电子电路分析与设计模拟电子技术答案第17章.docx
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电子电路分析与设计模拟电子技术答案第17章.docx
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电子电路分析与设计模拟电子技术答案第17章
Microelectronics:
CircuitAnalysisandDesign,4theditionChapter17ByD.A.NeamenProblemSolutions______________________________________________________________________________________
Chapter17
17.1
(a)RC=0−(−0.2)=2kΩ0.1
(b)(i)υ1=−1V,Q1off,Q2on
υO2=0−(0.2)
(2)=−0.4V
υO1=0
(ii)υ1=−0.4V,Q1on,Q2off
υO1=0−(0.2)
(2)=−0.4V
υO2=0
(c)For(i)and(ii)
P=IQ0−V−=(0.2)(1.8)=0.36mW()()
______________________________________________________________________________________
17.2
−1−0.7−(−2.5)=0.08mA,⇒RE=10kΩ(a)iE=RE
RC=0−(−0.25)=6.25kΩ0.04
(b)(i)υ1=−1.3V,Q1off,Q2on
iE=
υO2
υO1=0
(ii)υ1=−0.7V,Q1on,Q2off
iE=−1−0.7−(−2.5)=0.08mA10=0−(0.08)(6.25)=−0.50V
υO1
υO2=0
(c)(i)iE=0.08mA,P=(0.08)(2.5)=0.2mW
(ii)iE=0.11mA,P=(0.11)(2.5)=0.275mW
______________________________________________________________________________________
17.3(a)
(b)
iC2=IQ=0.5=iC1=IQ=0.5=3−0⇒RC2=6KRC23−1⇒RC1=4KRC1−0.7−0.7−(−2.5)=0.11mA10=0−(0.11)(6.25)=−0.6875V
Microelectronics:
CircuitAnalysisandDesign,4theditionChapter17ByD.A.NeamenProblemSolutions______________________________________________________________________________________⎛V⎞ISexp⎜BE1⎟iC1⎝VT⎠=IQ⎡⎛V⎞⎛V⎞⎤IS⎢exp⎜BE1⎟+exp⎜BE2⎟⎥⎝VT⎠⎝VT⎠⎦⎣
⎛V−VBE1⎞1+exp⎜BE2⎟VT⎝⎠
vI=VBE1−VBE2
1
⎛−v⎞1+exp⎜I⎟⎝VT⎠
0.11==0.20.5⎛−vI⎞1+exp⎜⎟⎝VT⎠
⎛−v⎞1−1=4exp⎜I⎟=⎝VT⎠0.2
(−vI)=(0.026)ln(4)=1(c)SoiC1=IQI
______________________________________________________________________________________
17.4
(a)vI=0.5V,Q1on,Q2off=v02=3Vv=−0.0360V
v01=3−
(1)(0.5)=2.5V
(b)vI=−0.5VQ1off,Q2on⇒v01=3V
v02=3−
(1)(0.5)=2.5V______________________________________________________________________________________
17.5
(a)Q2on,vE=−1.2−0.7=−1.9V
−1.9−(−5.2)iE=iC2==1.32mA2.5
v2=−1V=−iC2RC2=−(1.32)(RC2)
C2
Qon,v=−0.7−0.7=−1.40V1E(b)
iE=iC1==1.52mA2.5
v1=−1V=−iC1RC1=−(1.52)(RC1)
R=0.658kΩ−1.4−(−5.2)R=0.758kΩC1
Forv=−0.7V,Qon,Qoffin12(c)
⇒vO1=−0.70V
vO2O2
Forvin=−1.7V,Q1off,Q2on=−1−0.7⇒v=−1.7V
⇒vO2=−0.7V
vO1=−1−0.7⇒vO1=−1.7V
Microelectronics:
CircuitAnalysisandDesign,4theditionChapter17ByD.A.NeamenProblemSolutions______________________________________________________________________________________
iE=mA(d)(i)Forvin=−0.7V,1.52
iC4=
iC3=−1.7−(−5.2)3−0.7−(−5.2)=1.17mA
or=1.5mA3P=(iE+iC4+iC3)(5.2)=(1.52+1.17+1.5)(5.2)P=21.8mW
=1.5mA(ii)Forvin=−1.7V,iE=1.32mAiC4=
iC3=−0.7−(−5.2)3−1.7−(−5.2)
=1.17mA3P=(1.32+1.5+1.17)(5.2)
or
______________________________________________________________________________________
17.6a.
or
b.I3=3.7−0.7=1.5mA0.67+1.33P=20.7mWVR=I3R4+Vγ=(1.5)(1.33)+0.7VR=2.70Vlogic1level
iE=
vB3=3.7−0.7⇒3.0VForvX=vY=logic1.3−0.7=2.875mA=iRC10.8=3.7−(2.875)(0.21)=3.10V
⇒v01(logic0)=2.4V
2.7−0.7=2.5mA=iRC20.8
=3.7−(2.5)(0.24)=3.1V
⇒v02(logic0)=2.4VForvX=vY=logic0,QRoniE=vB4
______________________________________________________________________________________
17.7
0.7−0.7−(−2.1)(a)R1==10.5kΩ0.20
0−(−2.1)(b)R5=R6==17.5kΩ0.12
(c)IQ=IREF=0.20mA
υO1=−0.7V,υC2=0RC1=
0.7−0=3.5kΩ0.2
Microelectronics:
CircuitAnalysisandDesign,4theditionChapter17ByD.A.NeamenProblemSolutions______________________________________________________________________________________
(d)IQ=IREF=0.20mA
υO2=−0.7V,υCR=00.7−0=3.5kΩ0.20
______________________________________________________________________________________
17.8
VR=−0.5VRC2=
iE=
R5=−0.5−0.7−(−3)=0.4mA,RE=4.5kΩRE−0.5−(−3)=6.25kΩ0.4
0.7−0.2=1.25kΩ0.4VB2=−0.5+0.7=0.2V,R1=
R2=0.2−0.7−0.7−(−3)=4.5kΩ0.4
0−(−3)R3=R4==3.75kΩ0.8
υOR=−1V,⇒υCR=−0.3V
0.7−(−0.3)=2.5kΩ0.4
−0.7−(−3)υI=0,iE==0.511mA4.5
υC1=−1+0.7=−0.3VRC2=
0.7−(−0.3)=1.957kΩ0.511
______________________________________________________________________________________
17.9
υO=logic1=1.8V,logic0=1.2VRC1=
ForυI=logic1=1.8ViE=0.8=1.8−0.7⇒RE=1.375kΩRE
2.5−1.9=0.75kΩ0.8
1.5−0.70.8ForυI=logic0,QRon;iE===0.5818mA1.375REυC1=1.2+0.7=1.9V,⇒RC1=
υCR=1.2+0.7=1.9V2.5−1.9=1.031kΩ0.5818
1.8R2=R3==2.25kΩ0.8
______________________________________________________________________________________
RC2=
Microelectronics:
CircuitAnalysisandDesign,4theditionChapter17ByD.A.NeamenProblemSolutions______________________________________________________________________________________
17.10
Neglectingbasecurrents:
(a)IE1=0,IE3=0
IE5=5−0.7⇒IE5=1.72mA2.5Y=0.7V
IE1=
IE35−0.7⇒IE1=0.239mA18=0(b)
(c)
I5−0.7⇒IE5=1.72mA2.5Y=0.7VIE5=IE1=IE3=5−0.7⇒IE1=IE3=0.239mA18E5(d)Sameas(c).
______________________________________________________________________________________
17.11
VR=−
(1)
(1)−0.7⇒VR=−1.7V
(b)QRoff,thenvO1=Logic1=−0.7V
QRon,thenvO1=−
(1)
(2)−0.7⇒=0,Y=5V(a)
vO1=Logic0=−2.7V
QA/QB−off,thenvO2=Logic1=−0.7V
QA/QB−on,thenvO2=−
(1)
(2)−0.7⇒
vO2=Logic0=−2.7V
(c)
(d)A=B=Logic0=−2.7V,QRon,VE=−1.7−0.7⇒VE=−2.4VVE=−0.7−0.7⇒VE=−1.4VA=B=Logic1=−0.7V,QA/QBon,A=B=Logic1=−0.7V,QA/QBon,
−2.7−(−5.2)iC3==1.67mA1.5
−0.7−(−5.2)iC2==3mA1.5
P=(1.67+1+1+1+3)(5.2)⇒P=39.9mW
A=B=Logic0=−2.7V
iC3=3,mAiC2=1.67mA
P=39.9mW______________________________________________________________________________________
Microelectronics:
CircuitAnalysisandDesign,4theditionChapter17ByD.A.NeamenProblemSolutions______________________________________________________________________________________
17.12
a.ANDlogicfunctionb.logic0=0V
Q3on,i=5−(1.6+0.7)=2.25mA1.2V2=(2.25)(0.8)⇒logic1=1.8V
5−0.7iE1=⇒iE1=1.65mA2.6
5−(0.7+0.7)iE2=⇒iE2=3mA1.2
iC3=0,iC2=iE2=3mA
V2=0
iE1=
iE2c.5−(1.8+0.7)⇒iE1=0.962mA2.65−(1.6+0.7)=⇒iE2=2.25mA1.2
iC2=0,iC3=iE2=2.25mA
V=1.8V2d.
______________________________________________________________________________________
17.13
3.5+3.1−0.7=2.6V(a)υR=(logic0+logic1)/2−0.7=2
(b)ForυX=υY=logic1=3.5V
υE1=3.5−0.7−0.7=2.1ViE1=
iRC12.1−0=0.175mA120.41⇒RC1=6.86kΩ=⋅iE1=0.05833mA=3RC1
(c)ForυX=υY=logic0=3.1V
υE1=VR−0.7=2.6−0.7=1.9ViE2=
iRC21.9=0.1583mA120.41⇒RC2=7.58kΩ=⋅iE2=0.05277mA=3RC2
(d)ForυX=υY=logic0=3.1V
iE=0.1583mA
3.1−0.7=0.3mA8
P=(0.1583+0.3)(3.5)=1.60mW
______________________________________________________________________________________
iR1=
Microelectronics:
CircuitAnalysisandDesign,4theditionChapter17ByD.A.NeamenProblemSolutions______________________________________________________________________________________
17.14
AssumeVγ=0.4V
(a)Logic1=0.2V,Logic0=−0.2V
0−0.7−(−3.10)(b)iE==0.25⇒RE=9.6kΩRE
(c)iD1+iR1=iE
2iR1+iR1=3iR1=0.25⇒iR1=0.08333mA
0.4=4.8kΩ0.08333
0.2−0.7−(−3.10)(d)iE==0.2708mA9.6
0.40.4iR2===0.0833mAR24.8R1=
iD2=iE−iR2=0.2708−0.0833=0.1875mA
(e)iE=0.2708mA
−0.2−(−3.10)=0.8788mA3.3
0.2−(−3.10)iR3==1.0mA3.3
P=(iE+iR3+iR4)[0.9−(−3.10)]=(0.2708+1.0+0.8788)(4)=8.6mW
______________________________________________________________________________________
17.15
a.iR4=
i1=
i3=
i4=(−0.9−0.7)−(−3)1(−0.2−0.7)−(−3)15(−0.2−0.7)−(−3)⇒i1=1.4mA⇒i3=0.14mA
⇒i4=0.14mA15
i2+iD=i1+i3=1.4+0.14=1.54mA0.4i2=⇒i2=0.8mA0.5
iD=0.74mA
b.v0=−0.4Vi1=1.4mAi3=
(0−0.7)−(−3)⇒i3=0.153mA15i4=i3⇒i4=0.153mAi2+iD=i4⇒i2=0.153mAiD=0v0=−(0.153)(0.5)⇒v0=−0.0765V
Microelectronics:
CircuitAnalysisandDesign,4theditionChapter17ByD.A.NeamenProblemSolutions______________________________________________________________________________________
c.i1=i3=(0−0.7−0.7)−(−3)1⇒i1=1.6mA
(−0.2−0.7)−(−3)⇒i3=0.14mA15i4=i3⇒i4=0.14mAi2+iD=i3⇒i2=0.14mAiD=0.0v0=−(0.14)(0.5)⇒v0=−0.07V
i1=(0−0.7−0.7)−(−3)⇒i1=1.6mA1d.
(0−0.7)−(−3)i3=⇒i3=0.153mA15
i4=i3⇒i4=0.153mA
i2+iD=i1+i4=1.6+0.153=1.753mA
i2=0.4⇒i2=0.8mA0.5
iD=0.953mA
v=−0.40V0______________________________________________________________________________________
17.16
(a)(i)A=B=C=D=0V,Q1,Q2,Q3,Q4off;Q5,Q6on
2.5=iR1
(2)+0.7+
1.8=iR1(2+0.0824)⇒iR1=0.8644mA
Y=2.5−iR1
(2)⇒Y=0.771V
(ii)A=B=0V,C=D=2.5V;Q1,Q2,Q6off;Q3,Q4,Q5on
⎛iR1⎞2.5=iR1
(2)+0.7+⎜⎟(15)⎜91⎟⎠⎝
1.8=iR1(2+0.1648)⇒iR1=0.8315mA
Y=2.5−iR1
(2)⇒Y=0.837V
(iii)A=C=2.5V,B=D=0V;Q1,Q3on,Q5,Q6off
iR1=0⇒Y=2.5V
(b)Y=(AORB)AND(CORD)
(c)(i)P=iR1(2.5)=(0.8644)(2.5)=2.16mW1⎛iR1⎞⎜⎟⎟(15)2⎜⎝91⎠
(ii)iR3=2.5−0.7=0.12mA15
P=(iR1+iR3)(2.5)=(0.8315+0.12)(2.5)=2.38mW
2.5−0.7=0.12mA15
P=(iR2+iR3)(2.5)=(0.24)(2.5)=0.60mW
______________________________________________________________________________________
(iii)iR2=iR3=
Microelectronics:
CircuitAnalysisandDesign,4theditionChapter17ByD.A.NeamenProblemSolutions______________________________________________________________________________________
17.17
a.logic1=0V
logic0=−0.4V
b.v01=AORB
v02=CORDv03=v01ORv02or
v03=(AORB)AND(CORD)______________________________________________________________________________________
17.18
a.ForCLOCK=high,IDCflowsthroughtheleftsideofthecircuit..IfDishigh,IDCflowsthroughtheleftRresistorpullinglow.IfDislow.IDCflowsthroughtherightRresistorpullingQlow.ForCLOCK=low,IDCflowsthroughtherightsideofthecircuitmaintainingQandintheirpreviousstate.
b.
P=1.7IDC(3)=(1.7)(50)(3)⇒P=255μWP=(IDC+0.5IDC+0.1IDC+0.1IDC)(3)______________________________________________________________________________________
17.19
(a)(i)ForυI=0.1V
υ1=0.8V2.5−0.8=0.1417mA12
i2=i3=0,υO=2.5Vi1=
(ii)ForυI=2.5V
υ1=0.7+0.8=1.5V2.5−1.5=0.0833mA12
υO=0.1Vi1=i2=
i3=2.5−0.1=0.20mA12
(b)(i)υ1=0.7+0.7=1.4V
υI=υ1−0.7=0.7V
(ii)υ1=0.7+0.8=1.5V
υI=υ1−0.7=0.8V
______________________________________________________________________________________
Microelectronics:
CircuitAnalysisandDesign,4theditionChapter17ByD.A.NeamenProblemSolutions______________________________________________________________________________________
17.20
(a)vI=0⇒V1=0.7V
3.3−0.7=0.433mA6
iB=iC=0
vo=3.3Vi1=(b)____________________________________________________________________________________
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