Giao Trinh NLDT2_ThangVC

8/14/2019 Giao Trinh NLDT2_ThangVC

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MC LC

C LC .................................................................................................................................1CHNG 1: CC MCH TO QUAN H HM S DNG KHUCH I THUT TON41.1. Khi nim chung..................................................................................................................4

1.2. Cc mch tnh ton v u khin tuyn tnh ........................................................................41.2.1. Mch cng o..............................................................................................................41.2.2. Mch tr .......................................................................................................................51.2.3. Mch tch phn o.......................................................................................................61.2.4. Mch tch phn tng......................................................................................................71.2.5. Mch tch phn hiu......................................................................................................71.2.6. Mch vi phn ................................................................................................................8

1.3. Cc mch khuch i v tnh ton phi tuyn lin tc ............................................................91.3.1. To quan h hm s c dng c tuyn ca phn t phi tuyn. .....................................91.3.2. To quan h hm i ca dng c tuyn volt-ampe ca phn t phi tuyn.................. 101.3.3. Mch khuch i Loga................................................................................................101.3.4. Mch khuch i i Loga..........................................................................................121.3.5. Mch nhn dng nguyn tc khuch i Loga v i Loga..........................................121.3.6. Mch ly tha bc hai.................................................................................................131.3.7. Mch chia theo nguyn tc nhn o...........................................................................141.3.8. Mch chia dng nguyn tc khuch i Loga v i Loga:..........................................151.3.9. Mch khai cn............................................................................................................. 15

1.4. Cc mch phi tuyn khng lin tc.....................................................................................171.4.1. Mch to hm chnh lu hn ch:................................................................................171.4.2. Mch so snh tng t ................................................................................................ 22

Chng 2: U CH.............................................................................................................. 312.1. nh ngha ......................................................................................................................... 312.2. u bin ...........................................................................................................................31

2.2.1. nh ngha .................................................................................................................. 312.2.2. u bin thng thng..............................................................................................312.2.3. Cc ch tiu cbn ca dao ng u bin............................................................ 342.2.4 Phng php tnh ton mch u bin : ......................................................................362.2.5. Cc mch u bin c th :........................................................................................402.2.6. u chn bin ....................................................................................................... 45

2.3. Gii u bin .................................................................................................................... 532.3.1. Gii u bin vi tn hiu u bin thng thng ...................................................... 532.3.2 Gii u chn bin: ...............................................................................................55

2.4. u tn v u pha .......................................................................................................... 562.5. Gii u tn. ..................................................................................................................... 60

2.5.1. Nguyn l chung.........................................................................................................60

2.5.2. Cc mch gii ch tn s ............................................................................................. 60Chng 3: I TN................................................................................................................633.1. nh ngha ......................................................................................................................... 63

3.1.1. t vn :..................................................................................................................633.1.2. Khi nim: .................................................................................................................. 64

3.2. Nguyn l i tn: .............................................................................................................653.2.1. S khi v nguyn l lm vic:...............................................................................653.2.2. Nhiu sinh ra ca b trn tn v cch khc phc.......................................................... 673.2.3 Thu i tn trong 1 di tn rng ................................................................................... 703.2.4 i tn thc hin i bng tn s............................................................................70


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3.3. Mch i tn dng Diode..................................................................................................713.3.1. Mch i tn n ........................................................................................................ 713.3.2. Mch i tn cn bng ................................................................................................714.4.3. Mch i tn vng ...................................................................................................... 723.3.2 Mch i tn dng Transistor....................................................................................... 73

CHNG 4: CHUYN I AD V DA ................................................................................ 764.1. Csl thuyt...................................................................................................................76

4.2. Cc thng s cbn...........................................................................................................784.3. Nguyn tc lm vic ca ADC...........................................................................................794.4. Cc phng php chuyn i tng t - s ........................................................................ 80

4.4.1 Phn loi ...................................................................................................................... 804.4.2. Chuyn i AD theo phng php song song..............................................................814.4.3. Chuyn i AD ni tip ..............................................................................................824.4.4. Chuyn i AD theo phng php kt hp..................................................................834.4.5. Chuyn i AD ni tip dng vng hi tip.................................................................844.4.6. Chuyn i AD theo phng php tich phn n gin:................................................ 854.4.7. Chuyn i AD theo phng php tch phn hai sn dc ..........................................88

4.5. Cc phng php chuyn i s - tng t ........................................................................924.5.1. Chuyn i DA bng phng php n trbc thang.................................................93

4.5.2. Phng php mng n tr: .......................................................................................944.5.3. Phng php Shanon Rack:...................................................................................... 97

Chng 5: NGUN CUNG CP ........................................................................................... 1015.1. Khi nim chung..............................................................................................................101

5.1.1. t vn ................................................................................................................. 1015.1.2. Thng s ngun cung cp.......................................................................................... 101

5.2. Ngun chnh lu n gin................................................................................................ 1035.2.1. S khi.................................................................................................................1035.2.2. Bin p...................................................................................................................... 1035.2.3. Chnh lu.................................................................................................................. 1045.2.4. Lc ........................................................................................................................... 1055.2.5. Cc b chnh lu bi p............................................................................................. 106

5.3. n p:.............................................................................................................................. 1075.3.1. t vn :................................................................................................................ 1075.3.2. n p thng s: ......................................................................................................... 1075.3.3. n p so snh:........................................................................................................... 108

5.4. n p xung: ..................................................................................................................... 1135.5. Cc vi mch n p: ..........................................................................................................114TI LIU THAM KHO ...................................................................................................... 118


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CHNG 1: CC MCH TO QUAN H HM S DNG KHUCH ITHUT TON

Chng ny nhm gii thiu vic ng dng mch khuch i thut ton (KTT)

trong cc mch to quan h hm s. Kho st cc mch khuch i loga v i loga, khai

n, bnh phng, mch nhn, mch chia, mch chnh lu hn ch, mch to hm so snh

1.1. Khi nim chungHin nay, cc b khuch i thut ton (KTT) ng vai tr quan trng v c ng

dng rng ri trong k thut khuch i, tnh ton, u khin, to hm, to tn hiu hnh

sin v xung, s dng trong n p v cc b lc tch cc... Trong k thut mch tng t,

cc mch tnh ton v iu khinc xy dng ch yu da trn b TT. Khi thay

i cc linh kin mc trong mch hi tip ta s c c cc mch tnh ton v iu khin

khc nhau.

C 2 dng mch tnh ton v iu khin: Tuyn tnh v phi tuyn.

Tuyn tnh: c trong mch hi tip cc linh kin c hm truyn t tuyn tnh.

Phi tuyn: c trong mch hi tip cc linh kin c hm truyn phi tuyn tnh.

V mt k thut, to hm phi tuyn c th da vo mt trong cc nguyn tc sau y :

1. Quan h phi tuyn Volt - Ampe ca mt ghp pn ca diode hoc BJT khi phn cc

thun (mch khuch i loga)

2. Quan h phi tuyn gia dc ca c tuyn BJT lng cc v dng Emit (mchnhn tng t).

3. Lm gn ng c tuyn phi tuyn bng nhng n thng gp khc (cc mch to

hm dng diode).

4. Thay i cc tnh ca n p t vo phn t tch cc lm cho dng in ra thay i

(kho diode, kho transistor).

1.2. Cc mch tnh ton v iu khin tuyn tnh

1.2.1. Mch cng o


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Hnh 1.1. S mch cng o

p dng quy tc dng in nt cho N ta c:

1.2.2. Mch tr

Hnh 1.2. S mch tr

R1 = RN/a

R2 = RP/a

in p ca vo thun:

in p ca vo o:

V:

Nu RN = RP:


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1.2.3. Mch tch phn o

Hnh 1.3.a. S mch tch phn o

Phng trnh dng in nt ti N:

i1 + iC = 0

Hay:

Suy ra:

=> in p ra t l vi tch phn n p vo.

Thng chn hng s thi gian = RC = 1s. Vout(t=0) l iu kin u, khng ph thuc

vo in p vo Vin1.

Nu Vin1 l in p xoay chiu hnh sin: vin1 = Vin1 sint th:

=> Bin n p ra t l nghch vi tn s. c tuyn bin - tn s ca mch tch

phn f() c dc -20dB/decade:

Mch c gi l mch tch phn trong mt phm vi tn s no nu trong phm vi tn

s c tuyn bin - tn ca n gim vi dc 20dB/decade. gim nh hng ca

dng tnh It v in p lch khng c th gy sai sng k cho mch tch phn, ca

thun ca b KTT ngi ta mc thm mt n tr thay i c R1 v ni xung

masse.


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Hnh 1.3.b Mch tch phn o c bin trR1 b dng lch khng

iu chnh R1 sao cho R1 R th gim c tc dng ca dng in lch khng I0 = IP

IN v in p lch khng V0 = VP VN (khi Vout = 0)

1.2.4. Mch tch phn tng

Hnh 1.4. S mch tch phn tngDng phng php xp chng v vit phng trnh dng in nt i vi nt N ta tm

c:

1.2.5. Mch tch phn hiu

Hnh 1.5. Mch tch phn hiu


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Vit phng trnh i vi nt N:

i vi nt P:

Bin i v cho VN = VP, R1CN = R2CP = RC

Suy ra:

1.2.6. Mch vi phn

Hnh 1.6. S mch vi phn

Ta c:

Gi thit: vin1 = Vin1 sint


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H s khuch i ca mach:

K tng theo tn s v th bode c dc 20dB/decade.

Vy : Mch c gi l mch vi phn trong mt phm vi tn s no nu trong phm vi

tn s c tuyn bin - tn ca n tng vi dc 20dB/decade.

1.3. Cc mch khuch i v tnh ton phi tuyn lin tc

1.3.1. To quan h hm s c dng c tuyn ca phn tphi tuyn.Gi s ta c mt phn t tuyn tnh trong n c c tuyn:

( ) I f v=Dng in qua diode v in p t ln diode c quan h:

D0T

vexp vDi I =

Hnh 1.7. Mch to quan h hm s c dng c tuyn ca phn t phi tuyn

Ta c: Vh = VP VN

M do KTT l tng nn: VP = VN

VP = 0

VP = VN = 0Lc in trvo n v cng ln, v dng vo KTT v cng b: IN = 0.

M: VN = Vht +Vout = 0

( )

( )

.

.

.

out ht ht ht

out ht

out ht

out in ht

V V I R

V I R

V f v R

V f V R

= = =

=

=


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1.3.2. To quan h hm i ca dng c tuyn volt-ampe ca phn tphi tuyn.Gi s ta c mt phn t tuyn tnh trong n c c tuyn:

( )v f I= Ta c: h P NV V V=

M do KTT l tng nn:

P NV V=

0

0P

P N

V

V V

=

= =Lc in trvo n v cng ln, v dng vo KTT v cng b: 0NI = .Do : ( )ht I I f v= =

( )10 N ht out out V V V V V f I = + = = =

M: in N in

N N

V V VI

R R

= =

1 inout

N

VV f

R

=

1.3.3. Mch khuch i Loga

Hnh 1.8.a S mch khuch i Loga dng Diode

to mch khuch i loga, mc diode hoc BJT mch hi tip ca b KTT. Mch

in dng diode (1.8.a.) c th lm vic tt vi dng in I nm trong khong nA mA.

Dng in qua diode v in p t ln diode c quan h:

Trong :

ID, vD : dng in qua diode v ip p t ln diode.


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1.3.4. Mch khuch i i Loga

Hnh 1.9.a S mch khuch i i Loga dng Diode

V :

Hnh 1.9.b S mch khuch i i Loga dng Transitor

Do VBE = -Vin :

1.3.5. Mch nhn dng nguyn tc khuch i Loga v i Loga

a, S khi

Hnh 1.10. Mch nhn dng nguyn tc khuch i Loga v i Loga


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Cc mch khuch i loga v i loga c th dng mch nh xt mc trn.

Coi mch tng c th dng mt khuch i tng KTT. Mch ch lm vic c vi cc

tn hiu vx, vy > 0 (do tnh cht hm loga). Mch nhn 4200 l mt trong nhng mch tiu

biu c ch to theo nguyn tc ny.

A = K1ln(Vx/K2)

B = K1ln(Vy/K2)

C = A+ B = K1ln(VxVy/K22)

C = ln(VxVy/K22)

zV = K3VxVy/K2

2

b)S nguyn l

Vout

Vin2

Vin1

D3

D2

D1

R6

R5

R4

R3

R2

R1

Hnh 1.11. S nguyn l mch nhn

1.3.6. Mch ly tha bc haiu hai u vo ca mch nhn vi nhau ta s c mch ly tha:

Hnh 1.12. S mch ly tha bc hai

Lc ny vx = vy => vz = K.vx2

Gi sn p vo c dng sin: vx = Vcost

Th in p ra:


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=> C th dng mch ly tha bc hai thc hin nhni tn s.

1.3.7. Mch chia theo nguyn tc nhn oa, Mch chia thun

Hnh 1.13. S mch chia thun

Ta c:

b, Mch chia o

Hnh 1.14. S mch chia o

Phng trnh cn bng dng ti N:


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1.3.8. Mch chia dng nguyn tc khuch i Loga v i Loga:

Hnh 1.15. Mch chia dng nguyn t Loga v i Loga

Hnh 1.18 Mch chia tng t dng nguyn tc khuch i Loga v i Loga

b) S nguyn l:

R7

Vout

Vin2

Vin1

D3

D2

D1

R6

R5

R4

R3

R2

R1

Hnh 1.16. nguyn l mch chia dng nguyn tc loga v i loga

1.3.9. Mch khai cnMch khai cn c thc hin bng cch mc vo mch hi tip ca b KTT mt mch

ly tha.


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Hnh 1.17.a Mch khai cn o

Dng phng php xp chng ta c:

Do:

Hnh 1.17.b Mch khai cn thun

Ta c: Vz = VN

M:

Mch n hnh 1.17.a ch lm vic vi n p vo vZ < 0, cn mch n hnh 1.17.b th

vZ > 0. Trong trng hp ngc li th mch s c hi tip dng lm mch b kt.


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ngn nga ngi ta mc thm diode (mi mch mt diode) u ra ca b TT nh

hnh v.

1.4. Cc mch phi tuyn khng lin tc

1.4.1. Mch to hm chnh lu hn ch:

1.4.1.1. t vn :Xt mch chnh lu dng diode :

VoutVin

R

D

Hnh 1.18. Mch chnh lu dng D

Mch ny c hn ch l nu nh yu cu chnh lu tn hiu vo cmV th mch khng

thc hin c. V vy ta kt hp vi mch KTT c ng c tuyn nh sau:

Hnh 1.19. Chnh lu vi tn hiu nh

Khi ng c tuyn qua gc ta gi l mch hn ch mc 0. Cn khng qua gcta gi l mch hn ch khc mc 0.

1.4.1.2. Chnh lu hn ch mt Diode:


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Hnh 1.20. S mch chnh lu hn ch 1 Diode

- Khi Vin < 0 Va 0 Va >0 th D thng Vout = Va VDthng

Va= K0Vh

M: Vh = VP VN

Va = K0 ( VP VN )

Va = K0 ( Vin Vout )

K0 ( Vin Vout ) = Va + VDthng

0out0 01 1

Dthong

in

VKV V

K K=

+ +

Vi 0K>> v DthongV


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- Mch trn l mch hn ch di. c mch hn ch trn ta o chiu Diode.

1.4.1.3. Mch chnh lu hn ch 2 Diode.

Hnh 1.22. S mch chnh lu hn ch 2 Diode

- Khi Vin > 0 Va0 th D1 tt, D2 thng

Va =K0(VP VN)= - K0VN = VDthong + Vout .

p dng nguyn l xp chng :

2 1

1 2 1 2 N in out

R RV V V

R R R R= +

+ +

2 1

1 2 1 2 0

Dthong out

in out

V VR RV V

R R R R K

+ + =

+ +

1 2

1 2 0 1 2 0

1 Dthongout in

VR RV V

R R K R R K

+ = + +

Vi 0K>> , DthongV


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Hnh 1.23. ng dng ca mch chnh lu hn chmc 0 dng 2 D.

Nhn xt:- ng c tuyn truyn t i qua gc ta nn ta c th thc hin chnh lu

c cc tn hiu nh

- Mch chnh lu nh hnh v trn l chnh lu hn ch trn mc 0. Mun c mch

chnh lu hn ch di mc 0 ta ch vic i chiu 2diode.

- Nu R1 = R2 th bin tn hiu ra = bin tn hiu vo.

- Nu R1 > R2 th bin tn hiu vo > bin tn hiu ra.

- Nu R1 < R2 th bin tn hiu vo < bin tn hiu ra.

1.4.1.4. Mch chnh lu hn ch khc mc 0:

Hnh 1.24. S mch chnh lu hn ch khc mc 0

- K h i Vin > E0 Vh < 0 Va = K0Vh < 0 D1 thng, D2 tt

- K h i Vin < E0 Vh > 0 Va = K0Vh > 0 D1 tt, D2 thng


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(0 0a Dthong out N V V V K E V = + =

2

1 2 N in

RV V

R R=

++ 1

1 2out

RV

R R+

( 2 10 01 2 1 2

Dthong out in out

R RV V K E V V

R R R R

+ = + +

2 10

0 1 2 1 2

Dthong outin out

o

V V R R E V V

K K R R R R+ =

+ +

Vi K0 c4 510 10

1 2 20

1 1

2 20

1 1

1

out in

in

R R RV E V

R R

R RV E

R R

+=

= + +

Hnh 1.25. ng dng ca mch chnh lu hn ch khc mc 0.

1.4.1.5. ng dng ca mch chnh lu hn ch

Mch chnh lu hn ch 1 hoc 2 diode chnh lu hn ch 2 na chu ki vi tn

hiu nh.

Vi mch hn ch 1 diode


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Hnh 1.26. Mch chnh lu vi tn hiu vo nh dng mch hn ch 1D

Vi mch hn ch 2 diode

Hnh 1.27. Mch chnh lu vi tn hiu vo nh dng mch hn ch 2D

1.4.2. Mch so snh tng tMch so snh tng t c nhim v so snh mt in p vo Vin vi mt n p chun

Vch. Tn hiu vo dng tng t sc bin thnh tn hiu ra di dng m nh phn.

Ngha l u ra hoc mc thp (L) hoc mc cao (H). N l mch ghp ni gia

ANALOG v DIGITAL.

c m: Phn bit gia b KTT thng thng vi b so snh chuyn dng (m thc

cht cng l mt b KTT).

- B so snh c tc p ng cao hn thi gian xc lp v phc hi nh.

- L KTT lm vic trng thi bo ha nn mc ra thp (L) v mc ra cao (H) ca n

l mc dng v mc m ca ngun. Cc mc ny phi tng ng vi mc logic.

Hnh 1.28. Mch so snh v c tuyn vo ra


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1.4.2.1. B so snh khng c tr

a) Vin, E0a vo 2 ca khc nhau ca KTT

Vina vo ca thun:

(a) (b)Hnh 1.29. Vin a vo ca thun, Eo a vo ca o ca KTT

- Khi Vin < E0 Vh < 0 Vout = VrL

- Khi Vin > E0 Vh > 0

Vout = VrH

Vina vo ca o:

Hnh 1.30. Vin a vo ca o, Eo a vo ca thun ca KTT

- Khi Vin < E0 Vh < 0

Vout = VrH

- Khi Vin > E0 Vh > 0

Vout = VrL

b) Vin, E0a vo cng 1 ca ca KTT:

Vina vo ca thun:


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Eo

R2Vout

Vin

R1

Hnh 1.31. Vin, E0 cnga vo ca thun ca KTT

- Ti ca o

NV = 0- Ti ca thun

20

1 2

P

RV E

R R=

+

+1

1 2

in

RV

R R

+Vi KTT l tng

Ta cn bng: in SSV V=

2

01 2

RE

R R+1

1 2SS

RV

R R+

+ =0

SSV =0 2

1

0E R

R Vss Vh > 0 Vout = VrH- Khi Vin < Vss Vh < 0 Vout = VrL

- Khi Vin = Vss Vh = 0 (chuyn trng thi)

Hnh 1.32. ng c tuyn so snh

Vina vo ca o:


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Hnh 1.35. Vin v Eo a vo 2 ca ca KTT

- Ti ca o

VN = E02- Ti ca thun

201

1 2

R

RPV E

R=

+1

1 2in

RV

R R+

+Vi KTT l tng

VP = VN

Ta cn bng: VN = VP, vi Vin = Vss

2

02 011 2

RE E

R R=

+1

1 2SS

RV

R R+

+

Vss =1

201

1

202 )1(

R

RE

R

RE +

- Khi Vin > Vss Vh > 0 Vout = VrH

- Khi Vin < Vss Vh < 0 Vout = VrL- Khi Vin = Vss Vh = 0 (chuyn trng thi)

Nhn xt:

- K h i E02 = 0

201

1SS

RV E

R=

=> Chuyn v trng hp b

- K h i R1 = Vss = E01

=> Chuyn v trng hp a

1.4.2.2. Mch so snh c tr:

to mch so snh c tr ta dng hi tip dng

a) Vinc a vo ca thun:


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Hnh 1.36. Vin a vo ca thun- Ti ca o

VN = E0

- Ti ca thun

2 1

1 2 1 2P in out

R RV V V

R R R R= +

+ +Gi s ban u Vin nh lm cho 0P N hV V V< < Vout = VrL

Khi Vin ng dn VP tng dn

Khi Vin = VSS1 VP = VN Vh = 0 (chuyn trng thi)

2 10 1

1 2 1 2SS rL

R R E V V

R R R R= +

+ +

1 1

1 02 2

1SS rL

R RV E V

R R

= +

Hnh 1.37. c tuyn so snh

Khi Vin tip tc tng Vout = VrH

Gi s Vin ln VP > VN Vh > 0 Vout = VrH

Khi Vin gim dn VP gim dn, cho n khi Vin = VSS2

VP = VN (chuyn trng thi )

2 10 2

1 2 1 2SS rH

R R E V V

R R R R= + =

+ +


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1 12 0

2 2

1SS rH

R RV E V

R R

= +

rH rLV V=

Khi Vin tip tc gim Vh < 0 Vout = VrL

11 2

2

2tre SS SS rH R

V V V V R

= =

Nhn xt: Nu 1R = th mt hi tip dng Vtre = 0

b) Vinc a vo ca o:

R2

R1

Vout

Vin

Hnh 1.38. Mch so snh c tr khi Vin a vo ca o.

Ti ca thun

1

1 2P out

RV V

R R=

+Ti ca o

N inV V=

Gi s Vin rt nh sao cho:

VN < VP (Vh > 0) => Vout = VrH

Khi Vin tng dn 1in SSV V= sao cho ( 0)P N hV V V= = chuyn trng thi ( rH rLV V

Ta cn bng P NV V= sao cho 1,in SS out rH V V V V = =

1

1 1 2SS rH

RV V

R R =

+Gi s inV ln ( )N PV V>

0h P N

out rL

V V V

V V

= < =

Khi Vin gim 2in SSV V = sao cho 0P N hV V V= = chuyn trng thi (VrH VrL)

Ta cn bng P NV V= sao cho 2 ,in SS out rLV V V V = =


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1 12 SS1

1 2 1 2

0SS rL rH R R

V V V V R R R R

= = = > S


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= Vt costt +1

2Vs cos (t-s) t +

1

2Vs cos (ts) t

Hnh 2.2. Ph tn hiu u bin

Ph ca tn hiu u bin c dng nh hnh 3.2.

Trong trng hp u bin thng thng hay cn gi l iu ch hai bin tn th

ngi ta s pht i c thnh phn bin tn trn, bin tn di v ti tin.

b, Tn hiu iu ch l tn hiu phi u ha.

Khi tn hiu u ch c ph bin thin t Smin Smax th ph ca tn hiu u bin

c dng nh hnh 3.3

Hnh 2.3. Ph tn hiu u bin

c, Xt nng lng cc thnh phn ca tn hiu u bin thng thng.

Cng sut ra trn ti:2V

PR

=

Xt vi R=1 th P = 2V

Cng sut ti tin l cng sut bnh qun trong 1 chu k ca ti tin:2 2

22mt mt

U UP

= =

Cng sut vi mt bin tn:


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22 22 2

1

1

2

2

db

.2 2 42 2 2 2

2

2

12

ms ms mt mt bt t

mt

t bt

t t

t

U U U U m mP P

U

P P P

mP P P

mP P

= = = =

= +

= +

= +

Xt vi: m = 1:

2

1

1

1

1

3

21

4 44

6

6

db t

bt t t

t bt

db bt

dbbt

P P

mP P P

P P

P PP

P

=

= =

=

= =

Nhn xt:

- Cng sut mt bin tn ch =1/6 cng sut ca tn hiu u bin.

- Nu ta chn m


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Hnh 2.4.c tnh u ch tnh

I (t nS) (n 2 ): Bin dng in ng vi hi bc cao ca tn hiu u ch.

I (t s) : Bin cc thnh phn bin tn

Trong : It : bin tn hiu ra

V : gi tr tc thi ca tn hiu vo

A : gi tr cc i

B : ti tin cha iu ch

ng c tuyn thc khng thng to ra cc hi bc cao khng mong mun. Trong

ng lu nht l cc hi (t 2 S) c th lt vo cc bin tn m khng th lc c.

gim K th phi hn ch phm vi lm vic ca bu ch trong n thng ca ctuyn. Lc buc phi gim h su ch m.

2.2.3.2. H s mo tn s

Hnh 2.5.c tnh bin tn s

Gi : mo : h su ch ln nht

m : H su ch ti tn sang xt.


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H s mo tn sc xc nh theo biu thc :

Hoc : MdB = 20logM

2.2.4 Phng php tnh ton mch u bin :Hai nguyn tc xy dng mch u bin :

- Dng phn t phi tuyn : cng ti tin v tn hiu u ch trn c tuyn ca phn t phi

tuyn .

- Dng phn t tuyn tnh c tham su khin c : Nhn ti tin v tn hiu iu ch

nhphn t tuyn tnh .

2.2.4.1. iu bin dng phn tphi tuyn

Phn t phi tuyn c dng u bin c th l n in t, bn dn, cc n c kh,

cuc cm c li st hoc n trc tr s bin i theo n p t vo.

Ty thuc vo im lm vic c chn trn c tuyn phi tuyn, hm sc trng ca

phn t phi tuyn c th biu din gn ng theo chui Taylo khi ch lm vic ca

mch l ch A ( = 1800) hoc phn tch theo chui Fourrier khi ch lm vic ca

mch c gc ct < 1800 ( ch AB, B, C). Phng php tnh ton cho 2 trng hp

nh sau :

a, Trng hp 1 : iu bin ch A ( = 1800)

Hnh 2.6. Mch u ch dng Diode

Mch lm vic ch A nu tha mn iu kin:

Khai trin dng iD theo chui Taylor:


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Vi VD: in p trn Diode D v trn ti Rt

Vi:

Hnh 2.7.c tuyn ca Diode v th thi gian ca tn hiu vo ra

Hnh 2.8. Ph tn hiu u bin khi lm vic ch A

Thay UD vo biu thc (1) ta nhn c :

Khai trin (2) v b qua cc s hng bc cao n 4 s c kt qu m ph ca n c

biu din nh hnh 3.8.

Khi a3 = a4 = a4 =.....a2n+1 = 0 (n = 1,2,3) ngha l ng c tnh ca phn t phi tuyn l

1 ng cong bc 2 th tn hiu iu bin khng b mo phi tuyn.

tha mn iu kin (*) mch lm vic ch A th m phi nh v hn ch cng sut

ra. Chnh v vy m ngi ta rt t khi dng iu bin ch A.


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b, Trng hp 2 : iu bin ch AB, B hoc C ( < 1800)

Khi < 1800, nu bin n p c vo diode ln th c th coi c tuyn ca n l

mt ng gp khc.

Phng trnh biu din c tuyn ca diode lc :

ID = 0 khi VD 0

SVD khi vD > 0, S : H dn ca c tuyn

Chn m lm vic ban u trong khu tt ca Diode (ch C).

Hnh 2.9. Mch u ch dng Diode

Hnh 2.10.c tuyn ca Diode v th ca tn hiu vo ra khi lm vic ch C

Dng qua diode l 1 dy xung hnh sin, nn c th biu din iD theo chui Fourier nh

sau:


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I0 : thnh phn dng in mt chiu.

I1: bin thnh phn dng in cbn i vi ti tin

I2, I3.....In : bin thnh phn dng in bc cao i vi ti tinI0, I1, I2.....In : c tnh ton theo biu thc ca chui Fourrier:

Theo biu thc (*) ta c th vit:

Khi tt = th iD = 0:

Ly (3) (4) =>

y c xc nh t biu thc (4) :


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T biu thc (6) v (7) bin ca thnh phn dng in cbn bin thin theo tn hiu

iu ch (VS).

2.2.4.2. iu bin dng phn ttuyn tnh c tham s thay i

Thc cht qu trnh iu bin ny l qu trnh nhn tn hiu tng t. y l qu trnh

nhn tn hiu dng b nhn tng t. Trong mch n ny,quan h gia n p ra Vb

v in p vo Vt l quan h tuyn tnh. Tuy nhin, khi Vs bin thin th im lm vic

chuyn tc tuyn ny sang c tuyn khc lm cho bin tn hiu ra thay i c

iu bin.

Hnh 2.11. Mch u bin dng phn t tuyn tnh

n c vo tnh cht ca mch nhn,ta vit c biu thc ca n p ra sau y:

2.2.5. Cc mch u bin c th :2.2.5.1. Mch u bin dng 1 Diode

Hnh 2.12. Mch u bin dng 1 Diode


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Khai trin dng iD theo chui Taylor:

Vi VD: in p trn Diode D v trn ti Rt

VD = Vtcostt + Vscosst

Hnh 2.13.c tuyn ca Diode v th thi gian ca tn hiu vo ra

Hnh 2.14. Ph tn hiu u bin khi lm vic ch A

Thay UD vo biu thc (1) ta nhn c :

ID = a0 + a1(Vscosst + Vtcostt ) + a2(Vscosst + Vtcostt)2 + a3(Vscosst + Vtcostt)

3

+.= a1Vscosst + a1Vtcostt + a2Vs

2cos2st + 2VscosstVtcostt + a2Vt2cos2tt +

a3Vs3cos3st + 3 a3Vs

2cos2stVtcostt + 3a3Vt2cos2tt Vscosst + a3Vt

3cos3tt +

ID = a1Vscosst + a1Vtcostt + a2Vs2 1 os2 t

2sc + + VsVt (cos (t + s )t + cos (t - s )t )

+ a2Vt2 1 os2 t

2tc + + a3Vs

3cos3st +3

2a3Vs

2Vt (cos (t + s )t + cos (t - s )t )


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42

+3

2a3Vt

2Vs(cos (t + s )t + cos (t - s )t ) + a3Vt3cos3tt +.

Bin i tip tc ta s c kt qu sau:

ID = A costt + Bcosst + Ccos2st + D cos2tt + E(cos (t + s )t + cos (t - s )t )

+ F(cos (t + 2 s )t + cos (t - 2 s )t ) + G(cos (2t + s )t + cos (2t - s )t ) +.

Ta thy dng ra gm cc thnh phn hi: t, s, 2t, 2s

t + s, t - s, t + 2 s, t - 2 s .

2t + s, 2t + s .

Nh vy lc ly 3 thnh phn ca tn hiu u bin thng thng t-s, t, t+s th

chng ta s dng mch lc l tng. Nhng mch lc thc t li c dng hnh chung v

tn s cng hng cao nn b rng ph ca mch l rng nn ta thu c c cc thnh

phn hi khc na (t-2s, t-3s ).iu ny gy ra hin tng mo tn hiu u bin.

Nhn xt:- Vi mch u bin dng 1D c nhc m l mo tn hiu u bin l ln v khng

loi bc cc thnh phn hi bc cao ca tn hiu u ch

2.2.5.2. Mch u bin dng 2 Diode

Hnh 2.15. Mch u ch cn bng dng Diode

in p t ln D1,D2:

Dng in qua Diode c biu din theo chui Taylor:


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Dng in ra: i = i1 i2..

Thay (1), (2) vo (3) v ch ly 4 s hngu.

Bin i tng t nh mch u bin dng 1 Diode ta nhnc biu thc dng in ra:

Trong :

Hnh 2.16. Ph tn hiu u bin cn bng.

Theo hnh v ta thy cc thnh phn chn s b loi b. Nh vy khng c thnh phn hi

bc 2n ca s ( nh 2s, 4s, 6s ) v khng c thnh phn ti tin t, 2t,3tNh

vy trong trng hp u bin dng 2D th mo iu bin gim i hn so vi u bin

dng 1D.

2.2.5.3. Mch u bin vng

Hnh 2.17. Mch u bin vng.

Gi : iI l dng in ra ca mch u ch cn bng gm D1, D2


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44

iII l dng in ra ca mch u ch cn bng gm D3, D4

Hnh 2.18. Ph tn hiu u bin vng

Theo cng thc (4) mc u bin cn bng dng diode, ta c c biu thc tnh iI :

Ta c:

Trong :

Vi v3, v4 l in p t ln D3, D4 v c xc nh nh sau :

Thay (3) vo (2) v sau thay vo (1), ng thi ly 4 s hngu ta c kt qu :

Nhn xt:

- Mch u ch vng c th khc cc hm bc l ca S v cc bin tn ca

St, tn hiu ra ch c 2 bin tn trn v bin tn di do mo phi tuyn rt nh,

nh hn iu bin dng 1 diot v iu bin cn bng.

- Thnh phn ti tin b loi b.Mun c thnh phn ti tin ta phi cng thm thnh

phn ti tin trc khi pht i.

2.2.5.4. iu bin dng mch nhn.


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Hnh 2.19. S u bin dng mch nhn .

Ta c:

vs = Vscosst

vt = Vtcostt

Ta thy :Vb = k Vs Vt

= k (Vscosst + E0 ) ( Vtcostt )

=

1

2 k Vs Vt[cos (t + s )t + cos (t - s )t ] +k E0 Vtcostt.Nh vy Vb gm cc thnh phn ti tin, hai bin tn trn v di.

Nu nh E0=0 th suy ra:

Vb =1

2k Vs Vt[cos (t + s )t + cos (t - s )t ]

Vy ch c 2 thnh phn bin tn trn v bin tn di. Mun c ti tin ta cn phi thm

vo trc khi pht.

Hnh 2.20. Ph tn hiu u bin dung mch nhn

2.2.6. iu chn bin2.2.6.1. Khi nim

Ph tn hiu iu bin gm ti tn v hai di bin tn, trong ch c cc bin tnmang tin tc. V hai di bin tn mang tin tc nh nhau (v bin v tn s) nn ch cn

truyn i mt bin tn l thng tin v tin tc, cn ti tn th c nn trc khi truyn

i. Qu trnh gi l iu chn bin.

u im ca u ch dn bin so vi u ch hai bin :

- rng di tn gim i mt na.


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46

- Cng sut pht x yu cu thp hn vi cng mt c ly thng tin.

- Tp m u thu gim do di tn ca tn hiu hp hn

2.2.6.2. Cc phng php iu chn bin.

Vi u chn bin ta ch pht i mt bin tn. Ta ch pht i bin tn (t + s ). Vy

ta vn dng mch u ch 1D, 2D, 4D, iu ch vng nu trn nhng ch cn thay

mch cng hng ra thnh (ts ).

a, Mch u chn bin 1 Diode:

Theo iu bin thng thng ta c kt qu:

ID = A costt + Bcosst + Ccos2st + D cos2tt + E(cos (t + s )t + cos (t - s )t )

+ F(cos (t + 2 s )t + cos (t - 2 s )t ) + G(cos (2t + s )t + cos (2t - s )t ) +.

Nhng do iu chn bin ch pht i bin tn trn nn ta c kt qu:

ID = Acostt + Bcos2tt + C(cos (t + s )t ) + D(cos (t + 2 s )t ) + E(cos (2t + s )t

)+.

Phu bin ca tn hiu l:

Hnh 2.21. Ph tn hiu ca u chn bin 1 Diode

b, Mch u chn bin 2 Diode:


Nhng do iu chn bin ch pht i bin tn trn nn ta c kt qu:

iD

= Acos(t+

s)t + Bcos(2

t+

s)t

Phu bin ca tn hiu l:


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47


c, Mch u chn bin 4 Diode:


Nhng do iu chn bin ch pht i bin tn trn nn ta c:

idB = iI + iII = 2Acos(s + t)t

Phu bin ca tn hiu:


d, iu chn bin theo phng php lc

Hnh 2.24. S khi mch u ch theo phng php lc

t:

ft1: Tn s ca ti tn th nht

ft2: Tn s ca ti tn th hai


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x: H s lc ca b lc

Trong s khi trn y, trc tin ta dng mt tn s dao ng ft1 kh nh so vi di

tn yu cu ft2 tin hnh iu ch cn bng tn hiu vo VS(t). Lc h s lc tngln c th lc bc mt bin tn d dng. Trn u ra b lc th nht s nhn c

mt tn hiu c di ph bng di ph ca tn hiu vo.

fs = fSmax fSmin , nhng dch mt lng bng ft1 trn thang tn s, sau a n b

iu ch cn bng th hai m trn u ra ca n l tn hiu ph gm hai bin tn cch

nhau mt khong f = 2 (ft1 + fSmin ) sao cho vic lc ly mt di bin tn nhb lc th

hai thc hin mt cch d dng.

Lc ln 1:

Hnh 2.25. Ph tn hiu trong iu bin vng ln 1

Lc ln 2:


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Hnh 2.26. Ph tn hiu trong u bin vng ln 2

Nhn xt:

- u bin vng ln 1, do tn s cng hng thp nn di thng mch cng hng hp

nn d dng loi b bin tn di.Khong cch gia hai bin tn l 2s.

- u bin vng ln 2, tn s cng hng cao. Khong cch gia hai bin tn ln =

2(t1+ s). Do vy ta d dng loi bc bin tn di.

e, iu chn bin theo phng php quay pha

Tn hiu ra ca hai bu ch cn bng:

Hnh 2.27. S mch u chn bin theo phng php quay phaTa c:

Vs = Vscosst

Vt = Vtcostt

Ta thy :Vb1 = k Vs Vt

= k Vs Vt cosst costt


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Hnh 2.28. khi mch u chn bin theo phng php lc v quay pha kt hp

Vs = Vscosst

Vt1 = Vtcost1t

Tn hiu ra ca hai bu ch cn bng1:Vb1 = k Vs Vt1

= k Vs Vt1 cosst cost1t

=12

k Vs Vt1[cos (t1s )t + cos (t1-s )t ]

Vb1= k Vs Vt1

= k Vs Vt1 cosst sint1t

=

1

2 k Vs Vt1[sin (t1+s )t + sin (t1-s )t ]

Sau b lc 1, cn li bin tn trn ca hai bu bin dng mch nhn 1 lch pha nhau

90 . C th coi y l tn hiu u ch quay pha. iu ch ny cng vi ti tin t2

c a n b bu bin dng mch nhn 2 lch pha nhau 90 . in p ra sau hai

bu bin dng mch nhn 2:

Ta c:

Vt2 = Vt2cost2t

Vb1 = 12

k Vs Vt[cos (t1+s )t

T ta suy ra:

Vb2 = k Vb1 Vt2

= k1

2k Vs Vt1cos (t1+s )t Vt2cost2t


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=1

4k2 Vs Vt1Vt2 [ cos (t2+t1+s )t + cos (t2-t1-s )t]

Li c:

Vt2 = Vt2 sint2t

Vb1

=1

2k V

sV

t[sin (

t1+

s)t

Vb2 = k Vb1 Vt2

= k1

2k Vs Vt1sin (t1+s )t Vt2cost2t

= 14

k2 Vs Vt1Vt2 [- cos (t2+t1+s )t + cos (t2-t1-s )t]

Qua mch hiu ta c:

Vb2 = Vb2- V

b2

= 14

k2 Vs Vt1Vt2 [ cos (t2+t1+s )t + cos (t2-t1-s )t] -14

k2 Vs Vt1Vt2 [- cos

(t2+t1+s )t + cos (t2-t1-s )t]

=1

2k2 Vs Vt1Vt2 [ cos (t2+t1+s )t]

Phn bin tn hiu theo phng php lc quay pha kt hp c biu din nh sau:


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Hnh 2.29. Ph ca dao ng u chn bin theo phng php lc quay pha kt hp

(a) Ph tn hiu u ch

(b) Ph tn hiu ra trn bCCB1

(c) Ph tn hiu ra b lc

(d) Ph tn hiu ra mch hiu

2.3. Gii u bin

2.3.1. Gii u bin vi tn hiu u bin thng thnga. khi:

-Gii u bin c nhim v ly ra theo quy lut thay i bin ca tn hiu u

bin ngha l ly ra tn hiuiu ch Vs(t).

- Khi thc hin gii u bin ngi ta thc hin loi b ti tin v gi li tin tc.


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Hnh 2.30. S khi qu trnh gii u bin vi tn hiu u bin thng thng

Hnh 2.31. S nguyn l gii u bin vi tn hiu u bin thng thng

- Dng sng:


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Hnh 2.32. Dng sng gii u bin vi tn hiu u bin thng thng

b . Nguyn l hot ng :

- Tn hiu u tin khi qua b chnh lu s ct i mt na tn hiu v n c mt thnh

phn ngc chiu khc 0. Ty theo cc tnh ca phn t chnh lu m ta c thnh phn

ngc chiu m hoc dng. Tn hiu u bin khi m n pha thu do nh hng ca

tn hiu phadinh tc l cng trng pha thu thay i lm cho thnh phn mt

chiu U0 cng thay i theo. Do vy ngi ta cn phi lc thnh phn mt chiu U0

thc hin hi tip m v gi cho mc tn hiu ra l khng i.

- lc tn hiu u ch ngi ta dng b lc thng thp. u ra b lc thng thp

ngoi thnh phn tn hiu u ch cn c thnh phn 1 chiu U0. V loi b thnh

phn mt chiu ny ngi ta dng b lc xoay chiu.

2.3.2 Gii u chn bin:Vi u chn bin ngi ta ch pht i bin tn trn (t + s ). my thu ngi ta thu

ly bin tn trn sau cho vo mch u bin vng cng vi tn s dao ng to ra

trong my thu c tn s bng tn s ti tin v ng b vi ti tin ca i pht. u ra b

iu bin vng c 2 bin tn trn v di


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Cch to ra ti tin ng b vi ti tin ca i pht nh sau:

i vi tn hiu u chn bin ti tin b nn. Tc l n rt l nh so vi bin tn

trn. Nhim v pha thu l chng ta phi thu li v khuych i ti tin v dng n ng

b vi dao ng ti tin ca my thu.

2.4. iu tn v iu pha2.4.1. iu tn dng Diode bin dung

Hnh 2.33. Mch u tn dng Diode bin dung

iu tn tc l thay i tn s ti tin theo tn hiu u ch. Nh vy tn s ti tin thay

i t fminn fmax, f = fmax - fmin c gi l di tn cc i.

Da vo c m ca D bin dung, ngi ta c th thc hin u tn. n nh tn s

trung tm ca dao ng u tn ngi ta mc ni tip D bin dung vi thch anh. Khi

in p ngc ca D bin dung thay i theo tn hiu u ch s lm cho CD thay i

theo v lm f thay i theo tn hiu u ch => Nh vy ta thc hin c u tn.

Us(t) = - UD E0 => UD = - (E0 + Us(t))

=> UD = - (E0 + Umscosst)

Nhn xt:

- u im: C tn s trung tm n nh do dng thch anh.

- Nhc m: di tn cc i

f = fmax - fmin nh

2.4.2. iu pha theo Amstrong


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Hnh 2.34. Mch u pha theo Amstrong v th vectca tn hiu

Ti tin t thch anh a n bu bin 1 (B1) v iu bin 2 (B2) lch pha 900, cn

tn hiu u ch vSa n hai mch u bin ngc pha. in p ra trn hai bu

pha:

th vc t ca tn hiu

1dbV v

2dbV v vc t tng ca chng

V =

1dbV +

2dbV l mt

dao ng c u ch pha v bin . u bin y l iu bin k sinh.

hn chu bin k sinh => chn nh ( < 0,35)

2.4.3.iu tn dng Transistor in khng

Phn tn khng: Dung tch hoc cm tnh c tr s bin thin theo in p u ch

t trn n c mc song song vi h dao ng ca b dao ng lm cho tn s dao

ng thayi theo tn hiu u ch. Phn tn khng c thc hin nhmt mch di

pha trong mch hi tip ca BJT. C 4 cch mc phn tn khng nh hnh v.


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Vi mch phn p RC ta tnh c:

IC = S.VBE => IC lun lun cng pha vi VBE.

Nu chn Cj1


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Tham s ca n khng tng ng ph thuc vo h dn S ca BJT.

iu tn dng phn tn khng c tht c lng di tn tng i lt

f

fkhong

2%

S b to dao ng u tn bng phn tn khng phn p RC :

Hnh 2.35. S mch to dao ng u tn phn tn khng phn p RC

T1 : BJT in khng; T2 : BJT dao ng

2.4.4. iu tn bng dao ng xung.

Hnh 2.36. iu tn bng dao ng xung

EB thay i theo tn hiu tin tc nn lm thay i thi gian ng mca Transistor


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Thay i tn s xung ra.

2.5. Gii u tn.

2.5.1. Nguyn l chunga. Nhim v:

- Ly ra quy lut thay i tn s ca ti tin thu c tin tc ban u.

- thc hin c tch sng u tn th cn to ra c tnh truyn t nh sau:

Hnh 2.37. Nguyn l gii u tn

B = fmax fmin ; Yu cu ABC l ng thng

2.5.2. Cc mch gii ch tn sa, Tch sng bng mch cng hng.


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Hnh 2.40. c tuyn lc ca mch cng hng lch


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63

Chng 3: I TN

3.1. nh ngha

3.1.1. t vn :

Tn hiu (ting ni, hnh nh) mun truyn c i xa, ta phi thc hin u chtn scao. Ty theo phng thc u ch, mi tn hiu u ch c mt b mt rng ph nht

nh.

VD:m thanh: b rng ph tn hiu u bin f < 20KHz, v iu tn f


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thu c 1 knh theo yu cu ta khng dng my thu trc tip m dng qu trnh

i tn.

f0 < f0

B = f0 / Q < B

f0 = ftg (tn s trung gian)

VD: m thanh iu bin ftg = 455KHz

Kthut truyn hnh : ftg = 4.5MHz hoc 5.5MHz

my pht: thc hin u ch tn hiu tn s thp ln tn s sng mang cao tn,

ngi ta dng tnh cht khng ng thng ca phn t phi tuyn.

Hnh 3.2. Smy pht

Do tnh cht khng ng thng ca cc phn t phi tuyn nn tn hiu ra ngoi tn hiu

iu ch cn c cc thnh phn hi. Nu chng ta thc hin qu trnh pht trc tip th

cc thnh phn hi ny s gy nhiu cho my thu. khc phc nhc m ny ngi ta

tin hnh iu ch tn hiu tn s trung gian sau thc hin qu trnh i tn a ra

tn hiu sng mang. My pht thc hin theo nguyn l trn gi l my pht i tn.

Ngoi ra trong cc h thng thng tin (v tinh, viba) chng ta s dng cc b chuyn tip

chnh l cc bi tn.

3.1.2. Khi nim:i tn l qu trnh tc ng ln hai tn hiu sao cho trn u ra ca b trn nhn c tn

hiu tng hoc hiu ca hai tn hiu .


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+an [ Uthmcos( tht ) + Unsmcos( nst ) n] +

i = a0 + a1 [Uthmcos( tht ) + Unsmcos( nst ) ]

+ a2 [ Uthm cos( tht ) + Unsm cos( nst ) + 2UthmUnsmcos( tht )cos( nst ) ] +..+

+ an [ Uthmcos( tht ) + Unsmcos( nst ) ]n +

i = a0 + a1 [Uthmcos( tht ) + Unsmcos( nst ) ]( (2 2

2

1 cos 2 1 cos 2

2 2th ns

thm nsm

t ta U U

+ + + +

+a2 [2UthmUnsmcos(tht)cos( nst ) ++ an [ Uthmcos(tht) + Unsmcos(nst) n] +

i = a0 + a1 [ Uthmcos( tht ) + Unsmcos( nst ) ] + 22

a( Uthm + Unsm)

+ 22

a[ Uthmcos( 2tht ) + Unsmcos( 2nst )

+ a2UthmUnsm [cos( ns + ns )t + cos( ns th)t ] +...

+ an [ Uthmcos( tht ) +Unsm( nst)n] +

Vy dng in ra ca b trn tn c nhiu thnh phn. Trong c thnh phn 1 chiu,

thnh phn bc nht cath ns thnh phn bc 2 cath ns thnh phn tn s tng hiu

cath ,ns.

Ngoi ra cn c cc thnh phn bc cao

Khi: m, n = 1 => = ns th : B trn tn n gin

m, n > 1 => B trn tn t hp.

Trong cc thnh phn th thnh phn ns th l thnh phn tn s thp nht. Ngi ta

chn tn s ny lm tn s trung gian

T ta c biu thc i tn tg = ns th

Biu thc i tn

tg = ns th (Nuns > th )

tg = th ns (Nu th > ns)

ftg = fns fth

ftg = fth fns

Sau khi ra khi b trn ta c 1 b lc tn s trung gian. V vy di thng B ca mch

cng hng c tnh chn lc cao.


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3.2.2. Nhiu sinh ra ca b trn tn v cch khc phc3.2.2.1. Nhiu tn s trung gian

Nu u vo b trn c tn s trung gian th n s i qua b trn tn gi l nhiu tn s

trung gian:

t hf

B Q=

B tng tn hiu chn lc km

Cch khc phc:

Trc b trn tn ta phi lc b ftg ( lc b nhiu)

Hnh 3.4. S trn tn mc cng hng cc i

XL = XC

L =1C

=1

LC =

1

LC= 2ftg

ftg =1

2 LC

1

1AB

J LJ CZ

J LC

= =

Loi bc ftg

Mc cng hng cc tiu


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Hnh 3.5. S trn tn mc cng hng cc tiu

( )1

AB L C

JZ J L J L J X X

J C C

= + = =

ZAB = 0 ftgi xung t nn ta loi bc ftg

3.2.2.2 Nhiu tn snh

Trong thc t s c 1 thnh phn tn s cao hn thnh phn fns cng tho mn c i

tn v cng c i thnh ftg .Tn s c gi l tn snh

Gi si tn theo ftg = fns fth ( fns > fth )

Hnh 3.6. th nhiu tn snh

ftg = fanh fns

Mch cng hng fth c di thng B ln c th thu fanh (nhiu)

Cch khc phc

Thc hin i tn 2 ln s c 2 tn s trung gian: ftg1,ftg2 ( ftg1 > ftg2 )


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Hnh 3.7. Khc phc nhiu tn snh bng i tn 2 ln

Trong ln i tn ln 1:

ftg1 = fns fth

Hnh 3.8. Qu trnh i tn ln 1

thc hin khc phc nhiu tn snh ta s thc hin i tn 2 ln.

Hnh 3.9. Qu trnh i tn ln 2

Do tn s ftg1 ln . Do vy tn snh fanh1 nm ngoi di thng ca mch cng hng vo

v qua bi tn ln 1 ta loi bc fanh ta ch thu c fth. Sau ta thc hin i tn

ln 2 v thu c ftg2. Do ftg2 thp nn di thng ca mch lc hp v ta cng loi b

c thnh phn tn s fanh2


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uns1 = uns2 = uns = Unsmcos( nst )

Do dng in tn s trung gian qua cc diode:

itg1 = Itgmcos [ ( ns th)t ]

itg2 = Itgmcos { [( ns th)t ] + } = Itgmcos( ns th)t

Trn mch cng hng ta nhn c:

itg = itg1 + itg2 = 2Itgmcos [( ns th) t]

Nhn xt:

- Mch trn tn cn bng lm tng dng in trung gian u ra so vi i tn n th

i tn cn bng bin tn hiu ra ca dng in trung gian tng gp i.

- Nu c nhiu tn s trung gian to nn do b dao ng ngoi sai th diode tn s trung

gian ny s a vo 2 diode l cng pha. Tuy nhin do o pha BA2 nn nhiu ny b

trit tiu.

4.4.3. Mch i tn vng

Hnh 3.15. Mch i tn vng

Mch i tn vng gm 2 mch i tn cn bng mc ni tip.

in p tn hiu t ln 4 diode c xc nh:

uth1 = Uthmcos( tht )

uth2 = Uthmcos( tht )uth3 = Uthmcos( tht )

uth4 = Uthmcos( tht )

in p ngoi sai t ln 4 diode ng pha nhau:

uns1 = uns2 = uns3 = uns4 = uns = Unsmcos( nst )

Do dng in tn s trung gian qua cc diode ( do Uth to ra )


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( )

( ) ( )

( ) ( )

( )

1 ns

2 ns ns

3 ns ns

4 ns

os

os os

os os

os

tg tgm th

tg tgm th tgm th

tg tgm th tgm th

tg tgm th

I I c t

I I c t I c t

I I c t I c t

I I c t

=

= + = = + = =

Trn mch cng hng ta nhn c:

(1 2 3 4 ns4 ostg tg tg tg tg tgm th I I I I I I c t = + + + =

Nhn xt

- Mch i tn vng ch cha cc thnh phn tn s ns th cc thnh phn khc bkh do d tch c thnh phn tn s trung gian nh mong mun

- Mch i tn vng lm tng dng in trung gian u ra so vi i tn cn bng th

i tn vng c bin tn hiu ra ca dng in trung gian tng gp i.

3.3.2 Mch i tn dng TransistorMch i tn dng transistor c th mc theo s Bazo chung hoc Emito chung. Cc

ny khc nhau cch t n p ngoi sai vo transistor. Trn css nguyn

l, ngi ta thit k nhiu loi thc t khc nhau nh di y :

a, S Utha vo B, Unsa vo E.

Hnh 3.16. S Utha vo B, Unsa vo E.

b, Utha vo E, Unsa vo B.


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C E

T 4

C B

E c

C B

C E

T 3

C E

C 2

C 1

T 2

C

C B

T 1

C B

R B 1

R E

R B 2

R B 1 R B 2

R E

R E

R B 1

R B 2

R ER B 2 R B 1

Hnh 3.19. V di tn bng Transistor

Trong :

ftg1,ftg2 to mch cng hng kp.

T1: B trn vi ftha vo bazo, fnsa vo emito

T2 : B trn dao ng ngoi sai 3 m n dung

T3,T4,T5 : Tng khuch i trung tn cng hng


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CHNG 4: CHUYN I AD V DA

4.1. Csl thuyt.K thut PCM.

- khi:

Hnh 4.1. S khi k thut PCM

Tn hiu vo l tn hiu tng t bin i lin tc theo thi gian. Tn hiu ny c

a vo b ly mu ri rc ha tn hiu. u ra b ly mu l tn hiu M l mtdy xung c bin thay i theo quy lut ca tn hiu tng t.

Hnh 4.2. Minh ha vic ly mu tn hiu

- Do tn hiu tng t l bt k nn cc mc ly xung cng l bt k (v hn mc).

- Khi lng t ha c nhim v lm trn tn hiu xung c mc bt k thnh xung

c mc tng ng vi 2n mc. Nh vy sai su tin m chng ta gp phi khi

chuyn i tng t - s l do khi lng t ha gy nn.

- Tn hiu ra khi khi lng t (M) sc a vo khi m ha v to ra cckhi nh phn SD (n bit).

- V d:Gi s tn hiu tng t c tn s ln nht fAmax v tn s ly mu fLM

Ta c ting ni nm trong vng tn s 0,3 3,4 (KHz)

fLM tho mn iu kin khng xy ra hin tng chng cht tn hiu

fLM 2fAMax

Trong ting ni fLM = 2.3,4 = 6,8KHz

Nhng thng thng ta ly fLM = 8KHz

Mi mu ngi ta thng m ho bi 8 bit. Nh vy tc ly mu v = 64Kbit/s

Ly mu ng t ha M haSA M

M SD


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*) Lng t ho

+)Lng t ho u (tuyn tnh)

Cc mc lng t ho l bng nhau

Hnh 4.3. Lng t ha u.

Sai s ca lng t ha u l Q/2

+) Lng t ha phi tuyn:

Hnh 4.4. Lng t ha phi tuyn

phi ghp gia ngun tn hiu c dng tng t vi cc h thng x l s ngi ta

dng cc mch chuyn i tng t - s (ADC : Analog-Digial Converter) v cc mch

chuyn i s - tng t (DAC : Digial- Analog Converter).

Hnh v (6.1) biu din qu trnh bin i tn hiu dng tng t sang dng s.

Tn hiu tng t UAc chuyn thnh dng bc thang u. Vi 1 phm vi ca gi tr

UAc biu din bi 1 gi tri din thch hp.


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Chng hn gi tr UAc chuyn thnh dng bc thang 7 bc v mi bc, ta gn cho

UA mt gi tr ri rc. V d khi UA bin thin trong mt khong nh 3,5 4,5 ta gn cho

n mt gi tr l 100.

Mt cch tng qu, gi tn hiu tng t l SA (UA), tn hiu s l SD (UD). SD c biu

din di dng m nh phn nh sau :

Trong : bk = 0 hoc bk = 1 (vi k = 0 k = n - 1) v c gi l bit.

+ bn-1 : bit c ngha ln nht (MSB : Most significant bit). Mi bin i ca MSB tng

ng vi s bin i na di lm vic.

+ bo : bit c ngha nh nht (LSB : Least significant bit). Mi bin ca LSB tng ng

vi s bin i mt mc lng t. Mt mc lng t bng mt nc ca hnh bc thang

V d : vi mt mch bin i N bit vi l N s hng trong dy m nh phn. (Trong v dtrn hnh v 6.1 : N = 3) th mi nc trn hnh bc thang chim mt gi tr.

12AM

LSB n

UQ U = =

UAM : l gi tr cc i cho php ca n p tng t.

ULSB = Q : gi l mc lng t.

Sai s lng t ha c xc nh nh sau :

2QQU =

Khi chuyn i AD phi thc hin vic ly mu tn hiu tng t. m bo khi phc

li tn hiu mt cch trung thc, tn s ly mu fM phi tha mn iu kin :

Fthmax : tn s cc i ca tn hiu

B : di tn s ca tn hiu.

4.2. Cc thng s bna) Di chuyn i

Di chuyn i ca n p tng tu vo l khong n p m b chuyn i AD

c th thc hin chuyn i c.

UAmin UAmax


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b) Sai s chuyn i

- Lin quan n sai s lng t ho

Q =12 n

AMaxS

c)Tc chuyn i

Lin quan n thi gian chuyn i 1 gi tr no v thi gian hi phc mch

chuyn v trng thi ban u trc khi thc hin chuyn i trng thi khc nhau. Tc

chuyn i t l nghch vi thi gian

f =phuchoichuyendoi

TT +1

Mt ADC c tc chuyn i cao th chnh xc gim v ngc li. Ngha l yu cu

v chnh xc v tc chuyn i mu thun vi nhau. Ty theo yu cu s dng,

phi tm cch dung ha cc yu cu mt cch hp l nht.

4.3. Nguyn tc lm vic ca ADCNguyn tc lm vic ca ADC c minh ha theo s :

Hnh 4.5. th thi gian ca n p vo v ra mch ly mu.

Trc ht tn hiu tng t UAc a n mch ly mu. Mch ny c 2 nhim v:

- Ly mu tn hiu tng t ti nhng thi m khc nhau v cch u nhau (ri rc ha

tn hiu v mt thi gian).


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- Gi cho bin n p ti cc thi m ly mu khng i trong qu trnh chuyn i

tip theo (tc l trong qu trnh lng t ha v m ha).

Tn hiu ra ca mch ly mu c a n mch lng t ha thc hin lm trn vi

chnh xc bng Q/ 2.

Vy qu trnh lng t ha thc cht l qu trnh lm trn s. Lng t hac thc

hin theo nguyn tc so snh, tn hiu cn chuyn i c so snh vi mt lot cc n

v chun Q.

Sau mch lng t ha l mch m ha. Trong mch m ha, kt qu lng t ha c

sp xp li theo mt trt t nht nh ph thuc vo loi m yu cu trn u ra b

chuyn i .

Php lng t ha v m ha gi chung l php bin i AD.

4.4. Cc phng php chuyn i tng t- s4.4.1 Phn loiC nhiu cch phn loi ADC. Cch phn loi hay dng hn c l phn loi theo qu

trnh chuyn i v mt thi gian. N cho php phn on mt cch tng qut tc

chuyn i. C 3 phng php chuyn i sau:

+ Chuyn i song song: Tn hiu tng tc so snh cng mt lc vi nhiu gi tr

chun. Do tt c cc bit c xc nh ng thi v a n u ra.

+ Chuyn i ni tip theo m m: Qu trnh so snh c thc hin tng bc theo

quy lut m m. Kt qu chuyn i c xc nh bng cch m s lng gi tr

chun c th cha c trong gi tr tn hiu tng t cn chuyn i.

+ Chuyn i song song- ni tip kt hp: Qua mi bc so snh c th xc nh c ti

thiu 2 bit ng thi.


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4.4.2. Chuyn i AD theo phng php song song

Hnh 4.6. S nguyn l b chuyn i AD theo phng php song song

S dng 2n 1 b so snh, UAa vo ca P, ca N s dng b phn p. Tn hiu sau khi

ra khi 2n 1 b so snh c a vo cc trigger D v u ra ca trigger D

Tn hiu tng t UAc a ng thi n cc b so snh t cc b so snh c n p

so snh km nhau bng mc lng t ho. to ra cc n p so snh ny ngi ta

dng cc b phn p n trnh hnh v

Nu lng t ha l lng t ha u th tt c cc gi trn trtrng nhau.

V d: UAmax = 7,2V to ra mc n p so snh

USS1 = 1V, USS2 = 2V, USS3 = 3V, USS4 = 4V

USS5 = 5V, USS6 = 6V, USS7 = 7V

Nhn xt:

+)Chuyn i song song c u im: Tc chuyn i nhanh (cc bit to ra

ng thi). Sai s bin i thp v c th to ra dng m theo mun.

+)Chuyn i song song c nhc m: Kt cu phc tp do c s linh kin ln.

Nn vic ng dng ch c gii hn vi chuyn i AD c s bit nh v tc cao.

+)Cch khc phc: khc phc nhc m ny ngi ta s dng b chuyn i

ni tip.


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4.4.3. Chuyn i AD ni tipSD = bn-1bn-2b1b0

SA= bn-12n-1 +b12

1 + b020

Hnh 4.7. S khi chuyn i AD ni tip

Cho in p vo UA so snh ni tip vi cc mc n p vo so snh. Vi mc n p

so snh ny gim dn i 2 ln v u ra ca mi 1 tng so snh c 1 bit. Tng no c in

p so snh ln ng vi bit c trng s cao v ngc li.Phng php chuyn i nh sau:

- Gi s UA > USS1 khi bit ra b = 1 ng thi qua b chuyn i DA n c mc UA

=USS. in p ny c tr vi n p UAa vo tng chuyn i tip theo.

- Gi s UA < USS1 khi bit ra b = 0 ng thi qua b chuyn i DA th UA = 0. in

p ny c tr vi UA to ra mc n p a tng chuyn i tip theo


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V d: UAmax = 7,2V

USS1 =2

1 UAmax =2

2,7 = 3,6V

USS2 =2

1USS1 =

2

6,3= 1,8V

USS3 =21 USS2 =

28,1 = 0,9V

UA1 = 5,2V b2 = 1V, UA1 = 3,6V

UA2 = 5,2 3,6 = 1,6V USS3 b0 = 1

b2b1b0 = 101

Nhn xt: Phng php chuyn i ni tip c

-u im: Kt cu n gin v c bao nhiu bit cn by nhiu b so snh

- Nhc m: Tc chuyn i chm

4.4.4. Chuyn i AD theo phng php kt hp

Hnh 4.8. B chuyn i AD theo phng php song song ni tip kt hp

y l s kt hp phng php song song v phng php ni tip nhm dung ha u

khuyt m ca hai phng php ny : gim bt phc tp ca phng php song

song v tng tc chuyn i so vi phng php ni tip.

B chuyn i ADC u tin l b chuyn i song song n1bit . Trong bc so snh th

nht => xc nh c N1 bit. T B1 => BN1. N u s bit l n th s t ng chuyn i l:

l =1n

n

Mi tng dng 2n - 1 b so snh. Nh vy chuyn i n bit phi dng s b so snh l:

s b so snh = l(2n1 1) =1n

n(2n1 1)b


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V d N = 9; N1 = 3

Phng php song song-ni tip kt hp => S b so snh:

l(2n1 1) =1n

n(2n1 1) = 3.7 = 21

Phng php song song => s b SS:

(2n - 1) = 29 1 = 512 1 = 511

4.4.5. Chuyn i AD ni tip dng vng hi tip tng ca phng php :

Ta bin tn hiu tng t thnh s xung. Sau m s xung bng bm nh phn.

Trng thi ra ca bm nh phn chnh l trng thi ra ca tn hiu s m ta cn chuyn

i

Gi s n = 3 bit

Hnh 4.9. Minh ha bm 3 bit

khi

Hnh 4.10. S chuyn i AD ni tip dng vng hi tip

Mc ch ca chuyn i vng hi tip l bin UA thnh s xung trc khi chuyn i

trng thi ra ca bm thun nghch = 0 sau khi qua b chuyn i D_A th ta thu c

UM = 0. UA,U M cng a vo b tr


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t Uh = UA UM, gi thit UA> 0 Uh > 0 qua mch th A+ = 1, A- = 0. Vy bm

thun s lm vic

Trng thi ra ca bm s tng dn sau qua chuyn i D_A th UM tng dn Uh

gim dn cho n khi UM = UA th Uh =0 A+ = 0, A- = 0 bm thun nghch ngng

m trng thi ra ca tn hiu l tn hiu s cn chuyn i

Gi thit nu UA gim Uh > 0 A+ = 0, A- = 1 bm nghch s lm vic trng

thi ra ca bm s gim dn cho n khi UA =UM (U

h = UA UM =0 ). Lc bm

li ngng lm vic trng thi ra ca bm chnh l trng thi ra ca tn hiu s cn

chuyn i. Phng php ny l phng php xp x lin tip

4.4.6. Chuyn i AD theo phng php tich phn n gin:- tng: Ngi ta tp ra mt khong thi gian tx t l vi n p cn chuyn i dng

mca mt dy xung chun to nn mt n p xung t l vi n p cn chuyni. Sau cho s xung vo bm nh phn. Trng thi ra ca s nh phn chnh l

trng thi ca tn hiu s cn chuyn i

- S khi:

Hnh 4.11. Chuyn i AD theo phng php m n gin


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Hnh 4.12. Gin thi gian minh ha phng php tch phn n gin

in p qut ng thng c ng thi c a vo 2 b so snh 1 v so snh 2.

b so snh 1 n p Uqueta vo ca thun cn b so snh 2 th Uquetc a vo

ca o v so snh vi UA. Sau 2 tn hiu M,N c a vo cng AND. P l kt qu

ca php and gia M v N.

Trong khong thi gian t1,t2 m ca cho cc xung chun i vo bm. Ngoi khong

khng c xung a vo bm nh phn m s xung

Nh vy khi m

B so snh 1 c: t < t1 : Uq(t) < 0 N = 0

t > t1 : Uq(t) > 0 N = 1

B so snh 2 c : t < t2 : UA > Uq(t) M = 1

t > t2 : UA > Uq(t) M = 0

P =M.N

To ra khong mm s xung tx = t2 t1

Khi a P vo cng and 2 mca cho 1 dy xung chun fx to ra n xung. Cho s

xung vo bm nh phn u ra ca bm nh phn l tn hiu s chng ta cn

chuyn i

Xt quan h n v UA


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Gi s ta c Uq(t) =RC

1 0U dt

Uq(t) =RC

1 U0t + C

Uq(t1) = 0 =

RC

1U0t1 + C

C = -RC

1U0t1

Uqm = Uq(t2) =RC

1U0t2 + C -

RC

1U0t2

Uqm =RC

1U0(t2 t1) =

RC

1U0tx

Do Uqm = UA

UA = RC1

U0tx

tx =o

A

U

URC

t: k1 =0U

RC tx = k1UA

Vy s xung n l:n = fx.tx = fx.k1.U A = k2UA

Khi UA tng lm cho s xung tng ln nn trng thi ra tng theo

Xt trn th UA ng U

A ng vi khong thi gian t2


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Hnh 4.13. Gin thi gian minh ha khi UA tng

Nhn xt

- Phng php ny c chnh xc ph thuc vo R,C,U 0,fx. Chnh v vy yu cu cc

thng s ny phi n nh

- Phng php ny gp sai s 1 xung do s khng ng b ca thi gian mca vi

thi gian tn s xung nhp. V vy gim sai s ngi ta tng tn s xung nhp fx hocgim dc qut ng thng. Tuy nhin time chuyn i ko di.Ta thng s dng

phng php tng tn s xung chun.

4.4.7. Chuyn i AD theo phng php tch phn hai sn dcPhng php ny ngi ta chuyn tn hiu tng t thnh khong thi gian sau m

ca cho dy xung chun a vo bm nh phn


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Hnh 4.14. Phng php tch phn hai sn dc

Hnh 4.15. Gin thi gian minh ha phng php tch phn 2 sn dc


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Hnh 4.16. S xung m ctrong khong thi gian tx2

IC1 : mch tch phn to Uq(t)

IC2: so snh Uq(t) vi 0

Trc trng thi chuyn i ca bm = 0 bu khin nhn xung chun t khi to

xung chun u khin kho K v tr 1. Nh vy IC1 s lm nhim v tch phn UA

n trc ca Uq. ng vi sn dc ny khi cha a vo b so snh IC2 ta c tn

hiu M = 1 ( mc cao). Khi m t s xung no n1 th b iu khin u kin

kha K chuyn sang v tr 2. IC1 lm nhim v tch phn Uch to ra sn sau Uq(t).

ng thi bu khin tn hiu ra mc cao. P l kt qu php and M v N. Ta c s

xung ra c rng tx2 = t2 t1.Tn hiu P a vo cng and tip theo cng vi dy

xung chun a vo bm nh phn l tn hiu s cn chuyn i

Xt mi quan h gia s xung n2 v tn hiu chuyn i UA s xung n2 l tn hiu UA cn

chuyn iKhi kho K v tr 1:

Uq(t) = 0 = -RC

1A

U dt = RC1

UAt + C

Uq(0) = 0 = -RC

1UA0 + C

C = 0


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91

U+qm = Uq(t1) = -RC

1UAt1 = -

RC

1UAtx1

Khi kho K v tr 2 :

Uq(t) = -RC

1ch

U dt

Uq(t) = -RC

1 Ucht+ C

Uq(t2) = 0 = -RC

1Ucht2 + C

C =RC

1Ucht2

U-qm = Uq(t1) = -RC

1 Ucht1 +RC

1 Ucht2

U-qm = RC

1Uchtx2 (tx2 = t2 t1)

V U+qm = U-qm

-RC

1UA tx1 =

RC

1Uchtx2

UA tx1 = Uchtx2

tx2 = Ach

U

Utx1

Xt s xung n1 = fx.tx1 tx1 =x

f

n1

N2 = fx.tx2 tx2 =x

f

n2

x

f

n2 = - Ach

U

Ux

f

n1

n2 = - Ach

U

Un1 n2 = k2UA (vi k2 = 1

ch

n

U)

Gi s UA tngT th ta thy dc xung khng thay i t2 tng n t

2

Nhn xt:

- So vi cc phng php trc th phng php ny c chnh xc cao hn

- chuyn i chnh xc th Uch phi n nh


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92

4.4.8. Phng php chuyn i ni tip theo m nh phn

Hnh 4.17. B chuyn i AD ni tip theo m nh phn

Mi tng bao gm mt b so snh, mt kha u khin v mt mch tr.

Mt u vo ca cc b so snh l mc n p ngng. Mc n p ngng ln nht l

max

2AU tng u tin v tng ng vi bit ln nht. nhng tng sau, in p

ngng s l : max4

AU

, max8

AU

ty theo s tng s dng trong mch.

Mch chuyn i theo phng php ny c s tng bng s bit cn xc nh. Mi tng

cho ra mt bit. Gi x tn hiu vo bin thin trong phm vi 0 UAmax. Tn hiu vo s

c so snh vi n p chun Uch1 = max2

AU

. Nu UA > max2

AU

th ng ra ca b so snh

(SS) s cho ra mc logic 1 v lc ny kha K sc ni ti mc n p chun Uch1

mch tr tn hiu. Khi tr sc thc hin ly UA = max2

AU

(UA Uch1). Kt qu ca

php tr sc tip tc a vo so snh tng 2 vi Uch2 = max4

AU

. Ngc li nu UA 00

.2

ch ht R n

U RU

R=

+ Khi b1 =1; b0 = b2 = = bn-1 = 0

ng t nh trn:

11 11

0 0. . .2 .2 2

ch ht ch ht R n nU R U RU bR R= =

+ Khi bn-1 =1; b0 = b1 = = bn-2 = 0

1

11

0

. .2 .2n

nch ht R nn

U RU b

R

=

p dng nguyn l xp chng:

( )0 10 10

. . .2 ... .22

nch ht R nn

U RU b b

R

= + +

VD: SD = 101 ; n=3 bit

=> ( )0 1 2

30 0

5. . 1.2 0.2 1.2 . .

2 8ch ht ht

R ch

U R RU U

R R= + + =


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98

t

C MU U e

=

Trong : = RC

UM : Gi trn p ti thi m C phng.

Hnh 4.24. Gin thi gian m t s phng n trn t C

Ngi ta chn RC sao cho sau khong T/2 th in p trn t UC ch cn l mt na so

vi gi tr ban u.

2

2

ln1 ln 22

ln 2

T

MM

UU e

T

TRC

=

=

=

VD: Gi s SD = 1101


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99

Hnh 4.25. Gin thi gian gii thch nguyn l chuyn i Shanon-Rack

- Trong khong thi gian t 0 t1: K1 ng, K2 m, t C c np ng thng ti

gi tr U0

- T t1 t2 K1 m, K2 ng, t C phng qua n tR di dng hm m ti gi tr

U0/2- T t2 t3 do bt b1 = 0 nn K1 m, K2 ng, UC khng i.

- T t3 t4 K1 m, K2 ng, t C phng qua n trR di dng hm m ti gi tr

U0/4

- T t4 t5 : b2 = 1, t C np ng thng n gi tr U0 + U0/4


U0/2 + U0/8

- T t6 t7 K1 ng, K2 m, t C c np ng thng ti gi tr U0 + U0/2 + U0/8


cn li U0/2 + U0/4 + U0/16

Vy:


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100

( )

( )

0 0 0

0

0 1 2 30

2 4 16

1 4 816

1.2 0.2 1.2 1.216

r

U U UU

U

U

= + +

= + +

= + + + Tng qut: n bt

( )1 1 00 1 1 02 ....... 2 22n

r nn

UU b b b= + + +

Nhn xt:

+ Nhc m: Chm hn so vi phng php trc.

+ u im: Mch n gin hn v thc hin chnh xc hn.


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101

Chng 5: NGUN CUNG CP

Ngun nui mt chiu l cn thit cho mi thit bn t. Tr mt s trng hp cc

thit bn tc thit k ch dng cc ngun n ho nh pin, c-quy; trong nhiu

trng hp ngun nui mt chiu c to ra bng cch bin i v chnh lu dng in

xoay chiu 50 Hz t mng n cng nghip thnh ph. Nh ti chng 1 v mch chnh

u dng diode bn dn ni, do c gn sng bin gy ra bi s bin i gi tr tc

thi ca ngun n p xoay chiu nn cn c b lc thng thp san bng gn sng.

Cng do thng ging ca ngun n vo v thng ging ca ti cng cc bin ng khc

nn mun c c n p ra b ngun n nh th phi thit k thm cc mch n p

(hoc n dng) b tr cc bin ng ny. S khi ca mt ngun nui mt chiu

ni chung c biu din trn hnh 5.1.

Hnh 5.1. S khi ca ngun nui c n p.

Bin p l thit b bin i n p xoay chiu li vo (th d, 220V~) thnh in p xoay

chiu li ra c bin cn thit.

Mch chnh lu c nhim v chuyn n p xoay chiu bn th cp bin p thnh in

p mt chiu c bin bin i mp m.

Mch lc thng thp san bng cc mp m, chn cc thnh phn sng xoay chiu v ch

cho thnh phn mt chiu c bin khng i i n ti.

Bn p (hoc n dng) c nhim v lm n nh n p (hoc dng in) li ra trn

hai u ti cho d cc in p trc hay trti thay i trong mt gii hn no .

i y sm qua mt s loi ngun nui v mch n p.

5.1. Khi nim chung

5.1.1. t vn Mch n t mun lm vic c cn phi c ngun mt chiu cung cp. Ngun mt

chiu y c th l: Pin, acquy, bi n

5.1.2. Thng s ngun cung cpa, H s gn sng


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102

Hnh 5.2. Dng sng ngun tn hiu ra

H s gn sng c trng cho mp m ca ngun. H s gn sng c xc nh bng

biu thc:

Kgs =tb

r

U

U.100%

Trong : Kgs l h s gn sng

Url chnh lch gia n p ra ln nht v in p ra nh nht

Utb l in p ra trung bnh.

H s gn sng cng nh cng tt.

b, in trra ca ngun cung cp:

Hnh 5.3. in trra ca ngun

Ur = E Ir.Rr

E: L in p ra khi hmch Rt =

Khi Rr ln => Ur gim. V vy, yu cu n trra cng nh cng tt.


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103

5.2. Ngun chnh lu n gin

5.2.1. S khi

Hnh 5.4. S khi ngun chnh lu n gin.

Ui : 110V/220V 50 Hz

U2 : T l vi Ui . U2 c th ln hn Ui hoc c th nh hn Ui

Nu U2 > Ui : Bin p l tng p.

Nu U2 < Ui : Bin p l h p.

Khi chnh lu c nhim v chuyn t ngun xoay chiu v ngun mt chiu.

Khi lc c nhim v lm gim gn sng ca ngun tn hiu.

5.2.2. Bin p

Hnh 5.5. Bin p

Bin p gm c 2 cun scp ( c s vng 1n ) v th cp ( c s vng 2n )

Cun scp: P1 = U1.I1

Cun th cp: P2 = U2.I2

Hiu sut bin p: =1

2

P

P. 100% = 1 1

2 2

U I

U I.100%

Bin p c hiu sut tng i ln (thng trn 90%). Thc t ngi ta hay dng bin p

truyn ti nng lng.

Ta c:

1 1

2 2

U n

U n= = n

Vi bin p l tng: = 100%

2 1 2 2 1 1P P U I U I = =


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104

1 2

2 1

U In

U I = =

=> Bin p dng chuyn i n p

5.2.3. Chnh lu

a, Chnh lu na chu k nguyn l v gin thi gian

Hnh 5.6. S nguyn l v gin thi gian chnh lu na chu kb, Chnh lu hai na chu k

Cch 1: Dng hai b chnh lu na chu k c chung Rt v cng chiu D nhng in p

t vo hai D ngc pha nhau. to ra hai n p ny ngi ta dng bin p o pha.

nguyn l v gin thi gian nh hnh v:

Hnh 5.7. S nguyn l v gin thi gian chnh lu 2 na chu k

Nhn xt: Mch chnh lu hai na chu k nh trn c Ur > 0. c Ur < 0 ta ch cn o

chiu hai D.

Cch 2: Dng chnh lu cu. S nguyn l nh hnh v. Gin gn sng Ur tng

t nh trn.


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106/118

106

mun n p ra ny hon ton l mt chiu th phi dng mt mch lc tn thp bao gm

cc phn t r, C v R lm suy gim ht cc thnh phn xoay chiu trn.

Hnh 5.9. Lc gn sng trn ti

5.2.5. Cc b chnh lu bi pTrong cc b chnh lu ni trn, in p ra mt chiu khng ti cc i cng ch bng

bin th li vo xoay chiu. Trong mt s trng hp, khi cn mt n p ra c gi

tr cao hn m vn ch dng in p vo xoay chiu c bin thp th phi cn dng b

chnh lu bi p.

Hnh 5.10. B chnh lu nhn i th

Hnh 5.10 l s b chnh lu nhn i th. Trong na chu k m, dng in s np cho

t C1 qua diode D1 vi th phn cc trn t nh hnh v. ln ca th ny n gin

bng bin n p vo xoay chiu. Trong na chu k dng tip theo, dng in i s

i qua C2, D2 v np n cho C2. Nh vy n p np cho C2 lc ny s bng tng bin

n p vo xoay chiu trong na chu k m ca ngun n vo. Hay ni cch khc

th trn t C2 , tc l th cp cho ti bng 2 ln bin n p vp xoay chiu.


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107

Hnh 5.11. S b chnh lu bi p N ln

Hnh 5.11 l s b chnh lu nhn th bi p nhiu ln. Trong na chu k m th nht,

D1 thng, D2 v D3 cm; t C1 c np n th UC1 U2 ~. Na chu k dng tip theo,

D2 thng, D1 v D3 cm, dng qua D2 np cho C2in p gp i UC2 = 2U2 ~. Na chu

k m tip theo na, D3 thng, D1 v D2 cm; dng qua D3 np cho cc t C1 mc ni tip

vi C3 vi th bng U2~ + UC2 = 3 U2~. Tuy nhin do C2 c np n th U2~ nn

C3 sc np n th UC2 = 2U2~. L lun tng t cho mt th 4 v cc mt tip theo.

5.3. n p:

5.3.1. t vn :Ngun chnh lu n gin c nhc m: Khi n p li (U1) thay i, lm cho U2

thay i theo, ko theo Ur thay i => Ur khng n nh.

Khi Rt thay i cng lm cho Ur thay i theo. V vy ngun chnh lu n gin ch dngcho cc thit b c n p ra khng cn n nh.

n nh n p ra ta lm nh sau:

Hnh 5.12. S khi ngun

5.3.2. n p thng s:Ta s dng diode Zenner n p thng s

Hnh 5.13. c tuyn V-A ca Diode Zenner


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108

U = UZ , IDmin - IDmax

Hnh 5.14. n p thng s dng Diode Zenner

in trR to m lm vic trn on AB

Ur = UZ = const

5.3.3. n p so snh:a) S khi:

Hnh 5.15. S khi n p so snh

n p so snh lm vic da trn hi tip m n nh n p ra. Gm cc khi: khuch

i hiu chnh vi h s khuch i k, ton b phn cn li l khi hi tip.

c hi tip m cn tha mn iu kin:

o180k ht + =

C 2 loi n p so snh:

k = 0o,

ht =180o

ht = 0o,

k =180o

b) Mch n cc khi:


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109

* To n p chun:

Hnh 5.16. To n p chun.

*Khi phn p:

Hnh 5.17. Khi phn p

Vi IUss


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110

Hnh 5.18. Khi khuch i so snh

- Nu dng KTT th Upa v Uch c tha vo cc thun v o

ca KTT. Ty theo cch a ta c :

o

o

180

0

ht

ht

=

=* Khi khuch i hiu chnh:

Dng transistor cng sut c h s khuch i ln v tn nhit tt.

Hnh 5.19. Khi khuch i hiu chnh.

123

c) n p loi 1:k = 0

o, ht = 180o

Uk


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111

Hnh 5.20. n p loi 1

+) Nguyn l:

R8 v Dz c nhim v to ra Uch

R5, R6, R7 to ra Upa

T4 dng so snh

T1, T2, T3, R1, R2, R3: l khi K hiu chnh mc theo kiu CC k = 0

Gi s Ur ngII

pa r

I II

RU U

R R =

+tng

4BETU ng

4CTU gim.

Mc kiu EC (ht = 180o):

4CTU gim UBK Darlington gim Ur gim

Ur const

d) n p loi 2:

k = 180o, ht = 0

o


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112

Hnh 5.21. n p loi 2Trong T1, T2, T3, R1, R2, R3 l khi K hiu chnh

Tn hiu vo: chn B

Tn hiu ra: chn C

DZ, R4 to Uch

R5 R6 R7 to Upa

T4: K so snh.

Ta c:

Uch = Upa + UBE

Upa = Ir I II

RU

R R+

UBE = Uch Ir I II

RU

R R+

Khi Ur tng Upa tng UBE gim4CT

U ng Ur gim

*) Hai n p trn gi l n p so snh (hay cn gi l n p lin tc)

Trn th trng ta c IC 78XX l IC n p XX(V)Tuy nhin hai n p trn c nhc m: Khi Uv ng th s lm cho hiu sut gim, v

vy n p lin tc c nhc m l gy lng ph nng lng v chi ph cho h thng

tn nhit tn km.

Ta khc phc bng cch cho Tranzitor hot ng ch xung.


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113

5.4. n p xung:*Nguyn ln p xung:

Hnh 5.22. S khi n p xung

L nguyn l hi tip m u khin Ur thng qua vic thay i thi gian ng mca

tranzitor iu chnh lm vic ch kha.

Upa = kpa.Ur

- Khi to Uch: To n p chun Uch v Upa a KSS1 c Uo.

- Khi to n p rng ca: To n p rng ca Uo n KSS2

Hnh 5.23. Gin thi gian ca cc tn hiu


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114

Khi Tranzitor T thng c dng in qua cun L np cho tng thi tch t

ng lng trong cun cm. Khi tranzitor ngt, ng lng ny to nn 1 dy xung

ngc rt ln t ln Tranzitor v c th lm hng n. V vy ta cn mc thm mt

diode. Diode ny gi l diode hm, mc vo mch thng dng cho xung ngc,qua

diode v ti nhm bo v qu p.

- Trong khong thi gian tx, tranzitor thng bo ha.

1L I U dt

L=

- Khi UB = 0, Tranzitor tt

Pv = Uv.ILtb.tx

Pr = Urtb.ILtb.Tx

Pv Pr Uv.ILtb.tx = Urtb.ILtb.Tx

x rtb

x v

t U

T U =

xrtb v

x

tU U

T =

Nhvy ta c 3 cch n p xung:

Cch 1: Gi Tx khng i ( Tx = const ), tx v Uv thay i t l nghch vi nhau ( tng

tx bao nhiu th gim Uv by nhiu ln v ngc li )Cch 2: Gi tx = const, Tx v Uv thay i nh nhau

Cch 3: Tx, tx cng thay i theo Uv Urtb = const.

5.5. Cc vi mch n p:Ngy nay thng ngi ta ch to cc vi mch c chc nng n p vi cc tham s

chun. Cc vi mch ny do c ch to hng lot nn gi thnh cng rt r v thng

dng. Cc vi mch n p gm 2 loi: loi th nht c n p li ra cnh v loi th

hai c in p ra c thc u chnh trong mt di no .Cu to bn trong ca cc vi mch n p ny c sn hnh nh hnh 5.24


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116

n nh cao, cn phn cng sut c n u khin l cc transistor hoc vi mch

cng sut ln. khi v hnh dng ca vi mch ny cho trn hnh 5.26

.

Hnh 5.26. khi v hnh dng vi mch n p A-723.

Hnh 5.26 l mt s sng dng ca cc vi mch n p loi 7805 v 7905. Cc

loi vi thn p khc v nguyn tc cng c dng tng t. Nhn chung vic s

dng chng rt thun tin v d dng. Hnh 5.26.a l s to thn p cnh +5Vt vi mch 7805. Khi cn nng cao dng ra ca b n p c thu thm mt

transistor cng sut ph nh hnh 5.26.b. Cng vi cc transistor bn trong vi mch,

n to ra mt s Darlington cho php dng ra bn p tng ln.

Hnh 5.26. Mt s s ngun n p dng vi mch.

Trong trng hp cn th ra n p cao hn +5V nhng ch c vi mch 7805, c th dng

hnh 5.26.c trong diode n p (zener) c th UZc mc trong mch gia chn

3 ca vi mch vi t. Cch ny cho php in p rac tng ln mt lng bng UZ.

in trR dng u chnh dng ca diode n p n mt gi tr gn cnh


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117

I = (U2 UZ ) / R .

Khi cn c ngun n p lng cc, c th dng s nh hnh 5.26.d, trong s dng

bin p c cun th cp c m gia c ni t v hai vi mch n p tri du l 7805

cho li ra +5V v loi 7905 cho li ra l -5V.


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TI LIU THAM KHO

1. K thut mch n t, Phm Minh H, NXB KHKT, 1999.2. K thut mch n t 2, Nguyn Vn Tun.

3. K thut mch n t, o Thanh Ton, Phm Thanh Huyn, V Quang Sn, HGiao thng vn ti.4. Electronics Circuits, Ghausi ISBN Editor, 1982

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