电工学答案（秦曾煌）：dz15

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?15.2?bvè
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?15.2.35.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,3
?15.3?bvè
^￥?
s.,,,,,,,,,,,,,,,,,,,,,3
?15.3.25.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,3
?15.3.35.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,4
?15.3.45.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,5
?15.3.55.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,6
?15.4?á
yT?￥×?.,,,,,,,,,,,,,,,,,,,,,,,6
?15.4.25.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,6
?15.4.55.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,7
?15.6?

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?15.6.15.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,9
?15.6.25.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,9
?15.7?μsbvè
^.,,,,,,,,,,,,,,,,,,,,,,,,,11
?15.7.35.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,11
?15.9??r?85#bvè
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?15.9.25.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,13
?15.9.35.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,14
1
List of Figures
1515.2.3m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,3
2515.3.3m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,4
3515.3.5m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,6
4515.4.5m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,8
5515.6.1m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,9
6515.6.2m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,10
7515.7.3m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,11
8515.9.2m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,13
9515.9.2m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,14
2
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15.2bbbvvvèèè
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15.2.3
m1?
?UCC = 10V
?1pUCE = 5V
IC = 2mA

kpRCRB
￥E′b
!?85￥fl = 40b
[333]
m1:515.2.3m
?UCE = UCC?RCIC Vp
RC = UCC?UCEI
C
= 10?52￡10?3? = 2:5k?
IB … ICfl = 240mA = 0:05mA
RB … UCCI
B
= 100:05k? = 200k?
15.3bbbvvvèèè
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sss
15.3.2
51m
U￥%?
ê?bvè
^?
UCC = 9V
?85￥fl = 20
IC =
1mAb?1pjAuj? 100

k9
RC
RB#UCEb
3
[333]
IB … ICfl = 120mA = 0:05mA
RB … UCCI
B
= 90:05k? = 180k?
rbe = [200+(20+1)￡ 261:05]? = 720? = 0:72k?
jAuj = flRCr
be
( b
HjAujKv)
RC = jAujrbefl = 100￡0:7220 k? = 3:6k?
UCE = UCC?RCIC = (9?3:6￡1)V = 5:4V
15.3.3
μBbvè
^?51m
U
?85￥
{+?[#bvè
^￥?a°
@μL?m2
Ub
kù
(1)RB,RC,RLò1

$(2)?á
3
>?￥Kv
{?
èaUiM1

$(3)??Fv
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v
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@μLá
3
I
1?Y
$(5)?è
^?
e?
?M
o|?8
5DB?fl′lB?￥50
?
HIB
IC
UCE#jAuj|?M
$
[333]
m2:515.3.3m
?m2 V?
á
′1
IC = 2mA; IB = 40?A; UCE = 5V
è÷èa1
UCC = 10V
è
@bv"
1
fl = ICI
B
= 20:04 = 50
4
(1)
RB … UCCI
B
= 100:04k? = 250k?
RC = UCC?UCEI
C
= 10?52 k? = 2:5k?
??
@μL V¤
tanfi0 = 1R0
L; 28?5 = 1R0
L; R0L = 1:5k?
?N¤
R0L = RCRLR
C +RL
RL = RCR
0
L
RC?R0L =
2:5￡1:5
2:5?1:5k? = 3:75k?
(2)?m2 V?
8?UCEQ = (8?5)V = 3V UCEQ?UCES = (5?0:3)V = 4:7V
?
>?￥Kv
{èa?1UoM = 3V
5pjAuja
p?á
3
>?
￥Kv
{?èaUiM
rbe = 200(?)+(1+fl)26(mV)I
E(mA)
= [200+(1+50)￡ 262 ]? = 0:86k?
jAuj = flR
0
L
rbe =
50￡1:5
0:86 = 87
?
^
UiM = UoMjA
uj
= 387V = 34:5mV
(3)
n5á
3??
>?
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H V??RB
hlE′[9vIB
|á
yT
?Q
MB?b
(4)|RLE′9v
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YVQ?￥?
@μLDà
￥fi0?|μ
hlb
(5) IB?M
IC?hlB?
UCE9v
jAuj|hlB?b
15.3.4
X?
bvè
^￥
{èE13:3k?

{
7
^èa￥μr′Uo0 = 2V

kù?bvè
^¤μμèERL = 5:1k?
H

{èa|/??

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[333]
UoL = RLr
o +RL
Eo = RLr
o +RL
Uo0

UoL = RLr
o +RL
Uo0 = 5:13:3+5:1 ￡2V = 1:2V
5
15.3.5
m3?
UCC = 12V
RC = 2k?
RE = 2k?
RB = 300k?
?85￥fl =
50bè
^μ
?
{
b
kp
(1)èabv
Au1 =
_Uo1
_UiAu2 =
_Uo2
_Ui
(2)
{
èEro1ro2b
[333]
m3:515.3.5m
IB = UCC?UBER
B +(1+fl)RE
= 12?0:6300+(1+50)￡2mA = 0:028mA
IE = (1+fl)IB = (1+50)￡0:028mA = 1:43mA
rbe = [200+(1+50)￡ 261:43]? = 1127? … 1:13k?
V"è
{

Au1 =
_Uo1
_Ui =?
flRC
rbe +(1+fl)RE =?
50￡2
1:13+(1+50)￡2 …?1
ro1 … RC = 2k?
V?

{

Au2 =
_Uo2
_Ui =
(1+fl)RE
rbe +(1+fl)RE … 1
ro2 … rbe +R
0
S
fl …
rbe
fl =
1130
50? = 22:6?
T?
R0S = RS==RB

!?|÷
=ERS … 0
5R0S … 0b
15.4ááá
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15.4.2
??m15.4.1
U￥sa
T
ê?bvè
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X?UCC = 24V
RC =
3:3k?
RE = 1:5k?
RB1 = 33k?
RB2 = 10k?
RL = 5:1k?
fl = 66
i
6
!RS … 0b(1)
kpá
′IB
ICUCE
(2)±M?rè
^
(3)9
?85
￥
{?èErbe
(4)9
èabv
Au
(5)9
bvè
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7
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abv

i
a
üμèERLèabv
￥?Y
(6)
bvè
^￥
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{èEb
[333]
(1)
VB = UCCR
B1 +RB2
RB2 = 2433+10 ￡10V = 5:58V
IC … IE = VB?UBER
E
= 5:58?0:61:5 mA = 3:32mA
IB … ICfl = 3:3260 mA = 0:05mA
UCE = UCC?(RC +RE)IC = [24?(3:3+1:5)￡3:32]V = 8:06V
(2)
rbe = 200(?)+(1+fl)26(mV)I
E(mA)
= [200+(1+66)￡ 263:32]? = 0:72k?
(3)
Au =?flR
0
L
rbe =?66￡
3:3￡5:1
3:3+5:1
￡ 10:72 =?183:7
(4)
Au =?flRCr
be
=?66￡ 3:30:72 =?302:5
(5)
ri = rbe==RB1==RB2 … rbe = 0:72k?
ro … RC = 3:3k?
15.4.5
!9B?5?85bvè
^
X?RL = 3k?b1pjAuj? 60
ri?
1k?
ro < 3k?
yT?×?bypê¨ú
l?
q53GD100
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cC
fl′ Vê50? 100-WbKa°á
yT?
^?
abp
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cH
b
[333]
7
m4:515.4.5m
(1)ê4bvè
^?85
1pyT?×?
 Vê¨sa
ê?bvè
^

??m15.4.1

êUCC =
12V
?yp�853GD100

!fl = 50
i
!jAuj = 60
ri =
1k?b
(2)?
9
?
Trbe … [200+(1+fl)26I
E
]? … ri Vp
IC … IE … 26(1+fl)r
i?200
= 26￡511000?200mA = 1:66mA
?
TjAuj = flR
0
L
rbe Vp
R0L = 60￡150 k? = 1:2k?
R0L = RCRLR
C +RL
'
RC = R
0
LRL
RL?R0L =
1:2￡3
3?1:2k? = 2k?
!VB = 4V
RE = VB?UBEI
E
= 4?0:61:66 k? … 2k?
è
@IB … ICfl = 1:6650 mA = 0:033mA
!I2 = 10IB
'
I2 = 10￡0:033mA = 0:33mA … I1
¤
RB2 = VBI
2
= 40:33k? = 12:12k?(|12k?)
RB1 = UCC?VBI
1
= 12?40:33 k? = 24:24k?(|24k?)
8
(3)°á
yT?
?UCE = UCC?(RC +RE)ICS°
@μL(m4)
IC = 0 UCE = UCC = 12V
UCE = 0 IC = UCCR
C +RE
= 122+2mA = 3mA
UCE = [12?(2+2)￡1:66]V = 5:4V
á
yT?
a
l?|f ?/
?á
3
>?b
15.6



{
{
{
15.6.1
m5
U￥

{?
X?RS = 50?
RB1 = 100k?
RB2 = 30k?
RE =
1k?
?85￥fl = 50
rbe = 1k?

kpAu
rirob
[333]
m5:515.6.1m
Au = (1+fl)REr
be +(1+fl)RE
= (1+50)￡11+(1+50)￡1 = 0:98
ri = RB1==RB2==[rbe +(1+fl)RE] = 16k?
ro … rbe +R
0
S
fl =
1000+50
50? = 21?
T?
R0S = RS==RB1==RB2 … 50?
15.6.2
)bvè
^?m6
U
?85￥fl1 = fl2 = 40
rbe1 = 1:37k?
rbe2
= 0:89k?b(1)°
@Y
^
i
ò)è
^￥á
′(9
UCE1
H-
{IB2)

9
(2)±M?rè
^
i9
Au1
Au2Au
(3)9
rirob
[333]
m6:515.6.2m
(1)-á
′
VB1 = 2033+8:2 ￡8:2V = 4V
IC1 … IE1 = 4?0:63+0:39mA = 1mA
IB1 … 140mA = 25?A
UCE1 … 20?(10+3+0:39)￡1 = 6:6V
aá
′
IC2 … IE2 = UC1?UBE2R
E2
= (20?10￡1)?0:65:1 mA = 1:8mA
IB2 = 1:840 mA = 45?A
UCE2 = (20?5:1￡1:8)V = 10:8V
(2)-)èabv
Au1 =?fl1 R
0
L1
rbe +(1+fl1)R00E1 =?40￡
9:1
1:37+(1+40)￡0:39 =?21
T?
R0L1 = RC1==
rbe2 +(1+fl2)￠ RE2RLR
E2 +RL
a)èabv
Au2 = (1+fl2)R
0
L
rbe2 +(1+fl2)R0L =
(1+40)￡2:5
0:89+(1+40)￡2:5 = 0:99
)èabv
Au = Au1 ￠Au2 =?21￡0:99 =?20:8
10
(3)
ri = ri1 = RB1==RB2==[rbe1 +(1+fl1)R00E1] = 4:77k?
ro = ro2 … rbe2 +RC1fl
2
= 0:89+1040 k? = 272?
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V' è￥
)bvè
^ A
4ú

{?èE
??

{èEb
15.7μμμsssbbbvvvèèè
^
^
^
15.7.3
m7
U￥μsbvè
^?
fl = 50
UBE = 0:7V

{?èaui1 =
7mV
ui2 = 3mVb
(1)9
bvè
^￥á
′IB
IC#òè￥èêVE
VCVB

(2)ü
{?èaUi1
ui2s31
s
uic1
uic2μ
s
uid1
uid2

(3)p?

{uoc1uoc2

(4)p?
μ

{uod1uod2

(5)p?
9
{uo1uo2

(6)p


{uoc


μ

{uod

9
{uob
[333]
m7:515.7.3m
(1)á
H
ui1 = ui2 = 0
???m15.7.5￥?5°
@Y
^ V¤
RBIB +UBE +2REIE = UEE
IB = UEE?UBER
B +2(1+fl)RE
11
?
^
IB = 6?0:710￡103 +2￡(1+50)￡5:1￡103A
= 0:01￡10?3A = 0:01mA
IC = flIB = 50￡0:01mA = 0:5mA
IE = (1+fl)IB = 51￡0:01mA = 0:51mA
VC = UCC?RCIC = [6?5:1￡103 ￡0:5￡10?3]V = 3:45V
VE =?6+2REIE = [?6+2￡5:1￡103 ￡0:51￡10?3]V =?0:798V
VB =?RBIB =?10￡103 ￡0:01￡10?3V =?0:1V
(2)
uic1 = uic2 = ui1 +ui22 = 7+32 mV = 5mV
uid1 =?uid2 = ui1?ui22 = 7?32 mV = 2mV
(3)?515.7.2
￡
ü￥
T¤
uoc1 = uoc2 =?fl RCR
B +rbe +2(1+fl)RE
uic1
T?
rbe = [200+(1+50)￡ 260:51]? = 2:8k?
?
^
uoc1 = uoc2 =?50￡ 5:110+2:8+2(1+50)￡5:1 ￡5mV
=?2:39mV
(4)
uod1 =? flRCR
B +rbe
uid1 =?50￡5:110+2:8 ￡2mV =?39:8mV
uod2 =? flRCR
B +rbe
uid2 =?50￡5:110+2:8 ￡(?2)mV = +39:8mV
(5)
uo1 = uoc1 +uod1 = [(?2:39)+(?39:8)]mV =?42:2mV
uo2 = uoc2 +uod2 = [(?2:39)+39:8]mV = +37:4mV
(6)
uoc = uoc1?uoc2 = 0
uod = uod1?uod2 = (?39:8?39:8)mV =?79:6mV
uo = uo1?uo2 = (?42:2?37:4)mV =?79:6mV = uod
12
15.9???rrr???888555###bbbvvvèèè
^
^
^
15.9.2
m8
U￥÷
{?
X?UDD = 12V
RS = 12k?
RG1 = 1M?
RG2 =
500k?
RG = 1M?b
kpá
′aèabv
a
{?èE
{èEb
!VG … VS
gm = 0:9mA=Vb
[333]
m8:515.9.2m
VS … VG = RG2UDDR
G1 +RG2
= 500￡121000+500V = 4V
UDS = UDD?VS = (12?4)V = 8V
ID = VSR
S
= 412￡103A = 0:33￡10?3A = 0:33mA
Au =
_Uo
_Ui =
gm _UgsRS
_Ugs +gm _UgsRS =
gmRS
1+gmRS … 1
ri = RG +(RG1==RG2) =
1+ 1￡0:51+0:5
M? = 1:33M?
p
{èEro

m9


|
{?

^

!?|÷
=El

{?


{
FB?
@èa_Uo

?¤μ
μèERL
5"?

 Vpè
@
_Io = _Uo
RS?
_Id = _Uo
RS?gm
_Ugs
7
_Ugs =? _Uo
5
_Io = _Uo
RS +gm
_Uo = (gm + 1
RS)
_Uo
13
m9:515.9.2m
#
ro =
_Uo
_Io =
1
(gm + 1R
S
)
= 1g
m
==RS
?RS 1g
m
H
ro … 1g
m
'5?
RS = 12k? 1g
m
= 10:9 = 1:1k?
#
ro … 1:1k?
15.9.3
?r?85μsbvè
^???m15.11
U
X?gm = 1:5mA=V
pèa
bv
Au = uou
i
b
[333]
uoDuiQM

[
Au = uou
i
=?12gmRD =?12 ￡1:5￡15 =?11:25
14
15