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Khoa ien ien T. Ky thuat mach ien T I
-1- Tom Tat Bai Giang. -1-
BAI GIANG TOM TAT MON:BAI GIANG TOM TAT MON:BAI GIANG TOM TAT MON:BAI GIANG TOM TAT MON: IEN T INgi soan: TS. Pham Hong Lien.
Giao trnh chnh: Mach ien T 1 Le Tien Thng, HBK Tp.HCM.
Chng 1: Diode ban dan.Chng 1: Diode ban dan.Chng 1: Diode ban dan.Chng 1: Diode ban dan.I.I.I.I. Diode chnh lu:Diode chnh lu:Diode chnh lu:Diode chnh lu:
1111---- Quan he gia ien ap va dong ien cua Diode (H2Quan he gia ien ap va dong ien cua Diode (H2Quan he gia ien ap va dong ien cua Diode (H2Quan he gia ien ap va dong ien cua Diode (H2----1):1):1):1):
= 1
nKTqVexpIi D0D (1-1)
iD : Dong ien trong Diode (A).
VD : Hieu ien the hai au Diode (V).I0 : Dong ien bao hoa ngc (A).
q : ien tch electron 1,6.10-19 J/V.
K : Hang so Bolzman 1,38.10-23 J/0K.
N : Hang so co gia tr trong khoang (12) phu thuoc vao loai ban dan.
Goi ien the nhiet:q
KTVT = (1-2)
T (1-1) ta co:
=
T
D0
T
D0D nV
VexpI1nVVexpIi (1-3)
nhiet o T=3000K, tng ng T=270C, ta co VT2526mV. Khi o ien trong cua Diode c tnh bi phng trnh:
)(i
nVIi
nVrD
T
CD
Td =
+= (1-4)
ac tuyen Volt-Ampere cua Diode tren (H2-2)Kieu mau mach tng ng cua Diode tren (H2-3a,b,c).
2222---- Phng trnh ng tai cua Diode (H2Phng trnh ng tai cua Diode (H2Phng trnh ng tai cua Diode (H2Phng trnh ng tai cua Diode (H2----5).5).5).5). Phng trnh ng tai mot chieu cua Diode (DCLL)
1DDS RIVV += (1-5)
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Phng trnh ng tai xoay chieu cua Diode (ACLL)
)R//R(ivv L1dds += (1-6)
T (1-5) va (1-6) tren he toa o tong quat ta co:
DQdDDQdD Iii&Vvv +=+= (1-7)
Vi:
VD va iD la thanh phan tc thi cua ien ap va dong ien. VDQ va IDQ la cac gia tr mot chieu cua ien ap va dong ien. vd va id la cac gia tr xoay chieu cua ien ap va dong ien.
Vay phng trnh ng tai xoay chieu ACLL trong he toa o tong quat se
la:
sDQDL1DQD v)Ii)(R//R(Vv += (1-8)
3333---- Chnh lu ien ap xoay chieu:Chnh lu ien ap xoay chieu:Chnh lu ien ap xoay chieu:Chnh lu ien ap xoay chieu:a- Chnh lu ban song: (H2-6)
ien ap au vao: tsinvv maxs =
ien ap trung bnh DC tren tai:
=
+= maxL
LS
LmaxDC
VRR
RVV (1-9)
b- Chnh lu toan song: (H2-8a,b,c)
ien ap trung bnh Dc tren tai:
= maxLDCV2V (1-10)
4444---- Mach loc:Mach loc:Mach loc:Mach loc: (H2(H2(H2(H2----9a,b)9a,b)9a,b)9a,b)
Khi co tu C mac song song vi RL trong cac mach chnh lu ta co quan hegia ien ap trung bnh tren tai vi bien o ien ap au vao va ien tr R L va tuien C nh sau:
maxL
LDCmaxDC V1CfR4
CfR4fC4
IVV
+== (1-11)
5555---- Mach nhan oi ien ap:Mach nhan oi ien ap:Mach nhan oi ien ap:Mach nhan oi ien ap: (H2(H2(H2(H2----11a,b)11a,b)11a,b)11a,b)
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ien ap ra gan gap oi ien ap vao.
II.II.II.II. Diode on ap Zener:Diode on ap Zener:Diode on ap Zener:Diode on ap Zener:1111---- Cac tham so c ban cua diode ZenerCac tham so c ban cua diode ZenerCac tham so c ban cua diode ZenerCac tham so c ban cua diode Zener:::: (H3(H3(H3(H3----1)1)1)1)
ien ap on nh VZ khi dong ien qua zener thay oi trong khoang I zminIzmax. Thc te maxzminz I10
1I . (1-12)
ien tr ong tai iem lam viec.
dIdVr Zd = (1-13)
Diode Zener ly tng c coi co rd 0.
ien tr tnh:Z
Zt IVR = (1-14)
He so on nh:
d
t
Z
Z
Z
Z
ZZ
ZZ
rR
IV
dVdI
V/dVI/dIS === (1-15)
2222---- Mach on ap dung Diode Zener:Mach on ap dung Diode Zener:Mach on ap dung Diode Zener:Mach on ap dung Diode Zener: (H3(H3(H3(H3----2)2)2)2)
Mach tren hnh 3-2 luon thoa man he phng trnh:
+=
+=
ZiRS
LZR
VRIVIII
(1-16, 1-17)
Trong o ch co VZ const, con cac ai lng khac co the bien oi nhngphai thoa man ieu kien:
IZmin khi ILmax va VSmin IZmax khi ILmin va VSmax
T (1-16) va (1-17) tuy tng trng hp cu the ma ta co the suy ra cac hephng trnh khac nhau.
V du neu Ri = const th ta co he phng trnh:
(VSmin VZ)(ILmin + IZmax) = (VSmax VZ)(ILmax IZmin) (1-18)
V du neu Ri =const va RL = const ngha la IL = const th ta co he phngtrnh:
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VSmin = (IZmin + IL)Ri + VZ = IminRi + VZ (1-19)
VSmax = (IZmax + IL)Ri + VZ = ImaxRi + VZ (1-20)
Chu y v VL = VZ const nen khi IL thay oi ta co:
maxL
ZminL IVR
= (1-21)
minL
ZmaxL I
VR = (1-22)
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Chng II: Transistor hai lp tiep giap (BJT)Chng II: Transistor hai lp tiep giap (BJT)Chng II: Transistor hai lp tiep giap (BJT)Chng II: Transistor hai lp tiep giap (BJT)
che o tn hieu ln. che o tn hieu ln. che o tn hieu ln. che o tn hieu ln.
I.I.I.I. Cac tham so c ban cua Transistor.Cac tham so c ban cua Transistor.Cac tham so c ban cua Transistor.Cac tham so c ban cua Transistor. (H2(H2(H2(H2----1)1)1)1)+ He so truyen at dong ien phat khi mac Base chung
Thong thng = 0,95 0,99, ly tng = 1.
+ He so truyen at dong ien khi mac Emitter chung:
=
1(vai chuc vai tram lan).
+ Dong ien ra cc Collector:IC = IE + ICBO (2-1)
Trong o ICBO la dong ien phan cc ngc hay con goi la dong nhiet,thng rat nho.
+IE = IC + IB (2-2)suy ra IB = (1-)IE ICBO (2-3)
=
= CCBO
CCBOCB
IIIII1I (2-4)
tan so thap (H2-1) ta co: hfe = = hFE (2-5)
II.II.II.II. Mach phan cc cho Transistor:Mach phan cc cho Transistor:Mach phan cc cho Transistor:Mach phan cc cho Transistor:1111---- Mach phan cc Collector:Mach phan cc Collector:Mach phan cc Collector:Mach phan cc Collector:
Ta co phng trnh tai mot chieu:
VCC = VCEQ + ICQRC + IEQRE
VCEQ + ICQ(RC + RE) (2-6)
So lng ien t ti c Collector
So lng ien t phat i t cc Emitter =
VCEQ+
-
RE
ICQ
+VCCRC
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CQ
CEQCCEC I
VVRR
=+ (2-7)
RE thng c tnh theo cong thc thc nghiem:
CQCC
EQREE I
V)3,01,0(IV
R
= (2-8)
Thay vao (2-7) de dang tnh c RC.
Neu RE = 0 t (2-7) ta co:CQ
CEQCCC I
VVR
= (2-7)
2222---- Mach phan cc Base:Mach phan cc Base:Mach phan cc Base:Mach phan cc Base:a- Mach nh dong Base:
Ta co: RbIBQ + VBE + IEQRE = VCC (2-8)
VBE la ien ap m cua Transistor, con ky hieu la Vnh H2-2 chng 1. VBESi 0,7v va VBEGe 0,2v. Ngaynay chu yeu dung Transistor Silic nen t (2-8) ta co :
CCEEQEQ
b VRI7,01I
R =+++
Suy ra:
1RR
V
1RR
7,0VIb
E
CC
bE
CCEQ
++
++
= v VCC >>0,7v (2-9)
Phng phap nay t c dung do dong IBQ phu thuoc nhieu vao nhiet o.Phng phap nay ch c dung oi vi mach mac Collector chung e nang cao trkhang vao.
b- Mach nh ap Base: (H2-3)
IBQ
RbICQ
+VCC
RC
RE
iiRb
IBQ
IEQ VCC
RL
REVBB
R2
IBQI1
+VCCRC
RE
I2
R1
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Ta co: CC21
1BB VRR
RV
+= (2-10)
21
21b RR
RRR+
= (2-11)
BBCC
CCb
CC
BBb1 VV
VR
VV1
1RR
=
= (2-12)
BB
CCb2 V
VRR = (2-13)
Phng trnh tai DC: VCC = VCEQ + ICQ(RC + RE) (2-14)
Ap dung nh luat KII ta co: Vkn = 0, suy ra:
BBEEQBEBQb VRIVIR =++ (2-15)
++
=
1RR
7,0VIIb
E
BBEQCQ coi VBE = 0,7v (2-16)
Thay vao (2-14) ta tnh c VCEQ.
Thong thng khi thiet ke ta thng chon RE >> (1-)Rb e on nh dong IEQ.V vay neu cha biet Rb ta thng chon:
EEb R101R)1(
101R += (2-17)
Phng phap phan cc Base nay hay c dung nhat.
c- Mach nh dong Emitter:Ap dung nh luat KII Vkn = 0 ta co:
EEEEQBEBQb VRIVIR =++ (2-18)
Suy ra:
1RR
7,0VIb
E
EEEQ
++
= vi VBE = 0,7v (2-19)
Rb
IBQ
+VCC
RC
RE
-VEE
IEQ
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Phng trnh tai DC trong trng hp nay se la:
VCC + VEE = VCEQ + ICQ(RC + RE) (2-20)
Phng phap phan cc Base nay ch c dung khi mach yeu cau chat lngcao nh mach khuech ai vi sai, mach khuech ai thuat toan (KTT) v no phai ton
them mot nguon cung cap.
III.III.III.III. Giai tch mach TransistGiai tch mach TransistGiai tch mach TransistGiai tch mach Transistor bang o th:or bang o th:or bang o th:or bang o th:1111---- Bo khuech ai mac Emitter chung:Bo khuech ai mac Emitter chung:Bo khuech ai mac Emitter chung:Bo khuech ai mac Emitter chung:
Ta co the chia thanh 4 loai mach c ban nh sau:
a- Khong co CE, khong co CC: (H2-3)Bo khuech ai co the c thiet ke che o toi u (song ra tot nhat) hoac
che o bat ky.Che o toi u:Che o toi u:Che o toi u:Che o toi u: Thiet ke sao cho song ra ln nhat va khong b meo (I cmmax
hoac VLmax), thng cha biet cac ien tr phan cc R1, R2.
T o th (H3-2), ta thay song ra se ln nhat khi:
ACDC
CCCQTmaxcm RR
VII+
== (2-21)
ACTCQTCEmaxcm RIVV == (2-22)
Vi s o (H3-1) ta co: RAC = RDC = RC + RE nen t (2-21) va (2-22) ta suy ra:
)RR(2V
IEC
CCTCQ
+= (2-23)
2VV CCTCEQ = (2-24)
Che obat ky:Che obat ky:Che obat ky:Che obat ky: Thng cho trc R1, R2 hoac VCEQ hoac ICQ. Ap dung caccong thc (2-10, 11, 14, 16) se xac nh c (ICQ, VCEQ)
Neu ICQ < ICQTth Icm = ICQ. Neu ICQ > ICQTth Icm = iCQmax ICQ.
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b- Co CE, khong co CC (Tu Bypass Emitter) (H2-5)Che o tChe o tChe o tChe o toi u:oi u:oi u:oi u:
RDC = RC + RE va RAC = RC thay vao (2-21) ta c:
EC
CC
ACDC
CCTCQmaxcm RR2
VRR
VII+
=+
== (2-25)
C
E
CC
EC
CCCACTCQTCEQmaxcm
RR2
VRR2
RVRIVV+
=+
=== (2-26)
Che o bat ky:Che o bat ky:Che o bat ky:Che o bat ky: c tnh toan theo cac cong thc (2-10, 11, 14, 16) va actuyen tai AC c ve nh sau:
( )CEQCEAC
CQC VvR1Ii = (2-27)
Cho VCEQ = 0 ACCEQ
CQmaxC R
V
Ii += (2-28)
Cho iC = 0 vCEmax = VCEQ + ICQRAC (2-29)
Phng trnh (2-28) va (2-29) e xac nh iCmax va vCEmax trong cac trng hpiem tnh Q bat ky
Q2
VCE(V)
iC(mA)iCmax1
ACDC
CC
RRV+
ACT R
1ACLL
0
ICQT
DCR1DCLL
Q1
QT
iCmax2
VCET
VCEmax1
VCC
VCEmax2
2ICQT
2VCET
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c- Khong co CE, co CC:Che o toi u:Che o toi u:Che o toi u:Che o toi u:
ECDC RRR +=
LC
LCEAC RR
RRRR+
+=
Thay vao (2-21) ta c:
LC
LCEC
CCTCQmaxCm
RR
RR
R2R
VII
+++
== (2-30)
ACTCQTCEmaxCE RIVV == (2-31)
LC
LCEC
CC
LC
CmaxCm
LC
CmaxLm
RRRRR2R
VRR
RIRR
RI
++++
=+
= (2-32)
LC
LCEC
CC
LC
LCLmaxLmmaxLm
RR
RRR2R
VRR
RRRIV
+
+++== (2-33)
Che o bat ky nh tren nhng chu y:
ICQ < ICQT : ICm = ICQ. ICQ > ICQT : ICm = iCmax ICQ. Cm
LC
CLm IRR
RI+
= (2-34)
VLm = ILmRL. (2-35)
d- Co CE, co Cc: (tu ghep vo han) (H2-6)ECDC RRR +=
LC
LCAC RR
RRR+
=
CC
RLii
iL
VBB
Rb
+VCC
RC
RE
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thay vao (2-21) ta c:
LC
LCEC
CCTCQmaxCm
RRRRRR
VII
+++
== (2-36)
LC
LC
LC
LCEC
CCTCEQmaxCm RR
RR
RRRRRR
VVV+
+++
== (2-37)
LC
LCEC
CC
LC
CmaxCm
LC
CmaxLm
RRRRRR
VRR
RIRR
RI
+++
+
=+
= (2-38)
LC
LCEC
CC
LC
LCLmaxLmmaxLm
RR
RRRR
VRR
RRIV
+
+++== (2-39)
Che o bat ky xac nh nh tren.
So sanh 4 trng hp tren ta nhan thay tac dung cua cac tu C E va CC la lamtang bien o dong ien ra va ien ap ra (so sanh cac cong thc (2-21), (2-25), (2-30)va (2-36)).
e- Tnh toan cong suat: Cong suat nguon cung cap:
PCC = VCCICQ (2-40)
Cong suat trung bnh tieu tan tren tai:
L
2Lm
L2LmL R
V21PI
21P == (2-41)
che o toi u: ICmmax = ICQT nen oi vi trng hp a va b taco:
C
2
TCQC
2
maxCmmaxL RI2
1
RI2
1
P == v RL = RC (2-42)
Con oi vi trng hp c va d th do RC RL nen ta co:
L2
maxLmmaxL RI21P = (2-43)
Cong suat tieu tan tren Collector:
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2IRI)RR(PP
2Cm
AC2CQECCCC += (2-44)
Hieu suat:CC
L
PP
= (2-45)
CC
maxLmax P
P= (2-46)
che o lp A hieu suat cc ai %25max =
He so pham chat: 2PP
maxL
maxC = (2-47)
2222---- BoBoBoBo khuech ai mac Collector chung:khuech ai mac Collector chung:khuech ai mac Collector chung:khuech ai mac Collector chung:
Che o toi u:Che o toi u:Che o toi u:Che o toi u: Trong ca 3 hnh neu khong co CC ta co:
ECDC RRR +=
LE
LECAC RR
RRRR+
+=
thay vao cong thc (2-21), (2-22) ta se co:
LE
LE
EC
CCTCQmaxEm
RR
RRRR2
VII
+++
= (2-48)
++
+++
==LE
LEC
LE
LEEC
CCACTCQTCEQ RR
RRR
RRRRRR2
VRIV (2-49)
VL
R2 CC
CE
RL
+VCC
RC
RER1
-VEE
VLRb
CC
CE
RL
+VCC
RC
REVL
Rb CC
CE
RL
+VCC
RC
RE
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Trong ca 3 hnh neu co CC ta co:ECDC RRR +=
LE
LEAC
RR
RRR+
=
thay vao cong thc (2-21), (2-22) ta se co:
LE
LEEC
CCTCQmaxEm
RRRRRR
VII
+++
= (2-50)
+
+++
==LE
LE
LE
LEEC
CCACTCQTCEQ RR
RR
RRRRRR
VRIV (2-51)
Ta luon co:
maxCmLE
EmaxLm IRR
RI+
= (2-52)
LE
LEmaxCmLmaxLmmaxLm RR
RRIRIV+
== (2-53)
Che o bat ky:Che o bat ky:Che o bat ky:Che o bat ky: Tnh theo cac cong thc c xay dng trong phan machphan cc cho Transistor.
ac tuyen tai mot chieu DCLL va ac tuyen tai xoay chieu ACLLc ve tng t nh trong mach khuech ai Emitter Common.
3333---- Bo khuech ai mac Base chung:Bo khuech ai mac Base chung:Bo khuech ai mac Base chung:Bo khuech ai mac Base chung:
Che o toi u:CDC RR =
LC
LCAC RR
RRR+
=
Thay vao (2-21), (2-22) ta c:
Vi+
-
Ri
VL
R1
CC
RL
+VCC
RC
R2
Cb
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LC
LCC
CCTCQmaxCm
RRRRR
VII
++
== (2-54)
+
++
===LC
LC
LC
LCC
CCACTCQTCEQmaxCm
RR
RR
RR RRR
VRIVV (2-55)
maxCmLC
CmaxLm IRR
RI
+= (2-56)
LmaxLmmaxLm RIV = (2-57)
Che o bat ky: c tnh trc tiep t mach
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Chng III:Chng III:Chng III:Chng III: On nh phan cc cho Transistor BJTOn nh phan cc cho Transistor BJTOn nh phan cc cho Transistor BJTOn nh phan cc cho Transistor BJT
Chng nay nham nghien cu s dch chuyen iem Q theo ICBO, VBE khi thayoi nhiet o va theo khi b lao hoa. Coi gan ung cac ai lng VCC, VBB khongthay oi.
Neu s thay oi ICBO, VBE va la nho th bien xet ICQ se la ham tuyen tnhtheo cac bien khac.
ICQ = ICQ(ICBO, VBE, ) (3-1)
Tha so on nh dong ien:
E
b
CBO
CQI R
R1II
S +
= (3-2)
Tha so on nh ien ap:EBE
CQV R
1VI
S
= (3-3)
Tha so on nh he so khuech ai:
( )
++
+
=
=
E2b
Eb
1
1CQCQ
R1RRRII
S (3-4)
Vi 1CQ2CQCQ12 III& == (3-5)
Khi o s thay oi cua ICQ se c tnh bang cong thc:
++= SVSISI EVCBOICQ
( )
++
+
+
+=
E2b
Eb
1
1CQBE
ECBO
E
b
R1RRRIV
R1I
RR1 (3-6)
Chu y VBE thng co gia tr am.
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Chng IV. Thiet ke va phan tch tn hieu nho tan so thap.Chng IV. Thiet ke va phan tch tn hieu nho tan so thap.Chng IV. Thiet ke va phan tch tn hieu nho tan so thap.Chng IV. Thiet ke va phan tch tn hieu nho tan so thap.
I. Cac thong so Hybrid:Cac thong so Hybrid:Cac thong so Hybrid:Cac thong so Hybrid: Tr khang vao khi ngan mach tai:
0vi
vh21
1i
=
=
o li ien ap ngc khi h mach nguon:0iv
vh12
1r
==
o li dong ien thuan khi ngan mach tai:0vi
ih21
2f
==
Tong dan ngo ra khi h mach nguon:0iv
ih12
2o
==
ng vi cac cach mac khac nhau EC, BC hay CC ma ch th hai c chnh. V du: hie, hib, hic, ...
II. Cach mac Emitter chung:Cach mac Emitter chung:Cach mac Emitter chung:Cach mac Emitter chung:Tr khang vao khi ngan mach tai:
EQ
Tfeie I
Vmhh = (4-1)
Trong o: - VT =25mV 3000K (270C) (4-2)
- m = 1 2 phu thuoc vao chat ban dan. V du BJT Silic co m = 1,4khi o:
EQ
Tfeie I
Vh4,1h = (4-3)
* oi vi H4-1a, co mach tng ng rut gon H4-4, ta co:
He so khuech ai dong ien:( )
b
ie
fe
ieb
bfe
i
b
b
L
i
Li
Rh1
hhR
Rhii
ii
iiA
+
=+
=== (4-4)
Neu hie
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Khoa ien ien T. Ky thuat mach ien T I
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* oi vi H4-5, co mach tng ng H4-6 ta co:
He so khuech ai dong ien:( )
( )
bi
ieLC
Cfe
iebi
bife
LC
C
i
b
b
L
i
Li
R//rh
1
1RR
RhhR//r
R//rhRR
Rii
ii
iiA
++
=+
+=== (4-8)
Neu RC >> RL & Rb//ri >> hie ta co: Aimax = -hfe (4-9)
Tr khang vao: iebii h//R//rZ = (4-10) Tr khang ra: CC
oeo RR//h
1Z
= (4-11)
* oi vi H4-17 ta co:
He so khuech ai dong ien:( )
( ) ( ) Efeiebi
bife
RC
C
i
b
b
L
i
Li Rh1hR//r
R//rhR
Rii
ii
iiA
L+++
===
+
(4-12)
Tr khang vao: ( )[ ]Efeiebii Rh1h//R//rZ ++= (4-13) Tr khang ra: Co RZ = (4-14)S o EC hay c dung nhat do co A i, Av ln
III.Cach mac Base chung:Cach mac Base chung:Cach mac Base chung:Cach mac Base chung:T mach H4-9, H4-10, va H4-11 cac tham so cua cach mac Base chung (BC)
co the a ve cac tham so cua cach mac Emitter chung (EC) nh sau:
Tr khang vao khi ngan mach tai:
fe
ieib h1
hh+
= (4-15)
o li dong ien thuan khi ngan mach tai:fe
fefb h1
hh+
= (4-16)
Tong dan ra khi h mach nguon:
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fe
oeob h1
hh+
= (4-17)
Nh vay e tnh cac tham so cua s o B.C ch can biet cac tham so cuas o E.C. V hfb 1 nen s o B.C t c dung pham vi tan so thap,
nhng c dung rat nhieu pham vi tan so cao e giam anh hng cuacac ien dung ky sinh.
IV.Cach mac Collector chung:Cach mac Collector chung:Cach mac Collector chung:Cach mac Collector chung:T H4-14 va H4-15 ta co:
Vb = Vbe + iERE (4-18)
Vbe = ibhie (4-19)
VE = iERE = ib(1 + hfe)RE (4-20)
Suy ra:
Vb = ibhie + ib(1 + hfe)RE = ib[hie + (1 + hfe)RE] (4-21)
i
b
b
b
b
E
i
Ev V
VVi
iV
VVA == (4-22)
Efeb
E R)h1(iV
+= (4-23)
Efeieb
bR)h1(h 1Vi ++
= (4-24)
Goi: [ ]Efeieb'b R)h1(h//RR ++= (4-25)
'bi
'b
'bi
i'b
ii
b
RrR
RrVR
V1
VV
+=
+= (4-26)
Thay (4-23, 24, 26) vao (2-22) ta c:
'bi
'b
Efeie
Efe
i
Ev Rr
R
R)h1(h
R)h1(
V
V
A ++++
== (4-27)
vi Rb theo (4-25).
Tr khang vao cua s o H3-7 : Zi = hie + (1 + hfe)RE (4-28)
T s o H3-8 ta xac nh c tr khang ra:
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Khoa ien ien T. Ky thuat mach ien T I
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fe
'bi
ibiE
Eo h1
R//rh0Vi
VZ+
+==
= (4-29)
Nh vay e tnh cac tham so cua mach C.C ta cung ch can biet tham so cuamach E.C. S o C.C co Av
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Khoa ien ien T. Ky thuat mach ien T I
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Chng V:Chng V:Chng V:Chng V: Transistor hieu ng trng.Transistor hieu ng trng.Transistor hieu ng trng.Transistor hieu ng trng.
I. Ly thuyet hoat ong cua JFET:Ly thuyet hoat ong cua JFET:Ly thuyet hoat ong cua JFET:Ly thuyet hoat ong cua JFET:VDS: ien ap gia cc mang va cc nguon.
VDS (tai iem nghen) = Vp = Vpo + VGS (5-1)
Vpo: ien the nghen c tra tren o th. VGS: ien ap gia cc cong va cc nguon.
ien the anh thung Breakdown la mot ham cua ien ap GS:GSDSSDSX VBVBV += (5-2)
Trong o BVDSS la ien the Breakdown ng vi VGS = 0.
Tai vung bao hoa, dong dien mang c tnh gan ung:
++=
23
po
GS
po
GSpoD V
V2VV31II vi VGS < 0 (5-3)
Ipo: dong ien nghen c tra tren o th.
T (5-3) ta thay:
Khi VGS = 0 ta co: ID = Ipo (5-4) Khi VGS = -Vpo ta co: ID = 0 (5-5)
II. Ly thuyet hoat ong cua IGFET:Ly thuyet hoat ong cua IGFET:Ly thuyet hoat ong cua IGFET:Ly thuyet hoat ong cua IGFET:ien ap gia cc mang va cc nguon:
VDS(tai iem nghen) = Vp = Vpo + VGS (5-6)
BVDSX = BVDSS + VGS (5-7)
2
po
DSpoD VV1II
+= (5-8)
Khi VGS = 0 ta co: ID = Ipo (5-9) Khi VGS = -Vpo ta co ID = 0 (5-10)
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III. Giai tch o th va phan cc:Giai tch o th va phan cc:Giai tch o th va phan cc:Giai tch o th va phan cc:1111---- Phan cc JFET:Phan cc JFET:Phan cc JFET:Phan cc JFET:
Phng trnh tai DC (DCLL): VDD = VDS + ID(Rd + RS) (5-11)
Do IG 0 nen mach t phan cc: VGS = -IDRS (5-12)
2222---- Phan cc JGFET:Phan cc JGFET:Phan cc JGFET:Phan cc JGFET: H5-14Phng trnh tai DC (DCLL): VDD = VDS + ID(Rd + RS) (5-13)
nh ngha nguon ap cung cap cho cc cong la:
DD21
1GG VRR
RV+
= (5-14)
Theo nh luat KII Vkn = 0 ta co:
IG(Rb + R3) + VGS + IDRS = VGG (5-15)
Do IG 0 nen ta co:
SDDD21
1SDGGGS RIVRR
RRIVV +
== (5-16)
e ID t thay oi theo nhiet o ta phai chon:
2 VVVVGGpo
GGGSQ += (5-17)
Suy ra: poGSQGG VV2V += (5-18)
DQ
poGSQS I
VVR
+= (5-19)
Cac gia tr cho bi phng trnh (5-17, 18, 19) se xac nh iem tnh Q va cctieu hoa s phu thuoc vao nhiet o cua tnh iem.
IV. Giai tch tn hieu ln, s sai dang:Giai tch tn hieu ln, s sai dang:Giai tch tn hieu ln, s sai dang:Giai tch tn hieu ln, s sai dang:oi vi IGJET:
2
po
GSpoD V
V1II
+= (5-20)
a tn hieu AC vao cc cong:
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tcosVVvVV 0imGSQiGSQGS +=+= (5-21)
Thay (5-21) vao (5-20) ta c:
44 344 2144444 344444 214444 34444 21
haibachaiphanThanh
0po
impo
nhatbachaiphanThanh
0po
im
po
GSQ
po
DCbnhtrungphanThanh
2
po
im
2
po
GSQ
Dt2cos
V
V
2
Itcos
V
V
V
V1I2
V
V
2
1
V
V1I +
++
+
+= (5-22)
Khi Vim = 0 ta co:
2
po
GSQpoDQ V
V1II
+= (5-23)
Khi Vim
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-23- Tom Tat Bai Giang. -23-
1)1)1)1) Bo khuech ai cc nguon chung:Bo khuech ai cc nguon chung:Bo khuech ai cc nguon chung:Bo khuech ai cc nguon chung: H5-17Tr khang vao nhn t nguon: Zi = R3 + (R1//R2) (5-30)
Tr khang ra nhn t tai: Zo = Rd//rds (5-31)
( )
( )
Lm
213
ioLm
i
gs
gs
L
i
LV Rg
R//RRr1
1Z//RgVV
VV
VVA
++
=== (5-32)
vi ri
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Khoa ien ien T. Ky thuat mach ien T I
-24- Tom Tat Bai Giang. -24-
Tr khang vao:1Rr
iV
R ddsi
sgsg
+
+== (5-42)
He so khuech ai:
1Rrr
RVV
A ddsi
didV
+
++
== (5-43)
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Chng VI: Mach Transisrtor ghep lien tangChng VI: Mach Transisrtor ghep lien tangChng VI: Mach Transisrtor ghep lien tangChng VI: Mach Transisrtor ghep lien tang
1)1)1)1) Transistor ghep Cascading:Transistor ghep Cascading:Transistor ghep Cascading:Transistor ghep Cascading:a) Tang E.C-E.C: H6-1
o li dong ien:
+
+
+
===
1e'
1b
'1b
2ie'
2b
'2b1fe
L2C
2C2fe
i
1b
1b
2b
2b
L
i
Li hR
RhRRh
RRRh
ii
ii
ii
iiA (6-1)
vi Rb1 = ri//Rb1 va Rb2 = Rc1//Rb2
Tr khang vao: Zi = ri//Rb1//hie1 (6-2)
Tr khang ra: Zo = Rc2 (6-3)
Neu trong H1-1 khong co CE1 va CE2 th:
++
++
+
=
1E1fe1e'
1b
'1b
2E2fe2e'
2b
'2b1fe
L2C
2C2fei RhhR
RRhhR
RhRR
RhA (6-4)
Zi = ri//Rb1//[hie1 + (1 + hfe1)RE1] (6-5)
Zo = Rc2
b) Tang E.C-C.C hnh bai tap 6-3
o li dong ien:
( ) ( )
+
++
+=
==
1e1b
1b
L2E2fe2ie2b1C
2b1C
L2E
2E2fe
i
1b
1b
L
L
L
i
Li
hRR
R//RhhR//RR//R
RRRh
i
i
i
'i
'i
i
i
iA
(6-6)
Zi = Rb1//hie1 (6-7)
ZoZi
hie1
iC1
hfe1ib1
RE2hfe2
RLhfe2
hie2
ib1
ii
iL
Rb2 VLRb1 RC1
ib2
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+=
2fe
2b1C2ib2Eo h
R//Rh//RZ (6-8)
* Neu khong co CE1 khi o:
( ) ( )
++
++
+=
==
1E1fe1e1b
1b
L2E2fe2ie2b1C
2b1C
L2E
2E2fe
i1b
1bL
LL
iLi
RhhRR
R//RhhR//RR//R
RRRh
iii'i'iiiiA(6-9)
Zi = Rb1//[hie1 + hfe1RE1] (6-10)
+=
2fe
2b1C2ib2Eo h
R//Rh//RZ (6-11)
2)2)2)2) Mach khuech ai vi sai:Mach khuech ai vi sai:Mach khuech ai vi sai:Mach khuech ai vi sai:T H6-4 ta co:
fe
bE
EE2EQ1EQ
hRR2
7,0VII+
== (6-12)
VCEQ1 = VCEQ2 VCC + VEE ICQ(RC + 2RE) (6-13)
Cac dong ien co the c phan thanh hai thanh phan:
+ Mot chung2
iii 210+
= (6-14)
+ Mot vi sai 12 iii = (6-15)o li dong ien mot chung:
++
+=
fe
bibE
1b
LC
CC
hRhR2
RRR
RA (6-16)
o li dong ien mot vi sai:
+
+=
fe
bib
b
LC
Cd
hRh2
RRR
RA (6-17)
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v vay iL = ACi0 + Adi (6-18)
Ty so nen tn hieu ong pha:
fe
b
ib
E
h
R
h
RCMRR
+
(6-19)
e tang CMRR ta dung nguon dong cc phat.
3)3)3)3) Cach ghep Darlington:Cach ghep Darlington:Cach ghep Darlington:Cach ghep Darlington:T H6-10, ap dung nh luat K II. Vkin = 0 ta co:
IB1Rb + VBE1 + VBE2 + ICQ2RE = VBB1 (6-20)
1BBE2CQb2fe1fe
2CQ VRI7,07,0Rhh
I=+++ (6-21)
Tong quat dong ien ra ICQ2 c tnh:
2fe1fe
bE
1BB2CQ
hhRR
4,1VI+
= (6-22)
Neu2fe1fe
bE hh
RR >> th (6-22) tr thanh:
E
1BB2CQ R
4,1VI (6-23)
Trong o
CC21
11BB VRR
RV+
= (6-24)
21
21b RR
RRR+
= (6-25)
Phng trnh tai mot chieu au ra:
( )EC2CQ2CEE2C2fe
2CQ2CC2CECC RRIVRIh
IIRVV +++
++= (6-26)
VCE1 = VCE2 VBE2 = VCE2 0,7 (6-27)
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2fe
2CQ1CQ h
II = (6-28)
T H3-2 ta co:
b
1ieLC
C2fe1fe
2ie1ib1fe
b
b
LC
C2fe
i2b
2bL
iLi
Rh21
1RR
Rhh
hhhR
RRR
Rh
ii
ii
ii
A
++=
++
+=
==
(6-29)
Chu y:
1ib1fe
1ie
2fe1EQ
3
2fe1EQ
3
2fe2ie hhh
hI10.25h4,1
I10.25h4,1h ====
(6-30)
Tr khang vao: Zi = hie1 + hfe1hie2 2hie1 (6-31)
Tr khang ra: Zo = RC
* Mach Darlington cung co the lam viec che o toi u khi o:
RDC = RC + RE (6-32)
RAC = RC//RL (6-33)
Dong ien ra che o toi u:
LC
LCEC
CC
ACDC
CCCQmaxcm
RRRRRR
VRR
VIITU22
+++
=+
== (6-34)
ACCEQCEQmaxcm R.IVV TU2TU22 == (6-35)
maxcmLC
CmaxLm 2
IRR
RI+
= (6-36)
LmaxLmmaxLm RIV = (6-37)
* Mach Darlington cung co the mac theo kieu C.C.
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He so khuech ai:
( ) ( ) 2fe1feLE1iebi
bi
LE
E2fe1fe
i
1b
1b
L
L
L
i
Li
hhR//Rh2R//RR//R
RRRhh
ii
i'i
'ii
iiA
+++=
==
(6-38)
( ) ( ) 2fe1feLEiebibi
LE
EL2fe1fe
i
LL
i
LT hhR//Rh2R//R
R//RRR
RRhhiiR
iVA
1+++
=== (6-39)
ii
L
ii
LL
i
LV AR
RiRiR
VVA === (6-40)
Tr khang vao:
( )[ ]2fe1feLE1iebii hhR//Rh2//R//RZ += (6-41)
Tr khang ra:
+=
2fe1fe
bi2ibEo hh
R//Rh2//RZ (6-42)
VL
+VCC
T1
Ri
C1
RL
ZoZi
IC1
IE1=IB2ii
R2
RE
RC
R1 C2
T2
hie1
RLhfe1hfe2
ib1
iL
VLRi
Zo
hfe1hie2
ii REhfe1hfe2Rb
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4)4)4)4) Mach khuech ai Cascode:Mach khuech ai Cascode:Mach khuech ai Cascode:Mach khuech ai Cascode: H6H6H6H6----13131313T1 mac theo kieu EC, T2 mac theo kieu BC. Do o ay la bo khuech ai lien
tang E.C-B.C.
Che o DC:Che o DC:Che o DC:Che o DC: T hnh ve neu I2 >> IB1 va I2 >> IB2 ta co:CC
321
212BB VRRR
RRV++
+= (6-43)
2BB21
11BB VRR
RV+
= (6-44)
1fe
21E
1BB1E1C
hR//RR
7,0VII+
= (6-45)
VCE1 = VBB2 VBE2 RCIC1 REIC1 = VBB2 0,7 IC1(RC + RE) (6-46)
VCE1 = VCC IC1RL VBB2 + 0,7 (6-47)
Che o AC:Che o AC:Che o AC:Che o AC:
( )( )( )
+=
==
1ie21
211fe2fb
i
1b
1b
2e
2e
L
i
Li
hR//RR//Rhh
ii
ii
ii
iiA
(6-48)
iLi
LL
i
LT ARi
Rii
VA === (6-49)
Tr khang vao: Zi = R1//R2//hie1 (6-50)
Tr khang ra: Zo = (6-51)
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Chng VIIChng VIIChng VIIChng VII: Mach khuech ai hoi tiep.: Mach khuech ai hoi tiep.: Mach khuech ai hoi tiep.: Mach khuech ai hoi tiep.
1. Khai niem c ban ve mach hoi tiep:Khai niem c ban ve mach hoi tiep:Khai niem c ban ve mach hoi tiep:Khai niem c ban ve mach hoi tiep:
Hoi tiep la hien tng a tn hieu t ngo ra cua bo khuech ai ngc tr vengo vao. Hoi tiep gom 3 loai:
+ Hoi tiep noi bo sinh ra do tnh chat vat ly cua Transistor. V du: hoi tiep quaCbc khi mac E.C.+ Hoi tiep k sinh: sinh ra do cac phan t ghep mach: ien cam, bien ap, tcam ...
Hai loai hoi tiep tren la khong mong muon v no lam xau i cac ch tieuky thuat cua bo khuech ai.
+ Hoi tiep ben ngoai: do ta mac vao e cai thien cac ch tieu ky thuat cua bokhuech ai. Khi cac phan t hoi tiep la thuan tr ta co hoi tiep khong phu thuoc
tan so va khong ao pha, khi hoi tiep bao gom cac khau RC, L, bien ap,Transistor, ... th hoi tiep phu thuoc tan so va ao pha.
Neu hoi tiep a ve ong pha vi tn hieu vao, ta goi la hoi tiep dng va chdung trong cac bo dao ong. Neu hoi tiep a ve ngc pha vi tn hieu vao ta goi lahoi tiep am va chung c s dung rat nhieu trong cac bo khuech ai e cai thiencac ch tieu ky thuat. Trong chng nay ta ch xet hoi tiep am.
Phan loai hoi tiep:Phan loai hoi tiep:Phan loai hoi tiep:Phan loai hoi tiep:
+ Tuy theo ien ap hoi tiep (V f) ty le vi ien ap ra (Vo), dong ien ra(Io) hay ty le vi ca hai ma hoi tiep thuoc loai hoi tiep ien ap, hoitiep dong ien hay hoi tiep hon hp. e phan biet ba loai hoi tiep nayta dung phep th:
Ngan mach tai ma mat hoi tiep (Vf= 0) th o la hoi tiep ien ap. H mach tai ma mat hoi tiep (Vf= 0) th o la hoi tiep dong ien. Ca khi ngan mach va h mach tai ma van con hoi tiep (V f 0) tho la hoi tiep hon hp.
Bo khuech ai
AV
Mach hoi tiep
TaiNguon
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-32- Tom Tat Bai Giang. -32-
+ Tuy theo ien ap hoi tiep a ve ngo vao mac noi tiep hoac mac songsong vi nguon tn hieu vao ma ta co hoi tiep noi tiep (sai lech ap)hoac hoi tiep song song (sai lech dong).
+ Nh vay ta co bon loai hoi tiep hay dung Hoi tiep ap, sai lech ap. Hoi tiep ap, sai lech dong. Hoi tiep dong, sai lech ap. Hoi tiep dong, sai lech dong.
2. Anh hng cua hoi tiep am en cac tham so cua bo khuech ai:Anh hng cua hoi tiep am en cac tham so cua bo khuech ai:Anh hng cua hoi tiep am en cac tham so cua bo khuech ai:Anh hng cua hoi tiep am en cac tham so cua bo khuech ai:a) oi vi o li ap khi co hoi tiep am AVf:
V
VVf A1
A
A += (7-1)
+ AV: o li ap khi cha co hoi tiep.+ : he so hoi tiep.
b) oi vi s mat on nh cua o li ap:+ He so bat on nh cua AV:
V
V
AdAq = (7-2)
+ He so bat on nh cua AV khi co hoi tiep am:V
f A1 qq += (7-3)
c) Anh hng en meo tan so va meo pha:+ He so meo tan so: 1
AAM
f
0 = (7-4)
A0: o li day gia. Af: o li tai f.
+ He so meo tan so khi co hoi tiep am f la:V
f A11M1M
+
= (7-5)
V
f A1MM+
= (7-5)
+ o dch pha khi cha co hoi tiep la , o dch pha khi co hoi tiep am f:
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Vf A1 +
= (7-6)
d) Anh hng en he so meo phi tuyen va tap am He so meo phi tuyen:
m1
2nm
2m3
2m2
m1
2nm
2m3
2m2
IIII
VVVV +++
=+++
=LL
(7-7)
Trong o Vnm (Inm) la thanh phan hai bac th n (n = 1, 2, 3, ...)
He so meo phi tuyen khi co hoi tiep am:V
f A1 +
= (7-8)
Tap am khi co hoi tiep am:V
TATAf
A1
VV
+
= (7-9)
e) Anh hng en tr khang vao: Hoi tiep noi tiep: ( )Viif A1ZZ += (7-10) Hoi tiep song song:
V
iif A1
ZZ+
= (7-11)
f) Anh hng en tr khang ra:
Hoi tiep dong ien: ( )Voof A1ZZ += (7-12)
Hoi tiep ien ap:V
oof A1
ZZ+
= (7-13)
3. Hoi tiep ien ap, sai lech dong ien:Hoi tiep ien ap, sai lech dong ien:Hoi tiep ien ap, sai lech dong ien:Hoi tiep ien ap, sai lech dong ien:T H7-2 ta co o li dong thuan:
'i
iifi A0G
AA ==
= (7-14)
o li dong khi co hoi tiep am:
T1A
RGA1AA i
Lii
iif
=
+= (7-15)
o li vong c nh ngha:
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Liii
'L
L RGA0iV
VT ==
= (7-16)
Neu T >> 1 ta co:L
f
i
Lif R
RiiA == (7-17)
i
f
i
Lvf r
RVVA == (7-18)
Tr khang vao khi co hoi tiep:T1
ZZ iif
= (7-19)
Tr khang ra khi co hoi tiep:T1
ZZ oof
= (7-20)
4. Hoi tiep ien ap, sai lech ien ap:Hoi tiep ien ap, sai lech ien ap:Hoi tiep ien ap, sai lech ien ap:Hoi tiep ien ap, sai lech ien ap: o li ien ap khi khong co hoi tiep:
'v
vvfV A0K
AA ==
= (7-21)
o li ien ap khi co hoi tiep:
T1A
AK1A
VVA V
VV
V
i
Lvf
=
+== (7-22)
o li vong T:VV
'VV
i'L
L AKAK0VV
VT ===
= (7-23)
Tr khang vao khi co hoi tiep:)T1(ZZ iif = (7-24)
Tr khang ra khi co hoi tiep:
T1ZZ oof
= (7-25)
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Khoa ien ien t Ky thuat mach ien T I
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MOT SO BAI TAP MAU CHO QUYENMOT SO BAI TAP MAU CHO QUYENMOT SO BAI TAP MAU CHO QUYENMOT SO BAI TAP MAU CHO QUYENGiao trnh mach ien t IGiao trnh mach ien t IGiao trnh mach ien t IGiao trnh mach ien t I
Chng I: DIODE BAN DAN.Chng I: DIODE BAN DAN.Chng I: DIODE BAN DAN.Chng I: DIODE BAN DAN.
I.I.I.I. Diode ban dan thong thng:Diode ban dan thong thng:Diode ban dan thong thng:Diode ban dan thong thng:1) Ve dang song chnh luVe dang song chnh luVe dang song chnh luVe dang song chnh lu: (Bai 1-1 trang 29)
Cong thc tong quat tnh VL:
LLi
DSL RRR
VVV+=
VD = 0,7V (Si) va VD = 0,2V (Ge)
aaaa---- Ve VVe VVe VVe VLLLL(t) vi V(t) vi V(t) vi V(t) vi VSSSS(t) dang song vuong co bien o 10 va 1V(t) dang song vuong co bien o 10 va 1V(t) dang song vuong co bien o 10 va 1V(t) dang song vuong co bien o 10 va 1V
Ket qua vi gia thiet: Ri = 1, RL = 9, VD = 0,7V.V Diode chnh lu ch dan ien theo mot chieu nen:
Trong 0T21 > , Diode dan iD 0 iL 0 VL 0.
V37,8991
7,010V 1L =+
= va V27,09
917,01V 2L =
+
=
Trong 0T21
< , Diode tat iD = 0 iL = 0 VL = 0.
bbbb---- Ve VVe VVe VVe VLLLL(t) vi V(t) vi V(t) vi V(t) vi VSSSS(t) dang song sin co bien o 10 va 1V(t) dang song sin co bien o 10 va 1V(t) dang song sin co bien o 10 va 1V(t) dang song sin co bien o 10 va 1V....
iL
iD
RL
RiVLVs
-
-+VD
10
-10
0 1 - -
++
VS
2 3 4 t(ms)
1
-1
0 1 - -
++
VS
2 3 4 t(ms)
8,37
0 1
VL1
2 3 4 t(ms)0,27
0 1
VL2
2 3 4 t(ms)
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Khi VS = 10sinot ngha la VSm = 10V >> VD =0,7V ta co:99
9110R
RRVV L
Li
Sm1L =
+
+
tsin9V 01L
(Ta giai thch theo 0T21
> va 0T21
< ) Khi VS = 1sin0tngha la VSm = 1V so sanh c vi 0,7V:
+ VS > 0,7V, Diode dan, iD 0, iL 0, VL 0.6,0tsin9,09
917,0tsin1
V 00
2L =+
=
Tai sin0t = 1, |VL2| = 0,27V.+ VS < 0,7V, Diode tat, iD = 0, iL = 0, VL = 0.
Vi dang song tam giac ta co ket qua tng t nh song sin.
2) Bai 1Bai 1Bai 1Bai 1----3:3:3:3: e co cac ket qua ro rang ta cho them cac gia tr ien tr: R1 =1K, Rb = 10K, RL = 9K.
aaaa---- Ve VVe VVe VVe VLLLL(t) vi dang song vuong co bien o 10V va 1 V.(t) vi dang song vuong co bien o 10V va 1 V.(t) vi dang song vuong co bien o 10V va 1 V.(t) vi dang song vuong co bien o 10V va 1 V. 0T21 > , Diode dan, RthD 0, dong iL chay qua Ri, D, RL nen ta co:
V37,810.9.10.9107,010R
RRVV
V 333LLi
DS1L =
+
=
+
=
V27,010.9.10.9107,01R
RRVV
V 333LLi
DS2L =
+
=
+
=
0T21
< , Diode tat, Rng = , dong iL chay qua Ri, Rb, RL nen ta co.
iLRL9K
Ri=1K
VLVs+
-
-+ VD
Rb=10K
10
0-10
9
- -+ +
12 3
4 t(ms)
VS
VL1
0 1 2 3 4 t(ms)
1
0-1
12 3
4 t(ms)
VS
VL2
0 1 2 3 4 t(ms)
0,7
0,27
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V5,410.9.10.91010
10RRRR
VV 3343L
Lbi
S1L =
++=
++=
V45,010.9.10.91010
1RRRR
VV 3343L
Lbi
S1L =
++=
++=
bbbb---- Ve VVe VVe VVe VLLLL(t) vi dang song sin co bien o 10V va 1 V.(t) vi dang song sin co bien o 10V va 1 V.(t) vi dang song sin co bien o 10V va 1 V.(t) vi dang song sin co bien o 10V va 1 V. e n gian khi VSm = 10V (>>VD = 0,7V) ta bo qua VD. Khi o:
+ 0T21
> , Diode dan, RthD 0, dong iL chay qua Ri, D, RL nen ta
co:
)V(tsin910.9.10.910
tsin10R
RRV
V 03
330
LLi
S1L =
+
=
+=
+ 0T21
< , Diode tat, Rng = , dong iL chay qua Ri, Rb, RL nen ta co.
)V(tsin5,410.9.10.91010
tsin10R
RRRV
V 03
3430
LLbi
S1L =
++
=
++=
Khi VS = 1sin0t so sanh c vi VD ta se co:+ 0T
21
> , khi VSm 0,7, Diode dan, RthD 0, dong iL chay qua Ri,
D, RL nen ta co:
)V(63,0tsin9,010.9.10.910
7,0tsin1R
RR7,0tsin1
V 03
330
LLi
02L =
+
=
+
=
Tai 2t0 = , sin0t = 1, ta co VL2m = 0,9 - 0,63 = 0,27V
+ 0T21
> , khi VSm < 0,7, Diode tat, RngD = , dong iL chay qua Ri,
Rb, RL nen ta co:
tsin315,010.9.10.91010
tsin7,0R
RRRtsin7,0
V 03
3430
LLbi
02L =
++
=
++
=
10
-100 1 - -
++
VS
2 3 4 t(ms)
1
-10 1 - -
++
VS
2 3 4 t(ms)
8,370 1
VL1
2 3 4t(ms)
0,270 1
VL2
2 3 4 t(ms)-4,5
-0,45
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+ 0T21
< , Diode tat, Rng = , dong iL chay qua Ri, Rb, RL nen ta co.
tsin45,010.9.10.91010
tsin1R
RRRtsin1
V 03
3430
LLbi
02L =
++
=
++
=
2)2)2)2) Dang mach Thevenin ap dunDang mach Thevenin ap dunDang mach Thevenin ap dunDang mach Thevenin ap dung nguyen ly chong chap:g nguyen ly chong chap:g nguyen ly chong chap:g nguyen ly chong chap:Bai 1-20 vi Vi(t) = 10sin0t
a- Ve mach Thevenin:Ap dung nguyen ly xep chong oi vi hai nguon ien ap VDC va Vi: Khi ch co VDC, con Vi = 0 th ien ap gia hai iem A-K:
V3
10.5,110
10.5,15
rR
rVV 33
3
ii
iDCAK =
+
=
+
=
Khi ch co Vi, con VDC = 0 th ien ap gia hai iem A-K la:)V(tsin4
10.5,11010tsin.10
rRRVV 033
3
0ii
iiAK =
+=
+=
Vay khi tac ong ong thi ca VDC va Vi th sc ien ong tngng Thevenin gia hai iem A-K la:
VL
+
-Vi
+
-
iD
RL1,4K
Ri=1K
VDC=5v
KA
ri=1,5K
RT id
VTK
A
RLRi//ri iL
VT
KA
10
0-10
9
- -
+ +t(ms)
VS
VL1
t(ms)
1
0-1
t(ms)
VS
VL2
t(ms)
0,7
0,315+ +
- --4,5 -4,5
0,585
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)V(tsin43rR
RVrR
rVV 0ii
ii
ii
iDCT +=
++
+=
ien tr tng ng Thevenin chnh la ien tr tng ng cu aphan mach khi Diode h mach la:
=++
=++
= K210.4,110.5,110
10.5,1.10R
rR
r.RR 3
33
33
Lii
ii
T
b- Ve ng tai DC khi2
,3
,2
,3
,0t0
= .
Tai V3V0t T0 == Tai )V(46,6
2343V
3t T0 =+=
=
Tai )V(71.43V2
t T0 =+=
=
Tai )V(46,023
43V3t T0 ==
= Tai )V(11.43V
2t T0 ==
=
Theo nh luat Ohm cho toan mach ta co.
T
TD
TT
DT
RV
V.R1
RVV
i +=
=
Tai )mA(15,110.237,0.
10.21i0t 330 =+==
Tai )mA(88,210.246,67,0.
10.21i
3t 330 =+=
=
Tai )mA(15,310.277,0.
10.21i
2t 330 =+=
=
iD (mA)
3,152,88
1,15
3 6,46 7-1
VT
t
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Tai )mA(58,010.246,07,0.
10.21i
3t 330 ==
=
Tai )mA(85,010.217,0.
10.21i
2t 330 ==
=
c- Ve
( )
( ) )V(tsin8,21,2tsin437,0V7,0
10.2V10.4,1Rr//R VRRV.Ri.R)t(V
00T
3T3Lii
TLTTLDLL
+=+==
=+
===
II.II.II.II. Diode Zenner:Diode Zenner:Diode Zenner:Diode Zenner:1) Dang dong IL = const (bai 1-40); 200mA IZ 2A, rZ = 0
a- Tm Ri e VL = 18V = const.Imin = IZmin + IL = 0,2 + 1 = 1,2 A.Imax = IZmax + IL = 1 + 2 = 3 A.Mat khac ta co: Vimin = 22V = IZmin.Ri + VZ.Suy ra:
==
=
= 3,32,1
42,11822
IVV
RminZ
Zminii
Vimax = 28V = IZmaxRi + VZSuy ra
==
=
= 3,33
103
1828I
VVR
maxZ
Zmaxii
Vay Ri = 3,3.b- Tm cong suat tieu thu ln nhat cua Diode Zenner:
PZmzx = IZmax.VZ = 2.18 = 36W.
2) Dang dong IL const: (bai 1-41), 10mA IL 85mA.IZmin = 15mA.
VL
0-0,7
2,1
4,9V
t
RLVZ=10v13v
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a- Tnh gia tr ln nhat cua R imaxLminZ
Zii
minLmaxZ
Zi
IIVV
RIIVV
+
+
Khi VDC = 13V ta co=
+
30
085,0015,01013R maxi
Khi VDC = 16V ta co=
+
60
085,0015,01016R maxi
Vay ta lay Rimax = 30.b- Tm cong suat tieu thu ln nhat cua Diode Zenner.
PZmax = IZmax.VZ.Mat khac: Vimax = IZmaxRi + VZ
mA20030
1016R
VVIi
Zmaximax =
=
=
mA19019,001,02,0III minLmaxmaxz ==== W9,11019,0P maxz ==
3)
Dang IZ const; IL const (Bai 1-42)30 IL 50mA, IZmin = 10mA.rZ = 10 khi IZ = 30mA; Pzmax =800mW.
a- Tm Ri e Diode on nh lien tuc:mA80
108,0
VPI
Z
maxZmaxZ ===
Vay 10mA IZ 80mATa co: Imin = IZmin + ILmax = 60mAImax = IZmax + ILmin = 110mAMat khac: Vimin = Imin.Ri + VZ = 20V
RLVZ=10v20v
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== 7,16606,0
1020R maxi
Vimax = Imax.Ri + VZ = 25V
== 36,13611,0
1025R mini
Suy ra: 136,4 Ri 166,7Vay ta chon Ri =150
b- Ve ac tuyen tai:Ta co: VZ + IZRi = VDC ILRi Vi VDC = 20V ta co:
==
===+
mA50IkhiV5,1215005,020mA30IkhiV5,1515003,020
150IVL
LZZ
Vi DC = 25V ta co:
==
===+
mA50IkhiV5,1715005,025
mA30IkhiV5,2015003,025150IV
L
LZZ
Tng ng ta tnh c cac dong IZ:
mA7,36150
105,15I 1Z =
= ; mA7,16150
105,12I 2Z =
=
mA70150
105,20I 3Z =
= ; mA50150
105,17I 4Z =
= ;
IZ(mA)
VZ
36,7
50
30
80
70
10
20,5 17,5
15,5
VZ =10V 0
rZ =10
16,7
12,5
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Chng IIChng IIChng IIChng II: TRANSISTOR HAI LP TIEP GIAP: TRANSISTOR HAI LP TIEP GIAP: TRANSISTOR HAI LP TIEP GIAP: TRANSISTOR HAI LP TIEP GIAP
I.I.I.I. Bo khuech ai RBo khuech ai RBo khuech ai RBo khuech ai R----C khC khC khC khong co Cong co Cong co Cong co CCCCC va khong co Cva khong co Cva khong co Cva khong co CEEEE (E.C).(E.C).(E.C).(E.C).1) Bai 2-10: 20 60, suy ra ICQ khong thay oi qua 10%.
Phng trnh tai mot chieu:VCC = VCEQ + ICQ(RC + RE).
mA81010.5,1
525RRVV
I 33EC
CEQCCCQ =
+
=
+
=
Neu coi ay la dong ien ban au khi = 60 sao cho sau mot thi gian ch con = 20 th yeu cau ICQ 7,2mA.
Ta giai bai toan bai toan mot cach tong quat coi 1 = 20; 2 = 60.E22bbE11b R10
1RRR101R ==
==== K610.60.101RRK210.20.
101R 32bb31b
Vay 2K Rb 6K
Mat khac
+
=
bE
BBCQ R
R
7,0VI , neu coi VBB const th ta co:
9,0R
R
RR
II
1
bE
2
bE
2CQ
1CQ
+
+
= (1)
Co the tnh trc tiep t bat phng trnh (1):
+
+
+
12bE
1
bE
2
bE
9,01RR1,0R
R9,0R
R
==
+
=
+
K53,310.3,28
100
209,0
601
10.1,09,01
R1,0R 3
3
12
Eb
Chon Rb = 3,5K.
VCEQ = 5V+
-
+25V
R2
R1
RC=1,5K
RE=1K
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Neu bo qua IBQ ta co VBB VBE + IEQRE = 0,7 + 8.10-3.103 = 8,7V. Suyra:
==
=
= K4,55368652,010.5,3
257,81
110.5,3
VV
1
1RR3
3
CC
BBb1
=== K06,10100577,8
2510.5,3VVRR 3
BB
CCb2
Ta co the tnh tong quat: Chon Rb = 4K thay vao (1):%9,88
12001067
2010.410
6010.410
II
33
33
2CQ
1CQ==
+
+
= , b loai do khong thoa man (1).
Chon Rb =3K thay vao (1): 91,011501050
20
10.310
6010.310
II
33
33
2CQ
1CQ==
+
+= thoa
man bat phng trnh (1), ta tnh tiep nh tren.
2) Bai 2-11: Vi hnh ve bai (2-10) tm gia tr cho R1, R2 sao cho dong iC xoaychieu co gia tr cc ai. iem Q toi u c xac nh nh sau:
ACCQTTCEQ
ACDC
CCTCQmaxCm
R.IVRR
VII
=
+==
T hnh ve: RDC = RC + RE = 1,5.103 + 103 = 2,5K.
RAC = RC + RE = 1,5.103 + 103 = 2,5K.Suy ra: mA510.5,210.5,2
25I 33TCQ =+=
VCEQT= 5.10-3.2,5.103 = 12,5V
Chon === K1010.100.101R
101R 3Eb (bo qua IBQ)
VBB VBE + ICQT.RE = 0,7 + 5.10-3.103 = 5,7V
VCE(V)
iC(mA)
VCEQT= 12,5 25
10RV
DC
CC =
( )5
RR2V
EC
CC =+
310.5,2
1ACLLDCLL
QT
0
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==
=
= K13K95,12772,0
10
257,51
110.10
VV
1
1RR4
3
CC
BBb1
=== K44K85,437,5
2510VV
RR 4BB
CCb2
V RDC = RAC nen phng trng tai DC va AC trung nhau.
3) Bai 2-14: iem Qbat ky v biet VBB = 1,2V; = 20. Tm gia tr toi a cua daoong co the co c C va tnh .
Biet = 20, VBEQ = 0,7V.
Ta co: mA3,3501007,02,1
RR
VVI
bE
BEQBBCQ =
+
=
+
=
e tm gia tr toi a cua dao ong co the co c C ta phai vephng trnh tai DC, AC
VCEQ = VCC ICQ(RC + RE) = 6 3,3.10-3.1,1.103 = 2,37V Vay gia tr toi a cua dao ong la:
ICmmax = iCmax ICQ = 5,45 3,3 = 2,15mASuy ra VLmax = ICmmax.RC = 2,15.103.10-3 = 2,15V
PCC = ICQ.VCC = 3,3.10-3.6 = 19,8mW( ) ( ) mW31,210.10.15,2
21R.I
21P 3
23C
2maxCmL ===
+6V
Rb = 1K
RC = 1K
RE = 100VBB = 1,2V
45,5RV
DC
CC =
ICQ = 3,3
iC (mA)
VCE(V)2,37 3 60
2,725 QTQbk
=
11001ACLLDCLL
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Hieu suat: %7,1110.8,1910.31,2
PP
3
3
CC
L ===
II.II.II.II. Bo KRC khong co CBo KRC khong co CBo KRC khong co CBo KRC khong co CCCCC, C, C, C, CEEEE (tu bypass Emitter) (EC)(tu bypass Emitter) (EC)(tu bypass Emitter) (EC)(tu bypass Emitter) (EC)1) Bai 2-15: iem Q bat ky.
a- Tm R1, R2 e ICQ = 01mA (Rb
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T hnh ve ta nhan thay e ICm ln nhat va khong b meo th ICmmax =10mA.Ta co the tm iCmax va VCemax theo phng trnh
( )CEQCEC
CQC VVR1Ii =
Cho VCE = 0 mA60150
5,710R
VIi 2C
CEQCQmaxC =+=+=
Cho iC = 0 V95,7150.10VR.IV 1CEQCCQmaxCE =+=+=
2) Bai 2-16: iem Q toi u (hnh ve nh hnh 2-15).e co dao ong Collector cc ai ta co:
ACDC
CCCQTmaxCm RR
VII
+== (1)
VCEQT = RAC.ICQT (2)RDC = RC + RE = 150 + 100 = 250RAC = RC = 150Thay vao (1) ta c: mA25
15025010I CQT =+
=
V75,310.25.150V 3CEQT ==
VBB 0,7 + ICQT.RE = 3,2V.=== K1100.100.
101R
101R Eb
=
=
= K47,168,0
10
102,31
10
VV
1
1RR33
CC
BBb1
=== K1,331252,3
1010VV
RR 3BB
CCb2
VCE(V)
iC(mA)
VCEQT= 3,75
2ICQT= 50
40RR
V
EC
CC
=+
150
1ACLL
2VCEQT=7
10
ICQT= 25
2501DCLL
-
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e ve ACLL, rat n gian ta ch can xac nh:iCmax = 2ICQTva VCemax = 2VCEQT.
III.III.III.III. Bo K RBo K RBo K RBo K R----C co CC co CC co CC co CCCCC va Cva Cva Cva CEEEE (E.C).(E.C).(E.C).(E.C).1) Bai 2-20: iem Q toi u
RDC = RC + RE = 900 + 100 =1K=
+=
+= 450
900900900.900
RRRR
RLC
LCAC
mA9,6RR
VII
DCAC
CCCQTmaxCm
+==
VCEQT = ICQT.RAC = 6,9.10-3.450 = 3,1VVBB = 0,7 + RE.ICQT= 0,7 + 100.6,9.10-3 = 1,4V
=== K1100.100.10
1
R10
1
R Eb
=
=
= 116386,0
10
104,11
10
VV1
1RR33
CC
BBb1
CE
Vcc=10V
R2
R1
RC=900
RE100
CC
RL=900K
VCE(V)
iC(mA)
VCEQT= 3,1
2ICQT = 13,8
10RR
V
EC
CC =+
4501ACLL
6,2 100
ICT= 6,9
10001DCLL
-
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=== 71434,1
1010VV
RR 3BB
CCb2
Ta co dong xoay chieu:
V1,3V
mA45,39,6900900
900I.RR
RI
Lm
CmLC
CLm
=
=+
=+
=
2) Van bai 2-20 neu ta bo tu CE th ta se co bo khuech ai R.C co CC ma khongco CE. Khi o ket qua tnh toan se khac rat t v RE
-
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(V
>> bER
R nen co the tnh gan ung theo cong thcE
BBCQ R
7,0VI
= )
VCEQ = VCC ICQ(RC + RE) = 25 2,1.10-3.3.103 = 18,7V
T hnh ve ta thay: ICQ < ICQT nen ICm = ICQ = 2,1mA
mA05,110.1,2.10.210.2
10.2IRR
RI 333
3
CmLE
LLm =
+=
+=
VLmmax = RL.ILm = 2.103.1,05.10-3 = 2,1V* Cach ve DCLL va ACLL cua bo K R.C mac C.C tng t nh cach mac E.C
( )CEQCEAC
CQC VVR1Ii =
vi =+
+= k2RR
RRRR
LE
LE
CAC
Cho VCE = 0 suy ra mA45,1110.2
7,1810.1,2RV
Ii 33
AC
CEQCQC =+=+=
iC = 0 suy ra V9,2210.1,2.10.27,18IRVV 33CQACCEQmaxCEQ =+=+=
* Vi bai toan tren neu cha biet R1 va R2 ta co the thiet ke e dong ien raln nhat: RDC = RC + RE = 103 + 2.103 = 3K.
Q
VCE(V)
iC(mA)
VCEQ
= 18,7
ICmax = 11,45
3,8RV
DC
CC =
310.2
1ACLL
100
ICQT= 5
310.3
1DCLL
22,9 25
ICQ = 2,1
-
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Ta co: mA510.210.3
25RR
VI 33
ACDC
CCCQT =
+=
+=
VCEQT = ICQT.RAC = 10V.
2) Bai 2-24: Mach c nh dong Emitter.Theo nh luat K.II: Vkn = 0 ta coRbIBQ + VBEQ + RE.IEQ VEE = 0
Suy ra mA93100
7,010R
R
7,0VI
bE
BBEQ =
+
=
VCEQ = VCC + VEE ICQ(RC + RE)= 10 + 10 93.10-3.150 = 6,05V
mA5,4610.93.100100
100IRR
RI 3Em
LE
ELm =
+=
+=
VLm = ILmRL = 46,5.10-3.102 = 4,65V ay la iem Q bat ky nen ta co:
( )CEQCEAC
CQC VVR1Ii =
+ Cho VCE = 0 suy ra mA214RV
IiAC
CEQCQmaxC =+=
+Cho iC = 0 suy ra V675,1050.10.9305,6RIVV 3ACCQCEQCE =+=+=
iL
I.
CC
VL
CE
VEE=-10v
Rb
-
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Neu bai nay c tnh che o toi u th:RDC = RC + RE = 150
=+
= 50RR
RRR
LE
LEAC khi o
mA100A1,05015020
RRV
I DCACCCCQT ==+=+=
VCEQT= ICQT.RAC = 5V
-
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Chng IVChng IVChng IVChng IV:THIET KE VA PHAN TCH TN HIEU NHO TAN SO THAP.THIET KE VA PHAN TCH TN HIEU NHO TAN SO THAP.THIET KE VA PHAN TCH TN HIEU NHO TAN SO THAP.THIET KE VA PHAN TCH TN HIEU NHO TAN SO THAP.
I.I.I.I. S o mac Emitter chung E.C:S o mac Emitter chung E.C:S o mac Emitter chung E.C:S o mac Emitter chung E.C:1) Bai 4-7: Q bat ky.
a-
Che o DC
K3205,3
20.5,3RR
RRR21
21b
+=
+=
V320.205,3
5,3VRR
RV CC21
1BB
+=
+=
mA6,4
10010.3500
7,03I3CQ
+
=
VCEQ = VCC ICQ(RC + RE) = 20 4,6.10-3.2.103 = 10,8V
==
76010.6,410.25h.4,1h 3
3
feie
b- Che o AC:
i
b
b
L
i
Li i
iii
ii
A == (1)Zo
iC
Zi
Ri2K
ibRb3Kii
RC1,5K
iL
RL=1,5Khie100ib
1,2K
RL=1,5Kii
RC=1,5K
CC2-+
+VCC=20V
CE+
-
R13,5K
iL
R2=20K
Ri=2K RE1,5K
CC1-+
-
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50100.10.5,110.5,1
10.5,1h.RR
Rii
ii
ii
33
3
feLC
C
b
C
C
L
b
L =+
=+
==
( )61,0
76010.2,110.2,1
hR//RR//R
ii
3
3
iebi
bi
i
b =+
=+
=
Thay vao (1) ta co: Ai = -50.0,61 = -30,6Zi = Ri//Rb//hie = 1200//760 = 465Zo = RC = 1,5K.
2) Bai 4-11: Q bat ky va hfe thay oi.a- Che o DC:
100R5010.50.101R
101R bE11b ==== , bo qua IBQ.
mA83
5010010
7,07,1RR
7,0VI
1
bE
BB1EQ =
+
=
+
=
mA10010
7,07,1R
7,0VIE
BB2EQ =
=
=
2110.8310.25.50.4,1h 3
3
1ie
==
5,5210.10010.25.150.4,1h 3
3
2ie
suy ra 21 hie 52,5b- Che o AC:
RL=100Rb=100
VBB=1,7vii
RC=100
CC-+
+VCC=20v
CE+-
iL
RE10
-
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ieb
bfe
LC
C
i
b
b
L
i
Li hR
R.h.
RRR
ii
ii
ii
A++
===
66,2021100
100.50.100100
100A 1i =++
=
1,495,52100
100.150.100100
100A 2i =++
=
Zi = Rb//hie suy ra Zi1 = 100//21 = 17,36Zi2 = 100//52,5 = 34,43
Vay 20,66 Ai 49,1817,36 Zi 34,43
3) Bai 4-12: Dang khong co tu CEa- Che o DC:
mA5,4
1001010
7,07,5
hRR
7,0VI 43
fe
bE
BBCQ =
+
=
+
=
(co the tnh ICQ = 5 mA)VCEQ = VCC ICQ(RC + RE) = 20 4,5.10-3.(3.103) = 6,5V
==
77810.5,4
10.25
.100.4,1h 3
3
ie
b- Che o AC:
Rb=10K
VBB=5,7Vii
RC=2K
CC-+
+VCC=20V
iL
RE=1K
RL=100
RL=100
ib
Rb10KiiRC2K
iLhfeRE
100ib
hie=778
iC
ibRb100
ii
RC100
iLhie hfeib RL = 100
-
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i
b
b
L
i
Li i
iii
ii
A == (1)
24,95h.RR
Rii
ii
ii
feLC
C
b
C
C
L
b
L =+
==
09,01077810
10RhhR
Rii
54
4
Efeieb
b
i
b =++
=++
=
Thay vao (1) ta c Ai = -95,24.0,09 = -8,6
[ ] =+= K1,910//10Rhh//RZ 54Efeiebi
II. S o mac B.C:S o mac B.C:S o mac B.C:S o mac B.C: Bai 4-21, hoe = 4101) Che o DC:
91,01110
h1hh
fe
fefb ==
+=
==+
=
3210
10.25.10.4,1.111
h1hh 3
3
fe
ieib
54
fe
oe
ob
1011
10
h1
hh
==+
=
2) Che o AC:
VCC
R2
Vi+
- R1 Cb
ri=50 RL=10K
RL10K
iL1/hob
105
iC
hfbie0,91ib
hib
32
ieRi 50
Vi
+
-
-
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i
e
e
L
i
LV V
iiV
VV
A == (1)
82791,0.1010
10.10h.
h1R
h1R
ii
.iRi
iV
54
54
fb
obL
obL
e
C
C
LL
e
L =+
=
+
==
012,03250
1hR
1hR
V.V1
Vi
ibiibi
i
ii
e =+
=+
=+
=
Thay vao (1) ta c AV = (-827).(-0,012) = 10,085 10
III. S o mac C.C:S o mac C.C:S o mac C.C:S o mac C.C: Bai 4-231) Che o DC
VCC = IBQRb + VBEQ + REIEQ
mA65,4
1001010
7,010
RR
7,0VI53b
E
CC
EQ=
+
=
+
=
VCEQ = VCC REIEQ = 10 4,65.10-3.103 = 5,35 V
2) Che o AC
RL1K
Vi+
-RE
1K
ZoZi
Cc2ri 500
100KRb
Cc1
+VCC =10V
iLri 500
Vi+
-
hie 753ib
Rb100K
Re.hfe100K
RL.hfe100K
Vb VL
-
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=
75310.65,4
10.25h4,1h 33
feie
ib
bL
iLv V
VVV
VV
A == (1)( )
( )[ ]
985,0000.50753
500.100R//Rhhi
R//Rh.iVV
LEfeieb
LEfeb
b
L
=+
=
+=
(2)
Rb = Rb//[hie + hfe(RE//RL)] = 33,3
994,010.3,33500
K3,33Rr
RRr
V.R.
V1
VV
3'bi
'b
'bi
i'b
ii
b =+
=
+=
+= (3)
Thay (2), (3) vao (1) ta co: AV = 0,985.0,994 = 0,979 0,98
[ ] +
+= 37,12553,7//10
hR//rh//RZ 3
fe
biibEo
( )[ ] ==+= K3,33RR//Rhh//RZ 'bLEfeiebi
hie/hfe 7,53ie
RE1K
ri/hfe5
Rb/hfe1K
Zo
-
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Chng VI:Chng VI:Chng VI:Chng VI: MACH TRANSISTOR GHEP LIEN TANG.MACH TRANSISTOR GHEP LIEN TANG.MACH TRANSISTOR GHEP LIEN TANG.MACH TRANSISTOR GHEP LIEN TANG.
I. Transistor ghep Cascading:I. Transistor ghep Cascading:I. Transistor ghep Cascading:I. Transistor ghep Cascading:1) E.C1) E.C1) E.C1) E.C C.EC.EC.EC.E
Bai 6Bai 6Bai 6Bai 6----1111: iem Q bat ky, 2 tang hoan toan oc lap vi nhau.aaaa ---- Che o DCChe o DCChe o DCChe o DC
==>=+
=+
= 500R.h.101RK1,2
10.710.310.7.10.3
RRR.RR Efeb33
33
2111
21111b
suy ra, khong c bo qua IBQ1;
V310.10.710.3
10.3V.RR
RV 33
3
CC2111
111BB =
+=
+=
mA2,16
502100100
7,03
hR
R
7,0VI
1fe
bE
1BB1EQ
1
1
=
+
=
+
=
VCEQ1 = VCC IEQ1(RC1 + RE1) = 10 16,2.10-3.300 = 5,14V
===
10810.2,16
10.25.50.4,1I
10.25.h4,1h 33
1EQ
3
1fe1ie
==
-
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50h.1ii.
ii
ii
2fe2b
2C
2C
L
2b
L === (2)
( )
06,550.1458164
164
h.hR//R
R//Rii.
ii
ii
1fe2ie2b1C
2b1C
1b
1C
1C
2b
1b
2b
+
+
==
(3)
951,01082100
2100hR
Rii
1ieb
b
i
1b =+
=+
= (4)
Thay (2), (3), (4) vao (1) ta coAi = (-50).(5,06).(0,951) 241
Zi = Rb//hie1 = 2,1.103//108 103Zo =
e tm bien o nh oi xng cc ai ta ve DCLL va ACLL.
T ( )CEQCEAC
CQC VvR1Ii =
vCE = 0 suy ra, ICmax = ICQ +VCEQ/RAC= 1,2.10-3 + 7,3/2.103 = 4,85mA
iC = 0 suy ra, vCemax = VCEQ.RAC= 7,3 + 1,2.10-3.2.103 = 9,7V
T ac tuyen DCLL va ACLL ta co ICmmax = 1,2mA
Bai 6Bai 6Bai 6Bai 6----2222: iem Q toi u nen phai tnh tang th hai trc, tang 1 sau.
aaaa---- Che o DC:Che o DC:Che o DC:Che o DC:RDC2 = RC2 + RE2 = 2250; RAC2 = RC = 2K.
mA35,220002250
10RR
VI
2AC2DC
CCT2CQ =
+=
+=
VCEQ2T = ICQ2T.RAC2 = 2,35010-3.2.103 = 4,7V
Q
VCE(V)
iC(mA)
7,3
iCmax = 4,85
4,4RV
DC
CC =
310.2
1ACLL
0
2501DCLL
9,7
10
ICQ = 1,2ICmmax
-
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===
74510.35,2
10.25.50.4,1I
10.25.h4,1h 33
2EQ
3
2fe2ie
RDC1 = RC1 + RE1 = 200 + 100 = 300;RAC1 = RC1//Rb2//hie2 = 200//900//745 134,4
mA234,134300
10
RR
V
I 1AC1DCCC
T1CQ =+=+=
===
7610.2310.25.50.4,1
I10.25.h4,1h 3
3
1EQ
3
1fe1ie
b- Che o AC:Che o AC:Che o AC:Che o AC: S o tng ng tn hieu nho nh tren ch co hie1 va hie2 cogia tr khac. Ta ap dung luon cong thc (1) tren:
( )434
7621002100.
745164164.2500
hRR
.hR//R
R//Rh.hA
1ieb
b
2ie2b1C
2b1C2fe1fei
++=
++=
2) E.C2) E.C2) E.C2) E.C C.C:C.C:C.C:C.C:
Bai 6Bai 6Bai 6Bai 6----3333 iem Q toi uaaaa---- Che o DC:Che o DC:Che o DC:Che o DC:
Tang 2:RDC2 = RE2 = 1K; RAC2 = RE2//RL = 500.
mA7,650010
10RR
VI 3
2AC2DC
CCT2CQ =
+=
+=
VCEQ2T= ICQ2T.RAC2 = 6,7.10-3.500 = 3,35V
===
52210.7,610.25.100.4,1
I10.25.h4,1h 3
3
2EQ
3
2fe2ie
==== K101010.100.101R.h101R432E2fe2b
VBB = 0,7 + ICQ2T.RE = 0,7 + 6,7.10-3.103 = 7,4V
==
=
= K46,3826,0
10
104,71
10
VV
1
RR
44
CC
2BB
b12
=== K5,134,7
1010VV
RR 42BB
CCb22
Tang 1:
RDC1 = RC1 + RE1 = 400 + 100 = 500;RAC1 = RC1//Rb2//[hie2 + hfe(RL//RE)]= 400//104//[261 + 100.500] = 400//8333 382.
mA34,11382500
10RR
VI
1AC1DC
CCT1CQ =
+=
+=
VCEQ1T= ICQ1T.RAC1 = 11,34.10-3.382 = 4,33V
-
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===
30934,11
10.25.100.4,1I
10.25.h4,1h3
2EQ
3
1fe1ie
=== K1100.100.101R.h
101R 1E1fe1b
VBB1 = 0,7 + ICQ1T.RE1 = 0,7 + 11,34.10-3.100 = 1,834V
==
=
= K25,128166,010
10834,11
10
VV
1RR 33
CC
1BB
1b11
=== K45,5834,11010
VV
RR 31BB
CC1b21
bbbb---- Che o AC:Che o AC:Che o AC:Che o AC:
i
1b
1b
2b
2b
LT i
i.
ii
.iV
A = (1)
( )( ) 3LEfe2b
L 10.5,50R//Rh1iV
=+= (2)
( ) ( )
75,051407
10.385100.50500522385
385
h.R//Rh1hR//R
R//Rii
.ii
ii
2
1feLEfe2ie2b1C
2b1C
1b
1C
1C
2b
1b
2b
=++
=
+++
==
(3)
764,030910
10hR
Rii 3
1ie1b
1b
i
1b =+
=+
= (4)
Thay (2), (3), (4) vao (1) ta coAT = (50,5.103).(-0,75).(0,764) -29000AT = -29000V/A = -29V/mA.
Zi = Rb//hie1 = 103//309 = 236
[ ] =+=
+= 907,9//1085,322,5//10
hR//R
h//RZ 332fe
2b1C2ibEo
iL
VL
ZoZi
Rb12,1Kiihie1309 100ib R
C1400 Rb210K
hie2 522
(1+hfe)RL101K
ib1 ib2iC1
(1+hfe)RE101K
RE1K
Zo
hib2ie2
Rc1//Rb2hfe2
-
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3)3)3)3) Dang bai hon hp E.CDang bai hon hp E.CDang bai hon hp E.CDang bai hon hp E.C C.C:C.C:C.C:C.C: Bai 6-4
Tm R ei
02
i
01
iV
iV
= (1)
i
1b
1b
2b
2b
01
i
01
ii
.ii
.iV
iV
= (2)
i
1b
1b
2b2E2fe
i
02
ii
.ii
.R)h1(i
V+= (3)
suy ra2E2fe2b
01 R).h1(i
V+= (4)
2b
2C
2C
3b
3b
01
2b
01
ii
.ii
.iV
iV
= (5)
T (5) suy ra
2E2fe2fe3E3fe3ie2C
2C3E3fe
2b
01 R)h1(hR)h1(hRR
R.R)h1(
iV
+=++++
+=
5050100.505010R10
10.5050 333
=+++
(6)
105 = 7050 + RR = 100K - 7,05K 93K
Tmi
01
iV
T (2) ta co:
505050501010.9310
10.10.5050
R)h1(hRRh.R.R)h1(
ii
ii
iV
iV
333
23
3E3fe3ie2C
2fe2C3E3fe
2b
2C
2C
3b
3b
01
2b
01
+++
=
++++
+==
(7)
( )
( )63,7
655010.5
50501000500100.500
h.
R)h1(hR//R
R//R
i
i.
i
i
i
i
4
1fe
2E2fe2ie2b1C
2b1C
1b
1C
1C
2b
1b
2b
==++
=
+++
==
(8)
( )
( )5,0
101010
hR//RR//R
ii
33
3
1ie1bi
1bi
i
1b =+
=+
= (9)
Thay (7), (8), (9) vao (2) ta c:
iC2
hie11K
Ri
100Kii
hie2 1K
100ib1
RC1
1K
Rb2
1Khfe2RE2
5050
ib2ib1 iC1
Rb1
1K
Vo2100ib2
hie3 1K
RC2
1Khfe3RE3
5050
ib3
-
8/2/2019 Bai Tap Mach Dien Tu
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Khoa ien ien t Ky thuat mach ien T I
- 64 - Mot so bai tap mau - 64 -
( ) ( )mAV27,195,0.63,7.5050
iV
i
01 =
Zi = Ri//Rb1//hie1 500
[ ] ++=
++= 5,471093010//50
hR
hRh//RZ
3fe
2C
3fe3ib3Eo
II.II.II.II. Transistor mac vi sai va DarlingtnTransistor mac vi sai va DarlingtnTransistor mac vi sai va DarlingtnTransistor mac vi sai va Darlingtn1) Bai 61) Bai 61) Bai 61) Bai 6----23: E.C23: E.C23: E.C23: E.C E.C.E.C.E.C.E.C.
aaaa---- Che o DCChe o DCChe o DCChe o DC
V25,29.10.310
10V.RR
RVV 33
3
CC2111
112BB1BB =
+=
+==
mA55,142,7107,025,2
hR
R2
7,0VII 3
fe
bE
1BB2EQ1EQ =
+
=
+
==
IE = 2IE1 = 3,1mAVCEQ1 = VCC 2RE.ICQ1 = 9 2.500.1,55.10-3 = 7,45VVCEQ2 = VCC 2RE.ICQ1 RC2.ICQ2
= 9 103.1,55.10-3 2,5.103.1,55.10-3 = 3,575V
RE3
50
Zo
hib3 10R/hfe3 930
Rc2
hfe3 =10
+VCC =9V
ii
R213K
iL
ib3
R11K
RE1500
T1 T2
RC22,5K
R223K
R121K
T3
T4
RC4=RL60
RE460
hfe=100
iC4hie1
hfe2ib2100ib2
RL(1+hfe)RE1
(1+hfe)ib4ii
hie3
RC22,5K
Rb2750
h2feRE4
ib1 ib2 hfe3hie4
Rb1750
hie2
-
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Khoa ien ien t Ky thuat mach ien T I
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V725,34,1875,39VVI.RVV 4BE3BE2CQ2CCCR 4E ===
mA6260725,3
RV
II4E
R4EQ4CQ
4E ===
mA62,010
10.62h
IIII 2
3
fe
4CQ4BQ3EQ3CQ ===
VCEQ4 = VCC ICQ4(RC4 + RE4) = 9 62.10-3.120 = 9 7,44 = 1,56VVCEQ3 = VCEQ4 VBE4 = 1,56 0,7 = 0,86V
===
225810.55,1
10.25.100.4,1I
10.25.h4,1h 33
1EQ
3
1fe1ie
===
564510.62,0
10.25.100.4,1
I10.25
.h4,1h 33
1EQ
3
3fe3ie
===
45,5610.6210.25.100.4,1
I10.25.h4,1h 3
3
1EQ
3
4fe4ie
bbbb---- Che o ACChe o ACChe o ACChe o AC
( ) 50500101.500h1RRfe1E
'
E==+=
( )[ ]( ) [ ] =+=+++= K76,617101.606045,56h1Rh1hR 3fe4E4fe4ie' 4E
i
1b
1b
2c
2c
3b
3b
L
i
Li i
i.
ii
.ii
.ii
ii
A == (1)
( )( ) 10201101.101h1h1ii
4fe3fe3b
L ==++= (2)
( )
3
33
3
'4E4iefe3ie2C
2C
2c
3b
10,4605,631 5,2
10.76,6175700564510.5,210.5,2
Rhh1hRR
ii
=
+++
=
++++
=
(3)
( ) 1001.hii
ii
ii
2fe1b
2b
2b
2c
1b
2c === (4)
(V RE rat ln nen coi ib2 ib1)3
''E1ieb
b
i
1b 10.4,12828322258750
750RhR
Rii =
++=
++= (5)
iC2hie1 2258
100ib2
RC60
RE50,5K
Zo
(1+hfe)2ib310201ib3
ii
hie35645
RC22,5K
Rb750
RE4618K
ib1 ib2hie4(1+hfe)5700
Rb1750
hie2 2258 ib3 iL
Zo
-
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Khoa ien ien t Ky thuat mach ien T I
- 66 - Mot so bai tap mau - 66 -
vi [ ] =+= 28323008//5050Rh//RR b2ie'E
''E
Thay (2), (3), (4), (5) vao (1) ta co:Ai = 10201.(-4.10-3).(-100).128,4.10-3 = 522,3 (lan)
Zi = Rb1//(hie1 +RE) 750//(2258 + 2832) = 654Zo = Zo = Zo//RC = RC = 60
2) Bai 62) Bai 62) Bai 62) Bai 6----24:24:24:24: E.C C.C
aaaa---- Che o DCChe o DCChe o DCChe o DCAp dung nh luat K.II Vkn = 0 cho vong 2 ta co:
VBE3 + IEQ3RE3 VEE = 0 (1)
mA3,210
7,03R
VVI 3
3E
3BEEE3EQ =
=
=
mA15,12
III 3EQ2EQ1EQ ===
VCE1 = VCE2 = VCC RC1ICQ1 VE1 (2)Mat khac ap dung nh luat K.II Vkn = 0 cho vong 1 ta co:
-VBB1 + RbIBQ1 + VBE1 + VE1 = 0 (3) VE = VBB1 RbIBQ1 VBE1 = 1 104.1,15.10-5 0,7 =0,185V
Thay vao (2) ta c:VCE1 = VCE2 = 6 103.1,15. 10-3 0,185 = 4,665V 4,67V
Ta co VE1 = VCE3 + RE3.IEQ3 - VEE (4)Suy ra VCE3 = VEE + VE1 RE3IEQ3
= 3 + 0,185 103
.2,3.10-3
= 0,885VVRE6 = VCC RC2ICQ2 VBE4 - VBE5 - VBE6= 6 103.1,15.10-3 2,1 = 2,75V
mA2751075,2
RV
I6E
6RE6EQ ===
VCE6 = VCC VRE6 = 6 2,75 = 3,25VVCE5 = VCE6 VBE6 = 3,25 0,7 = 2,55VVCE4 = VCE5 VBE5 = 2,55 0,7 = 1,85V
VL
ZoZi
ii
-3V
Rb110K
RC11K
Rb210K
RE1K
VBB2
1V
T4T5
RC21K
VBB11V
T3 RE6
10
T6
-
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===
304310.15,110.25.100.4,1
10.15,110.25.h4,1h 3
3
3
3
1fe1ie
mA75,2hI
I6fe
6EQ5EQ == ; A10.75,2h
II 5
5fe
5EQ4EQ
==
===
72,1210.27510.25
.100.4,1I10.25
.h4,1h 33
6EQ
3
6fe6ie hie5 = 1272; hie4 = 127.200
bbbb---- Che o ACChe o ACChe o ACChe o AC
( ) += 766E3
fe6E'
6E 1010.Rh1RR
i
2b
2b
4b
4b
L
i
LT i
i.
ii
.iV
iV
A == (1)
= 7' 6E4b
L 10RiV
(2)
4
4
2
733
23
'6E4ie2C
2fe2C
2b
2C
2C
4b
2b
4b
10.3,96106,3811
10
1010.6,38110
10.10
Rh3Rh.R
ii
.ii
ii
=++
=++
=
++==
(3)
485,01010.086,610
10
Rh2RR
ii
ii
434
4
2b1ie1b
1b
i
1b
i
2b
=+
=
++==
(4)
Thay (2), (3), (4) vao (1) ta coAT = 107.(-96,3).10-4.(-0,485) = 46728V/A = 46,7V/mA
Zi = Rb1//[2hie1 + Rb2] 6,15K
( ) =
+= 37,0382,0//10
hh3
hR
//RZ 3fe
4ie3fe
2C6Eo
Chng VII:Chng VII:Chng VII:Chng VII: MACH KHUECH AI HOI TIEP.MACH KHUECH AI HOI TIEP.MACH KHUECH AI HOI TIEP.MACH KHUECH AI HOI TIEP.
iC2hie1 3043
VLhfe2ib2100ib2
ZoZi
ii
hie4127,2K
RC21K
Rb210K
RE6
ib1
Rb110K
hie2 3043 ib4
ib2
hie5.hfe4127,2K
hie5.hfe4.hfe5127,2K
Zo
RE63hie4h3fe
=0,381Rc2
h3fe 10
-
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I.I.I.I. Hoi tiep ap, sai lech dong.Hoi tiep ap, sai lech dong.Hoi tiep ap, sai lech dong.Hoi tiep ap, sai lech dong.1)1)1)1) Bai 7Bai 7Bai 7Bai 7----4444
=
=
40hK1h
GTfe
ie ;
==
=
?ii
A?T
KLi
Li
ay la dang hoi tiep ap, sai lech dong.
a- Tnh o li dong T: cho ii = 0'
1
1b
1b
2b
2b
1
i'
1
1
Vi
.ii
.iV
0iVV
T ==
= (1)
( ) 32fe22E2b
1 10.41h1RiV
=+= (2)
( ) ( )
941,08580
10.4110.411010.240.10.2
h1Rh1RhRh.R
ii.
ii
ii
3333
3
2fe22E2fe21E2ie1C
1fe1C
1b
1C
1C
2b
1b
2b
==+++
=
+++++==
(3)
634
1ief1ief
'1
'1
'1
1b 10.911010
1hR
1hR
V.
V1
Vi =
+=
+=
+= (4)
Thay (2), (3), (4) vao (1) ta co:
V1
iiRE221K
Rf10K
RC12K
RC22K
iL
RL100
+VCC
RE211K
T1VL
iC2
hie1 hfe1ib140ib1
RE21(hfe2+1)41K
ZoZi
iiRC12K
RC22K
ib1 ic1
40ib2Rf V1
hie2 1Kib2
V1+
-
RE22(hfe2+1)41K RL
100
-
8/2/2019 Bai Tap Mach Dien Tu
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Khoa ien ien t Ky thuat mach ien T I
- 69 - Mot so bai tap mau - 69 -
T = 41.103.(-0,941).91.10-6 = -3,51
b- Tnhi
L
ii
iA = cho V1 = 0
i
1b
1b
2b
2b
L
i
Li
i
i.
i
i.
i
i
i
iA == (1)
6,3940.1010.2
10.2h.
RRR
ii
.ii
ii
3
3
2feL2C
2C
2b
2C
2C
L
2b
L =+
=+
== (2)
941,0ii
1b
2b = (nh (3) phan tren) (3)
234
4
1ief
f
i
1b 10.911010
10hR
Rii =
+=
+= (4)
Thay (2), (3), (4) vao (1) ta co:Ai = (-39,6).(-0,941).91.10-2 = 33,9 34
c- Tnh Aif, Zif, Zof.54,7
51,3134
T1A
A iif =+
=
=
Zi = Rf//hie1 = 104//103 910
=+
=
= 20251,31
910T1
ZZ iif
Zo = RC2 = 2K
=+
=
= 44351,31
10.2T1
ZZ
3o
of
2)2)2)2) Bai 7Bai 7Bai 7Bai 7----11111111
=
=
C10h100h
GT ibfe
;
=
==
?TZ;Z
?ii
A
KL oii
Li
hie = hib.hfe = 10.100 = 1K
ii
Rf= Rb =10KRC 2K
C iL
RL
+VCC
RE100
C
-
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Khoa ien ien t Ky thuat mach ien T I
- 70 - Mot so bai tap mau - 70 -
a- o li vong T: cho ii = 0'L
b
b
L
i'L
L
Vi
.iV
0iVV
T ==
= (1)
433
233
feLC
CL
b
C
C
L
b
L 10.5101010.10.10h.
RRR
.Rii
.iV
iV
=+
=+
== (2)
634
iefief
'L
'L
'L
b 10.911010
1hR
1hR
V.
V1
Vi =
+=
+=
+= (3)
Thay (2), (3) vao (1) ta co:T =(-5.104).91.10-6 = -4,55
b- Tnh Ai, Zi, Zo.i
b
b
L'Li
Li i
i.
ii
0Vii
A ==
= (1)
50100.1010
10h.
RRR
ii
.ii
ii
3
3
feLC
C
b
C
C
L
b
L =+
=+
== (2)
234
4
ief
f
i
b 10.91
1010
10
hR
R
i
i =+
=
+
= (3)
Thay (2), (3) vao (1) ta co:Ai = (-50).91.10-2 = -45,45Zi = Rf//hie = 104//103 = 910Zo = RC = 103 = 1K
c- Tnh Aif, Zif, Zof.2,8
55,4150
T1A
A iif =+
=
=
=
+
=
= 164
55,41
910
T1
ZZ iif
=+
=
= 18055,41
10T1
ZZ
4o
of
II.II.II.II. Hoi tiep ap, sai leHoi tiep ap, sai leHoi tiep ap, sai leHoi tiep ap, sai lech ap:ch ap:ch ap:ch ap:
hie1iL
VLhfeib100ib
Zo
iiRC1K
ib ic
Rf Rf10KVL
+
-
RL
-
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Khoa ien ien t Ky thuat mach ien T I
- 71 - Mot so bai tap mau - 71 -
1)1)1)1) Bai 7Bai 7Bai 7Bai 7----10101010
=
=
C
50h50h
GT ibfe
;
=
==
?T
Z;Z
?ii
A
KL oii
Li
a- Tnh o li vong T (cho ii = 0)'L
1b
1b
2b
2b
L
i'L
L
Vi
.ii
.iV
0iVV
T ==
= (1)
3334
34
2fe2Cf
2Cf
2b
2C
2C
L
2b
L
10.5,8350.10.67,150.10.210
10.2.10
h.RR
R.Rii
.iV
iV
==+
=
+==
(2)
( )
46,1210.5,210.83,0 50.10.83,0
hhR//R
R//Rii
.ii
ii
33
3
1fe2ie2b1C
2b1C
1b
1C
1C
2b
1b
2b
=+
=
+==
(3)
63533
3
1fef1E1fe1ie1b
1fe1E'L
1b
10.2,110.210.5
1.10.510.5,2890
10.5
'Rh.R1.
R.hhRh.R
Vi
=+++
=
+++
=
(4)
Thay (2), (3), (4) vao (1) ta co:
VL
ii
Rf=10K
C
+VCC
R218K
CR111K
RC11K
RE1100
R2210K
R1210K
RC22K
RE21K
iC1
hfe1ib150ib1
hie1 2,5K
hie22,5K
RC22K
Zo
iiRC11K
Rb25K
Rf10K
ib2ib1 iC2
hfe2ib250ib2
Rf.hfe15.105
RE1hfe15K
VL+
-
Rb1890
-
8/2/2019 Bai Tap Mach Dien Tu
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Khoa ien ien t Ky thuat mach ien T I
- 72 - Mot so bai tap mau - 72 -
T = (-83,5.103).(-12,46).(1,2.10-6)= -1,25
b- Tnh AT, Zi, Zo.i
1b
1b
2b
2b
L
i
LT i
i.
ii
.iV
iV
A == (1)
2b
L
ii va
1b
2b
ii tnh nh tren theo cong thc (2), (3)
( )107,0
10.95,410.5,2890890
hRRhRR
ii
331fef1E1ie1b
1b
i
1b =++
=+++
= (4)
Thay (2), (3), (4) vao (1) ta co:Ai = (-83,5.103).(-12,46).(0,107) = 111.103V/A = 111V/mAZi = Rb1//[hie1 + (RE11//Rf)(1 + hfe)] = 890 //[2500 + 4950] = 795Zo = Rf= 10K
c- Tnh AVf, Zof, Zif.mAV49
AV10.49
25,1110.111
T1AA 33TTf ==
+=
=
( ) =+== 1788)25,11(795T1ZZ iif
=+
=
= 444425,11
10T1
ZZ
4o
of
2)2)2)2) Bai 7Bai 7Bai 7Bai 7----12121212
=
=
C50h20h
GT ibfe
;
=
=
?TZ;Z
?AKL oi
V
VL
Vi+
-
ri 1K
Rf=1K
C
+VCC
R2110K
CR111K
RC1500
RE1282
R2210K
R121K
RC2500
RE2282
C
RE1122
C
RE2122
hfe1ib1
20ib1
Rf(hfe1+1)21K hie2 1050
Vi+
-Rb1910 R
Cb2910
ib1 iC2iC1
VL+
-
ri 1K
RC1500
RC2 500ib2ii hie1 1050
20ib2
-
8/2/2019 Bai Tap Mach Dien Tu
73/