Trang 1
PHAN I
LY THUYET
Trang 2
Chng 1: MACH LOC TCH CC
1-1 Ham truyen co ap ng phang toi a:
Con goi la ham Butterworth. Khi bac cua bo loc tang len, tan so cat khong thay oi, nhng o doc cua bo loc tang dan en ly tng. Khi thiet ke cac bo loc bac cao: 3, 4, 5 ta da vao bang cac ham Butterworth a chuan hoa.
1-2 Mach loc tch cc bac nhat
a- Mach loc thong thap bac nhat: LTT1
Ham truyen: SCR1
ASVSVSH
11
0V
1
2
(1)
3
20V R
R1A (2)
12C CR
1 (3)
C1 V2(S)
R1 +-
R2 R3
Bo khuech ai ao
C 0
AV0
H(S)
C1
R1
R2
+ -
Bo khuech ai khong ao
Trang 3
Ham truyen: SCR1
ASVSVSH
12
0V
1
2
(1)
1
20V R
RA (2)
12C CR
1 (3)
b- Mach loc thong cao bac nhat
Ham truyen:
SRC11
ASVSV
SH
11
0V
1
2
(1)
1
20V R
RA (2)
C1
V2(S)
R1
R2
+ -
111 RC
1
0
AV0
H(S)
Trang 4
11t CR
1 (3)
1-3 Mach loc tch cc bac hai
a- Mach LTT2
AV0 = 1 (1)
212
20 CCR
1 (2)
10C22R
(3)
122
02 CR
1C
(4)
1
20V R
RA (1)
3221
20 RRCC
1 (2)
C1
V2(S)
R + - V1(S)
R
R R
C2
C2
Mach hoi tiep am mot vong
C1
V2(S)
R1 V1(S) R3
R2
C2 + -
Mach hoi tiep am 2 vong
Trang 5
Neu chon: 21
0V2
1
2
bA1b4
CC
(3)
10
12 Cf4
bR
(4)
0V
21 A
RR (5)
22120
22
3 RCCf4bR
(6)
Trng hp 1: AV0 = 1 (R3 = 0).
Neu chon 21
2
1
2
bb4
CC
(1)
Th 10
121 Cf4
bRR
(2)
2121
20 CCRR
1 (3)
Trng hp 2: R1 = R2 = R; C1 = C2 = C; AV0 1.
RC1
0 (1)
4
30V R
R123A (2)
C1 V2(S) R1
R4
V1(S) R3
R2
C2 + -
Mach LTT2 dung hoi tiep dng
Trang 6
59,022RR
4
3 (3)
b- Mach LTC2
Trng hp 1: AV0 = 1 va C1 = C2 = C.
212
20 RRC
1 (1)
C2R
01 (2)
2R
R 12 (3)
Trng hp 2: C1 = C2 = C; R1 = R2 = R;
RC1
0 (1)
23RR
1A4
30V (2)
59,022RR
4
3 (3)
c- Mach LTD2:
C1 V2(S) R1
R4
V1(S) R3
R2
C2 + -
Bo LTC dung hoi tiep dng
C1
V2(S) R1 V1(S)
R3
R2
C2
+ -
Bo LTD hoi tiep am 2 vong
Trang 7
Xet trng hp C1 = C2 = C ta co:
3321
110 R'RC2
1RRRRR
C21f
(1)
1
3
010V R2
RCRQA
(2)
21
21330 RR
RRR21CR
21Q
(3)
CR1
Qf
D3
0
(4)
21
21
RRRR'R
(5)
ieu kien: A0L > 2Q2.
min max
D
Q
H
C1 R1
R2
C2
+ -
LTD bac 2
Trang 8
Ham truyen: 2211
21
1
2
RSC1RSC1RSC
ZZSH
(1)
111 CR2
1f
(2)
222 CR2
1f
(3)
213 RC2
1f
(4)
3
10V f
flg10dBA
f1 f 0
A
H(dB)
f2 f3
-40dB
Trang 9
Chng 2: KHUECH AI CONG SUAT CAO TAN
(KCSCT)
2-1 Goc cat cua bo KCSCT:
Goc cat tnh theo o: T
tT180 000
(1)
Cac thanh phan dong ien ra c tnh da theo he so phan giai xung dong ien ra cua Transistor:
- Thanh phan trung bnh mot chieu:
0m0max0 I.II
- Thanh phan hai bac nhat:
1m1max1 I.II
- Thanh phan hai bac n:
nmnmaxn I.II
2-2 Cac mode hoat ong cua bo KCSCT lp C dung Transistor
Vm
Vm
Vmin C A 0
ib
V B
D
Vmax
Vb
iC
Imax
T
t0 t Imax
Hnh 2-1 Dang ac tuyen ong va gian o thi gian cua dong ien che o C
hfe
fcao
0 = hFE 0,7070
f
1
0 0,3f 3f f fT ftrungbnh fthap
Trang 10
Dai tan so lam viec cua Transistor c chia lam 3 oan:
- f0 f: tan so thap, cac tham so c coi la khong thay oi; hfe = 0;
- 0,3f f0 3 f: tan so trung bnh, cac tham so cua Transistor thay oi va xuat hien ien tr ky sinh (rbb), ien dung ky sinh (Cbe, Cbc)
2
0
fe
2
0
0*
ff
1
h
1
(3)
- f0 3 f: tan so cao, cac tham so cua Transistor thay oi, xuat hien rbb, Cbe, Cbc va cac iem cam ky sinh Lks.
00
00 f
fjj
(4)
Trong giao trnh ien t thong tin chu yeu chung ta se nghien cu bo KCSCT tan so thap va tan so trung bnh va ch xet che o kem ap. (Transistor nh mo nguon dong)
2-3 Bo KCSCT dung Transistor
1. Bo KCSCT dung Transistor che o kem ap mac Emitter chung.
CC LC +
VBB -
Lch
Cng
Cng
Lch
Cng
RE en
Rb
Rn
Cng
Trang 11
Cac bc thiet ke bo KCSCT khi cha ke en anh hng cua mach ghep au vao va au ra (Chu y: cac bc thiet ke khong nhat thiet theo trnh t a ra)
0- Xac nh pham vi lam viec cua Transistor theo (2-2) e ve s o tng ng tn hieu nho cho ung.
1- VCC = (0,5 0,8)VCEmax cho phep
2- Chon goc cat: = 600 900
3- Chon he so li dung ien ap: 1 = 0,85 0,95 = VCm1/VCC.
4- Xac nh bien o hai bac nhat tren Collector: VCm1 = 1VCC.
5- Xac nh cac dong ien:
*
1
1Cmn
I'I
; 1t1c'bTnn RC1'II
BOnB Itcos'Ii ; bmn I'I ; *COBOI
I
; 1Cm1
0CO II
6- *M* e'b*b CC'C ;
1
e'b*e'b
CC ;
1
11tc'be'bT*M
RCCC
1
1tc'bTe'b*b
RC1C'C
*'b
iEC Cj1Z
;
1
*
c'b*
e'b 1CC
Neu ke ca rbe ta co: ZiEC = rbe//ZiEC
Neu rbe >> ZiEC ta co ZiEC ZiEC
LC
|hfe|ib CC
Cbe rbe C*M
Rt1 In
R1 C*be
Trang 12
Neu rbe so sanh c vi ZiEC ta co:
20
e'b
20e'be'b
e'biEC
1
r
Cr1
rZ
7- Bien o ien ap kch thch vao: Vbm1 = In|ZiEC|
8- Cong suat vao cua nguon kch thch:
iEC2ni Z.'I2
1P
9- Xac nh tr khang nguon tng ng
e'bnn CR
T
fee'be'b
h1Cr
e dong ien au vao khong b meo th: n
e'be'bT
fen C
1C
hR
2
0
n*
n
1
RZ
10- Thien ap Base
)(Z'I7,0
)(Z'IVV
0
*
nn
0
*
nnBEB
11- ien tr tai tng ng: 1Cm
1Cmt
*
L IV
RZ
12- Cong suat nguon cung cap: PCC = ICOVCC.
13- Cong suat hu ch tren tai
t
21Cm
t2
1Cm1Cm1CmL RV
21RI
21IV
21P
Trang 13
14- Cong suat tieu tan tren Collector: PC = PCC PL.
15- Hieu suat cua mach: )()(
21
PP
0
1
CC
L
Trong thc te thng cong suat ra tren tai c biet trc nen ta co the tnh cac bc 0 4, 13, 11, 5, . . .
2. Bo KCSCT dung Transistor che o kem ap mac Base chung.
2
0
0*
1
; e'b* e'b
CC ; *
e'biEC Cj
1Z
Cac bc thiet ke tng t nh tren.
16. CC
0 CL1f vi C* c'bC 'CCC
Rt1 = 0Q0LC 00
1tC Q
RL
vi Q0 = 50 100
C20
2C Lf41C
c'bCC CC'C
Neu au vao bo KCSCT co mach cong hng Lb, Cb th ta cung xac nh tng t nh tren vi:
bb0 CL
1f ; Rt1 = 0Q0LC; b
20
2b Lf41C
b*
'bb 'CCC vi C*b tnh theo bc 6 tren.
2-4 Bo nhan tan dung Transistor
ie LC |h*fb|ie
CC Cbc
fe
e'b
hr
Rt1 In
Rn C*
be
CC LC + VBB -
Lch
Cng
Cng
Lch
Cng
RE Lb Rb Cb
Cng
Trang 14
Muc ch cua bo nhan tan:
- Nang cao tan so song mang
- M rong thang tan so lam viec
- Nang cao ch so ieu che trong may phat FM
- Nang cao o on nh tan so v khong co hien tng hoi tiep ky sinh qua Cbc do tan so hoat ong au vao va au ra khac nhau.
Tan so cong hng au vao:
bb0V CL
1 vi * e'bbb C'CC
vi e'b* e'b
CC ; b0101t LQR
Tan so cong hng au ra:
CC0ra CL
1k ; C020tn LQkR
Goc cat toi u cua bo nhan tan dung Transistor
k180
T ; k: he so nhan tan cua bo nhan
Cac bc thiet ke cua bo nhan tan:
0- Xac nh pham vi lam viec cua Transistor theo (2-2)
1- VCC = (0,5 0,8)VCEmax cho phep
ib LC |h*fe|ib
CC Cb Lb Rt2 Rt1 rbe C*be
Trang 15
2- Chon goc cat toi u: k
180T
3- Chon he so li dung ien ap:
k = 1 =0,85 0,95 = VCm1/VCC = VCmk/VCC
VCmk = kVCC
4- Xac nh xung dong hai bac k
1bm*
km
*
kCmk I.'I.I
5- Xac nh cong suat hu ch tren tai ng vi hai bac k
6- ien tr cong hng tng ng cua mach ra ng vi hai bac
k: 1tk
1
1Cm1
k
1Cm
Cmk
Cmktk R
I
VIV
R
7- Hieu suat cua bo nhan tan:
CC
Lkk P
P ; vi PCC = ICO.VCC
8- Do khong co hien tng hoi tiep qua Cbc nen
*
k
Cmk*
1
1Cmbnnn
III'II
iB = Incost IBO vi Cmkk
0**
COBO I.
1II
9- Tr khang vao cua tang
*'b
iEC Cj1Z
; vi
1
*'b
e'CbC
Neu ke ca rbe ta co ZiEC = rbe//ZiEC (tnh nh tren)
1 1 1 11 1 1
1 1
2 2k k k
Lk Cmk Cmk Cm Cm L LP I V I V P P
Trang 16
10- Bien o ien ap kch thch vao: Vbm1 = In|ZiEC|
11- Cong suat cua nguon kch thch: iEC2ni ZI21P
12- CBn P,V,Z tnh nh bo KCSCT
13- Tnh mach cong hng vao:
bb0 CL
1 ; vi
010
1tbb
*e'bb Q
RL;'CCC
14- Tnh mach cong hng ra
CC0 CL
1k ; vi 020
tnC Qk
RL
Trang 17
Chng 3: CAC MACH TAO DAO ONG
3-1 Cac van e chung ve mach tao dao ong
- Bo tao dao ong tan so thap, trung bnh: dung bo khuech ai thuat toan + RC hoac dung Transistor + RC.
- Bo tao dao ong tan so cao: 0,3f f0 3f dung Transistor + LC hoac dung Transistor + thach anh
- Bo tao dao ong tan so sieu cao: dung Diode Tunel, Diode Gunn.
- Cac tham so c ban cua mach dao ong: tan so dao ong, bien o ien ap ra, o on nh tan so, cong suat ra, hieu suat.
- Trong chng 3 ta ch xet mach dao ong LC, dao ong thach anh va ch xet ieu kien dao ong cua mach
- He so khuech ai cua bo khuech ai
*1
*
2A
**
V
VjexpAA
+ Modul he so khuech ai: *1
*
2*
V
VA
+ A goc di pha cua bo khuech ai.
- He so truyen at cua bo hoi tiep
ht**
jexp
Bo khuech ai A
Bo hoi tiep
V1 V2
Trang 18
+ Modul he so hoi tiep: *2
*
1*
V
V
+ B goc di pha cua bo hoi tiep
- ieu kien pha e mach dao ong: = B + B = 0,2
- ieu kien bien bo e mach dao ong: 1.A**
3-2 Bo dao ong LC dung Transistor
a- Mach tao dao ong 3 iem C kieu Colpits mac EC
Rb = R1//R2; Rb = Rb//rbe rbe (neu Rb >> rbe)
Cac bc thiet bo tao dao ong 3 iem C:
1- Xac nh pham vi tan so lam viec cua mach
2- Xac nh ieu kien pha
0C1XX
2BE1
; 0C1XX
1CE2
;
X3 = XCB = L > 0
Lch
R1
+VCC
Cng
L
R2
RE Cbe
Cng C1
C2 Vk
C E
B
L C1
C2 Vk
C
E
B
Cbe
Trang 19
3- Xac nh he so hoi tiep
nCC
VV
2
1
CE
BE (1)
- Ta thng biet f0, L t o suy ra:
21
2120
2t CCCC
Lf41C
(2)
- n co the tnh theo cong thc (3-45) nhng nhieu khi khong u d lieu e tnh
- Neu mach lam viec tan so thap ta co the chon
n = 0,01 0,05, t o tnh C1, C2.
- Neu mach lam viec tan so trung bnh, e mach hoat ong o nh ta chon: C2 = 10Cbe C2 = 11Cbe roi t (2) tnh C1, thay vao (1) tnh n
4- He so khuech ai cua s o mac EC
211
K2
e11
e21C n
h//RphhSZA (3)
- tan so thap: h21e = hfe, h11e = hie = rbe
- tan so trung bnh: 2
0
fe*
e21
1
h|h|
; e'b
e'b C.f21r
- Rk = 0LQ0 thng biet trc 0, L, Q0 (4)
- p: he so ghep au ra cua Transistor vi khung cong hng
n11
CCC
CCC
CC
VV
p21
2
1
21
21
K
CE
(5)
5- ieu kien bien o e mach dao ong: 1.A**
(6)
b- Mach tao dao ong 3 iem C kieu Colpits mac BC
Trang 20
Gia thiet RE >> hib
- Bc 1 va 2 lam nh tren, thng mach mac BC lam viec tan so thap.
- Bc 3: He so hoi tiep: 21
1
2
21
21
K
CE
CCC
CCC
CC
VV
p
Neu mach lam viec tan so thap ta co the chon n = 0,1 0,5; t o tnh C1, C2 v Ct thng tnh c.
Neu mach lam viec tan so trung bnh, tnh nh tren
- Bc 4:
2
11K
2
b11
b21C n
h//Rp
hh
SZA
tan so thap: h21b 1, h11b = hie/hfe
tan so trung bnh:
L C1
C2 Vk
Cbe
+VCC R1 R2
RE Cng
L C1
C2 Vk
Cbe
Trang 21
2
T
0
*
b21
1
1|h|
; e'bT
*
b11 Cf21h
Rk tnh nh tren.
He so ghep au ra cua Transistor vi khung cong hng:
1VV
VV
pBC
BC
K
BC
- Bc 5: ieu kien bien o e mach dao ong: 1.A**
c- Mach tao dao ong 3 iem C kieu Clapp mac EC
- Cac bc thiet ke tng t nh mach dao ong 3 iem C kieu Colpits mac EC, ch khac ve Ct va he so ghep p cua Transistor vi khung cong hng.
t0 LC2
1f
vi 021t C
1C1
C1
C1
Lch
R1
+VCC
Cng
L
R2
RE Cbe
Cng C1
C2
Vk
C
E
B
C0
L C1
C2
Vk
C
E
B
C0
Trang 22
Neu ta chon C1, C2 >> C0 th Ct Co khi o nhanh cong hng noi tiep L, C0 se quyet nh tan so cong hng cua mach va mach se on nh tan so hn
- He so ghep p: 1
0t
K
CE
CC
1CC
VV
p
d- Mach tao dao ong 3 iem C kieu Clapp mac BC
Cac bc thiet ke tng t nh mach dao ong 3 iem C kieu Colpits mac BC, ch khac ve Ct va he so ghep p.
- Khi biet f0, L ta tnh c Ct, ta se chon C0 ln hn Ct mot chut v du: Ct = 25pF th ta chon C0 = 30pF.
- He so ghep p: 21
21t
21
21
t
K
BC
CCCC
C
CCCC
CVV
p
3-3 Cac mach dao ong dung thach anh
a- S o tng ng cua thach anh
- Lq, Cq, rq la L, C, r cua thach anh (rq = 0)
- Cp: ien dung gia (Cp = 10 100pF) (Cq = 0,01 0,1pF)
Lch
R1
+VCC
L R2
RE Cng
C1
C2
Vk
C
E
B
C0
L
RE
C1
C2
Vk
C
E
B
C0
Trang 23
- Tan so cong hng noi tiep: qq
q CL1
(1)
- Tan so cong hng song song:
p
p
pq
p
q
pq
pqq
p C2C
1CC
1
CCC
CCCC
L
1
- Tr khang tng cua thach anh:
Zq = Xq = j0Lt (3)
Vi
p
2
q
0qp
20
2
q
0
t
CCC
1
L (4)
e thay oi tan so cong hng rieng cua thach anh ta mac CS noi tiep vi thach anh:
pqq
20qp
Spqq20Spq
St CCLCC
CCCLCCCCj1Z
(5)
khi o tan so cong hng noi tiep cua mach se la
Sp
qqq CC
C1f'f
(6)
rq
Lq Cp
Cq
CS TA
Trang 24
sp
q
q CCC
21
ff
e giam anh hng cua Cp ngi ta mac tu C0 song song vi
Cq p0qq0
qqp CCneufCC
C1ff
b- Mach tao dao ong dung thach anh vi tan so cong hng song song
e mach dao ong theo kieu 3 iem C kieu Colpits, thach anh phai tng ng nh cuon cam, ngha la: q < 0 < p
Thc te 0 p nhng e tnh toan n gian do p q ta coi
2qp
0
(1)
p
qqp C2
C1 (2)
Lch
+VCC
Cng R2
C1
C2 LTA
Cng
Cng R1 RE
C1
C2 LtTA
jXq
0 q p
0
Lt
Ct Ct
Trang 25
Biet 0 , Cq, Cp ta tnh c q
- ien cam rieng cua thach anh: q
2q
q C.1L
(3)
- ien cam tng ng cua thach anh:
pqq20qp20
2
q
0
t CCLCC
1
L
(4)
t0tTA LjZ (5)
- 21
21
t20
t CCCC
L1C
(6)
Cac phan con lai tnh toan tng t nh mach dao ong 3 iem C kieu Colpits mac EC.
Ap dung cac cong thc (1) (6) tren va cac cong thc trong mach dao ong 3 iem C kieu Colpits mac BC.
Khi tu CS mac noi tiep vi thach anh no ong vai tro nh tu C0 trong mach dao ong 3 iem C kieu Clapp. Khi tnh Lt, Ct ta se chon CS ln hn Ct mot chut, roi tnh C1, C2 nh cac mach tren.
c- Mach tao dao ong dung thach anh vi tan so cong hng noi tiep
C1
C2 TA
RE Mach B.C
Trang 26
Trong loai mach nay thach anh ong vai tro mach hoi tiep. Ch ung tai tan so cong hng noi tiep cua thach anh th Zq 0 khi o B B va mach se hoat ong nh 3 iem C kieu Colpits hoac Clapp va cung co the mac EC hay BC
3-4 Mach tao dao ong RC
n gian va thong dung nhat la mach dao ong cau Wiew
- Tan so dao ong: RC1
- ieu kien dao ong ve bien o: 1.A.A***
Ma 31
nen 3A*
- Mat khac 212
1*
R2R3RR1A
B
R
B
Z3
Z1
Z2
D2
C
R2
Vin R
R1
+ -
R
D1
Vout
Trang 27
Chng 4: IEU CHE TNG T
4-1 ieu bien
a- Pho cua tn hieu ieu bien va quan he nang lng trong ieu bien.
V(t) = Vcost (1)
V0(t) = V0cos0t (2)
VAM(t) = V0(1+mcost)cos0t (3)
)1(VVm
0
(4)
V
0 t
Vo
0 t
VAM
0 t
VAM
0 t
V0
2mV0
0 - 0 + 0
Trang 28
tcos2
mVtcos
2mV
tcosVtV 00
00
00AM
- Cong suat tai tin:L
20
0 R2V
P (6)
- Cong suat hai bien tan: 2
mPP2
0bt (7)
- Cong suat ieu bien:
2
m1PPPP2
0bt0AM (8)
- He so li dung cong suat: AM
bt
PP
k (9)
- Cong suat ieu bien ln nhat: 20maxAM m1PP (10)
ay la ieu kien e chon Transistor sao cho
PAMmax < PCmax
b- ieu bien Collector
ien ap Collector bien oi theo ien ap am tan:
tcosVVV CC*
CC (11)
vi 1VV
CC
1Cm
e am bao Transistor khong b anh thung, phai thoa man ieu kien: CEOmaxCE0 BVVVV (12)
maxCECCCC000 VV2m1Vm1VmVV2 oi vi ieu bien th maxCECC V5,0V (13)
Neu au ra cua mach ieu bien la mach loc co hieu suat CH th ieu bien Collector co cong suat nh la:
phepchomaxC
CH
20
maxAM Pm1P
'P
(14)
ay la ieu kien e chon Transistor co PCmax cho phep
Trang 29
e thiet ke bo ieu bien Collector ta se tien hanh theo hai phan nh sau:
- Cho trc
2m1
PPPPP2
AM0
CH
AAM
mA khi a biet 0P ta
tien hanh cac bc thiet ke nh oi vi mach KCSCT (muc 2-3)
- Thiet ke phan ieu bien: 1Cm0 mVmVV
- Pho cua ieu bien tVAM (theo 3) va ve pho
- Tnh cong suat hai bien tan (theo 7).
- Tnh he so li dung cong suat k (theo 9).
- Kiem tra ieu kien ien ap (theo 12)
- Kiem tra ieu kien cong suat (theo 14).
4-2 ieu tan va ieu pha
a- Quan he gia ieu tan va ieu pha
Dao ong ieu hoa song mang:
tcosVtcosVtV 00000 (1)
Tn hieu ieu che am tan: tcosVtV (2)
Tn hieu ieu tan
FM:
000FM tsintcosVtV (3)
Trong o: tcost 0 (4)
Vi : lng di tan cc ai
Ch so ieu tan:
V
km f ; he so ty le (5)
Tn hieu ieu pha PM:
Trang 30
000PM tcostcosVtV (6)
Trong o: tcost 0 (7)
Vi : lng di pha cc ai
Ch so ieu pha: kVm p (8)
vi k: he so ty le
Quan he gia o di tan va o di pha:
tsin..dt
d
(9)
T 3, 6, 9 ta nhan thay ch can biet tn hieu ieu tan FM se tm c tn hieu ieu pha PM va ngc lai.
b- Pho cua tn hieu ieu tan va ieu pha
Khi ch tnh cac thanh phan Im(mf) 0,01I0(mf) th be rong dai tan cua tn hieu ieu tan chiem la:
maxffFM 1mm2D (1)
- Khi mf > 1 ta co bieu thc gan ung:
DFM 2mfmax 2 (2)
goi la ieu tan bang rong
- Khi mf < 1 DFM 2max (3)
0 + 2
In(m)
0 t 0 -
0 + 0
I0 I1
I1
I2
0 - 2
Trang 31
Goi la ieu tan bang hep
e mf const khi tan so thay oi pha phat phai co mach pre-emphasis va pha thu co mach de-emphasis.
c- ieu tan bang Varicap
n
inV eCC
(1)
Cin: ien dung ban au khi e = 0
: hieu ien the tiep xuc si 0,7V
n: he so phu thuoc loai varicap .2,1,21,
31n
eVe pc (2)
vi Vpc: ien ap phan cc ban au cho varicap
tcosVtcosVe 00 (3)
RD
CD
CC1
CC2 CC0
CV
0 Vpc
V
V
Trang 32
Trong thc te ta phai tm moi cach e giam anh hng cua ien ap cao tan tren varicap, khi o: tcosVe (4)
Goi ien ap AC tren varicap a chuan hoa: pcV
ex
(5)
n
pcin0V V
CC
(6)
0Vn
V Cx1C (7)
Tuy theo cach mac varicap vao khung cong hng ta co the tnh gan ung o di tan do varicap gay ra theo ien ap ieu che V.
pc0a V
Vnf5,0f (8)
30V
0V
pc0b CC
CV
Vnf5,0f (9)
40V
0V
pc0c CC
CV
Vnf5,0f (10)
Mac Varicap n:
a b c
L CV0 L L CV0 CV0 C3
C4
C
+VCC
R2 LK Ra
Cng CV0
R Lch
RE R1
Trang 33
Neu chon 2
VV CCCEQ th 2
VV CCRE tao phan cc ngc cho varicap.
ien tr R thng c chon vai tram k. Do dong tren R bang 0 nen VR = 0V. e thiet ke mach ieu tan varicap can tien hanh 2 phan
- Phan th nhat: thiet ke e mach thoa man ieu kien dao ong ve pha va bien o (giong phan 3-2-b).
- Phan th hai: thiet ke mach ieu tan varicap. Tuy theo cach mac varicap vao khung cong hng theo s o a, b, c ma chon cong thc (8) hoac (9) hoac (10) e tnh ffV .
Tnh f1 = f0 - f Lf41C 2
121td
CV1
Tnh f2 = f0 + f Lf41C 2
222td
CV2
Ve ac tuyen CV = f(V).
Mac varicap ay keo:
C
E LK
B
CV0
C
R
RE
C1
+VCC
V
VR L
E
Cng CV1
C2
R2
Lch
RE R1
Cng CV2 Lch
Cng
Trang 34
S o mac varicap ay keo triet tieu c hoan toan song cao tan tren varicap nen cac cong thc 8, 9, 10 tren c tnh
chnh xac hn. 2
C2
CCC
CCC 2V1V
2V1V
2V1V0V neu 2V1V CC . Ve ly thuyet
pF1001C 0V , tren thc te gia tr hay gap pF5010C 0V ; v du pF50CC 2V1V pF25C 0V .
Cac bc thiet ke c tien hanh nh 2 phan tren
d- On nh tan so trung tam cua tn hieu ieu tan
Cac bien phap on nh tan so trung tam f0 c xep t n gian en phc tap:
- ieu tan trc tiep bang thach anh: o di tan hep, ch dung trong phat thoai quoc te.
- S dung thach anh lam bo dao ong: o di tan hep.
- On nh nguon cung cap, s dung cac ien tr bu nhiet.
- Ha thap tan so trung gian cua bo ieu tan e nang cao o on nh tan so.
- S dung he thong t ong ieu chnh tan so AFC-F: ch ieu chnh tho.
- S dung he thong t ong ieu chnh tan so hon hp AFC-F va AFC-P: AFC-F ieu chnh tho, con AFC-P ieu chnh tinh a 0f .
C1 V
E
B
CV1
C2 Lch
RE
C
CV2
Trang 35
Chng 5: VONG GI PHA PLL
5-1 Nhng u, khuyet iem cua vong gi pha PLL
u iem:
- Kha nang lam viec tan so cao.
- S oc lap ve kha nang chon loc va ieu hng tan so trung tam.
- Nhng linh kien ben ngoai t.
- De dang trong viec ieu hng
Khuyet iem:
- S thieu thon thong tin ve bien o tn hieu.
- T ong ieu chnh he so khuech ai kho.
5-2 S o khoi va nguyen ly hoat ong cua PLL
Vong ieu khien pha co nhiem vu phat hien va ieu chnh nhng sai sot ve tan so gia tn hieu vao va tn hieu ra, ngha la PLL lam cho tan so ra 0 cua tn hieu song song bam theo tan so vao i cua tn hieu vao.
Bo so pha LTT
VCO
K
S o khoi cua PLL
Vp(t) Vd(t)
Vi(t) = Visinit
Vo(t) = Vocos(ot + )
Trang 36
Khi tn hieu vao a lot vao dai bat cua PLL, th tan so f0 cua VCO se bam theo tan so vao i .
5-3 Mot so ng dung cua PLL.
- Tach song tn hieu ieu tan.
- Tach song tn hieu ieu bien.
- Tong hp tan so.
- Nhan tan so bang khoa hai PLL.
- ieu che tan so (FSK) va ieu che pha (PSK).
- ong bo tan so.
- Bo loc bam theo thong dai hoac loc chan.
Trang 37
Chng 6: MAY PHAT
6-1 nh ngha va phan loai may phat
Mot so ch tieu ky thuat c ban cua may phat:
- Cong suat ra cua may phat.
- o on nh tan so: 730
1010ff
- Ch so ieu che AM (m), ch so ieu tan FM (mf)
- Dai tan so ieu che.
6-2 S o khoi tong quat cua cac loai may phat
- S o khoi tong quat cua may phat ieu bien (AM).
- S o khoi tong quat cua may phat n bien (SSB).
- S o khoi tong quat cua may phat ieu tan (FM).
- S o khoi tong quat cua may phat FM stereo.
6-3 Cac mach ghep trong may phat
Yeu cau chung oi vi cac mach ghep.
- Phoi hp tr khang.
- am bao dai thong D.
- am bao he so loc hai cao.
- ieu chnh mach ghep.
Cac loai mach ghep c ban.
- Ghep bien ap.
- Ghep ho cam.
- Ghep hai mach cong hng.
Trang 38
6-4 Cac mach loc c ban trong may phat
a- Mach loc n.
He so pham chat cua mach vao:
i
0
i
Li R
LRXQ
(1)
He so pham chat cua mach ra:
L0C
L0 CRX
RQ (2)
He so pham chat tng ng cua mach:
0i
0it QQ
xQQQ
(3)
e truyen at cong suat ln nhat va ap tuyen tan so
bang phang nhat ta co 2
QQQQ it0i (4) vi tan so loc
cua mach: LC1
0 (5)
b- Mach loc n.
Vi C
Ri
RL
L
C1 Vi
C2
Ri
RL
L
Trang 39
Khi mach oi xng C1 = C2 =C.
C
i
C
L0i X
RXRQQ (1)
vi
2CL
10 (2)
Khi mach bat oi xng: C1 C2 ta co.
10t
i1C C
1Q
RRX
(1)
20t
L2C C
1Q
RRX
(2)
2C1CL XXX (3)
vi LiRRR (4)
0it QQQ (5)
c- Mach loc oi.
1QRQR
X 2t
ti1C
(1)
RXX
X 3C1C2C (2)
1QRQR
X 2t
tL3C
9 (3)
LiRRR (4)
C1 Vi
C2
Ri
RL
L1 L2
C3
Trang 40
2C1C1L XXX (5)
3C2C2L XXX (6)
2QQ
Q 0it
(7)
Trang 41
Chng 7: MAY THU
7-1 nh ngha va phan loai may thu:
Mot so ch tieu ky thuat c ban cua may thu:
o nhay: bieu th kha nang thu tn hieu yeu cua may thu ma van am bao:
- Cong suat ra danh nh PL.
- Ty so tn hieu tren nhieu (S/N).
o chon loc: la kha nang chen ep cac dang nhieu khong
phai la tn hieu can thu: 1AA
Sf
0E
Chat lng lap lai tin tc
7-2 S o khoi tong quat cua may thu:
S o khoi tong quat mach khuech ai trc tiep.
S o khoi tong quat mach oi tan AM va FM.
S o khoi tong quat mach n bien (SSB).
7-3 Mach vao may thu:
Mot so ch tieu ky thuat cua mach vao:
He so truyen at: A
0MV E
VA (1)
o chon loc f
0E A
AS (2)
Dai thong D (BW)
Tan oan lam viec.
a- Anh hng cua Anten hoac tang au en mach vao
Trang 42
Tan so cong hng cua mach vao: LC1
0 (1)
ien dan tng ng luc cong hng:
Ld
CdgR1
0
0000
t (2)
vi 0
0 Q1d he so ton hao (3)
Khi co anh hng cua ien dan anten:
A0a CjgY (4)
hoac ien dan cua mach vao tang khuech ai au:
A0i CjgY (5)
ien dan tong ng cua mach cong hng tr thanh:
i2
0t gmgg (6)
i2
t CmCC (7)
He so pham chat cua mach cong hng se giam t Q0 xuong Qi theo quan he:
0
i2
0
t0
0 ggm
1gg
dd
(8)
Neu i2CmC (9) kha nho th o sai lech tan so cong hng
se la: CCm
21
CC
21
ff i
2
0
0
Nguyen tac xac nh he so mac mach (m)
- Mc tang ton hao khong vt qua gii han cho phep.
m2Ci m2gi C g0 L
Trang 43
- S bien thien cua tham so anten va dan nap vao tang au khong gay anh hng en ch tieu cua mach vao.
Cac mach loc nhieu lot thang (nhieu tan so trung gian)
b- Cac mach ghep anten vi mach cong hng vao
f < 30MHz CA = 50 250pF; rA = 20 60.
f > 30MHz CA = 10 20pF; rA = 10.
7-4 Bo tron tan
Thc chat la mot tang khuech ai cao tang PF co hai tan so vao khac nhau va au ra bo oi tan ta co vo so tan so thns nfmf . V au ra bo oi tan ta at mot mach cong hng tai tan so trung gian: thnstg fff
bang song trung va ngan: KHz455ftg
bang song FM: MHz7,10ftg
7-5 Tach song
a- Nhng ch tieu ky thuat c ban cua bo tach song bien o.
He so truyen at: 0iTS
TS0TS mV
VVV
A (1)
Tr khang vao cua bo tach song: iTS
iTSiTS I
VZ (2)
Meo phi tuyen: %100.I
...II
m1
2m3
2m2 (3)
b- Tach song Diode
Tach song noi tiep: Da vao qua trnh phong nap cua tu ta se co dang ien ap am tan a tach song tren ien tr tai RL.
CK R LK C
Trang 44
ieu kien e tu C loc tan so trung gian au vao:
R1
C
(1)
ieu kien e RC khong gay meo tn hieu la:
mm1RC
2 (2)
V vay trong thc te 1m 8,07,0m
ien tr vao tach song: 2RR i (3)
e tnh R ta da vao bang sau:
SR 20 50 100 200 1000
ATS 0,75 0,84 0,89 0,93 0,98
Tach song song song t c dung do 3RR i
c- Tach song tan so
Yeu cau cua bo tach song tan so.
He so truyen at cao Smax ln: fddV
S TS0f
ac tuyen truyen at phang trong mot pham vi tan so rong.
ien ap vao khong can ln.
e ffV TS0 ma khong phu thuoc bien o ien ap vao nen trc bo tach song tan so can co bo han che bien o.
Trang 45
ac tuyen truyen at phai oi xng qua goc toa o ffff
Co 3 nguyen tac e thc hien tach song tan so.
- Bien tn hieu vao FM thanh tn hieu AM, roi dung tach song bien o e tach song thng dung IC chuyen dung nh IC e tach song.
- Bien tn hieu FM thanh tn hieu ieu che o rong xung roi thc hien tach song tn hieu o rong xung nh mot mach tch phan.
- Lam cho tan so cua tn hieu FM bam theo tan so VCO cua PLL, ien ap sai so chnh la ien ap can tach song.
Trang 46
PHAN II
BAI TAP
Trang 47
Chng 1: MACH LOC TCH CC
Bai 1.3 Thiet ke bo LLT bac 2 co tan so cat tren fc = 1Khz, C1 = 0,1F, |Av0| = 1 trong hai trng hp:
1. Bo LLT co hoi tiep dng.
2. Bo LLT co hoi tiep am nhieu vong.
Bo LLT co hoi tiep dng.
1b
22121
20
2b
0211
21
SRRCCSRRC11)S(H
Chon: 221.4
bb4
CC
21
2
1
2
C2 = 2C1 = 2x0,1F = 0,2F
11231010.14,3.4
2Cf4
bRR 73
10
121
Bo LLT hoi tiep am hai vong.
10
121
1
2v Cf4
bRR1RRA
0
R2 +-V1(S)
C2
C1 R1 V2(S)
V2(S) V1(S) C2
C1
R1
R2
+-R3
Trang 48
6,11221010.14,3.4
2RR 7321
4
22.1.4
b
A1b4
CC
21
v2
1
2 0
C2 = 4C1 = 4x0,1F = 0,4F
5576,1122.10.4.1010.10.4
1RCCf4
bR 776221
20
22
3
Bai 1.6 Thiet ke bo LLT bac 3 co tan so cat tren fc = 1Khz, C1 = 0,16F, Av0 = 3.
V cho moi C bang nhau nen ch co the dung bo LLT2 hoi tiep dng 1 vong.
Theo bang Butterworth ta co: B3(S) = (S+1)(S2 + S + 1), (b22 = 1; b12 = 1).
LLT1:
K110.16,0.10.28,6
1C.
1R 63C
1
K310x3R3R3RRA 312
1
2v0
LLT2: Chon
K110.16,0.10.28,6
1C
1RR 630
43
59,0RR
59,123RR
1A6
5
6
5v0
Chon R6 = 1K R5 = 590.
R4 V2(S) V1(S)
C
R1
R2
R3 + - +
- . V2(S) R5
R6
C
C
Trang 49
Bai 1.8 Thiet ke bo LLT bac 4 co tan so cat tren fc = 1Khz, tat ca cac C = 0,16F.
V tat ca cac tu ien bang nhau nen chon s o hoi tiep dng mot vong la n gian nhat. Khi o ta cung chon cac R bang nhau.
B4(S) = (S2 + 0,765S + 1)(S2 + 1,848S + 1)
Mat loc 1:
b21 = 1; b11 = 0,765
C1 = C2 = C3 = C4 = 0,16F
K110.16,0.10.28,6
1C
1RRRR 630
6521
235,1RR
RR
1765,03A4
3
4
3v01
Chon R4 = 1K R3 = 1235.
Mat loc 2:
152,0RR
RR
1848,13A8
7
8
7v02
Chon R8 = 1K R7 = 0,152. R8 = 152.
Bai 1.5 Thiet ke bo loc thong dai LTD bac 2 co tan so cong hng f0 = 10Khz; D = 2000Hz; Av0 = 10; C1 = C2 = 0,1F.
V1(S) R1 R2
R3
R4
R5 C1 +-
. +-
R6
R7
R8
C2
C3
C4 V2(S)
V2(S) V1(S) R1
R2
R3 C1 C2
+-
Trang 50
510.2
10Df
Q 34
00
K6,110.10.2.14,3
1DC1R 733
802010.6,1
A2R
R3
0V
31
21
21143822
320
'
RRRR16
4,6100
10.10.6,1.10.41
C.R.1R
R2 = 20.
Bai 1.11 Thiet ke bo LTD bac 2 co F = 1000 3000Hz; A0(dB) = 10dB; C1 = 0,1F f1 = 1000Hz; f2 = 3000Hz.
K6,110.1016,0
Cf21R 73
111
V2(S) V1(S) R1
R2
C1
C2
+-
1000 f 3000 0
1
H(dB)
Trang 51
10ffdB10
fflg10A
3
1
3
10V
Hz10010
100010f
f 13
K1610.10.28,6
1Cf2
1R7213
2
nF3,310.16.3
16,010.16.10.3.28,6
1Rf2
1C 63322
2
Trang 52
Chng 2:
KHUECH AI CONG SUAT CAO TAN
I. Khuech ai cong suat cao tan: (KCSCT)
Bai 2.2
PL = 0,1W; f0 = 10MHz; Q0 = 50;
f1 = 3500MHz; hfe = 100; Cbe = 10-9F;
Cbc = 1pF; PCmax = 2W; VCEmax = 40v;
maxiC = 1A; = 900; Zn = 1K.
1. MHz5,10f.3,0MHz10fMHz5,3100
10.3500hf
f 06
fe
T suy ra s o
lam viec tan so thap, ta co s o tng ng sau:
2. Chon VCC = 0,5VCEmax = 0,5x40 = 20V
3. Chon = 0,9.
4. Vcm1 = .VCC = 0,9x20 = 18V.
5. A1imaxmA11A011,018
1,0x2V
P2I C1Cm
L1Cm
6. ien tr cong hng tng ng:
LC CC Lch
Cng Rn
Rb en
Cng Lch
+ VBB -
+ VCC - Cng
LC CC 100ib
Rn
en hie Rt
ib
Trang 53
1636011,018
IV
R1Cm
1Cmt
7. H521,050.10.28,6
10.636,1Qf2
RL 7
3
00
tC
8. pF6,48655,20
1010.521,0.10.86,9x4
1Lf4
1C8
61420
2C
9. A10.22100.5,0
011,0).(
III 5
1
1Cmbmn
10. VBB = VBEO In.Zn.0( - ) = 0,7 22.10-5.103.0,3 = 0,634V
11. ICO = 0()..Ibm = 0,3x100x22.10-5 = 6,6mA
12.
53010.6,610.25.100.4,1
I10.25h4,1h 3
3
CQ
3
feie
13. Neu Rn = 1K th Zi = Rn//hie = 103//530 = 346
14. Bien o ien ap kch thch:
Vbm = Ibm.Zi = 22.10-5x346 = 76mV
15. PCC = ICO.VCC = 6,6.10-3x20 = 0,132W.
16. Hieu suat: %75132,0
1,0PP
CC
L
Bai 2.1: Gia thiet giong bai 2.2, ch khac fT = 350MHz
1. MHz5,10f3MHz10f;MHz5,3100
10.3500hf
f 06
fe
T
suy ra mach hoat ong tan so trung bnh co cac CKS. Cac bc 2 8 giong het nh tren
9. He so khuech ai dong ien dai tan so trung bnh:
Cbe Rt LC CM |hfe|ib
Rn
en rbe .
ib . -
CC
Trang 54
33027,3
100
10.5,3.210.21
100
1
hh
2
6
720
fe.
fe
.
10.
mA67,033x5,0
011,0.
II
1
1Cm'n
11. e dong vao Base khong b meo phai thoa man ieu kien:
e'bnn C.R1
5,4598,21
1010.10.5,3.28,6
1C.f2
1R3
96c'b
n
15027,3
5,45
10.5,3.210.21
5,45
1
RZ
2
6
720
n.
n
12. In = In[1 + TCbc.1().ZL] = 0,67.10-3[1 + 6,28.108.3,5.10-12.0,5.1,64.103] = 0,67.10-3[1 + 1,8] = 1,88mA
13. mV2,1810.88,1.15I.ZV 3n.
nnm
14. V7,05,0.15.10.63,07,0)(ZIVV 30.
n'nEO'BBB
15.
86,2j8,2.10.10.28,6.j
5,0)(ZC1Cj
)(Z 971Lc'bTe'b
1iEC
Mat khac:
86,2j'Cj
1Z'b
iEC
nF55,586,2.10.28,6
1'C 7'b
Chu y: Cb = Cbe + CM (khac vi ien t 2 la co them goc cat)
16. 51271.
c'bOEC 10.110j5,0.33110.10.28,6.j)(.1CjY
33
OECOEC 10.j10.j
1Y
1Z
17. mA7A10.710.67x33x32,0'I..I 35n.
0CO
18. PCC = ICO.VCC = 7.10-3.20 = 134mW = 0,134W
19. %6,74134,0
1,0PP
CC
L
Trang 55
Bai 2.8: Giong bai 2.1, ch khac au vao khung cong hng ch c ghep mot phan vao au vao e giam anh hng cua ien dung ky sinh va giam anh hng cua rbe.
Cac bc t 1 10 giong bai 2.2, nhng cach tnh mach vao
khac v au vao co khung cong hng. Cac bc t 15 19 giong bai 2.2.
Nh tren ta co: Cb = 5,55nF, khi o ien dung au vao: Cb = Cb + a2Cb = Cb + (1/25)5,55.10-9 = Cb + 222.10-12.
Coi he so pham chat cua khung cong hng au vao va au ra giong nhau: Q01 = Q02 = 50.
H517,04,3110.1624
50.10.21624
QR
L8
7010
1tb
pF51039,20
1010.517,0.10.86,9.4
1Lf4
1C8
614b
20
2b
Khi o ien dung can mac au vao: Cb = Cb a2Cb = 510pF 222pF = 288pF.
Tr khang vao:
e'b2e'b
tbi r25//1636ar
//RZ
16510.710.25..33.4,1
10.710.25h.4,1r 3
3
3
3.
fee'b
Suy ra: Zi = 1636//4125 = 1171,4 He so pham chat cua mach au vao:
Qi = 0RiCb = 6,28.107.1171,4.510.10-12 = 37,5 Neu au vao mac trc tiep vao Base, khi o ien dung ky
sinh rat ln (Cb = 5,55nF) >> Cb, nen mach lam viec khong on nh va khi o Zi rat nho (Zi = Rtb//rbe = 150) nen Qi = 0RiCb = 6,28.107.150.510.10-12 = 4,8
Trang 56
ngha la Lb, Cb cua tang KCSCT chnh la LC, CC cua tang tien KCSCT trc o.
II. Mach nhan tan
Bai 2.5
6
0
3500.1035 ; 10 0,3 10,5
100T
fe
ff MHz f MHz f MHz
h Suy ra mach hoat
ong tan so thap:
1. Chon goc cat toi u: 00
902
180
2. Chon nguon cung cap: VCC = 0,5VCEmax = 0,5x40 = 20V 3. Chon 1 = 2 = 0,9
Suy ra VCm1 = VCm2 = xVCC = 0,9x20 = 18V. 4. Cong suat ra cua bai 2.1 la cong suat hai bac nhat. Do o
cong suat ra do hai bac 2 gay ra se la:
W042,01,0x5,0
212,0P.)()(P.
)()(P 1L
1
21L
1
22L
5. Dong ien hai bac hai:
mA7,4180424,0x2
VP2I
2Cm
2L2Cm
6. ien tr tng ng cua mach cong hng ra:
383010.7,4
18IV
R 32Cm
2Cm2t
7. Neu coi gia thiet cua bai 2.1 trong o Q0 la he so cham chat cua khung cong hng au ra: Q02 =50 ta co:
LC CC Lch
Lb Rb Cb
Lch
+ VBB -
+ VCC - Cng
Rt2 LC Cb gmVbe
CC Lb Rt1 rbe
Trang 57
H61,08,62
10.38350.10.28,6.2
3830Q2
RL
7
7020
2tC
8.
pF10423,96
1010.61,0.10.86,9.4.4
1L.2
1C8
614C
20
C
9. Hieu suat bo nhan: %32318,075,0.5,0
212,0. 11
22
10. tan so thap: 30 22
( ) 0,318. 4,7.10 7,05
( ) 0,212CO CmI I x mA
11.
50049610.05,7
10.25.100.4,1I10.25.h.4,1hr 3
3
CO
3
feiee'b
12. tan so thap: A10.22100.212,010.7,4
h.90I
I'I 53
fe0
2
2Cm1bmn
13. Bien o ien ap kch thch au vao:
mV84V08425,0500//163610.22r//RIZ.IV 5e'b1t1bmi1bm1bm
14.
16243830x
5,0212,0R.
)()(R 2t
1
21t
15. Coi mach cong hng au vao va au ra co he so pham chat rieng bang nhau: Q01 = Q02 = 50. Thc te khi co rbe th he so pham chat Q01 < Q01.
16. F517,04,3110.1624
50.10.21624
QR
L8
7010
1tb
17. pF51039,20
1010.517,0.10.86,9x4
1Lf4
1C8
614b
20
2b
Bai 2.4
1. 6
0
350.103,5 ; 10 3 10,5
100T
fe
ff MHz f MHz f MHz
h Suy ra mach hoat
ong tan so trung bnh: Cac bc 2 en 10 giong nh bai 2.5.
12. tan so trung bnh:
. Rt2 LC Cb gmVbe
CC Lb Rt1 rbe
Cb
Trang 58
*
2 27
06
100 10033
3,0272 .1011
2 .3,5.10
fefe
hh
13.
1648,16310.05,7
10.25.33.4,1I10.25.h.4,1r 3
3
CO
3*
fee'b
14.
mA03,07,155
10.7,433.212,0
10.7,4
h90
II
33
*
fe0
2
2Cm'n
15. Vbm1 = Ibm1.Zi = Ibm1 = (Rt1//rbe) = 3.10-5(1636//164) = 4,47mV. Cac bc 15, 16, 17, 18 nh tren ch chu y: Cb = Cbmac + Cb*
20. Trong mach nhan tan do tan so cong hng au vao (0) va au ra (20) khac nhau, nen khong co anh hng cua CM*.
Khi o ta co: F10.25,0
10)(
CC 9
9
1
e'b*'b
Ta thay Cb* = 2000p >> Cb = 510pF. Do o mach nay lam viec khong on nh, ta phai cai tien mach nh bai 2.9
Bai 2.9:
Cac bc t 1 den 14 giong het nh bai 2.4 15. Vbm1 = Ibm1.Zi = Ibm1 = (Rt1//(rbe/a2) = 3.10-5(1636//4100) =
0,035mV = 35mV. Cac bc 16, 17, 18 giong bai 2.5
20.
F10.25,0
10CC 99
1
e'b*'b
suy ra pF802510.2C.a
9*
'b2
21. Khi o ien dung Cb = Cb Cb* = 510pF 80pF = 430pF.
Rt2 LC Cb gmVbe CC Lb Rt1
. Cb.a2 rbe
a2
VL
Trang 59
Chng 3: BO DAO ONG
I. Dao ong ba iem ien dung
Bai 3.2: Mach dao ong fthap. f0 = 10MHz, Q0 = 100, L = 1H, rbe = 500, fT = 3500MHz, hfe = 100, Cbe = 1000pF, PCmax = 4W, VCE0max = 40V, maxiC = 1A, CCE = 5pF.
Cac mach dao ong lam viec che o lp A nen tnh toan mach phan cc giong nh trong mon ien T I. Trong thc te: IEQ = 15mA va RE = 102103 (sinh vien t chon). V du: Chon IEQ = 5mA; RE = 103; VCC = 10V.
VRE = RE.IEQ = 103.5.10-3 = 5V Suy ra VCEQ = VCC VRE = 5V VBB 0,7 + RE.ICQ = 5,7V; Rb = (1/10).hfe.RE = 10K;
K26,237,510
1010VV
VRR 4BBCC
CCb1
K8,407,5
1010VVRR 4
BB
CCb2
50010.510.25100
I10.25hh 3
3
EQ
3
feie
1. ay la mach dao ong 3 iem ien dung nen thoa man ieu kien can bang pha v:
21 C
1X
; 1
2 C1X
; X3 = L.
R1
Cb
+VCC Lch
VK
R2
RE CE
C1
C2 E L
Trang 60
2. MHz35100
10.3500hf
f6
fe
T ;
MHz5,10f.3,0MHz10f0
S o lam viec tan so thap
3. He so hoi tiep: nCC
VV
2
1
CE
BE
n co the chon theo kinh nghiem: n = 0,010,05 (E.C) va n = 0,10,5 (B.C) hoac tnh theo cong thc. Chon n = 0,01 C2 = 100C1.
4. ien dung tnd ng cua khung cong hng:
pF5,25344,39
1010.10.86,9.4
1Lf4
1C8
61420
2t
pF5,253C100C
C100.CCC
CCC11
11
21
21t
21 CnF35,25C100 pF5,253C1
5. He so khuech ai cua s o mac E.C:
2
11K
2
e11
e21C n
h//Rp
hh
SZA
h21e = hfe = 100; h11e = rbe = 500; M510500
nh
Z 4211
ifa
6. RK = 0LQ0 = 6,28.107.10-6.100 = 6,28K. 7. He so ghep Transistor vi khung cong hng:
Cbe
C
Rb=R1//R2
B
CCE C1
C2 L
Trang 61
99,001,01
1n1
1CC
CC
CCCC
VV
p21
2
1
21
21
K
CE
8. 33632 10.23,110.155,6.5110.5//10.28,6.99,0
500100A
9. ieu kien bien o e mach t kch:
13,1201,0.10.23,1n.|A|||.|A| 3***
; Suy ra mach dao ong
Bai 3.1: Mach dao ong tan so trung bnh 1. Bc 1: nh tren.
2. MHz5,3100
10.350hf
f6
fe
T
MHz5,10f.3MHz10f0
Suy ra s o lam viec tan so trung bnh (co cac ien dung ky sinh)
3. He so hoi tiep co the chon nh tren, khi o C1, C2 giong het bai 3.2. Nhng e mach hoat ong on nh ta co the chon C2 = 10Cbe = 10.10-9 = 10-8F = 10nF, suy ra C2 = 9nF.
4. Nh tren ta co: Ct = 253,5pF
pF5,25310C
10.CCC
CCC 8
1
81
21
21t
20121
81 10.5,25310.5,253.C10.C
20418 10.5,25310.4,2531C10
121 10.5,253C.97465,0
pF255CpF260C '11
Chu y: C1 = C1 + CCE va C2 = C2 + Cbe trong o C1 va C2 mac ben ngoai, con CCE va Cbe la cac tu ien ky sinh.
Khi o: 026,010
10.260CC
n 812
2
1
5.
2
*
11K
2*
11
*
21C n
|h|//Rp
|h|
|h|SZA
tan so trung bnh ta co:
33
10.5,3.210.21
100
1
h|h|
2
6
722
fe*
21
Trang 62
5,4598,21
1010.10.5,3.28,6
1C.f2
1r|h|3
96e'b
e'b
*
11
K3,67
10.76,65,45
026,05,45
n|h|Z 422
*
11ifa
6. Nh tren RK = 6,28K.
7. 975,0026,01
1n1
1VV
pK
CE
8. 332 10.3,67//10.28,6.975,05,45
33A
397510.975,3
10.48,5.725,0K3,67//K97,5725,03
3
9. ieu kien bien o e mach t kch:
135,103026,0x3975n.|A|||.|A|***
Suy ra mach dao ong.
Dao ong 3 iem ien dung kieu Clapp Bai 3.3: Mach dao ong tan so trung bnh:
Bc 1, 2, 3 giong nh bai 3.1 tren 4. Nh bai tren ta co Ct = 253,5pF nhng
021t C1
C1
C1
C1
Chon C0 = 270pF e C0 quyet ng f0 trong mach
E
C0
VK
Cbe
C
B
CCE C1
C2
L
Trang 63
824
12
0
0
021
21
11
10.41,210.68445
10.5,16
.
11
.
11
t
t
t CC
CC
CCCC
CC
CC
18
188
18
1 C41,210C10.41,2.10.C10C
nF1,7C1
Khi o: 71,010.1010.1,7
CC
n 99
2
1
5. Giong bai 3.1 ch khac:
3,90
504,05,45
71,05,45
nhZ 22
11ifa
6. Nh tren: RK = 6,28K
7. 0357,01,7
2535,010.1,710.5,253
CC
VV
p 912
1
t
K
CE
8. 33,53,90//8725,03,90//10.28,6.0357,05,45
33A 32
9. ieu kien bien e mach t kch:
178,371,0x33,5n.|A|||.|A|***
Suy ra mach dao ong
10. Chu y: C2 = C2 Cbe = 10-8 10-9 = 9nF; C1 = C1 CCE = 7,095nF
Bai 3.4: Mach dao ong tan so cao. 1. Mach thoa ieu kien pha ve dao ong nh tren
2. MHz5,3100
10.350hf
f6
fe
T
f0 = 100MHz > 3f = 10,5MHz Vay mach hoat ong tan so cao. Ta khong the tnh cac Lk sinh, v vay phai dung s o dao ong mac Base chung. f = fT = 350MHz f0 = 100MHz < 0,3f = 105MHz Suy ra s o hoat ong tan so thap (khong can tnh Ck sinh)
Lch
RE
B R1
E
C0
VK
+VCC C1
Cb
L
R2
C2
+VCC
Trang 64
3. V s o lam viec tan so thap nen he so hoi tiep co the chon nh bai 3.2. ay la s o B.C nen chon n = 0,1.
1,0CC
CC
CCC.C
VV
21
1
2
21
21
BC
BE
4. Giong bai 3.3 ta co Ct = 253,5pF; chon C0 = 270pF va ta se
co: nF4,41CC1,01
C.n110.41,2
C.CCC
222
8
21
21
C1 = 0,1C1 + 0,1C2 = 0,1C1 + 4,14.10-9 0,9C1 = 4,14nF C1 = 4,6nF.
5. He so khuech ai cua s o mac B.C.
2
b11K
2
b11
b21C n
h//Rp
hh
SZA
h21b hfb 1;
5100500
hh
hfe
ieb11
500
1,05
nh
Z 22b11
ifa
6. Nh tren: RK = 0LQ0 = 6,28.108.10-6.100 = 62,8K
7. 3
9
12
8
12
21
21
10.6110.15,4
10.5,253
10.41,2
110.5,253
CC
CC
C
V
Vp
t
K
CE
8. 88,31500//2342,0500//10.8,62.10.372151A 36
9. ieu kien bien o e mach dao ong t kch:
1188,31,0x88,31n.|A|||.|A|***
Suy ra mach dao ong.
III. Dao ong thach anh:
Bai 3.6: Mach dao ong tan so thap kieu Colpits Chu y gia thiet giong bai 3.2 nhng cha biet L. Lch
C
Cng
C1
Cb
+VCC
Trang 65
1. ieu kien pha: e mach dao ong theo kieu 3 iem C Colpits,
thach anh phai tng ng nh cuon cam, ngha la: fq > f0 > fp. e n gian ta coi f0 nam gia fq va fp: Mach Colpits E.C
q11
13
qp
qqp f005,110.2
101fC2
C1ff
Mat khac: qqqpq
0 f0025,12f005,1f
2ff
f
Suy ra: MHz975,90025,110
0025,1f
f7
0q
MHz025,10f005,1f qp
2. MHz35100
10.3500hf
f6
fe
T
f0 = 10MHz < 0,3f = 10,5MHz Suy ra s o lam viec tan so thap.
3. ien cam rieng cua thach anh:
mH55,2392410
10.10.5,99.86,9.41
C.f21L 1312
q2
4. Tr khang tng tng cua thach anh tai tan so f0: Zt = j0Lt
nen: pqq20qp20qq
20
t CCLCC
1CLL
00572,110.55,2.10.10.86,9.4LCf4CL 31314qq20
2qq
20
fq f0 fp
f/2 f/2
Trang 66
H34H89,33
10.428,0.10.44,3910.572
10.00572,1101010.86,9.4100572,1L
1314
5
11131114t
5. pF5,710.34.10.86,9.4
1Lf4
1CC
CCC 614t
20
221
21t
6. Chon he so hoi tiep: n= 0,01.
nCC
VV
2
1
CE
BE
Suy ra C1 = 0,01C2 hoac C2 = 100C1.
Suy ra: pF5,757C100pF5,7C100C
C100.CC 111
11t
Vay C2 = 100C1 75,75nF va C1 7,75pF 7. He so khuech ai cua s o mac E.C
2
*
11K
2*
e11
*
e21C n
|h|//Rp
|h|
|h|SZA
h21e = hfe = 100; h11e = rbe = 500.
M510500
nh
Z 4211
ifa
8. RK = 0LtQ0 = 6,28.107.9,226.10-7.100 Suy ra RK = 57,94.100 = 5794
9. 99,001,01
1n1
1CC
CVV
p21
2
K
CE
10. 11355678.2,0M5//10.5794.99,0500100A 32
11. ieu kien bien o e mach t kch:
135,1101,0x1135n.|A|||.|A|***
Suy ra mach dao ong.
Bai 3.5: Mach lam viec ftrung bnh mac kieu Clapp.
1. ieu kien pha: tnh nh tren, nhng lu y khong nhng thach anh phai tng nh cuon cam, ma thach anh cung CS cung phai tng ng nh cuon cam ngha la: q < q < 0 < p.
VK
CS=C0
E
C
C1
C2
B
Trang 67
fq = 9,975MHz va fp = 10,025MHz
2. MHz5,3100
10.350hf
f6
fe
T
f0 = 10MHz < 3f = 10,5MHz Suy ra s o lam viec tan so trung bnh phai ke ca CKS. Cac bc 3, 4, 5 nh tren: Lq = 2,55mH; Lt = 34H va Ct = 7,5pF = C1 noi tiep C2 noi tiep CS.
6. e CS quyet nh tan so cong hng trong mach ta chon CS
Trang 68
0833,010.03,3.10.275
C.CCCxC
CCCC
CVV
p
812
21
21t
21
21
t
K
CE
10. 1,2910.455//5794x0833,05,45
33A 32
11. ieu kien bien o e mach t kch:
191,21,0x1,29n.|A|||.|A|***
Suy ra mach dao ong.
IV. Dao ong RC
Bai 3.8: Dao ong cau
e mach dao ong: 32
RRR
21
2
3R2 = 2R1 + 2R2 R2 = 2R1 Chon R1 = 1K, th R2 = 2K.
F16,010.16,010.10.28,6
1R
1C 6330
Bai 3.7: Dao ong cau Vien
VL
R1
C
+ -
R2
R
R
C
R1
-
R2
Trang 69
e mach dao ong: 2121
2 R2RRR
R31
Chon R2 = 1K R1 = 2K. F16,010.16,0
10.10.28,61
R1C 6330
Trang 70
Chng 4: IEU CHE TNG T
I. ieu bien Collector
Bai 4.4: PL = 100mW; f0 = 1MHz; f = 10KHz; Q0 = 50; m = 0,5; = 900. Tran sistor nh bai 2-1.
1. MHz5,3100
10.350hf
f6
fe
T
f0 = 1MHz < 0,3f = 1,05MHz Suy ra s o hoat ong tan so thap.
2. Chon nguon cung cap VCC = 0,5VCEmax = 0,5x40 = 20V. 3. Chon he so li dung ien ap = 0,9. 4. Bien o trung bnh phan hai bac nhat: VCm1 = V0 = VCC = 18V.
khi o V0 = 18cos(2.106t)
5. mA88,9
225,01
1,0x182
2m1
PxV
2V
P2I 2
AM
1Cm1Cm
01Cm
6. ien tr cong hng tng ng:
)(182210.88,9
18IV
R 31Cm
1Cmt
7. H8,510.5810.4,31
182250.10.28,6
10.822,1Qf2
RL 776
3
00
tC
8. nF37,475,228
1010.8,5.10.86,9.4
1Lf4
1C6
612C
20
2C
9. Bien o dong kch thch vao:
mA2,0A10.76,19100.5,010.88,9
h).(I
I 53
fe1
1Cmbm
10. mA610.2x100x3,0I.).(I 4bm0CO 11. Tr khang vao cua Transistor
LC CC Lch
V
Cng
+VCC
Trang 71
58310.610.25.100.4,1
I10.25h.4,1rZ 3
3
CQ
3
fee'bi
12. Bien o ien ap kch thch: Vbm = Zi.Ibm = 583.0,2.10-3 = 116mV
13. PCC = ICO.VCC = 6.10-3.20 = 120mW
14. %7012,0089,0
PP
CC
L
15. Bien o ien ap am tan: V = mV0 = 0,5x18 = 9V khi o: V(t) = 9cos(2.104t)
16. Pho cua tn hieu AM
t)1010(2cos5,4t)1010(2cos5,4)t10.2cos(18
t)cos(2
mVt)cos(
2mV
tcosVV
46466
00
00
00AM
17. mW8910.82,1.2
)18(R2V
P 32
L
20
0
18. mW125,112
)5,0(10.892
m.PP2
32
0bt
19. 1191
1m2
1PP
k
2AM
bt
20. Kiem tra: V0 + V = 18 + 9 = 27V < BVCEO = 40V PAMmax = P0(1 + m)2 = 89.10-3(1 + 0,5)2 = 200mW < PCmax.
Bai 4.5: Vtt = 10cos2.106t; V = 7cos2.104t; RL = 1K. 1. Giong nh bai 4-4: f0 = 1MHz < 0,3f = 1,05MHz Suy ra s o hoat ong tan so thap.
2. He so ieu che: 7,0107
VVm
0
3. Chon = 0,9 V11,119,0
10VV 1CmCC
106-104 106+104 106
VAM
4,5
18
0
Trang 72
4. Pho cua tn hieu ieu bien:
t)1010(2cos5,3t)1010(2cos5,3)t10.2cos(10
t)cos(2
mVt)cos(
2mV
tcosV)t(V
46466
00
00
00AM
5. mW500.2)10(
R2V
P 32
L
20
0
6. mW25,12249,010.5
2m.PP 2
2
0bt
7. Cong suat ieu bien: PAM = P0 + Pbt = 50 + 12,25 = 62,25mW
8. PAMmax = P0(1 + m)2 = 5.10-2(1 + 0,7)2 = 144,5mW < PCmax.
9. %201
5,02
1
1m2
1k
2
10. Kiem tra: V0 + V = 10 + 7 = 17V < BVCEO = 40V 11. Bien o hai bac nhat:
mA101010
RV
RV
I 3L
0
L
1Cm1Cm (chu y RL = Rt)
12. H18,3.4,31
1050.10.28,6
10Qf2
RL
4
6
3
00
L
13. nF97,74,125
1010.18,3.10.86,9.4
1Lf4
1C6
61220
2
14. Bien o dong kch vao:
mA2,0100.5,0
10h).(
II
2
fe1
1Cmbm
15. mA610.2x100x3,0I.).(I 4bm0CO
58310.610.25.100.4,1
I10.25h.4,1r 3
3
CQ
3
fee'b
16. V117,0583.10.2r.IV 4e'bbmbm
II. ieu tan Varicap
Bai 4.6: ieu tan dung Varicap n f0 = 100MHz; f = 75KHz; Q0 = 50; L = 0,1H;
106-104 106+104 106
VAM
3,5
10
0
Trang 73
Transistor nh bai 2-1
1. MHz5,3100
10.350hf
f6
fe
T
f0 = 100MHz > 3f = 10,5MHz Suy ra s o hoat ong tan so cao, nen phai chuyen sang dung mach dao ong B.C nh hnh ve
2. 21
219
71620
2t CCC.C
pF35,2544,39
1010.10.86,9.4
1Lf4
1C
Chon C1 = 30pF
pF5,16310)35,2530(10.35,25.10.3
CCxCC
C 121211
t1
t12
3. He so hoi tiep:
n155,010.5,16310.30
10.30CC
CC
CCCC
VV
1212
12
21
1
2
21
21
BC
BE
4. He so ghep Transistor vi khung cong hng:
1VV
VV
pCB
CB
k
CB
R1
L
C
Lch V
C
+VCC
R2
RE C2
C1
C
L
C
B
RE C2
C1
E
Trang 74
5. tan so thap nen: h21b 1; 1050500
hh
hfe
e11b11
6. Rt = 0LQ0 = 6,28.108.10-7.50 = 3140 7. He so khuech ai khi mac B.C:
8,36416//3140.1101
nh
//Rphh
SZA 22b11
K2
b11
b21C
8. ieu kien bien o e mach t kch: ongdaomach17,5155,0x37n.|A|||.|A|
9. Chon Vpc = 5V; n = ; = 0,7V; Suy ra ta co: 38
pc21
10 10.7557,0
V.5,16330
5,163.10.21.
21
VV.
CCC.nf
21f
suy ra
38
10.75V8,4411
5,163.10f suy ra V = 20mV
10. f1 = f0 - f = 99,925MHz;
f2 = f0 + f = 100,075MHz;
pF39,259381,3
1010.10.9985.86,9.4
1Lf4
1C10
71221
21t
pF317,259499,3
1010.10.10015.86,9.4
1Lf4
1C10
71222
22t
pF06,3010.11,138
10.39,25.10.5,163CC
xCCC 12
1212
1t2
1t211
pF955,2910.183,138
10.317,25.10.5,163CC
xCCC 12
1212
2t2
2t212
30,06pF
29,96pF CV0 = 30pF = C1
C
V(
Trang 75
Bai 4.7: ieu tan dung Varicap ay keo 1. Bc 1 va 2 nh tren
2. pF35,25CC
xCCCC
xCCC
21
21
20v10v
20v10vt
Chon pF10CC
xCCCpF20CC
20v10v
20v10v0v20v10v
Suy ra pF35,15CC
CCC21
213
3. He so hoi tiep: v s o mac B.C nen ta chon n = 0,1
21212
21
2
1
21
21
BC
BE
C9CC1,0C1,0C
1,0nCC
CC
CCCC
VV
pF35,151C9C
C9CCC
CCC1
11
21
213
9C1 = 153,5pF = C2 C1 = 17pF
4. Cac bc 4, 5, 6 nh tren
7. 5,6001,0
10//3140.1101
nh
//Rphh
SZA 22b11
K2
b11
b21C
8. ieu kien bien o e mach dao ong t kch:
ongdaomach16,81,0x26,86||.|A| 9. Chon Vpc = 5V; n = ; = 0,7V
R1
CV1 Ra VFM
+VCC
R2
L
C
VR
RE
C2
C1
R
CV2 -
Trang 76
38
pc30v
0v0 10.7557,0
V.
35,151010.10.
21.
21
VV
.CC
C.nf
21f
suy ra
39
10.75V57810f suy ra V = 43,4mV
10. Bc 10 nh tren ta co: Ct1 = 25,39pF; Ct2 = 25,317pF Suy ra: Cv01 = Ct1 C3 = 10,04pF; Cv02 = Ct2 C3 = 9,967pF
Trang 77
Chng 6: MAY PHAT
I. KCSCT + mach ghep au vao va au ra
Thiet ke bo KCSCT nh hnh ve co f0 = 10MHz. Transistor co cac thong so fT = 350MHz; hfe = 100; Cbc = 1pF; Cbe = 1000pF; PCmax = 10W; VCEmax = 40V; max iC = 1A. Cac bc thiet ke:
1. MHz5,3100
10.350hf
f6
fe
T
f0 = 10MHz < 3f = 10,5MHz Suy ra s o hoat ong tan so trung bnh
2. Chon VCC = 0,5VCEmax = 0,5x40 =20V 3. Chon = 0,9 4. Bien o hai bac nhat: PCm1 = .VCC = 0,9x20 = 18V.
5. Cong suat hai bac nhat: W5,28,0
2PP2
A1L
6. Bien o hai bac nhat cua dong ien ra:
mA27818
5,2x2V
P2I1Cm
1L1Cm
A1imaxmA556278x2I2 C1Cm 7. ien tr cong hng tng ng mach ra
65278,018
IV
R1Cm
1Cm1t
Rt1 chnh la tr khang vao cua mach ghep au ra. 8. Mach ghep au ra:
8,69xRRxRRR AtLi
Lch
+ VBB -
+ VCC -
-
L2
Rb
L1
C2
Cng
C1
Ri = 50 Pi = ?
Lch
Cng
C3
PA = 2W RA = 75
1=0,7 Qt1=5 =900 1=0,7 Qt2=5
Trang 78
275
8,6965Q
RRXt
i2C
pF59027.10.28,6
1Xf2
1C 72C0
2
295
8,6975Q
RRXt
L3C
pF55029.10.28,6
1C 73
pF5721CCCCC 1*
c'b'2
*c'b
'2
'2
662927XXX 3C2C2L
H110.28,6
66f2
XL 70
2L
9. He so khuech ai tan so trung bnh
33027,3
100
10.5,3101
100
1
hh
2
6
72
0
fe.
fe
.
10. Thanh phan trung bnh DC cua dong ra
mA178278x5,0
32,0I.)()(
I 1Cm1
0CO
11. Cong suat nguon cung cap PCC = ICO.VCC = 0,178.20 = 3,56W
12. Hieu suat cua mach KCSCT
%22,7056,35,2
PP
CC
L
13. Tr khang vao
'b*12
12697
1Lc'bTe'b
1iEC
Cj18j87,7j
35,6j50
0114,0110.28,6j5,0
5,0.6510.10.3501.10.10.28,6.j5,0
)(ZC1Cj)(
Z
L2
C2
Ri = Rt = 65 C3
RA = RL = 75
Trang 79
)CCC(nF023,287,7.10.28,6
1C *M*
e'b'b*
7'b*
14. Tr khang vao cua Transistor tan so trung bnh
5,6178,010.25..33.4,1
I10.25h.4,1r
3
CO
3.
fee'b
15. Mach ghep au vao: e s truyen at cong suat ln nhat va ap tuyen tan so bang phang nhat ta thiet ke sao cho Qi = Q0 = 2Qt.
65,010
5,6Q2r
QRX
t
e'b
0
LC
nF5,2465,0.10.28,6
1Xf2
1C 7C0
'1
Mat khac: nF5,22023,25,24CCCCCC * 'b
'11
*b1
'1
50050x5x2RQ2RQX itii2L
H810.28,6
500f2
XL 70
L1
16. e'b
*'b0
e'be'biEC
'iEC r.Cj1
rr//ZZ
5
3,15,6
25,42.1010.092,4.86,9.41
5,6rC1
rZ
1418
2e'b
2*'b
20
e'b'iEC
17. Cong suat au ra cua Transistor
mW75,010.5,75.10.3.21
5.533,0278,0
21Z
I21ZI
21P
44
2
i
2
1
1Cmi
2'nb
18. Cong suat au vao cua tang KCSCT
mW07,17,010.75,0PP
3
1
bi
II. KCSCT + ieu bien
L1
C'1 Ri = 50 Qt1 = 5
rbe Cb
Trang 80
Thiet ke mach ieu bien Collector co gia thiet nh bai tren biet m = 0,7 va f0 = 10MHz; f = 10KHz. e n gian ta khong tnh mach vao
1 4 nh bai tren: VCm1 = 18V = V0 5. Cong suat ieu bien:
W5,28,0
2PP2
AAM
Cong suat hai bac nhat:
2
m1PP2
0AM
W2
249,01
5,2
2m1
PPP2
AM1L0
6. Bien o hai bac nhat cua dong ien ra
mA18718
2x2V
P2I1Cm
1L1Cm
7. )(25,96187,018
IV
R1Cm
1Cm1t
Rt1 chnh la tr khang vao cua mach loc au ra 8. Mach loc au ra
8575x25,96xRRR Li
2,365
8596Q
RRXt
i1C
nF44,02,36.10.28,6
1Xf2
1C 71C0
1
325
8575Q
RRXt
L2C
nF5,032.10.28,6
1C 72
+VCC
L1
+ VBB -
Rb KCSAT
Lb C1
Cng
Cng
Lch
C2
RA = 75 PA = 2W
Trang 81
pF5,175,0.331101CC 121*
c'b*
c'b
pF5,4225,17440C '1 2,68322,36XXX 2C1C1L
H1,1H10.86,1010.28,62,68
f2XL 77
0
1L1
9. Nh tren 33*
10. A12,0187,0x5,0
32,0I.)()(
I 1Cm1
0CO
11. PCC = ICO.VCC = 0,12.20 = 2,4W
12. %704,2
68,1PP
CC
1L
13. Nh tren )25,96Zvi(83,7jZ LiEC
14.
625,912,010.25..33.4,1
I10.25h.4,1r
3
CO
3.
fee'b
15. 2
e'b2*'b
20
e'b'iEC
rC1
rZ
6587,1625,9
64,92.10.16,4.10.86,9.41
625,91814
16. Cong suat au vao
mW4,06.33.5,0
187,021Z
I21ZI
21P
2
i
2
11
1Cmi
2'nb
17. Bien o ien ap am tan V = mV0 = 0,7x18 = 12,6V khi o: V(t) = 12,6cos(2.104t)
18. Pho cua tn hieu am tan
t)1010(2cos3,6t)1010(2cos3,6)t10.2cos(18
t)cos(2
mVt)cos(
2mV
tcosVV
47477
00
00
00AM
L1
C1
Rt1 = Ri = 96 C2
RA = RL = 75
107-104 f 107+104 107
VAM
6,3
18
0
Trang 82
19. W41,0249,068,1
2m.PP
2
0bt
20. %4,165,241,0
PP
kAM
bt
21. Kiem tra: V0 + V = 18 + 12,6 = 30,6V < BVCEO = 40V PAMmax = P0(1 + m)2 = 1,68(1 + 0,7)2 = 4,85W< PCmax = 10W (Chu y: cac bc 9 16 co the khong phai tnh khi thi).