do an dien tu thong tin

n in t thng tin GVHH:TS.NGUYN L HNG

SVTH: Page 1

PH LC

PHN I

CHNG I: MY PHT

1.1.S khi ca h thng in t thng tin.

1.2.nh ngha v phn loi:

1.3.My pht iu tn FM

1.3.1.Nhim v cc khi.

1.3.2.Nguyn l hot ng :

1.3.3. Dng sng u ra cc khi:

1.4Cng sut pht

1.5. Kt lun chng.

CHNG II: MY THU FM

2.1. nh ngha my thu:

2.2.My thu i tn FM:

CHNG III: CC MCH THNG DNG TRONG MY PHT FM

3.1.Mch to dao ng .

3.1.1. Cc vn chung v mch to dao ng .

3.1.2. Mch to dao ng bng hi tip dng

3.1.2.1. iu kin mch to dao ng

3.1.2.2 . c im ca mch dao ng

3.1.2.3. Mch dao ng ba im in dung .

3.2. Mch khuych i thut ton.

3.3 Mch khuych i dng BJT.

3.3.1. Cc ch hot ng ca BJT.


SVTH: Page 2

PHN I

CHNG I: MY PHT

1.1. S khi ca h thng in t thng tin.

1.2. nh ngha v phn loi:

Mt h thng in t thng tin bao gm, my pht, my thu v mi trng

truyn sng. Trong my pht l 1 thit b pht ra tn hiu di dng sng

in t c biu din di 1 hnh thc no .


SVTH: Page 3

Hnh 1.2: s khi ca h thng thit b thu pht

Sng in t l gi l sng mang hay ti tin lm nhim v chuyn ti

thng tin cn pht ti my thu. Thng tin ny c gn vi ti tin theo 1

hnh thc iu ch thch hp. My pht phi pht i cng sut ln

cung cp t s tn hiu trn nhiu ln cho my thu.

My pht phi s dng iu ch chnh xc bo v cc thng tin c

pht i, khng b bin dng qu mc. Ngoi ra cc tn s hot ng ca

my pht phi c chn cn c vo knh v vng ph song theo qui nh

ca hip hi thng tin quc t. Cc tn s trung tm ca my pht phi c

n nh tn s cao. Do ch tiu ca my pht l: cng sut ra, tn s lm

vic , n nh tn s, di tn s iu ch.

C nhiu cch phn loi my pht.

Theo cng dng:

Theo tn s:

- Pht thanh:

+ 3 30 KHz (100 Km 10 Km) : i pht sng cc di: VLW

+ 30 300 Khz (10 Km 1 Km) : i pht sng di: LW

+ 300 3000 Khz (1Km 100m) : i pht sng trung : MW

+ 3 30 Mhz (100m 10m) : i pht sng ngn: SW

- Pht hnh:

+ 30 300 MHz (10m 1m): i pht sng met

+ 300 3000 MHz (1m 0.1m) : i pht sng dm


SVTH: Page 4

- Thng tin vi-ba, ra-da :

+ 3 30 Ghz (0.1m 0.01 m) : i pht sng cm

+ 30 300 Ghz (0.01m 0.001m) : i pht sng mm

Theo iu ch:

+ My pht iu bin AM

+ My pht n bin SSB

+ My pht iu tn (FM) v my pht iu tn m thanh ni (FM stereo)

Ngy nay my pht a s c nghin cu ng dng tt c vo cc loi

my pht thng tin s, pht thanh, pht hnh, v.v.

Theo cng sut:

+ My pht cng sut nh P ra < 100

+ My pht cng sut trung bnh 100 W P ra 10 KW

+ My pht cng sut ln 10 KW P ra


SVTH: Page 5

Hnh 1.3: s khi tng qut ca my thu pht FM

1.3.1. Nhim v cc khi.

Khi tin khuych i m tn:

Tng ny c tc dng khuych i tn hiu vo n mc cn thit a

vo tng in khng v iu ch FM.

Khi ch sng:

C nhim v to ra dao ng cao tn( sng mang) c bin v tn s n

nh, c tm bin i tn s rng. Mun vy ta c th dng mch dao ng

LC kt hp vi mch t ng iu chnh tn s AFC.

Tng in khng v iu ch FM:

Tng in khng s dng cc phn t in khng bin i tn hiu m

tn thnh tn s thay i(dung khng hoc cm khng bin i) thc

hin vic iu ch FM. Phn t in khng c th l transistor in khng,

n in khng hoc varicap.

Mch iu ch FM: mch iu ch tn s th sng mang c bin khng

i, nhng tn s thay i theo bin ca tn hiu m tn.


SVTH: Page 6

Tng nhn tng v khuych i cng sut cao tng:

C nhim v to ra cng sut cn thit theo yu cu cng sut ra ca my

pht. Cng sut ra cng ln th s tng khuych i trong khi ny cng

nhiu.

Mch ra:

phi hp tr khng gia tng khuych i cng sut v anten c cng

sut pht ln nht.

Anten:

bc x nng lng cao tn ca my pht thnh sng in t truyn i

trong khng gian.

Ngun DC:

Ngun cung cp phi c cng sut ln cho transistor hoc n in t

cng sut.

1.3.2.Nguyn l hot ng :

Tn hiu c a vo tng in khng, tng in khng s dng cc

phn t in khng bin i tn hiu m tn thnh in khng thay i ( dung

khng hoc cm khng bin thin ) thc hin vic iu ch FM . Phn t in

khng , n in khng hoc Varicap ( in dung bin i theo in p t vo

Varicap ) .

Sau tn hiu c trn vi sng mang c to ra t khi ch sng

iu ch FM . Tn hiu FM sau khi iu ch c nhn tn to ra tn s cao .

Cui cng tn hiu sau khi nhn tn c khuch i nh mch khuch i cng

sut cao tn ri pht ra anten .

1.3.3. Dng sng u ra cc khi:

- Tn hiu hnh sin miu t tin tc cn gi i c khuch i qua khi tin

khuch i.


SVTH: Page 7

- Tn hiu cao tn sau khi c iu tn thng qua tn in khng.

- Tn hiu ra sau khi nhn tn tn s c nhn gp vi ln nhng vn

gi nguyn c quy lut bin tn.

- Tn s dao ng chun TA bng tn s f0 so snh vi tn s f0 ca

dao ng ch sng.


SVTH: Page 8

1.4. Cng sut pht

L cng sut dng in cao tn a n AT ca my pht, l yu t

quan trng quyt nh c ly thng tin c hiu qu.

Hiu sut l t s gia cng sut ra AT trn ton b cng sut cung cp cho

my.

1.5. Kt lun chng.

Chng 1 l chng gii thiu v my pht FM c nhng u im ni bt

m my pht AM khng th c c v t mc ch yu cu nhim v nn em

chn n ny. S khi my pht FM, nguyn l lm vic, dng sng u ra

ca cc khi quan trng. Vy chng ny em khi qut my pht FM m

n ang thi cng.


SVTH: Page 9

CHNG II: MY THU FM

2.1. nh ngha my thu:

My thu l thit b u cui trong h thng thng tin v tuyn in.

My thu c nhim v tip nhn v lp li tin tc cha trong tn hiu chuyn

i t my pht di dng sng in t trng. My thu phi loi b c

cc loi nhiu khng mong mun, khuych i tn hiu sau gii iu ch

n nhn c tn hi ban u. My thu c rt nhiu tham s nhng

chng ta ch yu ch xt n cc ch tiu k thut c bn sau:

- nhy: biu th kh nng thu tn hiu yu ca my thu, c xc

nh bng sc in ng cm ng ti thiu, hoc cng sut ti thiu

ca tn hiu ra ca tn hiu ti anten m bo cho m thu lm vic

bnh thng.

- chn lc: l kh nng chn p cc dng nhiu khng mong mun,

khng phi l tn hiu cn thu. Ngha l chn lc l kh nng la

chn tn hiu ra khi cc loi nhiu tn t ti u my thu.

- Cht lng lp li tin tc: c nh gi bng mo ca tn hiu ,

ch yu l mo tng khuych i cng sut m tn cho tn hiu

ra loa khng b bin dng so vi tn hiu a vo b iu ch ca my

pht.

2.2.My thu i tn FM:

Hnh 2.1: s khi tng qut ca my thu i tn FM


SVTH: Page 10

Tn hiu cao tn t anten vo mch v c khuych i nh mch

khuych i cng sut . Sau tn hiu c a n b trn tn trn

vi tn hiu hnh sin ca dao ng ni. Tn hiu c KDTT v c a

vo b tch sng ti to li tn hiu m tn. Sau tn hiu m tn a

vo tng khuych i m tn a ra loa.

trnh hin tng iu bin nn gy mo tn hiu sau tch sng, ta

t b hn ch bin ngay trc b tch sng tn s hoc s dng b tch

sng t s v n c mch hn bin. i vi my thu FM, n nh tn s

yu cu rt cao nn bt buc phi c mch AFC.


SVTH: Page 11

CHNG III

CC MCH THNG DNG TRONG MY PHT FM

3.1.Mch to dao ng .

3.1.1. Cc vn chung v mch to dao ng .

Cc tham s c bn ca mch to dao ng :

+Tn s dao ng

+Bin in p ra

+ n nh tn s dao ng

+Cng sut ra

Nguyn tc c bn to mch dao ng iu ha :

+To dao ng bng hi tip dng

+To dao ng bng phng php tng hp mch

3.1.2. Mch to dao ng bng hi tip dng

3.1.2.1. iu kin mch to dao ng

Hnh 3.1 S khi tng qut ca mch to dao ng

(A): Khi khch i c h s khuch i l : K=K.

(B): Khi hi tip c h s truyn i l : =.


SVTH: Page 12

Ta c : =.

M : = K.

Suy ra : =K. .

Mch ch dao ng khi = ngha l lc ta c th ni im a va a v tn

hiu ly ra t mch hi tip c a tr li u vo ( mch khng c tn hiu

vo m vn c tn hiu ra )

Vy iu kin ca mch dao ng l : = K. =1

. . (+)=1

. .[ cos( + ) + j.sin( + ) ] =1 (*)

K : l h s khch i

: l h s hi tip m

: gc pha ca b khuch i

: gc pha ca mch hi tip

T (*) ta suy ra . =1

= + =2n vi n = 0, 1, 2, 3

Suy ra mun to dao ng phi tha mn 2 iu kin sau :

+ iu kin cn bng bin : Mch ch c th dao ng khi h s

khuch i ca b khuch i b c tn hao do mch hi tip gy ra .

+ iu kin cn bng pha : Mch ch c th dao ng khi tn hiu hi tip

v ng pha vi tn hiu vo .

3.1.2.2 . c im ca mch dao ng

- Mch dao ng l mt mch khuch i , nhng l mch khuch i t iu

chnh bng hi tip dng t u ra v u vo . Nng lng t dao ng ly t

ngun cung cp mt chiu .

- Mch phi tha mn iu kin cn bng v bin v pha .


SVTH: Page 13

- Mch phi cha t nht 1 phn t tch cc lm nhim v bin i nng lng

mt chiu thnh xoay chiu.

- Mch phi t nht 1 phn t phi tuyn hay mt khu iu chnh m bo

cho bin dao ng khng i trng thi xc lp .

3.1.2.3. Mch dao ng ba im in dung .

Phng php tnh ton mch dao ng ba im in dung : C nhiu phng php, nhng y ta xt phng php thng dng nht, l tnh ton mch dao ng theo phng php b khuch i c hi tip

a. Mch EC

Hnh 3.2 : S mch dao ng ba im in dung dng Transitor(mch EC)

Bc 1 : Tnh h s khuch i K

K= - . -

= -

21

11 .

Vi 21=

; 11= = 1//2//

Suy ra K = -S. = - 21

11 . ; S : h dn ca BJT

+ : Tr khng gia collector v t , nhng do C c ni vi khung

cng hng nn n l phn t tr khng ca khung cng hng .


SVTH: Page 14

Ta c cng thc sau : = (2. ) // (1)

: Tr khng vo phn nh sang nhnh collector-emitor

: l tr khng ca khung cng hng ti tn s cng hng.

=

.

L : in cm ca khung cng hng C : in dung ca khung cng hng r : in tr tn hao ca khung cng hng

Nu 1//2 11 ta c :

=

2 =

11

2

H s phn nh n :

n= -

= -

=

( 0< 1 )

Suy ra : n= 2

1 =

1

2 =

1

2 1 1 2

+ Tnh P : h s ghp ca Transistor vi khung cng hng :

P=

=

=

1

( 1+2) =

1

11

=

1 =

12

1(1+2) =

2

1+2 =

112

+1 =

1

+1

Thay vo (1) ta c :

= 2..

(2.)+

= = (

1

+1)2..

112

(1

+1)2.+

112

= .11

2.+11.(+1)2

Suy ra :

K = - 21

11.

.11

2.+11.(+1)2

Bc 2 : Xc nh h s hi tip


SVTH: Page 15

=

=

= -

2 :

1 = -

1

2 = -n

Bc 3 : Tnh tch K.Kht :

. = n..11

2.+11.(1+)2

Bc 4 : Xc nh iu kin dao ng ca mch :

. 1

(1 + )2 + 2.

11 -

21

11. . 0

Du = ng vi trng hp dao ng xc lp. Du < ng vi trng hp qu lc ng mch.

Bc 5 : Xc nh h s hi tip cn thit mch t dao ng c.

Thng n


SVTH: Page 16

= = 1

2..1.2

1+2

(4)

T (3),(4) ta tm c L , 1, 2

b. Mch BC:

Hnh 3.3: S mch dao ng ba im in dung dng

Transitor(mch BC)

3.2. Mch khuych i thut ton.

u im ca mch khuch i thut ton (OPAMP):

+Ngun in p cung cp thng nh 518V


SVTH: Page 17

+H s khuch i rt ln vo khong 103105 khi hot ng trong mng

+Tng tr vo ln nn d dng phi hp vi cc mch in v linh kin

ngoi vi

+Tng tr ra nh t 0,116

+ tri nhit nh 2mV/1oC.

tri ny nm trong phm vi cho php nn khng cn phi hiu chnh

dch im zero

+in p tn hiu ng ra gn bng gi tr Vcc nn hiu sut cao .

V nhng l do trn nn khi thit k ta chn Opamp khch i tn hiu ng vo .

Mch khuych i dng opamp c rt nhiu loi nhng y ta ch xt n

mch khuych i opam c tr khng vo ln.

Hnh 3.4: mch tr c tr khng vo ln.

Vit phng trnh dng in cho nt N1 v N2 ta c:

1Vin Vn

R

-

/

Vn

R n+

Vout Vn

KR

=0 ( m Vn= Vin2 )


SVTH: Page 18

Mch ny thng c dng ti tng khuych i u tin ca my pht.

S d n c dng v n c tr khng vo ln. Do n hn ch c nhiu

bin ln.

3.3 Mch khuych i dng BJT.

3.3.1. Cc ch hot ng ca BJT.

- Ch A: gc ct =1800

-Ch B: gc ct =900

-Ch C: gc ct


SVTH: Page 19

Av = Vout

V i

Ta c: Vo = -ibRc

Vi = reib + (1+re)RE ib

Av = -Vout

V i=-

(1 ) R

Rc

re E

Du - cho thy vo

v vi ngc pha.

* Tng tr vo:

Zi=Vi

Ii=Rb//Zb

* Tng tr ra:

Zout = Vout

Iout = Rc


SVTH: Page 20

PHN II: TNH TON

S MCH

4.1. Tng khuch i

R2 100k

R1 100k

R3 10k

R4 1.5k

R8 100k

R6 68k

R7 180

R5 10K

C1 1uF

C9 33p

C3 10u

C4 100n

C2 5n

C5

C

0J1

12

0

L1

1

2

L2

1

2

Q1 2SC1815Q2 2SC1815

C6 1n

C7 12p C10 10p

C11 33p

C12 100n

C13 100n

0

0

J3

12

J2

12

0

C8 5p

U3

OP-TL082

+3

-2

V+

7V

-4

OUT6

OS28

OS11


SVTH: Page 21

Tng ny c tc dng khuch i tn hiu sau khi c iu ch tng dao ng

My pht FM tn s pht l 92.5 Mhz, cng sut l 100 mW, ngun cung cp

9Vdc

4.1.1 .Tnh chn L2, C10

L2 C10 dao ng cng hng c tc dng chn lc tn s khuch i.

Tn s cng hng ca L2 C10 ng bng tn s ca mch iu ch

Tn s dao ng:

f=1

2210= 92.5Mhz

L2*C10= 1.5*10-18

Chn C10= 10pF, suy ra: L2=1.51018

101012 = 149*10-9 H =149 nH

Vy C10= 10pF, L2= 149 nH

4.1.2 .Tnh chn BJT Q2

tn hiu khng mo dng, ta chn BJT Q2 hot ng ch A.

S tng ng ca Q2

C9

R8

Q3

Vout

L2

Vin

Vcc

C10


SVTH: Page 22

Tr khng ra ca Anten l Ranten= 75

Tr khng ti Rt= XL3//XC10//Ranten

Xtd= XL2//XC10= 3

1

10

3+1

10

= 3

1310 =

292.5106149109

1(92.51062)2149109101012

Xtd= 174.1j

Rt= Xtd//Ranten=

+ =

174.175

174.1+75= 52.4

Gi VLP v ILP l in p nh

PL=Rt*I2Lhd= 1

2*

2

VLP= 2 = 2 0.1 52.4 = 3.23 V

ILP =

=

3.23

52.4 = 0.0616 A = 61.6 mA

Cng sut cung cp cho Q2 l:

Pcc= Vcc*Itb = Vcc*

= 9*

61.6

= 176,56 mW

Cng sut xoay chiu trn ti:

PAC= Rt*I2LPhd = Rt*2

2= 52.4*

0.06162

2 = 99.4mW

Cng sut tiu tn trn Q2:

Ptt=Pcc- PAC = 176.56 - 99.4= 77.16 mW

hf e*Ib

C10Ranten

hre.ib

L3

hie

Cce1/hoeCbe

R8

B CVin

E


SVTH: Page 23

Chn BJT Q2 l 2SC1815 c =100, f= 100 MHz, Icmax=150 mA,Pc=

400 mW tha mn:

{ > 2 >

> 2

4.1.3 .Tnh R8

Chn dng chy qua Q2 l IcQ2= 8.5 mA

IBQ2 = 2

=

8.5

100 = 0.085 mA

R8= 2

2=

90.6

0.085103 = 98.8*103 = 98.8 K

Chn R8= 100 K

4.2 .Tng dao ng

Tng dao ng c tc dng iu ch tn

hiu m tn to thnh tn hiu FM c

tn s t 88MHz n 108MHz.

Mch dao ng c thc hin theo

ch 3 im in dung.

R6

R7

Q1

C6

C7

C8

Vin

Vout

L1

Vcc


SVTH: Page 24

4.2.1 .Tnh chn Q1

Chn ch hot ng ca Q1 l ch A trnh mo tn hiu. Do vy, chn

VCE=80%*Vcc=0.8*9=7.2 V

Chn dng chy qua Q1 l Ic=10mA

Chn BJT Q1 l 2SC9018 c =100, tha:

{ = 400 > 2 = 2 = 2 10 7.2 = 144

= 150 > 2 = 20 = 15 > = 9

4.2.2. Tnh chn L1, C7, C8.

Ltd,Cbe C8 to thnh mch dao ng 3 im in dung

Gi Ltd= L1// C7 => jLtd = 1

1

7

1+ 1

7

Ltd= 1

1217 (*)

S mch dao ng 3 im in dung

Tn s dao ng ca mch l

fd = 1

28

8+

(**)

Khi in p vo BJT thay i th t Cbe thay i, do

tn s ca mch cng thay i => iu ch FM

Chn C8=5pF, Cbe= 1.3pF

T (**) => 92.5*106 = 1

2510121.31012

51012+1.31012

=> Ltd = 299*10-9 H

T (*) => 299*10-9= 1

1(292.5106)217

Chn C7= 10pF => L1=149 nH

Q1C8

Ltd

Cbe


SVTH: Page 25

4.2.3 .Tnh R6 R7 C6

VR7=Vcc-VCE = 9 7.2 = 1.8V

R7= 7

=

1.8

10= 180

Chn R7= 180

VR6= Vcc VBE VR7

R6*

= Vcc VBE VR7

R6*10

100 = 9 0.6 1.5

R6= 69 K

Chn R6 = 68 K

T C6 c tc dng lm cho transistor hot ng ng ch ti tn s cao tn.

Chn gi tr t C6= 103 ( 10nF) s hn ch c p ng cao tn n 15KHz

4.3 .Tng tin khuch i

Tng tin khuch i c tc dng khuch i bin tn hiu m tn ln

a vo tng dao ng.

Ta chn Opamp thay v BJT v

tr khng u vo ca Opamp l

v cng v tr khng ra thp,

li ln.

y ta chn Opamp TL082

4.3.1 .Chn R3 R4

Xt mch khuch i nh hnh.

ch mt chiu th ln

ca tng ny l bng 1 v Vout=

Vin

ch xoay chiu, ta mun c

li l Av=8 ln.

R1

-

+

U2B

TL084

5

67

411

R2

R3

R4

C2

Vin

Vcc

Vout


SVTH: Page 26

Av= 1 +3

4 = 8

R3= 7*R4

Chn R4= 1.5 K

R3 = 7*1.5 = 10.5 K

Chn R3 = 10 K

4.3.2 .Chn R1 R2

in tr R1, R2 to thnh cu phn p cho Opamp. Ta chn R1 v R2 khng

c nh v nh th s nh hng n tr khng u vo ca opamp.

Chn R1=R2= 100 K

4.4. Tnh chn cc thng s khc

4.4.1 .Tnh t C1

T C1 c tc dng lin lc, cch ly tn hiu mt chiu gia tn hiu u vo v

tng tin khuch i.

Ta chn gi tr C1 sao cho dung khng ca t nh hn tr khng u vo ca

Opamp.

Chn XC1= 1

10 Zinopamp=

1

10*

12

1+2

Tn hiu u vo c bin khong 775mV, tn s t 20Hz 20KHz. Chn tn

s ca tn hiu m thanh l 100 Hz.

XC1 = 1

21 =

1

21001 =

100103100103

10(100103+100103)

C1 = 0.3F

Chn C1= 1F

4.4.2 .Tnh t C2

T C2 c tc dng lm cho tng tin khuch i c li bng 1 khi c tn hiu

mt chiu v bng Av khi c tn hiu xoay chiu, vy th t C2 c gi tr nh

khi c tn hiu xoay chiu th t c xem nh ngn mch.

Chn C2= 5nF


SVTH: Page 27

4.4.3 .Tnh t C3 C4

T C3 C4 l cp t c tc dng lm n nh ngun, trnh nhiu tc ng vo.

Chn t C3 c gi tr ln v khi c tn hiu vi tn s thp tc ng vo ngun

th s b t a xung mass. Chn C3= 10F.

Chn t C4 c gi tr thp v khi c tn hiu tn s cao tc ng vo ngun th s

b t a xung mass. Chn C4= 100nF.

4.4.4 .Tnh C5 R5

T C5 l t lin lc, R5 l tr cch ly, hn dng gia tng tin khuch i v

tng dao ng.

V tr R5 cch ly v hn dng nn gi tr in tr ca R5 cn phi ln, ng

thi R5 ln tng tr khng u vo ca R5.

Chn R5= 10K

V khi ch xoay chiu th t C5 xem nh ngn mch nn tr khng u vo

ca mch chnh bng in tr vo ca Q1

Chn XC5 = ZinQ1 = R5//R6 // (re+R7)

1

21005 =68000 // 10000 //100(2.6 + 180) =5900

C5=270 nF

Chn C5=330nF

4.4.5. Tnh C9

C9 l t lin lc gia tng dao ng v tng khuch i. Chn gi tr ca t nh

hn gi tr in tr u vo ca tng khuch i.

ZinQ2= R8//rbe

M rbe=re = *25

= 100*

25

8.5 = 294

ZinQ2= 100000294

100000+294 = 293

Chn XC9 = 1

10ZinQ2 = 29.3


SVTH: Page 28

C9 = 29.3

290106 = 5.2*10-8 F =52 nF

Chn C9= 33nF

4.4.6 .Tnh C12

T C12 l t thot cao tn, trnh a tn s cao v ngun lm nhiu ngun.

Chn gi tr ca t nh. Chn C12= 100nF

do an dien tu thong tin

Documents