtain 2.75 = (0.47 IC3/α) + 0.7 + IC3 2.7 β RE + [RBB/(β + 1)] 5−0.7 Substituting α = 0.99 and β = 100 and solving for IC3 results in IC3 = 13.4 mA Now, VE3 and VC2 can be determined: IC 3 13.4 VE3 = α ×0.47= 0.99 ×0.47=+6.36V = 3 + (33.3/51) = 1.177 mA IC =αIE =0.98×1.177=1.15mA VC2 =VE3 +0.7=+7.06V Ex: 6.31 Thus the current is reduced by △IC = 1.28 − 1.15 = 0.13 mA which is a −10% change. Ex: 6.29 Refer to the circuit in Fig. 6.30(b). The total current drawn from the power supply is I =0.103+1.252+2.78=4.135mA Thus, the power dissipated in the circuit is P = 15 V × 4.135 mA = 62 mW 10 k􏰵 􏰒5 V IB 􏰀5 V Q1 off 0 I E V 􏰔 􏰒I 􏰁 1 E E Q2 on IE 1 k􏰵 􏰒5 V From the figure we see that Q1 will be off and Q2 will be on. Since the base of Q2 will be at a voltage higher than −5 V, transistor Q2 will be operating in the active mode. We can write a loop equation for the loop containing the 10-k􏱹 resistor, the EBJ of Q2 and the 1-k􏱹 resistor: −IE ×1−0.7−IB ×10=−5 Substituting IB = IE /(β + 1) = IE /101 and rearranging gives IE = 5−0.7 =3.9mA Thus, VE =−3.9V VB2 = −4.6 V IB = 0.039 mA Ex: 6.32 With the input at + 10 V, there is a strong possibility that the conducting transistor Q1 will be saturated. Assuming this to be the case, the analysis steps will be as follows: VCEsat|Q1 = 0.2 V VE =5V−VCEsat =+4.8V IE1 = 4.8V =4.8mA 1 k􏱹 VB1 =VE +VBE1 =4.8+0.7=+5.5V IB1 = 10−5.5 =0.45mA 10 IC1 =IE1 −IB1 =4.8−0.45=4.35mA IC 4.35 βforced = I = 0.45 = 9.7 B which is lower than βmin, verifying that Q1 is indeed saturated. Finally, since Q2 is off, IC2 =0 Ex:6.33 VO =+10−BVBCO =10−70 = −60 V Exercise 6–7 10 +1 101 This figure belongs to Exercise 6.32. 4 10 k􏰵 􏰀5.5 V 􏰀5 V 􏰀 Q1 on VCEsat 􏰔 0.2 V 􏰒 1 􏰀10 V 5 10􏰒5.5 10 􏰔 0.45 mA 􏰀4.8 V 2 34.8 mA Q2 off 􏰒5 V 1 k􏰵 6.1 1. Active 2. Saturation 3. Active 4. Saturation 5. Active 6. Cutoff 6.2 The EB junctions have a 4:1 area ratio. IC =ISeVBE/VT 0.5 × 10−3 = IS1 × e0.75/0.025 ⇒IS1 =4.7×10−17 A IS2 =4IS1 =1.87×10−16 A 6.3 IC =ISeVBE/VT 200×10−6 =ISe30 ⇒IS =1.87×10−17 A For the transistor that is 32 times larger, IS = 32×1.87×10−17 =6×10−16 A At VBE = 30 VT , the larger transistor conducts a current of IC =32×200μA=6.4mA At IC = 1 mA, the base–emitter voltage of the larger transistor can be found as 6.6 Old technology: 10−3 =2×10−15eVBE/VT 􏱾 10−3 􏱿 VBE = 0.025 ln 2 × 10−15 VBE =VT ln 6×10−16 =0.704V 6.4 IS1 = AE1 = 200×200 = 250,000 α= 6.9 6.10 B β = 80 =0.988 β+1 81 Chapter 6–1 New technology: 10−3 =2×10−18eVBE/VT 􏱾 10−3 VBE = 0.025 ln 2 × 10−18 6.7 5 × 10−3 = IS e0.76/0.025 IC =ISe0.70/0.025 Dividing Eq. (2) by Eq. (1) IC = 5 × 10−3e−0.06/0.025 = 0.45 mA ForIC =5μA, 5×10−6 =ISeVBE/0.025 Dividing Eq. (3) by Eq. (1) = 0.673 V 􏱿 = 0.846 V yields yields (1) (2) (3) 1×10−3 =6×10−16eVBE/VT 􏱾1×10−3 􏱿 10−3 =e(VBE−0.76)/0.025 VBE = 0.76 + 0.025 ln(10−3) = 0.587 V 6.8 IB =10μA IC =800μA IC β=I =80 IS2 AE2 0.4 × 0.4 IC1 = IS1eVBE1/VT IC2 = IS2eVBE2/VT For IC1 = IC2 we have e(VBE2−VBE1)/VT = IS1 = 250,000 IS2 VBE2 − VBE1 = 0.025 ln(250,000) = 0.31 V 6.5 IC1 = 10−13e700/25 = 0.145A = 145 mA IC2 = 10−18e700/25 = 1.45 μA For the first transistor 1 to conduct a current of 1.45 μA, its VBE must be 􏱾 1.45 × 10−6 􏱿 VBE1 = 0.025 ln 10−13 = 0.412 V α 6.11 β = 1 − α (1) α → α + △α β → β + △β β+△β= α+△α (2) 1 − α − △α α 0.5 0.8 0.9 0.95 0.98 0.99 0.995 0.999 β=α 1−α 1 4 9 19 49 99 199 999 β 1 2 10 20 50 100 200 500 1000 α=β β+1 0.5 0.67 0.91 0.95 0.98 0.99 0.995 0.998 0.999 Subtracting Eq. (1) from Eq. (2) gives △β= α+△α − α 1−α−△α 1−α △β = △α (1−α−△α)(1−α) Dividing Eq. (3) by Eq. (1) gives △β 􏱾△α􏱿􏱾 1 􏱿 (3) 6.14 For iB = 10 μA, iC =iE −iB =1000−10=990μA β=iC =990=99 iB 10 α= β = 99 =0.99 β + 1 100 ForiB =20μA, iC =iE −iB =1000−20=980μA β=iC =980=49 iB 20 α = β = 49 = 0.98 β+1 50 For iB = 50 μA, iC =iE −iB =1000−50=950μA β=iC =950=19 iB 50 α= β =19=0.95 β+1 20 6.15 See Table below. 6.16 First we determine IS , β, and α: 1×10−3 =ISe700/25 ⇒IS =6.91×10−16 A IC 1 mA β=I =10μA=100 Chapter 6–2 β = α 1−α−△α For △α ≪ 1, the second factor on the right-hand side is approximately equal to β. Thus △β 􏱾 △α 􏱿 β ≃β α Q.E.D. For △β = −10% and β = 100, β △α −10% α ≃ 100 =−0.1% 6.12 Transistor is operating in active region: β = 50 → 300 IB =10μA IC =βIB =0.5mA→3mA IE =(β+1)IB =0.51mA→3.01mA Maximum power dissipated in transistor is I ×0.7V+I ×V BCCB = 0.01×0.7+3×10 ≃ 30 mW 6.13 iC =ISevBE/VT = 5×10−15e0.7/0.025 = 7.2 mA iB willbeintherange 7.2 mAto 7.2 mA,thatis, α= β = 100 =0.99 β+1 101 ThenwecandetermineISE andISB: ISE = IS = 6.98×10−16 A α ISB = IS =6.91×10−18 A 50200 β 144 μA to 36 μA. iE will be in the range (7.2 + 0.144) mA to large-signal models, corresponding to Fig. 6.5(a) The figure on next page shows the four (7.2 + 0.036) mA, that is, 7.344 mA to 7.236 mA. to (d), together with their parameter values. This table belongs to Problem 6.15. Transistor a b c d e VBE (mV) 700 690 580 780 820 IC (mA) 1.000 1.000 0.230 10.10 73.95 IB (μA) 10 20 5 120 1050 IE (mA) 1.010 1.020 0.235 10.22 75 α 0.99 0.98 0.979 0.988 0.986 β 100 50 46 84 70 IS (A) 6.9 × 10−16 1.0 × 10−15 1.9 × 10−14 2.8 × 10−16 4.2 × 10−16 Chapter 6–3 6.17 􏰀5 V 2 k􏰵 C VC aiE B DE iE VE E 2 mA (b) C The figure shows the circuit, where α= β = 100 =0.99 β+1 101 IS 5×10−15 −15 ISE = α = 0.99 =5.05×100 A The voltage at the emitter VE is VE = −VDE = −VT ln(IE/ISE) 􏱾 2×10−3 􏱿 = −0.025 ln 5.05 × 10−15 = −0.668 V The voltage at the collector VC is found from VC = 5 − IC × 2 =5−αIE ×2 = 5−0.99×2×2 = 1.04 V 6.18 Refer to the circuit in Fig. 6.6(b). B iB 􏰀 vBE 􏰒 iC aiE DE ISE􏰔IS/a iE E a􏰔0.99 ISE 􏰔 6.98 􏰁 10􏰒16 A 5 × 10−15 50 0.5 × 10−3 IB=β= 50 =10A 􏱾I􏱿 VB = VBE = VT ln B IS ISB = β = −16 = 10 A IC −5 (d) iB iC ISB B􏰀 C 􏱾10−5􏱿 D vBE B IS bi = 0.025 ln 10−16 = 0.633 V ISB 􏰔 b E b􏰔100 ISB􏰔6.91􏰁10􏰒18A B 􏰒 We can determine RB from RB = VCC − VB IB = 15−0.633 =1.44M􏱹 10−5 iE To obtain VCE = 1 V, we select RC according to V−V 6.21 Dividing Eq. (6.14) by Eq. (6.15) and substituting iC /iB = βforced gives βforced = ISevBE/VT − ISCevBC/VT (IS /β)ev BE /VT + ISC ev BC /VT Dividing the numerator and denominator of the right-hand side by ISC ev BC /VT and replacing vBE − vBC by VCEsat gives 􏱾I􏱿 S eVCEsat/VT − 1 RC=CC CE IC = 15−1 =28k􏱹 0.5 6.19 IS = 10−15 A Thus, a forward-biased EBJ conducting a current Chapter 6–4 of 1 mA will have a forward voltage drop V 􏱾I􏱿 VBE =VT ln I S : ISC BE βforced = This equation can be used to obtain eVCEsat/VT and 􏱾 􏱿 􏱾 10−3 􏱿 = 0.025 ln 10−15 hence VCEsat as 􏱾I􏱿 1+β S eVCEsat/VT = forced = 0.691 V ISC = 100IS = 10−13 A 1 − βforced/β 􏱼I1+β􏱽 1 IS eVCEsat/VT +1 β ISC Thus, a forward-biased CBJ conducting a 1-mA current will have a forward voltage drop VBC : 􏱾1×10−3 􏱿 VBC =VT ln 1×10−13 =0.576V When forward-biased with 0.5 V, the emitter–base junction conducts I = IS e0.5/0.025 = 10−15e0.5/0.025 = 0.49 μA and the CBJ conducts I = ISC e0.5/0.025 = 10−13e0.5/0.025 = 48.5 μA 6.20 The equations utilized are vBC =vBE −vCE =0.7−vCE iBC = ISC ev BC /VT = 10−13 ev BC /0.025 iBE = ISBevBE/VT = 10−17e0.7/0.025 iB =iBC +iBE iC=ISevBE/VT −iBC=10−15e0.7/0.025−iBC Performing these calculations for vCE = 0.4 V, 0.3 V, and 0.2 V, we obtain the results shown in the table below. This table belongs to Problem 6.20. ISC ⇒ VCEsat = VT ln SC forced Q.E.D. Forβ=100andISC/IS =100,wecanusethis equation to obtain VCEsat corresponding to the given values of βforced. The results are as follows: 6.22 IS 1 − βforced/β βforced 50 10 5 1 VCEsat (V) 0.231 0.178 0.161 0.133 The emitter–base voltage VEB is found as the voltage drop across the diode DB, whose scale vCE (V) vBC (V) iBC (μA) iBE (μA) iB (μA) iC (mA) iC /iB 0.4 0.3 0.016 14.46 14.48 1.446 100 0.3 0.4 0.89 14.46 15.35 1.445 94 0.2 0.5 48.5 14.46 62.96 1.398 29 current is ISB = IS /β , it is conducting a 10-μA current. Thus, 􏱾10 μA􏱿 VEB = VT ln ISB where ISB = β = 50 =2×10−16 A Referring to the figure, we see that IE = IB + IC = IC + IC β Thus, IS 10−14 􏱾 10 × 10−6 􏱿 IC = IB = 0.091 mA Chapter 6–5 IE 1 1 = 1 =0.909mA VEB = 0.025 ln 1+β 1+10 For direction of flow, refer to the figure. 􏱾I􏱿 VEB=VTln B 2 × 10−16 = 0.616 V Thus, VB = −VEB = −0.616 V The collector current can be found as IC =βIB = 50 × 10 = 500 μA = 0.5 mA The collector voltage can now be obtained from VC = −5+IC ×8.2 = −5+0.5×8.2 = −0.9 V The emitter current can be found as IE =IB +IC =10+500=510μA = 0.51 mA 6.23 AtiC =1mA, vEB =0.7V ISB IS 10−15 −16 where ISB=β=10=10 A 􏱾 0.091 × 10−3 􏱿 VEB = 0.025 ln 10−16 = 0.688 V Thus, VE = VB + VEB = 0 + 0.688 = 0.688 V Ifatransistorwithβ=1000issubstituted, IC = IE = 1 =0.999mA 1+11+1 At iC = 10 mA, vEB=0.7+VTln 1 β 1000 Thus, IC changes by 0.999 − 0.909 = 0.09 mA, a 􏱾10􏱿 = 0.7 + 0.025 ln(10) = 0.758 V 9.9% increase. 6.25 = 5 20 + 1 α= 20 =0.95 21 IS =αIEe−VEB/VT = 0.95 × 5e−(0.8/0.025) IB = IC =ISeVEB/VT = 0.238 A = 238 mA IE AtiC =100mA, vEB =0.7+0.025ln 1 β + 1 αIE =ISeVEB/VT 􏱾100􏱿 Note that vEB increases by about 60 mV for every = 0.815 V decade increase in iC . 6.24 where =6×10−14 A A transistor that conducts IC = 1 mA with VEB = 0.70 V has a scale current IS = 1 × 10−3e−0.70/0.025 = 6.9 × 10−16 A The emitter–base junction areas of these two transistors will have the same ratio as that of their scale currents, thus EBJ area of first transistor 6 × 10−14 EBJ area of second transistor = 6.9 × 10−16 = 87 6.26 The two missing large-signal equivalent circuits for the pnp transistor are those corresponding to the npn equivalent circuits in Fig. 6.5(b) and 6.5(d). They are shown in the figure. E 􏰀 iE (b) Refer to Fig. P6.28(b). SinceVC =−4VislowerthanVB =−2.7V,the transistor is operating in the active mode. IC = −4−(−10) = 2.5 mA 2.4 k􏱹 IE = IC ≃ IC = 2.5 mA α V3 =+12−IE ×5.6=12−2.5×5.6=−2V (c) Refer to Fig. P6.28(c) and use IC = 0−(−10) =0.5mA 20 Assuming active-mode operation, and utilizing the fact that β is large, IB ≃ 0 and V4 ≃ 2 V Since VC < VB, the transistor is indeed operating in the active region. I5 =IE = IC ≃IC =0.5mA α (d) Refer to Fig. P6.28(d). Since the collector is connected to the base with a 10-k􏱹 resistor and β is assumed to be very high, the voltage drop across the 10-k􏱹 resistor will be close to zero and the base voltage will be equal to that of the collector: VB = V7 This also implies active-mode operation. Now, VE = VB − 0.7 Thus, VE = V7 − 0.7 I6 = VE − (−10) 3 = V7 −0.7+10 = V7 +9.3 (1) 33 Since IB = 0, the collector current will be equal to the current through the 9.1-k􏱹 resistor, IC=+10−V7 (2) 9.1 Sinceα1≃1,IC =IE =I6 resultingin 10−V7 = V7 +9.3 9.1 3 ⇒V7 =−4.5V and V7 + 9.3 I6= 3 = 3 =1.6mA Chapter 6–6 vEB iB 􏰒 B DE (IS /a) aiE iC C 6.27 6.28 (a) Refer to Fig. P6.28(a). 10.7 − 0.7 I1 = 5 k􏱹 = 2 mA Assuming operation in the active mode, IC =αI1 ≃I1 =2mA V2 =−10.7+IC ×5 = −10.7 + 2 × 5 = −0.7 V Since V2 is lower than VB, which is 0 V, the transistor is operating in the active mode, as assumed. −4.5 + 9.3 6.29 (a) Observe that the transistor is operating in the active region and note the analysis performed on the circuit diagram. Thus, IC =IE −IB =3−0.04=2.96mA and β≡IC =2.96=74 Chapter 6–7 IB 0.04 Since VC is lower than VB, the transistor is operating in the active region. From the figure corresponding to Fig. P6.29(a), we see that 6.30 VCC IE RC VC IC =1mA IB = 0.0215 mA Thus, 1 0.0215 IB IC β≡IC = IB (b) =46.5 VB IB IE ObservethatwithVC at3VandVB at4.3V,the transistor is operating in the active region. Refer to the analysis shown in the figure, which leads to β≡ IC = 3.952 =82.3 IB 0.048 (c) Since the meter resistance is small, VC ≃ VB and the transistor is operating in the active region. To obtain IE = 1 mA, we arrange that VBE = 0.7 V. Since VC ≃ VB, VC must be set to 0.7 by selecting RC according to VC =0.7=VCC −IERC Thus, 0.7=9−1×RC ⇒RC =8.3k􏱹 Since the meter reads full scale when the current flowing through it (in this case, IB is 50 μA), a full-scale reading corresponds to β ≡ IC ≃ 1 mA = 20 IB 50 μA If the meter reads 1/5 of full scale, then IB =10μAand β= 1mA =100 10 μA A meter reading of 1/10 full scale indicates that β = 1 mA = 200 5 μA 6.31 RC = VC − (−2.5) IC = −0.5+2.5 =4.04k􏱹≃4k􏱹 0.495 The transistor VEB can be found from 􏱾􏱿 VEB =0.64+VT ln 0.5mA 0.1 mA = 0.68 V Thus, VE =+0.68V and RE = 2.5−0.68 = 3.64 k􏱹 0.5 The maximum allowable value for RC while the transistor remains in the active mode corresponds to VC = +0.4 V. Thus, RCmax = 0.4 − (−2.5) = 5.86 k􏱹 0.495 6.33 Refer to Fig. 6.15(a) with RC = 5.1 k􏱹 and RE = 6.8 k􏱹. Assuming VBE ≃ 0.7 V, then VE =−0.7V,and −0.7 − (−15) IE = 6.8 =2.1mA IC =αIE ≃2.1mA VC = 15−2.1×5.1 ≃ 4.3 V 6.34 Refer to the circuit in Fig. P6.34. Since VC = 0.5 V is greater than VB, the transistor will be operating in the active mode. The transistor VBE can be found from 􏱾0.2 mA􏱿 VBE =0.8+0.025ln 1mA = 0.76 V Thus, VE =−0.76V Chapter 6–8 IC IB 􏰀2.5 V 5 k􏰵 b 􏰔 50 􏰒2.5 V I =VE−(−2.5) E 10 = −0.7+2.5 = 0.18 mA 10 IE VC VE 􏰔 􏰒0.7 V 10 k􏰵 Assuming the transistor is operating in the active mode, we obtain IE 0.18 IB = β+1 = 50+1 =3.5μA IC = β IE = 50 × 0.18 = 0.176 mA β+1 51 VC = +2.5 − ICRC = 2.5−0.176×5 = 1.62 V Since VC > VB, active-mode operation is verified.
􏱾􏱿
6.32
􏰀2.5 V
IE 􏰔 0.5 mA
RE
V C
IC RC
􏰒2.5 V
VE
􏰔􏰒0.5 V
I 􏱾β+1􏱿 101 IE=C=IC =0.2×
α β 100 = 0.202 mA
From the figure we see that VC = −0.5 V is lower than the base voltage (VB = 0 V); thus the transistor will be operating in the active mode.
􏱾 β 􏱿 100
IC =αIE = β+1 IE = 100+1 ×0.5
= 0.495 mA
The required value of RE can be found from RE = VE − (−1.5)
IE
−0.76 + 1.5
RE = 0.202 = 3.66 k􏱹
ToestablishVC =0.5V,weselectRC accordingto
RC = 1.5−0.5 =5k􏱹 0.2

6.35 (a) 4
IC 􏰔a􏰁0.26􏰔0.98􏰁0.26 􏰔 0.255 mA
􏰀1.5 V a 2.7 k􏰵
VC 􏰔 1.5 􏰒 0.255 􏰁 2.7 􏰔 0.81 V 5
VE 􏰔􏰒0.8 V 2 2.7 k􏰵
Chapter 6–9
6
IB 􏰔 IE 􏰒 IC 􏰔 0.005 mA
􏰀
0.8 V
1 􏰒
3 IE 􏰔􏰒0.8 􏰒 (􏰒1.5) 2.7
􏰔 0.26 mA
􏰒1.5 V 􏰀1.5 V
(b)
3
I 􏰔1.5􏰒0.8
E 2 2 k􏰵
􏰔 0.35 mA
2
VC 􏰔􏰒1.5􏰀2􏰁0.343
IB 􏰔 IE 􏰒 IC IB
􏰔 0.007 mA 􏰒
VE 􏰔 􏰀0.8 V 1􏰀
0.8 V
6
(c)
4 IC 􏰔 a 􏰁 0.35 2 k􏰵 􏰔 􏰒0.81 V 􏰔 0.98 􏰁 0.35
􏰔 0.343
􏰒1.5 V
5
2
􏰀3 V 3 IE 􏰔 3 􏰒 1.8 􏰔 0.12 mA
10
􏰀1 V 1 􏰀 0.8 V
􏰒
6 IB􏰔IC/50􏰔2.4􏱌A
10 k􏰵
VE 􏰔 􏰀1.8 V
4 IC􏰔a􏰁0.12
􏰔 0.98 􏰁 0.12
􏰔 0.118 mA
􏰀3 V
VC 􏰔0.118􏰁2􏰔0.236V 5 2 k􏰵
8.2 k􏰵
VC 􏰔3􏰒0.147􏰁8.2􏰔1.8V 5
VE􏰔1.5􏰒0.8􏰔0.7 2 4.7k􏰵
indicated on the corresponding circuit diagrams; the order of the steps is shown by the circled numbers.
(d)
4
IC 􏰔 a 􏰁 0.15 􏰔 0.147 mA 􏰀1.5 V
6 IB􏰔0.15􏰔3􏱌A
50 IB
􏰀
0.8 V 4.7
􏰒
􏰔 0.7 􏰔 0.15 mA 4.7
In all circuits shown in Fig. P6.35, we assume active-mode operation and verify that this is the case at the end of the solution. The solutions are
V 1 3 IE􏰔 E

6.36 ICBO approximately doubles for every 10◦C rise in temperature. A change in temperature from 25◦C to 125◦C—that is, an increase of 100◦C—results in 10 doublings or, equivalently, an increase by a factor of 210 = 1024. Thus ICBO becomes
Thus,
At20◦C, T =293KandV =
8.62 × 10−5 × 293 = 25.3 mV At50◦C, T =323KandVT =
8.62×10−5×323=27.8mV
If the transistor is operated at vBE = 700 mV,
then
(i) At 20◦C, iE becomes
iE = 0.5e(700−692)/25.3 = 0.69 mA (ii) At 50◦C, iE becomes
iE = 0.5e(700−632)/27.8 = 5.77 mA
6.40 vBE =0.7VatiC =10mA ForvBE =0.5V,
iC = 10e(0.5−0.7)/0.025 = 3.35 μA
At a current IC and a BE voltage VBE , the slope of the iC –v BE curve is IC /VT . Thus,
of700mV= 10mA =400mA/V 25 mV
ICBO = 10 nA×1024 = 10.24 μA 6.37
From the figure we can write
Chapter 6–10
T
􏱾I􏱿
IB = S evBE/VT −ICBO
(1)
When the base is left open-circuited, iB = 0 and Eq. (1) yields
β
IC =ISevBE/VT +ICBO
(2) IE=IS1+βevBE/VT (3)
SlopeatV
SlopeatVBE of500mV= 25mV
􏱾1􏱿
BE
6.38 Since the BJT is operating at a constant emitter current, its |VBE | decreases by 2 mV for every ◦C rise in temperature. Thus,
|VBE| at 0◦C = 0.7+0.002×25 = 0.75 V |VBE| at 100◦C = 0.7−0.002×75 = 0.55 V
6.39 (a) If the junction temperature rises to 50◦C, which is an increase of 30◦C, the EB voltage decreases to
vEB = 692−2×30 = 632 mV
(b) First, we evaluate VT at 20◦C and at 50◦C:
VT = kT q
where k = 8.62 × 10−5 eV/K.
To find the intercept of the straight-line characteristics on the ic axis, we substitute vCE = 0 and evaluate
iC = 10−15evBE/VT A
for the given value of v BE . The slope of each straight line is equal to this value divided by 100 V (VA ). Thus we obtain
3.35 μA
􏱾I􏱿
ICBO = S evBE/VT β
or equivalently,
ISevBE/VT =βICBO Substituting for IS ev BE /VT iC =iE =(β+1)ICBO
= 0.134 mA/V Ratio of slopes =
400 0.134
≃ 3000
(4) in Eqs. (2) and (3) gives
􏱾v􏱿 1+ CE
6.41 Use Eq. (6.18):
iC =ISevBE/VT
withIS =10−15 AandVA =100V,toget
−15 iC=10 e
VA vBE/0.025􏲃 vCE 􏲄
1+100
vBE
0.65 V
0.70 V
0.72 V
0.73 V
0.74 V
vCE (V)
iC
(mA)
iC
(mA)
iC
(mA)
iC
(mA)
iC
(mA)
0
0.196
1.45
3.21
4.81
7.16
15
0.225
1.67
3.70
5.52
8.24

vBE (V)
0.65
0.70
0.72
0.73
0.74
Intercept (mA)
0.2
1.45
3.22
4.80
7.16
Slope (mA/V)
0.002
0.015
0.032
0.048
0.072
6.42
I1 R1
6.8 k􏰵
􏰒
􏱾 6.8 􏱿
=−VBE 1+ 0.68 =−11VBE =−7.48V
which gives this circuit the name “VBE multiplier.” A more accurate value of VE can be obtained by taking IB into account:
VE = −(I1R1 + I2R2) 􏱾R􏱿
=− VBE+ 2VBE+IBR2 R1
􏱾R􏱿
=− 1+ 2 VBE−IBR2 (2)
R1
= −7.48 − 0.01 × 68 = −8.16 V
As temperature increases, an approximate estimate for the temperature coefficient of VE can be obtained by assuming that IE remains constant and ignoring the temperature variation of β. Thus, we would be neglecting the temperature change of the (IBR2) terms in Eq. (2). From Eq. (2) we can obtain the temperature coefficient of VE by utilizing the fact the VBE changes by – 2.2 mV/◦C. Thus,
Temperature coefficient of VE 􏱾R􏱿
=− 1+ 2 ×−2.2 R1
= −11 × −2.2 = +24.2 mV/◦C
At 75◦C, which is a temperature increase of 50◦C,
VE =−8.16+24.2×50=−6.95V
As a check on our assumption of constant IE , let us find the value of IE at 75◦C:
I1(75◦C) = VBE(75◦C) R1
0.68−2.2×10−3 ×50 = 6.8
= 0.084 mA
IE(75◦C) = I − I1(75◦C) = 1.1 − 0.084 = 1.016 mA
which is reasonably close to the assumed value of 1 mA.
6.43 ro = 1/slope = 1/(0.8 × 10−5) = 125 k􏱹
ro = VA IC
125k􏱹= VA ⇒VA =125V 1 mA
Chapter 6–11
R2 68 k􏰵
I2 IB
􏰀
VBE
IE VE
I 􏰔 1.1 mA
At25◦C,assumeIE =1mA.Thus, VBE = 0.68 V
I1 = VBE = 0.68V =0.1mA R1 6.8 k􏱹
IE =I−I1 =1.1−0.1=1mA which is the value assumed.
I2 =I1 +IB =I1 + IE β+1
=0.1+ 1 =0.11mA 101
Note that the currents in R1 and R2 differ only by the small base current, 0.01 mA. Had I1 and I2 been equal, then we would have had
(1)
I1R1 = VBE IR≃IR=V R2
22 12 BER1 VE =−(I1R1 +I2R2)
AtIC =10mA,
V 125V
􏱾 R􏱿
=−VBE 1+ 2 R1
ro= A= =12.5k􏱹 IC 10 mA

Chapter 6–12
6.44 ro = VA = 50 V IC IC
Thus,
AtIC =1mA, ro = 50V =50k􏱹 1 mA
AtIC =100μA, ro = 50V =500k􏱹 0.1 mA
6.45
􏰒VA
Slope of iC–vCE line corresponding to
(b)
E
DB (IS/b)
iC (mA) 1.3
1.1
iE
􏰀
vEB
ro
biB
vBE =710mVis
1.3 − 1.1 0.2 mA
△i 􏲀􏲀􏲀
βac=C􏲀 = =80
0 5 10 15 vCE(V) 0.3
B􏰒C iB iC
6.47 β = iC = 1 mA = 100 iB 10 μA
Slope= 15−5 = 10V =0.02mA/V
0.08mA △iB v CE constant 1.0 μA
Near saturation, VCE = 0.3 V, thus iC = 1.1−0.02×(5−0.3)
= 1.006 ≃ 1 mA
iC willbe1.2mAat,
△iC =△iB×βac+△vCE ro
where
ro = VA = 100 =100k􏱹
IC 1
0.02 23
vCE =5+ 1.2−1.1 =10V TheinterceptoftheiC–vCE straightlineontheiC
axis will be at
iC = 1.1−5×0.02 = 1 mA
Thus, the Early voltage is obtained as
Slope = iC(at vCE = 0) VA
Thus,
△iC =2×80+ 100 ×10 =180μA = 0.18 mA
6.48 Refer to the circuit in Fig. P6.48.
(a) For active-mode operation with VC = 2 V: IC=VCC−VC =10−2=8mA
1 RC1
⇒ VA = 0.02 = 50 V
ro = VA = 50V =50k􏱹
IC 1 mA
which is the inverse of the slope of the iC–vCE
line.
6.46 The equivalent circuits shown in the figure correspond to the circuits in Fig. 6.19.
IC 8
IB = β = 50 = 0.16 mA
VBB =IBRB +VBE
= 0.16×10+0.7 = 2.3 V
(b) For operation at the edge of saturation: VCE = 0.3 V
IC = VCC − VCE = 10 − 0.3 = 9.7 mA RC 1

IB = IC = 9.7 =0.194mA β 50
VBB =IBRB +VBE
= 0.194×10+0.7 = 2.64 V
(c) For operation deep in saturation with βforced = 10:
VCE = 0.2 V VCC − VCE
βforced 10 VBB =IBRB +VBE
= 0.98×10+0.7 = 10.5 V
6.49 Refer to the circuit in Fig. P6.48 (with VBB = VCC ) and to the BJT equivalent circuit of Fig. 6.21.
IC = VCC − 0.2 RC
IB = VCC −0.7 RB
βforced ≡ IC IB
RB =12.1k􏱹
Substituting in Eq. (3), we have
RC = 1.35 k􏱹
From the table of 1% resistors in Appendix J we select
RC = 1.37 k􏱹 For these values:
5−0.2
IC= 1.37 =3.5mA
5−0.7
IB = 12.1 = 0.36 mA
Thus,
βforced = 3.5 = 9.7 0.36
Pdissipated =VCC(IC +IB) = 5×3.86 = 19.2 mW
Chapter 6–13
=9.8mA IB = IC = 9.8 =0.98mA
IC = R C
=
10 − 0.2 1
6.50
􏰀5 V
Thus,
βforced = CC B
􏰀
VECsat 􏰔 0.2 V 􏰒
V 􏰔 􏰀4.8 V C
1 k􏰵
􏱾V −0.2􏱿􏱾R􏱿 VCC−0.7 RC
(1)
(2)
VB 􏰔 􏰀4.3 V RB 􏰔 10 k􏰵
I B
IC
Assume saturation-mode operation. From the figure we see that
Pdissipated =VCC(IC +IB)
=VCC(βforcedIB +IB)
= (βforced + 1)VCC IB
ForVCC =5Vandβforced =10and
Pdissipated ≤ 20 mW, we can proceed as follows.
Using Eq. (1) we can determine (RB /RC ):
􏱾5−0.2􏱿􏱾R 􏱿 B
V 4.8
10=
⇒ RB =8.96
C = 1 k􏱹
=4.8mA 1
5 − 0.7 RC
IC =
IB = VB = 4.3 =0.43mA
RB 10
Thus,
βforced ≡ IC = 4.8 = 11.2
IB 0.43
Since 11.2 is lower than the transistor β of 50, we have verified that the transistor is operating in saturation, as assumed.
VC =VCC −VECsat =5−0.2=4.8V
To operate at the edge of saturation,
VEC =0.3V and IC/IB =β=50
RC
Using Eq. (2), we can find IB: (10+1)×5×IB ≤ 20 mW ⇒IB ≤0.36mA
Thus,
VCC − 0.7 ≤ 0.36 mA RB
⇒RB ≥11.9k􏱹
(3)
From the table of 1% resistors in Appendix J we select

Thus,
5 − 0.3
The analysis and the results are given on the circuit diagrams of Figs. 1 through 3. The circled numbers indicate the order of the analysis steps.
6.52
􏰀3 V
1 k􏰵
IC 􏰔3􏰒(VB 􏰒0.4) VC 􏰔VB 􏰒0.4 1􏰀
IC = 1
β 50
= 4.7 mA
IB = IC = 4.7 =0.094mA
Chapter 6–14
RB = 4.3 = 4.3 =45.7k􏱹
IB
6.51
0.094
1 k􏰵
􏰀3 V
IC
IE 􏰔 1.3 mA 3 􏰔 3 􏰒 1.3 􏰁 1
VB
IE 􏰔 VB 􏰒 0.7
VCE 􏰔 0.3 V 􏰒VB 􏰒0.7
1 k􏰵
V C1
􏰔 􏰀1.7 V 4 VE 􏰔􏰀2􏰒0.7
􏰔 1.3 V 1 IE􏰔1.3􏰔1.3mA 2
1
(a)
􏰀3 V
􏰀2 V
1 k􏰵
Figure1
1 k􏰵
1 k􏰵
1 k􏰵
1 k􏰵
IC
3
VC 􏰔 3 􏰒 1 􏰁 1 􏰔 􏰀2 V
4
IE 􏰔 1 mA
Figure 1 shows the circuit with the value of VB that results in operation at the edge of saturation. Since β is very high,
IC ≃IE
3−(VB −0.4) = VB −0.7
11 ⇒ VB = 2.05 V
􏰀1.7 V
􏰀3 V
1 k􏰵
􏰀VB 􏰒 0.5 VCEsat 􏰔 0.2 V
􏰒VB 􏰒 0.7 1 k􏰵
Figure 2
IC 1 IB
VE 􏰔 􏰀1.0 V 1 IE 􏰔1.0
􏰔1 mA 2 (b)
VB
􏰀3 V
03
IE
VC 􏰔 3 􏰒 0 􏰔 3 V
VE 􏰔 0 V 1 02
(c)
4
0 V
Figure 2 shows the circuit with the value of VB that results in the transistor operating in saturation, with
IE = VB − 0.7 = VB − 0.7 1
IC = 3−(VB −0.5) = 3.5−VB 1
IB =IE −IC =2VB −4.2

For βforced = 2, IC =2
IB
3.5−VB =2
2VB − 4.2 ⇒VB =2.38V
6.53 Refer to the circuit in Fig. P6.53. (a) ForVB =−1V,
VE =VB −VBE =−1−0.7=−1.7V IE = VE −(−3) = −1.7+3 =1.3mA
11
Assuming active-mode operation, we have
IC =αIE ≃IE =1.3mA
VC =+3−IC ×1=3−1.3=+1.7V
SinceVC >VB−0.4,thetransistorisoperatingin the active mode as assumed.
(b) ForVB =0V,
VE =0−VBE =−0.7V
−0.7 − (−3)
IE = 1 =2.3mA
Assuming operation in the active mode, we have
IC =αIE ≃IE =2.3mA
VC =+3−IC ×1=3−2.3=+0.7V
Since VC > VB − 0.4, the BJT is operating in the active mode, as assumed.
(c) ForVB =+1V,
VE =1−0.7=+0.3V
I = 0.3 − (−3) = 3.3 mA E1
Assuming operation in the active mode, we have IC =αIE ≃IE =3.3mA
VC =3−3.3×1=−0.3V
Now VC < VB − 0.4, indicating that the transistor is operating in saturation, and our original assumption is incorrect. It follows that VC =VE +VCEsat = 0.3+0.2 = 0.5 V IC = 3−VC = 3−0.5 =2.5mA 11 IB =IE −IC =3.3−2.5=0.8mA βforced=IC=2.5=3.1 IB 0.8 (d) WhenVB =0V,IE =2.3mA.Theemitter current becomes 0.23 mA at VB =−3+0.23×1+0.7=−2.07V (e) The transistor will be at the edge of conduction when IE ≃ 0 and VBE = 0.5 V, that is, VB =−3+0.5=−2.5V Inthiscase, VE =−3V VC =+3V (f) The transistor reaches the edge of saturation whenVCE =0.3VbutIC =αIE ≃IE: VE = VB − 0.7 VB −0.7−(−3) Chapter 6–15 IE = 1 = VB + 2.3 VC =VE +0.3=VB −0.4 3−VC 3−VB +0.4 1 = 1 IC ≃IE 3.4−VB =VB +2.3 VB = 0.55 V For this value, VE =0.55−0.7=−0.15V VC =−0.15+0.3=+0.15V =3.4−VB IC = Since (g) For the transistor to operate in saturation with βforced = 2, VE = VB − 0.7 IE = VB −0.7−(−3) = VB +2.3 1 VC = VE +VCEsat = VB −0.7+0.2 = VB −0.5 IC = 3−(VB −0.5) = 3.5−VB 1 IB = IE − IC = 2 VB − 1.2 IC = 3.5−VB =2 IB 2 VB − 1.2 ⇒VB =+1.18V 6.54 (a) VB =0V 􏰀5 V (c) VB 􏰀5 V 4 mA 􏰀 1 k􏰵 0.3 V Chapter 6–16 4 mA 4 􏰒 1.3 􏰔 2.7 mA 5 VC 􏰔 VB 􏰒 0.4 VB 􏰒 0.4 VC 􏰔 2.7 􏰁 1 IC 4 3 1.3 mA 􏰔 2.7 V 6 1 k􏰵 2 􏰒 0.7 􏰔 1.3 mA IE 􏰒 1 VE 􏰔VB 􏰒0.7 0.7 mA 2 VE 􏰔 􏰒0.7 V 1 k􏰵 1 2 mA 1 k􏰵 Figure 3 2 mA 􏰒5 V Figure 1 VB 􏰒 0.7 1 The analysis is shown on the circuit diagram in Fig. 1. The circled numbers indicate the order of the analysis steps. (b) The transistor cuts off at the value of VB that causes the 2-mA current of the current source feeding the emitter to flow through the 1-k􏱹 resistor connected between the emitter and ground. The circuit under these conditions is shown in Fig. 2. Figure 3 shows the transistor at the edge of saturation. Here VCE = 0.3 V and IC =αIE ≃IE.Anodeequationattheemitter gives 􏰀5 V 4 mA 4 mA 0 0 2 mA 􏰒5 V Figure 2 VC 􏰔 􏰀4 V 1 k􏰵 VE 􏰔 􏰒2 V 1 k􏰵 IE = 2 + VB − 0.7 = VB + 1.3 mA A node equation at the collector gives IC =4−(VB −0.4)=4.4−VB mA Imposing the condition IC ≃ IE gives 4.4−VB =VB +1.3 ⇒VB =+1.55V Correspondingly, VE = +0.85 V VC =+1.15V VB 􏰔 􏰒1.5 V 2 mA 6.55 0.1 mA R1 VCC 􏰔 􏰀 3 V 0 Figure 1 RC 􏰔 1 k􏰵 RE 􏰔 1 k􏰵 ObservethatVE =−2mA×1k􏱹=−2V, IE =0,andVB =VE +0.5=−1.5V.Since IC = 0, all the 4 mA supplied by the current source feeding the collector flows through the collector 1-k􏱹 resistor, resulting in VC = +4 V. VB 􏰔 􏰀1.2 V R2 From Fig. 1 we see that R1+R2= VCC = 3 =30k􏱹 0.1 6.56 Refer to the circuit in Fig. P6.56. VE=1V IE = 3 − 1 = 0.4 mA 5 0.1 mA V R2 = 1.2 Chapter 6–17 CC R1 + R2 3×R2 =1.2 VB = VE − 0.7 = 0.3 V IB= VB =0.3=0.006mA 30 ⇒R2 =12k􏱹 50 k􏱹 50 IC = IE −IB = 0.4−0.006 = 0.394 mA R1 =30−12=18k􏱹 For β = 100, to obtain the collector current, we replace the voltage divider with its Thévenin equivalent, consisting of VC =−3+5×0.394=−1.03V Observe that VC < VB, confirming our implicit assumption that the transistor is operating in the active region. β=IC =0.394=66 IB 0.006 IC 0.394 α= I = 0.4 =0.985 R VBB =3× 2 12 18 + 12 =1.2V B12 E =3× R =R ∥R =12∥18=7.2k􏱹 R1 + R2 􏰀3 V 􏰀3 V 􏰀 VBC 􏰔 1 V 􏰒 IC 6.57 IE RE VBB R B IB IC IE RC 􏰔 1 k􏰵 VC RE 􏰔 1 k􏰵 VE 􏰔 􏰀0.7 V VC 􏰔 􏰒1 V RC Figure 2 􏰒3 V Refer to the figure. To obtain IE = 0.5 mA we select RE according to RE = 3 − 0.7 = 4.6 k􏱹 0.5 To obtain VC = −1 V, we select RC according to RC = −1−(−3) =4k􏱹 0.5 where we have utilized the fact that α ≃ 1 and thusIC ≃IE =0.5mA.Fromthetableof5% resistors in Appendix J we select RE =4.7k􏱹 and RC =3.9k􏱹 For these values, IE = 3−0.7 =0.49mA 4.7 IC ≃IE =0.49mA VBC =0−VC =−(−3+0.49×3.9)=−1.1V Refer to Fig. 2. Assuming active-mode operation, we can write a loop equation for the base–emitter loop: VBB =IBRB +VBE +IERE IE 1.2 = β +1 ×7.2+0.7+IE ×1 ⇒IE = 1.2−0.7 =0.47mA 1 + 7.2 101 IC =αIE =0.99×0.47=0.46mA VC =+3−0.46×1=+2.54V SinceVB =IERE +VBE =0.47+0.7=1.17V, we see that VC > VB − 0.4, and thus the transistor isoperatingintheactiveregion,asassumed.

6.58
􏰀3 V
3−0.7 = 3−0.7 2.2+20 2.2+ 100
Chapter 6–18
RE
IC
2.2 k􏰵
41 β+1 20 100
IE
V
⇒ 41 = β + 1 β+1= 410 =205
IB
RB
20 k􏰵
E
VB
2
VC
2.2 k􏰵
β = 204
􏰒3 V
Writing a loop equation for the EBJ loop, we have
􏰀3 V
RC
6.59
IE
RE
2.2 k􏰵 VE
VB RC
3=IERE +VEB +IBRB =IE ×2.2+0.7+ IE
(1)
×20 20 = 0.86 mA
VE =3−0.86×2.2=+1.11V
VB =VE −0.7=+0.41V
Assuming active-mode operation, we obtain
IC =αIE =40×0.86=0.84mA 41
VC =−3+0.84×2.2=−1.15V
VB
β+1 2.2 +
RB IB 20 k􏰵
IC
⇒ IE =
3−0.7 41
Since VC < VB + 0.4, the transistor is operating in the active mode, as assumed. Now, if RB is increased to 100 k􏱹, the loop equation [Eq. (1)] yields Assume active-mode operation: IE = 3−VEB RE + RB β+1 IE = 3 − 0.7 = 0.887 mA 2.2 + 20 2.2 k􏰵 􏰒3 V IE = 100 =0.5mA 2.2+41 IE 0.887 IB=β+1= 51 =0.017mA IC =IE −IB =0.887−0.017=0.870mA VB =IBRB =0.017×20=0.34V VE = VB + VEB = 0.34 + 0.7 = 1.04 V VC =−3+ICRC =−3+0.87×2.2=−1.09V Thus, VC < VB + 0.4, which means active-mode operation, as assumed. The maximum value of RC that still guarantees active-mode operation is that which causes VC to be 0.4 V above VB: that is, VC =0.34+0.4=0.74V. Correspondingly, 0.74 − (−3) RCmax = 0.87 = 4.3 k􏱹 3−0.7 51 VE =3−0.5×2.2=+1.9V VB =VE −VEB =1.9−0.7=+1.2V Assuming active-mode operation, we obtain IC =αIE =40×0.5=0.48mA 41 VC =−3+0.48×2.2=−1.9V SinceVC VB−0.4,thetransistorisoperatingin the active region, as assumed.
(b) For RB = 10 k􏱹,
IE = 4.3 =3.91mA
VE =3.91×1=3.91V
VB = 3.91+0.7 = 4.61 V
Assuming active-mode operation, we obtain
IC =αIE =0.99×3.91=3.87mA
VC =5−3.87=+1.13V
SinceVC −2.5,VC willbelowerthanVB and Q2 will be operating in the active region. Thus
IE = 2.5−0.7 = 1.5 mA 1 + 10
51
VE =−IE ×1=−1.5V
VB =−1.5−0.7=−2.2V
(d) ForvI =−5V,Q1 willbeoffandQ2 willbe
on,andthecircuitreducestothatinFig.3.
Chapter 6–29
􏰀2.5 V
10 k􏰵
􏰀2 V Q1
IB
VB
Figure1
VE
IE
1 k􏰵
Figure 3
Here we do not know whether Q2 is operating in the active mode or in saturation. Assuming active-mode operation, we obtain
IE = 5−0.7 =3.6mA 1 + 10
51 VE =−3.6V
VB = −4.3 V
which is impossible, indicating that our original
assumption is incorrect and that Q2 is saturated. Assuming saturation-mode operation, we obtain
Since VB will be lower than +2 V, VC will be higher than VB and the transistor will be operating in the active mode. Thus,
IE = 2−0.7 =1.1mA 1 + 10
51 VE =+1.1V
VB =1.8V
(c) For vI = −2.5, Q1 will be off and Q2 will be
on, and the circuit reduces to that in Fig. 2.
VE =VC +VECsat =−2.5+0.2=−2.3V −VE
IE = 1 k􏱹 = 2.3 mA
Figure 2
VB =VE −0.7=−3V
IB = −3−(−5) = 0.2 mA 10
IC =IE −IB =2.3−0.2=2.1mA βforced = IC = 2.1 = 10.5
IB 0.2
which is lower than β, verifying that Q2 is operating in saturation.

6.69 (a)
(b)
􏰀5 V
IC
􏰀5 V
Chapter 6–30
20 k􏰵 IB
VE + 0.7 V
IE
Assuming saturation-mode operation, the terminal voltages are interrelated as shown in the figure, which corresponds to Fig. P6.69(a). Thus we can write
10 k􏰵
VE + 0.2
IE
IC
(b)
1 k􏰵 VE
VE –0.2 1 k􏰵
VE – 0.7
VE 10k􏰵
IB
1 k􏰵
–5 V
(a)
IE = VE = VE
1 E1E
I = 5 − VE = 5 − V
IC = 5−(VE +0.2) = 0.5−0.1(VE +0.2) I = VE −0.2 = V −0.2
10 C1E
IB = 5−(VE +0.7) = 0.25−0.05(VE +0.7) I = VE −0.7−(−5) = 0.1 V +0.43
20 B10E
Assuming saturation-mode operation, the terminal voltages are interrelated as shown in the figure, which corresponds to Fig. P6.69(b). We can obtain the currents as follows:
Now, imposing the constraint
IE = IC + IB
results in
VE =0.5−0.1(VE+0.2)+0.25−0.05(VE+0.7) ⇒VE =0.6V
VC =0.8V
VB =1.3V
Imposing the constraint
IE = IB + IC
results in
5−VE = VE −0.2+0.1 VE +0.43 ⇒VE =+2.27V
VC =+2.07V
VB =1.57V
IC = 5−0.8 =0.42mA IB = 5−1.3 = 0.185 mA
20
βforced = 0.42 = 2.3 0.185
which is less than the value of β1 verifying saturation-mode operation.
IC = 2.07 =2.07mA 10 1
IB = 1.57−(−5) = 0.657 mA 10
βforced = IC = 2.07 = 3.2 IB 0.657
which is lower than the value of β, verifying saturation-mode operation.

This figure belongs to Problem 6.69, part (c).
Chapter 6–31
􏰀5 V
IE3 10 k􏰵 IC4
30 k􏰵
V + 0.2
V – 0.5 Q3
IB3 V Q4
Supernode
V – 0.5
V – 0.7 10 k􏰵
10 k􏰵
I
20 k􏰵
IE4
(c)
(c) We shall assume that both Q3 and Q4 are operating in saturation. To begin the analysis shown in the figure, which corresponds to Fig. P6.69(c), we denote the voltage at the emitter of Q3 as V and then obtain the voltages at all other nodes in terms of V , utilizing the fact that a saturated transistor has |VCE | = 0.2 V and of course |VBE | = 0.7 V. Note that the choice of the collector node to begin the analysis is arbitrary; we could have selected any other node and denoted its voltage as V . We next draw a circle around the two transistors to define a “supernode.” A node equation for the supernode will be
Thus
IE3 + IC4 = IB3 + I + IE4 where
IE3 = 5−(V +0.2) = 0.48−0.1V 10
IC4 = 5−(V −0.5) = 0.183−0.033V 30
IB3 = 5−(V −0.5) = 0.1V −0.05 10
I = V = 0.05V 20
I = V −0.7 = 0.1V −0.07 E4 10
(1)
(2)
(3)
(4)
(5)
VC3 = V = 2.044 V
VC4 = V − 0.5 = 1.54 V
Next we determine all currents utilizing Eqs. (2)–(6):
IE3 = 0.276 mA IC4 = 0.116 mA
IB3=0.154mAI=0.102mA
IE4 = 0.134
The base current of Q4 can be obtained from
IB4 = IE4 −IC4 = 0.134−0.116 = 0.018 mA
Finally, the collector current of Q3 can be found as
IC3 =I+IB4 =0.102+0.018=0.120 The forced β values can now be found as
(6) Substituting from Eqs. (2)–(6) into Eq. (1) gives
βforced3 = IC 3 IB3
βforced4 = IC 4 IB4
= 0.120 = 0.8 0.154
= 0.116 = 6.4 0.018
0.48 − 0.1V + 0.183 − 0.033V
= 0.1V −0.05+0.05V +0.1V −0.07
⇒V =2.044V
Both βforced values are well below the β value of 50, verifying that Q3 and Q4 are in deep saturation.