bg_ktdttt
TRANSCRIPT
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MC LC
CHNG 1: MCH KHUCH I............................................................................ 41.1 Khi nim chung....................................................................................................... 4
1.1.1 nh ngha.............................................................................................................. 41.1.2 Phn loi................................................................................................................. 41.1.3 Cc thng s mch khuch i ................................................................................ 5
1.2 Mch khuch i tn hiu nh.................................................................................. 81.2.1 Phn tch mch khuch i dng BJT ch tn hiu nh tn sthp.................... 81.2.2 Phn tch mch khuch i BJT bng s tng ng vt l ........................... 121.2.3 Phn gii theo tham s h: ...................................................................................... 181.2.4 Phn tch mch khuch i Transistor trng FET ................................................ 19
1.3 Ghp cc tng khuch i ...................................................................................... 261.3.1 Ghp lin tip. ...................................................................................................... 261.3.2 Mch khuch i Darlington ................................................................................ 301.3.3 Mch lin kt chng (Cascode) ............................................................................. 311.3.4 Mch khuch i vi sai ......................................................................................... 32
1.4 Mch khuch i cng sut .................................................................................... 361.4.1 nh ngha v phn loi ....................................................................................... 361.4.2 Mch khuch i ch A .................................................................................... 381.4.3 Mch khuch i ch B .................................................................................... 411.4.4 Mch khuch i ch C .................................................................................... 46
1.5 Mch khuch i c hi tip................................................................................... 471.5.1 Cc khi nim c bn: .......................................................................................... 471.5.2 Cc phng trnh c bn ...................................................................................... 481.5.3
nh hng ca hi tip m n cc thng sca mch ........................................ 49
1.5.4 Phng php phn tch b khuch i c hi tip............ .................... ............ ...... 55
1.6 Mch khuch i thut ton v mt sng dng c bn ..................................... 581.6.1 C bn vb khuch i thut ton ...................................................................... 581.6.2 Cc tham sc bn ca b khuch i thut ton................................................. 591.6.3 Cc s c bn ca b khuch i thut ton .................................................... 61
CHNG 2 : MCH LC TCH CC ..................................................................... 732.1. Khi nim mch lc tn s ..................................................................................... 73
2.1.1. Mch lc thng ............................................................................................ 742.1.2. Mch lc tch cc............................................................................................. 752.1.3. Tham sc trng mch lc tch cc ............................................................... 75
2.2. Mch lc thng thp ............................................................................................... 792.2.1. Mch lc thng thp bc 1................................................................................ 792.2.2. Mch lc thng thp bc 2................................................................................ 81
2.3. Mch lc thng cao ................................................................................................. 832.3.1. Mch lc thng cao bc 1 ................................................................................. 832.3.2. Mch lc thng cao bc 2 ................................................................................. 85
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2.4. Mch lc chn lc v lc thng di ........................................................................ 862.4.1. Mch lc chn lc............................................................................................. 862.4.2. Mch lc thng di bng tn rng .................................................................... 89
2.5. Mch lc chn di ................................................................................................... 902.6. Thit k mch lc tch cc bc cao ......................................................................... 91CHNG 3. MCH TO DAO NG V NH THI .......................................... 923.1. Mch to dao ng iu ha .................................................................................. 92
3.1.1. Mch dao ng di pha ...................................................................................... 933.1.2. Mch dao ng cu Wien.................................................................................. 963.1.3. Mch dao ng LC........................................................................................... 973.1.4. Dao ng Thch anh......................................................................................... 99
3.2. Mch to xung ...................................................................................................... 1023.2.1. Khi nim chung............................................................................................. 1023.2.2. Mch dao ng a hi lng n...................................................................... 103
3.3. Mch nh thi 555 ............................................................................................... 109CHNG 4: MCH IU CH V GII IU CH ........................................... 1144.1. Cc mch iu chtng t................................................................................. 114
4.1.1. Khi nim iu ch.............................................................................................. 1144.1.2. iu chbin .................................................................................................. 1144.1.3. iu chn bin (SSB)................................................................................. 1224.1.4. iu tn v iu pha ....................................................................................... 128
4.2. Cc mch tch sng tng t............................................................................... 1324.2.1. Tch sng iu bin ........................................................................................ 1324.2.1.1. Cc tham sc bn ........................................................................................ 1324.2.1.2. Mch tch sng bin .................................................................................. 1334.2.2. Tch sng tn hiu iu tn v iu pha .......................................................... 136
CHNG 5: MCH TRN TN V VNG KHA PHA...................................... 1405.1. Mch trn tn ....................................................................................................... 140
5.1.1. Nguyn l trn tn .......................................................................................... 1405.1.2. Mch trn tn n .......................................................................................... 1415.1.3. Mch trn tn cn bng .................................................................................. 1425.1.4. Mch i tn cn bng vng ........................................................................... 1435.1.5. Trn tn dng IC nhn 602. ............................................................................ 145
5.2. Vng kha pha (PLL) ........................................................................................... 1475.2.1. Cu to, nguyn tc hot ng........................................................................ 1475.2.2. ng dng PLL ................................................................................................ 151
CHNG 6. MCH N NH IN P MT CHIU......................................... 1546.1. Khi nim chung ................................................................................................... 154
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6.2. Mch n p tuyn tnh .......................................................................................... 1546.2.1. Mch n p ni tip ........................................................................................ 1556.2.2. Mch n p song song .................................................................................... 157
6.3. Mch n p xung .................................................................................................. 159CHNG 7: MCH CHUYN I TNG T-S V STNG T............. 1637.1. Mch chuyn i tng t-s (ADC) ................................................................... 163
7.1.1. Nguyn tc chuyn i .................................................................................... 1657.1.2. Tham sc bn ADC...................................................................................... 1677.1.3. Cc phng php chuyn i tng t-s....................................................... 167
7.2. Mch chuyn i s-tng t............................................................................... 1757.2.1. Nguyn tc chuyn i .................................................................................... 1757.2.2. c tnh kthut ca DAC............................................................................. 1767.2.3. Phng php chuyn i DAC........................................................................ 178
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CHNG 1: MCH KHUCH I
1.1 Khi nim chung1.1.1 nh ngha
Khuch i l mt qu trnh bin i nng lng c iu khin. Nng lng ca ngun
mt cung cp 1 chiu (khng cha ng thng tin) c bin i thnh dng nng lng xoaychiu (c quy lutbin i mang thng tin cn thit). Qu trnh khuch i s to ra trn ti
mt nng lng bin i theo quy lut ca ngun tn hiu nhng cao hn v d ng in, in
p hoc cng sut.
S khi mt b khuch i trn hnh (1.1)
Hnh(1-1): S khi b khuch i
Ngun tn hiu: ngun tn hiu xoay chiu cn khuch i.
Ngun cung cp: ngun mt chiu cp cho mch khuch i hot ng
(phn cc).
Khuch i: mch khuch i c nhim v chuyn nng lng t ngun
mt chiu cung cp sang tn hiu xoay chiu cn khuch i. Mch ny
c t nht mt phn t tch cc (BJT, FET hoc Op-Amp) .
Ti: ly tn hiu c khuch i.
1.1.2 Phn loi
Mch khuch i c phn loi theo mt s cch c bn sau:
Theo dng tn hiu: khuch i tn hiu lin tc (tng t) v khuch i
tn hiu ri rc (xung s).
Theo tn s: khuch i tn thp, tn cao, khuch i tn hiu 1 chiu,
xoay chiu. Theo phn t khuch i: khuch i bng transistor lng cc,
transistor trng hoc khuch i thut ton...
Theo thng s tn hiu: khuch i cng sut, khuch i in p, khuch
i dng in.
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Theo ln tn hiu: khuch i tn hiu nh, khuch i tn hiu ln
(khuch i cng sut).
1.1.3 Cc thng s mch khuch i
1.1.3.1 Cc thng svo ra Cc thng s vo mch khuch i gm: in p vo (Uv), dng in vo
(Iv), tr khng vo (Zv), cng sut vo (Pv).
Cc thng s ra mch khuch i gm: in p ra (Ur), dng in ra (Ir),
tr khng ra (Zr), cng sut ra (Pr)
1.1.3.2 H skhuch i H s khuch i p: tnh bng t l in p ra chia in p vo, c tnh theo cng
thc:
. jru u
v
UK K e
U
o uK
: ln h s khuch i
o : Gc lch pha gia Ur v Uv. Cc b khuch i ni chung thng c =
0 (khuch i ng pha) hoc = (khuch i ngc pha).
H s khuch i dng in: tnh bng t l dng in ra v dng in vo, c tnh
theo cng thc:
r
iv
IK
I
H s truyn t tn hiu u vo: tnh bng t l tn hiu thc t a vo b khuch
i v ngun tn hiu a vo:
nv
v
n
vn
RR
R
U
UK
Hnh(1-2): S tng ng ca b khuch i
H s khuch i cng sut: tnh bng t l cng sut ra chia cng sut vo:
+
-
Ur
Rr RtRv
Rn
+
-
Un
Iv Ir
Uv
Rn: in trngun tn hiu Rv: trkhng vo
Rr: trkhng ra Rt: in trti Un: in p ngun tn hiu
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.rP u iv
PK K K
P
1.1.3.3 Hiu sut:Hiu sut ca mch khuch i l t s ca cng sut ra v cng sut ngun tn hiu
1 chiu tiu th:
o
r
P
P
B khuch i nhiu tng th tng tiu th cng sut 1 chiu nhiu nht l tng cng
sut v hiu sut ca tng c ly xp x bng hiu sut ca tng khuch i cng sut.
1.1.3.4 c tuyn bin tn sc tuyn bin tn s th hin s bin thin ca bin tn hiu khi tn s tn hiu
vo thay i. Ta c h s khuch i in p ca mt mch khuch i:
v
rj
U
UeKK .||
| |K = 1 () : c tuyn bin tn s
= 2 () : c tuyn pha tn s
Vi tn s thp ( khng cao lm), = 0 hoc 1800 th khng xt v gc lch pha .
Cc b khuch i tn thp thng c dng c tuyn bin tn s nh trn hnh (0-3):
Hnh (1-3): c tuyn bin tn s ca b khuch i tn thp
Trong :
ft: l tn s gii hn di m ti h s khuch i gim 2 ln
fc: l tn s gii hn trn m ti h s khuch i gim 2 ln
Ko: H s khuch i ln nht t c
Di thng (hay bng thng) ca b khuch i c tnh bng: c tB f f
B khuch i thng thng (di rng) c bng thng B fc
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c tuyn bin tn s gim phm vi tn thp ch yu do t ni tng (C nt) kt hp vi tr
khng vo (Rv) to b lc thng caonh trn hnh (1-4).
Hnh (1-4): nh hng ca t ni tng tn s thp
Ti tn s cao, do t k sinh u ra kt hp vi in tr ra Rrv in tr ti Rt ca tng
khuch i (Rtd = Rr // Rt) nh trn hnh (1-5).
Hnh(1-5): nh hng ca t k sinh tn s cao
1.1.3.5 c tuyn truyn tBiu th mi quan h gia i lng u ra v i lng vo.
Hnh(1-5): c tuyn truyn t in p
V d c c tuyn truyn t in p nh trn hnh (1-6), ta c:
Khi Uv Uvmin th Ur=Urmin. y l tp m ca b khuch i.
Khi Uv Uvmax th Ur = Urmax, th hin tnh cht hn ch ca tng
khuch i.
Yu cu tn hiu a vo b khuch i phi tho mn iu kin:
Uvmin Uv Uvmax
Di ng ca b khuch i c nh ngha bng t s : max
min
vk
v
UD
U
Rv
Cnt
+
-
Ur 1
Rt d
Cks+
-
Ur
CksRr
Rt
f
|K|
f
|K|
Uv
UrB
A
Uvmin Uvmax
U rmax
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1.1.3.6 T stn hiu trn tp m (S/N)Tn hiu ra gm c 2 thnh phn: tn hiu (S) v tp m (N)
T s S/N cng ln th nh hng ca tp m cng t v ngc li.
Tp m sinh ra l do cm ng ca in t trng bn ngoi nh hng vo b
khuch i: Tp m do chnh cc phn t khuch i to nn (diode,
transistor ...)
Tp m do nh hng va in li thng qua ngun chnh lu
1.2 Mch khuch i tn hiu nh
1.2.1 Phn tch mch khuch i dng BJT ch tn hiu nh tn sthp gii cc bi ton c lin quan n cc linh kin tch cc, ta phi a chng v cc
dng mch in tng ng n gin.
S tng ng ca Transistor lng cc c ngha l ta biu din linh kin ny di
dng mt s mch in tng ng hoc mt hphng trnh ton m t hot ng can mt ch c thno . Yu cu ca cc s tng ng nh sau:
n gin, ch yu nn s dng quan h bc 1.
Tnh chnh xc.
Vi Transistor lng cc, thng s dng 2 kiu s tng ng khi phn tch
mch khuch i tn hiu nh, tn s thp:
S tng ng theo tham s vt l (re)
S tng ng theo tham s h (tham s hn hp)
1.2.1.1 S tng ngBJT theo tham sh:i vi tn hiu nh c th coi BJT l mt mng 4 cc tuyn tnh nn c th dng h
phng trnh ca mng 4 cc biu din dng v p vo v ra ca Transistor.
Hnh (1-6): S Transistor nh mt mng 4 cc.
Vi BJT, dng phng trnh hn hp theo tham s h l thun li hn c. Cp
phng trnh hn hp nh sau:
2221212
2121111
..
.
UhIhI
UhIhU
Transistor
I
U
Vo Ra
U
I
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S tng ng c minh ho trn hnh (1-8)
Hnh(1-7) . S tng ng mng 4 cc tuyn tnh theo tham s h.
Trong :
h11 l trkhng vo khi u ra ngn mch, cn c k hiu l hi, c tnh
theo cng thc
111 2
1
0U
h UI
h12 l h s hi tip in p khi u vo hmch, cn c k hiu l hr,c
tnh theo cng thc:
112 1
2
0U
h IU
h21 l h s khuch i dng in khi u ra ngn mch, cn c k hiu l hf,
c tnh theo cng thc:
2
21 21
0I
h UI
h22 l h s dn np khi u vo hmch, cn c k hiu l ho, c tnh theo
cng thc sau:
222 1
2
0I
h IU
phn bit thng s ny s dng cho cu hnh mch CE, CC hay BC ngi ta dng
thm k hiu ph nh bng sau:
Loi mch Thng sh
E chung (CE) hie, hre, hfe, hoe
C chung (CC) hic, hrc, hfc, hoc
B chung (BC) hib, hrb, hfb, hob
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V d, xt vi mch E chung, cc thng sc tnh nh trn hnh (1-9).
Hnh(1-8):Tnh cc tham s h ca mch E chung
Tng t ta c vi mi cu hnh mch cc thng sc xc nh theo bng sau:
Thng s h Mch E chung
(CE)
Mch B chung
(CB)
Mch C chung
(CC)
hi ie b bh V I ib e eh V I ic b bh V I
hr re b ch V V rb e ch V V
rc b eh V V
hf fe c bh I I fb c eh I I fc e bh I I
ho oe c ch I V ob c ch I V
oc e eh I V
Ghi ch Vi mch CE v CB c: hr0
v ho0
Cc thng sh ny c cho trong bng thng s k thut ca mi BJT. Cn ch rng vi
mt BJT xc nh, cc thng s h c th thay i ph thuc vo thng k ca nh sn xut,
nhit lm vicDo , nh sn xut sa ra gi tr ln nht v nh nht. Thng thng,
khi thit k chn gi tr trung bnh gia 2 gi tr ny.
Bng di l thng s ca s tng ng hn hp vi BJT thng dng m 2N3904:
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1.2.1.2 S tng ng vt lTs tng ng theo tham s hn hp ca BJT phn trn, nhn chung hr kh
nh v c th bqua cho vic phn tch tnh ton n gin. M hnh kiu ny c gi l
m hnh tng ng hay cn gi l s tng ng vt l, c dng rt ph bin khi
phn tch cc mch BJT. im tng ng gia m hnh tng ng vt l vi s tng
ng theo tham sh nh bng trn hnh (1-10).
Hnh(1-9) . S tng ng vt l ca BJT
Trong rel in trkhuch tn Emitter, c tnh theo cng thc sau:
T Te
EQ CQ
U Ur
I I
UT: in p nhit, c gi tr bng 26mV 250C
IEQ , ICQ : dng IC v IE ch tnh ca mch
Loi
mch
S tng ng
tham s h
S ng ng vt
l
S tng ng
gn ng
CC v
CE
hie =re; hfe=ac;
ro=1/hoe
CB
hib =re; hfe=-; ro=1/hob
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1.2.2 Phn tch mch khuch i BJT bng s tng ng vt l
1.2.2.1 Nguyn tc chungNhn chung, khi phn tch mt mch khuch i tn hiu dng BJT, chng ta lm theo
cc bc sau:
Phn tch ch mt chiu (DC): Coi ton b cc t in h mch v dng cc
phng php phn tch tm im cng tc tnh nh trong gio trnh Cu kin in t.
Phn tch ch xoay chiu (AC):
B tt c cc ngun 1 chiu (ngn mch ngun p v h mch ngun
dng).
Coi tt c cc tin ngn mch ti tn s tn hiu. Cc t ny sc
xt nh hng khi tnh tn s ct thp ca mch.
Thay BJT bng m hnh tng ng tn hiu nh tn s thp.
Gii cc phng trnh tm cc thng s ca mch khuch i nh h
s khuch i dng, h s khuch i p, trkhng vo, trkhng ra.
Tnh tn s ct thp (ft) ca mch khuch i.
Di y s phn tch mt s mch cn bn ca BJT dng s tng ng vt l
(re)
1.2.2.2 Mch Emitter chung (CE):Xt mch khuch i CE nh trn hnh (2-11a), p dng nguyn tc phn tch xoay
chiu ta c s
tng ng xoay chi
u tn hiu nh
nh trn h
nh (1-11b).
Hnh(1-10) : Mch Emitter chung
Trong UVl in p vo, iv l dng in vo, Ur l in p ra, ir l in p ra. Cc
thng s mch khuch i c tnh nh sau:
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H s khuch i in p: ruv
UK
U
H s khuch i dng in: riv
iK
i
Trkhng vo:v
v
v
U
Z I
Trkhng ra Zrc tnh khi u vo bng khng (Uv = 0):
0v
rr
r U
UZ
I
Ch : Zr ca mt mch khuch i khng ph thuc vo ti. Zrc tnh bng cch
ngn mch u vo (Uv = 0), thay th ti u ra bng mt ngun ginh Ur v ly Zr = Ur/Ir.
Nhn vo s tng ng hnh (b) ta c:
in p ra: Ur= -ir.RC = - ib.RC
in p vo: Uv = ib.re
Do , h s khuch i in p c tnh nh sau:
.
. .b C C
u
b e e
i R RK
i r r
Dng vo c tnh theo cng thc sau:.
i b ev b b
B B
V i ri i i
R R
Tng trvo tnh theo cng thc:
. .
//( )
v b e e Bv B e
v b v B B e
U i r r RZ R r
i i V R R r
Thng RB ln hn nhiu so vi re(thng RB > 10re) th ta c : Zvre
Tng trra tnh theo cng thc:
0v
rr C
r U
UZ R
I
H s khuch i dng in:.br B
ib ev B e
b
B
ii RK
i ri R ri
R
Hoc c th tnh Ki khi tnh c Kuv Zvtheo cng thc sau:
. .r C v vr ri u
v v v v C C
U R Z Z i UK K
i U Z U R R
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T cc kt qu tnh ton trn, ta rt ra mt s nhn xt sau v cc tnh cht ca mch
CE:
Ku < 0 nn in p vo v ra ngc pha nhau hay cn gi mch CE l mch khuch
i o.
Mch CE thng c h s khuch i dng v p ln, trkhng vo nhvi trm
(Zvre) v trkhng ra ln hng k (Zr= RC).
Mch CE c h s khuch i dng, p v cng sut ln thng c s dng
khuch i tn hiu nh trong x l v gia cng tn hiu tng ttn s thp v trung
bnh.
Xtnh hng ca t CE
xt nh hng ca t CE, tnh ton tng t vi mch hnh (1-12).
re ib
B C
E
RBRC
iv ib
ir
+ +
- -
UrUv
RB RC
RE
Vcc
C1
C2Uv
Ur
a) S mch khuch i CE khikhng c t CE
b) S tng ng xoay chiu
RE
Hnh(10-11): Mch CE khi khng c t CE
H s khuch i p ca mch:
)()(....
.
Ee
C
Ee
C
Ebeb
Cbu
Rr
R
Rr
R
Riri
RiK
Tng trvo:)(
).(....
EeB
BEe
B
ib
Ebeb
i
ii
RrR
RRr
R
Vi
Riri
i
VZ
Nhn xt:
in trREc tc nh n nh im cng tc cho mch phn cc BJT (xem trong
gio trnh Cu kin in t). Tuy nhin n li lm gim h s khuch i do hi tip
m thnh phn xoay chiu.
khc phc dng t CE c tc dng ngn mch thnh phn hi tip m xoay chiu
th trong mch tng ng xoay chiu c RE = 0 v h s khuch i ca mch s
tng ln.
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Dnh cho Sinh vin: Tng t hy tnh cc thng s khuch i ca mt s mch CE
trn hnh(1-13).
Hnh(1-12): Mt s dng mch CE
1.2.2.3 Mch Collector chung (CC)Xt mch collector chung v s tng ng xoay chiu trn hnh(1-14), trong
RB = (R1//R2).
Hnh(1-13): Mch Collector chung
Thc hin phn tch tnh ton cc thng s ca mch khuch i cho mch trn hnh (1-
14). Ta c:
in p vo: v b e r U i r U
in p ra: ( 1)r r E e E b E U i R i R i R
H s khuch i p:( 1)
1. . ( 1). .
b Er Eu
v b e b E e E
i RU RK
U i r i R r R
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Tng trvo:. . ( 1). .
// ( )v b e b E v B e E
vvb
B
U i r i RZ R r R
Uii
R
Tng tr ra: c tnh bng rr
r
UZ
i khi ni tt u vo (UV =0) v
mc ngun Uru ra:
re
ib
B
C
E
RB RE
ibir
+
+
-
-
Uv=0 Ur
Hnh(1-14): S tnh t khng ra
Dng in ra: ( 1)r b b bE E
Ur Ur i i i i
R R
M(1 )
b o
e E e
Ur Ur Ur i i
r R r
Vy trkhng ra:1
//1 1r e E
r
e E
UrZ r R
i
r R
H s khuch i dng in:( 1)
( 1)bri
v b
iiK
i i
hoc c th tnh Kikhi bit Ku v ZV theo cng thc sau:
. vr r Ei uv v v E
Zi U RK K
i U Z R
Nhn xt:
H s khuch i in p nhhn v gn bng 1 nn mch khng c kh
nng khuch i p. Mch CC c in p ra gn bng in p vo, tn hiu
vo v ra ng pha vi nhau (Ku >0) nn cn c gi l mch lp Emitter
(tn hiu ra lp li tn hiu vo cc Emitter.
Tng trvo rt ln v trkhng ra nh (ging bin th) nn thng c
s dng phi hp trkhng.
Tng khuch i C chung c trkhng vo ln, trkhng ra nh v h s
khuch i p 1, Ur v Uv ng pha nhau nn thng c dng lm tng
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lp hay tng m, c mc u vo cc my o v u ra cc my pht
phi hp trkhng.
Mch C chung c h s khuch i dng v h s khuch i cng sut ln
nn c s dng khuch i dng, cng sut.
Tng khuch i C chung c tn s gii hn lm vic cao tng ng tn
s gii hn ca transistor.
1.2.2.4 Mch Baz chung (BC)Xt mch Baz chung v s tng ng xoay chiu nh trn hnh (1-16).
Hnh(1-15): Mch Baz chung
Tng tnh tnh ton vi cc mch C chung v E chung ta c cc kt qu sau:
Trkhng vo: //vv E e
v
UZ R r
i
Trkhng ra: Zr= RC
H s khuch i p:.e c c cr
u
v e e e e
i R R RUK
U i r r r
(ln)
H s khuch i dng in:
o E. 1 ( do R r )r C v vr
i u e
v v v c e
U R Z Z iK K
i U Z R r
Nhn xt: c im ca mch khuch i Baz chung:
B khuch i l khng o (Ku > 0)
H s khuch i p cao t l vi RC(vi trm n vi nghn ln tng
ng mch CE). Tuy nhin do tr khng vo thp (in tr tip gip BE
phn cc thun re rt thp) Zvo= 30300 nn h s khuch i ton
phn thp hn nhiu so vi mch CE. Do phm vi tn s trung b nh
khng dng mch BC.
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H s khuch i dng gn bng nn khng dng mch CB khuch
i dng in hoc cng sut
Tr khng ra bng mch CE (Zra = RC = 100k1M.)
Tn s gii hn lm vic cao (cao hn so vi s mc cc pht chung)
do in dung hi tip nh. Mch CB thng lm vic tn s cao (ln
hn 100MHz).
ng dng: S dng trong di tn s v tuyn v tr khng vo nh khong vi chc
ph hp vi tr khng 50 ca antena v ng truyn.
1.2.3 Phn gii theo tham s h:
Dng phng php ny c th tnh c thng s khuch i ca tt c cc dng mc.
Ch cn ch l:
Vi mch CE th dng cc thng s h11e, h12e, h22e,h21e
Vi mch CB th dng cc thng s h11b, h12b, h22b,h21b
Vi mch CC th dng cc thng s h11c, h12c, h22c,h21c
S tng ng tng qut ca mch theo tham s h trn hnh (1-17).
HnHnh (1-16): S tng ng theo tham s h
Trong :
Un: Ngun tn hiu vo.
Rn: in tr trong ca ngun tn hiu vo.
Rt: in tr ti.
H s khuch i dng in:r
i
v
i
K i
m 21 21 22. ' . ' .v ri h i i h i h U
thay r r tU i R ta c: 21 22 21 22. ' . . ' .v r r t i h i h U h i h i R
Vy h s khuch i dng: 21221 .
ri
v t
i hK
i h R
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H s khuch i in p:
21 21 21
22 22 22
11 1211 12 21 11 12 21
22 22
1 1 1. ( // ) . ( // ) .( // )
1 1. . . . ( // ) ( // )
v t v t t
ru
v v rv v t t
h i R h i R h RU h h h
KU h i h U
h i h h i R h h h R
h h
Tng tr vo:
11 12 11 12 11 12
2111 12 11 12
22
. . . . . . .
. .1 .
v v r v t r v t i v
vv t i t
v t
u h i h u h i h R i h i h R K i
u hZ h h R K h h R
i h R
Tng tr ra l t s gia in p v dng in u ra khi u vo ni tt
(US = 0)12
11
.r
v
n
h ui
R h
m 21 1221 22 22
21 121122
11
1. r r
r v r r r r
n r
n
h h u ui h i h u i h u Z
h hR h ih
h R
1.2.4 Phn tch mch khuch i Transistor trng FET
C thdng FET khuch i tn hiu nhnh BJT. Vi BJT, s thay i dng
in u ra (IC) c iu khin bng dng in u vo (IB) nhng vi FET th s thay
i dng in cc mng (ID) c iu khin bng in th nhu vo (UGS). BJT c
h s khuch i dng in cn FET c h dn gm.
1.2.4.1 S tng ng ca FETVi tn hiu nh tn s thp th mch tng ng xoay chiu ca FET nh trn
hnh (1-18).
rgs gmugs rd
G D
S S
ugs uds
+
--
+
iD
Hnh(1-17): S tng ng xoay chiu FET
Trong :
rgs l trkhng vo ca FET. Vi JFET, rgs khonghng chc n hng
trm M v vi MOSFET thng hng trm n hng ngn M. Do
trong thc t c th b qua rgs trong mch tng ng.
rd l tng trra ca FET c nh ngha nh sau:
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GS
DSd U const
D
Ur
I
in trcc mng rd biu th snh hng ca in p cc mng UDS ti dng cc
mng ID khi in p trn cc cng khng i. Nh vy, in tr mng rd chnh l tr
khng ra ca FET ch xoay chiu trn cc mng. Gi tr ca rd ph thuc vo im
lm vic c ththay i t vi chc kn vi chc M. Nu trong mch thit k c RDkhng ln lm (vi k) th c th b qua rd trong mch tng ng.
gm l h dn ca FET, c trng cho siu khin ca in p UGS
n dng ID, tnh theo cng thc sau:
Dm
GS
Ig
U
Cch xc nh gmtrn c tuyn truyn t c cho trn hnh (1-19).
Dm
GS
Ig
U
Hnh(1-18): Xc nh h dn gm.
V mt ton hc, cch xc nh gmnh sau:
Vi JFET v D-MOSFET
Ta c phng trnh hm truyn t:
2
( )
( )
. 1 1GS DD DSS GS GS off
GS off DSS
U II I U U
U I
M Dm
GS
Ig
U
nn ta c :
( )( )
21DSS GS
m
GS off GS off
I Ug
UU
y c tr tuyt i |UGS(off)| biu th gi tr gmlun dng.
t 0( )
2DSS
m
GS off
Ig
U biu th gi tr gm ti UGS = 0V Ta c:
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0 0
( )
1 GS Dm m m
GS off DSS
U Ig g g
U I
Trong UGS v ID l cc bin thay i tu theo v tr im cng tc tnh Q, cn
dng bo ho IDSSv in p ngt UGS(off) l cc hng s. Hnh (1-20) minh ho v d tnh
gi tr gm ti cc im phn cc khc nhau.
2
8 14GS
D
UI mA
V
Hnh (1-19): Xc nh h dn gm ti cc gi tr im phn cc
Vi E-MOSFET
Vi E-MOSFET, phng rnh hm truyn t nh sau:
2
( )D GS GS ThI k U U
Trong kl hng sc xc nh tim lm vic c trng, UGS(Th)l in p UGSngng hnh thnh knh dn.
Gi tr h dn gm vn c xc nh theo cng thc DmGS
Ig
U
. Ta c biu thc
ton hc xc nh gm cho E-MOSFET nh sau:
( )2Dm GS GS ThGS
Ig k U U
U
Nguyn l xy dng tng khuch i dng FET cng ging nh tng dng BJT. C 3 s
c bn mc FET trong mch in l cc ngun chung (SC), mng chung (DC) v cachung.
1.2.4.2 Tng khuch i cc ngun chungS khuch i cc ngun chung dng D-MOSFET knh n v s tng ng
xoay chiu c cho trn hnh (1-21).
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Hnh
Hnh (1-20): Tng khuch i cc ngun chung
in trR1, RG, RD, RSdng xc nh ch lm vic tnh ca FET.
Cc thng s cn tnh cho mch khuch i dng FET cng tng t nh vi mch
khuch i dng BJT gm c h s khuch i in p (Ku), trkhng vo (Zv) v trkhng
ra (Zr). Cch chuyn i sang s tng ng xoay chiu cng tng tnh BJT ch thays tng ng BJT bng s tng ng FET.
T ni tng C1, C2 v Cs kh ln nn coi nh ngn mch xoay chiu nn trong s
tng ng b qua cc t.
Nhn vo s tng ng ta c:
in p ra: . .( // )r m gs D d
U g u R r
in p vo: v gsU u
Vy c h s khuch i p ca mch l:
. .( // )( // )
m gs d Dru m d D
v gs
g u r RUK g r R
U u
Du m ca biu thc Ku ch ra rng tn hiu vo Uv v tn hiu ra Ur lch pha nhau
1800. Nu rd> 10RD th c th ly Ku -gm.RD.
Trkhng vo ca mch : 1 2// //v
v gs
v
UZ R R r
i . Thng in trrgs rt ln so vi R1
v R2 nn c th ly Zv R1 //R2.
Trkhng ra ca mch:0v
rr
r U
UZ
i
= RD//rd. Nu rd> 10RD ta c th ly Zr RD.
Trng hp khng c t CS.
xt nh hng ca t CS ta tnh ton cho mch khng c t CSnh trn hnh (1-22).
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Hnh 1-21: S mch cc ngun chung khi khng c CS
Dng in chy qua in trrd :.
' r S r D S
d d
U U U i Ri
r r
Dng chy qua in trRD : ' .D m gsi i g u
M .gs G S v S v D S
u U U U U U i R
Vy c :.
' . .( . )r D SD m gs m v D S
d
U i Ri i g u g U i R
r
M rD
D
Ui
R . Thay vo ta c phng trnh :
.
.( . )
rr S
r D rm v S
D d D
UU R
U R Ug U R
R r R
Gii phng trnh ny ta c h s khuch i p :
.
1 .
m Dru
D Svm S
d
g RUK
R RUg R
r
Nu b qua rdtrong s mch tng ng (khi rd ln hn nhiu so vi RD v RS) th
ta c h s khuch i p c tnh theo cng thc sau:
.
1 .m D
um S
g R
K g R
Ta thy so vi trng hp c t CS th h s khuch i p gim (1+gm.RS) ln. Nh vy
khi dng RS phn cc cho cc S ca FET th s lm gim h s khuch i. T CS c vai
tr ngn mch in trRSch xoay chiu khc phc nhc im ny. T CS c vai tr
tng ng nh t CE trong mch E chung ca BJT.
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1.2.4.3 Tng khuch i cc mng chung.
Hnh(1-22): S mch cc mng chung
H s khuch i in p:
. .( // )m gs d S r
u
v v
g u r RUK
U U
M GSrv UUU
Nn. .( // ) .( // ) .
1 .( // ) 1 .m GS d S m d S m S r
u
v v m d S m S
g U r R g r R g RUK
U U g r R g R
H s khuch i KU ph thuc vo ti xoay chiu ca tng v h dn ca FET. H s
khuch i tin ti 1 khi tng gm v RS. V vy vi tng khuch i cc mng chung nn dng
transistor c h dn ln. H s khuch i mang du dng ngha l in p vo v in p
ra ng pha vi nhau. Do tn hiu ra ng pha vi tn hiu v gn bng tn hiu vo (Ku 1)
nn c th ni tn hiu ra lp li tn hiu vo cc ngun nn mch cc mng chung cn c
gi l mch lp cc ngun(tng t mch lp Emitter ca BJT).
Tng trvo ca mch: Zv= R1//R2
Tng trra: ta thy RS song song vi rd v song song vi ngun dng in gmugs.
Nu ta thay ngun dng din ny bng mt ngun in th ni tip vi in tr
1/gmv t ngun in th ny bng 0 trong cch tnh Zr ta tm c tng tr ra
ca mch: Zr= RS//rd//(1/gm).
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1.2.4.4 Tng khuch i cc ca chung
Hnh(1-23). Mch khuch i cc ca chung
S mch khuch i cc ca chung dng JFET knh n v s tng ng xoay
chiu c cho trn hnh().
Trc tin ta tnh h
s
khuch
i p :
r
uv
U
K U
Phng trnh dng in ti nt D: 1 2.m gsi g u i
M 1S D v r
d d
U U U U i
r r
; 2
r
D
Ui
R v
gs vu U .
Thay vo phng trnh trn c:
v r rm v
d D
U U Ug U
r R
Vy h s khuch i p:
//
//(1 )d Dr
u
v d m
r RUK
U r g
Nu b qua rdtrong s tng ng, ta c kt qu: .u m DK g R
tnh trkhng vo ta vit phng trnh dng in ti nt S:
v v rv m gs
S d
U U Ui g u
R r
M r u vU K U v gs vu U . Thay vo phng trnh trn ta c:
v v u vv m v
S d
U U K U i g U
R r
Do Ku >> 1 nn c th b qua s hng cui cng trong phng trnh trn nn ta c tr
khng vo:
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1//v
v S
v m
UZ R
i g
Tr khng ra ca mch c tnh bng cch ngn mch u vo (Uv =0) nn ngun
dng gm.ugs lc ny hmch, trkhng ra Zr= (rd//RD).
1.3 Ghp cc tng khuch iTrong cc phn trc chng ta kho st cc mch ring l dng BJT v FET. Cc mch
in t thc t thng ghp cc mch ring l ny li theo cc cch khc nhau nhm t c
mt mc ch no nh tng h s khuch i dng, h s khuch i p, bin i tr
khng...Di y chng ta s cng xem xt mt s cch ghp ni cn bn v c im ca
mi cch ghp ni ny.
1.3.1 Ghp lin tip.Mt b khuch i ghp lin tip gm nhiu tng khuch i mc lin tip nh trn hnh
(1-25). Mc ch ca vic ghp tng lin tip l lm tng h s khuch i ca b khuch i
nhm t yu cu cn thit. Trong s ny tn hiu ra ca tng trc l tn hiu vo ca tng
sau. Tuy nhin, vic mc lin tip ny cn m bo yu cu tn hiu ra khng b mo.
Hnh (1-24): S khi mch khuch i ghp lin tip
Tng khuch i ghp tn lin tip c mt sc im sau:
in p u ra tng trc l in p u vo ca tng sau: Ur(i)=Uv(i+1)
H s khuch i ca ton mch bng tch cc h s khuch i ca cc tng:
1 2 1 21 1 2
rN rN r ru u u uN
v v v vN
U UU UK K K K
U U U U
Nu tnh theo n v [dB] ta c:
1 2( ) ( ) ( ) ... ( )u u u uN K dB K dB K dB K dB
Tr
khng vo c
a t
ng khu
ch i c tnh b
ng tr
khng vo c
a t
ng u: Zv= Zv1.
Trkhng vo ca tng sau sng vai tr lm ti ca tng trc: Zvi= Rt(i-1).
Trkhng ra ca tng khuch i c tnh bng trkhng ra ca tng cui: Zr=
ZrN.
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Vic ghp gia cc tng c th dng tin, bin p hay ghp trc tip. Di y
ta xt c th tng cch ghp ny.
1.3.1.1 Ghp bng tin (ghp RC)Hnh (1-26) l s ghp 2 tng khuch i lin tip bng tin. Mc ch ca mch
ny l tng h s khuch i in p.
Gi h s khuch i p ca tng tng khuch i ln lt l Ku1 v Ku2, ta c h s
khuch i ca c mch l: 1 2.r
u u u
v
UK K K
U .
Hnh(1-25): Mch dng t in ghp 2 tng khuch i BJT
H s khuch i p ca tng thc nht l: 1 211
( // )C vu
e
R ZK
r
Trong Zv2 l trkhng vo ca tng th2, ng thi cng l ti ca tng th
nht c tnh bng cng thc: v2 3 4 2 2Z ( // // )eR R r .
H s khuch i p ca tng th 2 l: 222
Cu
e
RK
r
H s khuch i ca c tng khuch i l:
1 2 21 2
1 2
( // ). C v Cu u u
e e
R Z RK K K
r r
Trkhng vo ca ton mch l: 1 1 2 1 1( // // )v v eZ Z R R r .
Trkhng ra ca ton mch l: Zr = Zr2= RC2
C1, C2, C3 l cc t cch tng. Cc t cho thnh phn xoay chiu i qua v cn thnh phn
mt chiu nn phn cc mt chiu khng nh hng ln nhau gia cc tng, im lm vic
tnh cng c cch ly.
Khi dng phng php ghp tng ny cn ch n nh hng ca tin ti cc tn s
vo khc nhau c th lm bin i pha v bin ca tn hiu.
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Ti tn s trung bnh c th b qua nh hng ca tin (chn gi trin
dung sao cho XC = 0 ti tn s trung bnh ftb)
Ti tn s thp, dung khng XC=(1/2fC) ln nn dn in ca t gim, h
p trn t s lm gim bin b tn hiu u ra mi tng dn n lm gim h
s khuch i min tn s thp. T CE cng lm gim h s khuch i v
gim tc ng kh tn hiu hi tip m AC.
Ti tn s cao: cc tham s ca transistor nh h s khuch i dng in ,
in dung tip gip CCE s bnh hng. gim v nh hng ca t CCE lm
gim h s khuch i ca transistor v lm gim h s khuch i ca mch.
Hnh (1-26): c tuyn bin tn s ca b khuch i ghp t in
1.3.1.2 Ghp bng bin pCc tng khuch i c ghp vi nhau bng bin p. S ghp 2 tng khuch i
bng bin p nh trn hnh (1-28).
Hnh(1-27): Ghp tng khuch i bng bin p.
Cun s cp ca bin p ng vai tr thay cho in trti ca tng khuch i th nht.Bin p cch ly in p mt chiu gia cc tng khuch i v tng h s khuch i chung
tu thuc vo bin p tng hay gim.
Phng php ny c u im l khng c dng mt chiu trn ti v t hiu sut cao.
Tuy nhin nhc im l cng knh, c tuyn tn s khng bng phng trong di tn nn
khng s dng trong cc mch khuch i tn hiu nh tn s thp m dng trong tn s cao
iu chnh knh thu, bin p to mch cng hng.
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1.3.1.3 Ghp trc tipy l mt dng lin kt kh ph bin trong cc mch khuch i nht l trong k thut
ch to vi mch. S mch ghp trc tip dng BJT c cho trn hnh (1-29).
Hnh(1-28): Ghp tng khuch i trc tip
Mch ghp trc tip c cc u im:
Trnh c nh hng ca cc t ghp tng tn s thp.
Trnh c s cng knh cho mch.
in th tnh ca tng u cung cp in th tnh cho tng sau.
Nhc im:
Snh hng ch tnh ln nhau gia cc tng.
in p mt chiu cung cp thng c gi tr ln nu dng cng loi BJT. Vn
chnh trong ghp trc tip l n nh phn cc. Cch tnh phn cc thng
c p dng trn ton b mch m khng th tnh ring tng tng.
Khi tnh phn cc tnh cn ch rng in p mt chiu cung cp cho cc Baz ca BJT
th 2 (T2) l VB2do in p ti cc Collector ca T1(VC1) cung cp: VB2= VC1.
Thng s mch khuch i:
Tr khng vo ca mch bng tr khng vo ca tng khuch i th nht:
1 1 2 1 1( // // )v v eZ Z R R r .
H s khuch i in p : 1 2.u u uK K K
Trong Ku1 l h s khuch i p ca tng khuch i th nht c tnh theo cng
thc: 1 211
//C vu
e
R ZK
r .
Zv2 l trkhng vo ca tng khuch i th2 ng thi gi vai tr l ti ca tng th
nht c gi tr l:Zv2=2re2
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H s khuch i p ca tng khuch i th 2: 221
Cu
e
RK
r
Vy h s khuch i p ca c tng khuch i l:
1 2 21 2
1 2
//C v Cu u u
e e
R Z RK K K
r r
H s khuch i dng in:2
vi u
C
ZK K
R
Trkhng ra ca mch:Zr=RC2
1.3.2 Mch khuch i DarlingtonL dng lin kt rt thng dng gia 2 transistor (cng loi hoc khc loi)
Mch khuch i Darlington c tc dng tng trkhng vo v tng h s khuch i.
S mch khuch i Darlington dng Transistor cng loi v khc loi trn hnh(1-30).
c) S tng ng
T1
T2
B
C
E
TD
DB
C
E
T1
T2
B
C
E
a) Dng Transistor cng loi b) D ng Transistor kh c loi
Hnh(1-29): Mch khuch i Darlington
S lin kt gia 2 transistor nh trn tng ng vi mt transistor duy nht c h s
khuch i dng in l D= 12. Nu 2 transistor ng nht th D = 2.
V dng lin kt ny rt thng dng v thch hp cho vic nng cng sut nn ngy nay
ngi ta thng ch to cc lin kt ny di dng mt transistor duy nht gi l transistor
Darlington. Bng thng s transistor Darlington 2N999.
Thng s iu kin Min Max
UBE IC = 100mA 1,8V
hfc
(D)
IC = 10mA
IC = 100mA
4.000
7.000 70.000
tm hiu cc c tnh ca mch khuch i Darlington ta xt mch ng dng thng
dng c dng nh hnh (1-31).
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Hnh(1-30): Mch ng dng transistor Darlington
Cch tnh phn cc tnh cho mch ny cng tng tnh vi mch BJT thng thng,
ta c dng phn cc tnh:Q
BEB
B D E
Vcc U I
R R
Cn ch gi trD v UBE thng ln hn trng hp 2 transistor ri mc theo kiu
Darlington.
in p vo: 211211112211 )1()1( vvvvBvBvBvBBE rrRrIrIrIrIU =>
V mt xoay chiu transistor Darlington c mc theo kiu cc gp chung CC) nn c
tng trvo ln (hng trm k), tng trra nh (chng chc )v h s khuch i p xp
x bng 1, h s khuch i dng rt ln (hng nghn). Mch khuch i Darlington c in
trvo rt ln nn dng vo nh, mch Darlington c nhiu u im nh n nh cng tc
cao, mo tn hiu nh
1.3.3 Mch lin kt chng (Cascode)
lm vic tn scao thng dng mch khuch i Bazo chung. Tuy nhin tng
khuch i Baz chung c trkhng vo nh nn h s khuch i ton phn thp. Do ,
khc phc nhc im ny mc thm u vo mt tng khuch i Emitter chung to nn
tng khuch i Cascode.
Trong lin kt ny, mt transistor ghp chng ln mt transistor khc. S hnh (1-32)
l mch Cascode vi mt tng cc pht chung ghp chng ln mt tng cc gc chung.
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T2
T1
RC2
RE
RB1
RB2
RB3
CB
C1
CE
C2
+Vcc
Uv
Ur2
Ur1
re1 ib1
B1 C1
E1
RB re2 ie2
E2 C2
B2
RC2Uv
iv
Ur1 Ur2
a) S mch Cascode b) S tng ng xoay chiu
RB= RB2//RB3
Hnh(1-31): Mch khuch i xp chng Cascode
Ts tng ng xoay chiu ta tnh c cc thng s ca mch Cascode nh sau:
H s khuch i dng in: 2 1 1i c bK i i
Trkhng vo: 2 3 1 1/ // //v v v B B eZ U i R R r
H s khuch i p: 2 2 1 1 2 21 1 1 1 1 1 1
.
. .c C b C C r
u
v b e b e e
i R i R RUK
U i r i r r
Trkhng ra:Zr= RC2
T cc thng s ca mch khuch i Cascode ta c mt s nhn xt dau:
Trkhng vo ln (bng trkhng vo ca mch khuch i CE)
H s khuch i p ton phn ln do trkhng vo ln
Tn s lm vic ln (bng tn s gii hn ca mch CB)
1.3.4 Mch khuch i vi sai
Khuch i tn hiu 1 chiu:
Trong nhiu trng hp phi khuch i tn hiu bin thin chm (tn hiu 1 chiu), yu
cu bng tn ca b khuch i c tn s ct thp ft0. t c iu ny cn tng gi tr
t ni tng (Xc =1/C) gim suy hao tn hiu trn ttn s thp.
Trong thc t khng th tng gi tr t qu ln, gim ft cn ghp trc tip gia cc
tng khuch i khuch i tn hiu mt chiu.
Tuy nhin, ghp trc tip c mt snhc im sau:
Kh thc hin c ch cng tc cng khuch i (ch A) do snh
hng mt chiu ln nhau gia cc tng khuch i v ngun tn hiu.
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Tri im khng: l hin tng c tn hiu ra khi khng c tn hiu vo.
Nguyn nhn l do cc thng s ca transistor thay i theo nhit dn n
im lm vic ca transistor thay i.
i vi cc mch khuch i tn hiu xoay chiu th khng cn quan tm n hin
tng tri v qua cc tin tri skhng c khuch i. Tri chnh hng n h s
khuch i ca mch v c khc phc bng hi tip m.i vi cc mch khuch i tn hiu mt chiu tri cng c khuch i v a n
u ra nh tn hiu nn phi tm cch gim tri s dng mch khuch i vi sai.
Khuch i vi sai:
Cc mch khuch i xt khuch i trc tip tn hiu vo cn mch khuch i vi sai
ch khuch i sai lch gia 2 tn hiu vo nn in p ra ca n ch chu tc ng ca hiu
cc in p tri ca transistor. Do mc tri ca b khuch i vi sai rt thp. Trng hp
mch hon ton i xng th tri c kh hon ton.
1.3.4.1 Dng mch cn bnCu to ca mch gm 2 tng khuch i Emitter chung i xng c minh ho trn
hnh (1-33). Cc phn ttng ng ging nhau v mi c tnh: RC1 = RC2 = RC, RB1= RB2 =
RB, VCC = VEE, RE ni chung, hai transistor Q1 v Q2 ging nhau v mi c tnh.
Hnh(1-32): S mch vi sai cn bn
Mch c 2 u vo Uv1 v Uv2v 2 u ra l Ur1 v Ur2. C 2 phngphp ly tn hiu
ra:
u ra vi sai: Tn hiu c ly gia 2 cc Collector Ur= Ur1 Ur2 u ra n cc: Tn hiu c ly gia mt cc Collector v t
Mch c phn cc bng 2 ngun in thi xng c cc in thcc Baz bng
0V.
Tu thuc vo cch a tn hiu vo m mch khuch i vi sai c cc ch hot ng
khc nhau :
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Chn: a mt tn hiu vo (Uv1 hoc Uv2) cn tn hiu vo cn li ni
t.
Ch vi sai : a 2 tn hiu vo Uv1 Uv2.
Chng pha : a 2 tn hiu vo bng nhau Uv1 = Uv2.
1.3.4.2 Phn tch phn c
c tnh
Do mch c tnh cht i xng nn ta c cc dng phn cc tnh qua 2 transistor l nh
nhau :IC1 = IC2=IC; IE1 = IE2 = IE. S phn cc tnh mch khuch i vi sai trn hnh(2-
34).
Xc nh im phn cc tnh:
Phng trnh u vo: RBIB + UBE + 2REIE VEE = 0
2
Q Q
EE BEE C
BE
V UI I
RR
Khi mch hon ton i xng in th cc B bng 0 ta c RB.IB = 0 nn :
2Q QEE BE
E C
E
V UI I
R
Phng trnh ng ti tnh ca mch: VCC + VEE = RC.IC + UCE + 2RE.IE
Coi IC IE UCEQ = (VCC + VEE) (RC + 2RE) ICQ
Hnh(1-33): S phn cc tnh mch khuch i vi sai
1.3.4.3 Kho st thng sca mcha. Chng pha:
Trong chng pha, 2 tn hiu vo bng nhau Uv1 = Uv2. Do mch c tnh cht i
xng nn ta c ib1 = ib2v Ur1 = Ur2 = Ur.
Do mch hon ton i xng nn ta ch xt mt na mch. Ch dng chy qua RE l 2IE.
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Hnh(1-34): S tng ngxoay chiu mch khuch i vi sai
H s khuch i ng pha Kphc xc nh theo cng thc 11
rc
v
UK
U
Lm tng tnh cc phn trc ta tm c :
in p ra : 1 1 1 1 1r c C b C U i R i R
in p vo :1 1 1 1
2 2v b e e E b e b E
U i r i R i r i R
H s khuch i ng pha: 1
12 2
r C C
C
v e E E
U R RK
U r R R
Nu chn RE rt ln th h s khuch i ng pha Kc s tin n 0.
b. Ch vi saiVi ch vi sai ta c Uv1 Uv2. H s khuch i vi sai c tnh theo cng thc:
1 2
1 2
r rVS
v v
U UK
U U
Ta c:
1 1 1 2 1 1 2( ) ( )v b e e e E b e b b E U i r i i R i r i i R
2 2 1 2 2 1 2( ) ( )v b e e e E b e b b E U i r i i R i r i i R
Vy 2 2 1 2( )v v e b bU U r i i
Ta c :
1 1 1r c C b C
U i R i R
2 2 2r c C b C U i R i R
Vy 1 2 1 2 2 1( )r r b C b C b b C U U i R i R i i R
Vy h s khuch i vi sai :
1 2
1 2
Cr rVS
v v e
RU UK
U U r
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T s G = KVS/KC cng ln th thnh phn chung t nh hng n u ra.
Mun tng G phi gim KCv tng KVS.Nh vy phi dng RE ln. Tuy nhin iu ny
lm cho VCC v VEE phi ln. Phng php tt nht l dng ngun dng in. Ngun dng
in thay cho RE phi c 2 c tnh:
Cp dng in khng i
C tng trZS ln thay cho RE.
Mch khuch i vi sai c th mt cn bng do cc linh kin thng nh in tr, t
in hoc cc linh kin tch cc nh diode, transistor khng hon ton ging nhau. Khi mch
mt cn bng th khng cn duy tr c si xng dn n thnh phn chung c th to ra
tn hiu vi sai u ra.
Gii php khc phc hin tng mt cn bng:
Chn cc linh kin tht ging nhau
Gia dng in phn cc nh sai s vin trto ra in th vi sai nh
Thm bin trRE cn bng dng in phn cc.
Ch to theo phng php vi mch.
1.4 Mch khuch i cng sut1.4.1 nh ngha v phn loi
Tng khuch i cng sut to tn hiu ra c cng sut ln v p ng cc yu cu ca cc
ph ti nh cho loa, cho cc cun li tia (trong tivi), cho m t...Trong thc t, h thng
khuch i gm nhiu tng khuch i ghp vi nhau v tng cui cng thng l tngkhuch i cng sut. Tn hiu vo h thng c mc in p thp c khuch i thng qua
cc khuch i in p ri mi a vo tng khuch i cng sut. Tng khuch i cng sut
s dng cc mch khuch i cng sut.
Mch khuch i cng sut cn p ng cc yu cu sau:
Tn hiu khuch i phi t c cng sut yu cu.
mo tn hiu phi nh.
Hiu sut cao.
Cc mch cng sut c cng sut ra tvi trm mW n vi trm W, bin tn hiu ra
ln, transistor lm vic trong min khng tuyn tnh nn khng th dng mch tng ng
tn hiu nhphn tch nh chng trc m phi dng phng php th.
Da theo ch lm vic ca transistor ngi ta chia mch khuch i cng sut thnh
cc loi chnh sau:
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ChA: Transistor c phn cc cng tuyn tnh. Tn hiu c khuch i
gn nh tuyn tnh. Dng tn hiu ra c gi nguyn ch bin i vbin so
vi tn hiu vo. Ch ny c hiu sut thp (vi ti in trdi 25%) nhng
mo phi tuyn nh nht. im lm vic tnh trong ch A phi nm gia
ng ti tnh.
ChB: Transistor c phn cc vng ngt. Tn hiu ra ch c trong mt nachu k (m hoc dng). Ch ny c hiu sut ln (78%), tuy mo xuyn tm
ln nhng c th khc phc bng cch kt hp vi ch AB hoc dng hi tip
m.
Ch AB: C tnh cht chuyn tip gia ch A v chB. Transistor c
phn cc gn vng ngt tham gia vo vic gim mo khi tn hiu vo c bin
nh. Tn hiu ra c hn mt na chu k.
ChC: Transistor c phn cc di vng ngt. Tn hiu ra ch c trong nh
hn mt n
a chu k
, hi
u su
t l
n (>78%) nhng mo rt l
n. M
ch ch
Cc dng trong cc mch khuch i cao tn c ti l cc khung cng hng
chn lc tn s mong mun v t hiu sut cao.
Ch D: Transistor lm vic nh mt kho in tng m. Di tc dng ca
tn hiu vo iu khin transistor thng bo ho l kho ng, dng ICt cc i,
cn kho mkhi transistor ngt, dng IC = 0.
IC
UCE
A
AB
B
CVng ngt
Vng boha
Gii hncng sut
Hnh 1-36: Phn loi mch khuch i cng sut theo v tr im lm vic
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Hnh 1.37: Dng tn hiu ra ca cc mch khuch i cng sut
1.4.2 Mch khuch i ch AMch khuch i ch A hot ng vi tn hiu vo ln hay nhu khuch i tn hiu
vo mt cch tuyn tnh. Mch khuch i chny khng gy mo bin v pha ca tn
hiu. im lm vic c chn chnh gia ng ti tn hiu c khuch i ln nht
m khng b mo.
Transistor trong ch A lm vic trong min tch cc trong sut chu k tn hiu. Dng
collector trong mch ch A lun khc khng ngay c khi khng c tn hiu vo do dng
phn cc tnh ICQ khc khng. Transistor s tiu th cng sut ngay c khi khng lm vic
(khng khuch i tn hiu). Do hiu sut ca mch khuch i ch A rt thp, ti a l
25% (vi ti in tr), vi ti cun cm hoc bin p th khong 50%.
1.4.2.1 Mch khuch i ch A ti in tr:Xt mch khuch i cng sut ch A ti in trn gin trn hnh(). y l mch
khuch i E chung, n gin ta b qua phn tnh ton mch phn cc cho mch ny.Trong mch khuch i A chung c bn ny cng khng dng cun cm hoc my bin p.
im khc nhau gia mch ny vi mch khuch i tn hiu nh l tn hiu vo (Uv) c
bin ln (hng trm mV).
RC
ur
uv uCE
iC
+
-
Vcc
ti
Hnh 1.38.: Mch khuch i ch A ti in tr
Gi sim cng tc tnh Q nm chnh gia c ti tnh, ta c UCEQ = VCC/2 v ICQ =
VCC/2RC. Khi a tn hiu xoay chiu uv vo, dng iCv in thuCE(tn hiu u ra) s thay
i quanh im lm vic tnh Q. Khi tn hiu u vo ln, bin tn hiu ra sthay i ln
quanh im Q trong gii hn: iC= (0; 2ICQ
) v uCE= (0; Vcc).
Cng sut tiu th tc thi trn transistor l (b qua dng iB):
.Q CE C p u i
Vi tn hiu vo dng sin, dng in iCv in p uCEc vit di dng sau:
( ) sinC CQ pi t I I t
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V ( ) sinCE CEQ pu t U U t
Xt trong trng hp tn hiu ra t c bin ln nht ta c: Ip=ICQ v
Up=UCEQ=Vcc/2.Do , cng sut tiu th tc thi trn transistor l:
2(1 sin )2
CC CQ
Q
V Ip t
Ta thy khi khng c tn hiu vo transitor tiu th cng sut ln nht l (VCCICQ/2).
Hiu sutchuyn i cng sut ca tng khuch i c tnh theo cng thc sau:
L
S
P
P
Trong :
LP : l cng sut xoay chiu trung bnh trn ti
SP : l cng sut trung bnh ca ngun cp VCC.
Vi mch khuch i chA ang xt khi a tn hiu vo dng sin th cng sut ra
trung bnh trn ti l:
1 1(max)
2 2 2 4
CC CQCCL p p CQ
V IVP U I I
Cng sut trung bnh ca ngun cp VCC l:
.S CC CQP V I
Do hiu sut ln nht ca tng khuch i l:
1
4(max) 25%.
CC CQ
CC CQ
V I
V I
Cn ch rng hiu sut ca tng khuch i sthay i khi ta ni ti vo u ra ca
b khuch i. Hiu sut ny kh thp, do cc mch khuch i ch A kiu ny thng
khng c dng khi cn cng sut tn hiu ln hn 1W.
Trong thc t th in p v dng in ra ln nht s nhhn cc gi tr VCC/2 v ICQ
trnh a transistor vo trng thi bo ho hoc kho to mo phi tuyn. Do , hiu sutthc t ca cc mch khuch i chA thng nhhn 20% nn khng tn dng c kh
nng lm vic ca phn t khuch i, cng sut ra thp thng dng khuch i tn hiu
nh.
1.4.2.2 Mch khuch i ch A ghp bin pMch khuch i ch A ghp bin p s ti u ho c hiu qu ca ti. Hnh () l
mch khuch i ch A mc kiu E chung vi ti ghp bin p trn cc C.
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R1
RE
Vcc
C1
R2CE
RL
a:1 iL
uv
+
-u2
+
-u1
. .
Vcc 2Vcc
2ICQ
ICQ
ng ti tnh, dc =-1/RE
ng ti ng, dc =-1/RL
UCE
IC
(a) (b)
Hnh 1.38. Mch khuch i ch A ghp bin p
ng ti tnh v ng ti ng trn th hnh (1.38). Nu bqua in trca my
bin p v gi s RE rt nh, ta c in p UCEQ ca im cng tc tnh l:
CEQ CC U V
Trkhng ti my bin p l:
' 2L LR a R
Trong a l t l s vng dy ca cun s cp v cun th cp. tn hiu ra ln nht,
t l s vng dy phi tho mn biu thc sau :
' 22
2CC CC
L L
CQ CQ
V VR a R
I I
Cng sut trung bnh ti a a n ti bng cng sut trung bnh ti a a n cun s
cp trong trng hp bin p l tng, do c:1
(max)2
L CC CQP V I
Vi VCC vICQ l bin ln nht ca tn hiu dng sin. Nu b qua cng sut tiu th
trn cc in trphn cc R1 v R2 (do dng IB nh), ta c cng sut trung bnh ca ngun
cp VCC l :
S CC CQP V I
Do hiu sut chuyn i cng sut max ca mch khuch i ch A ghp bin p
ny l :
(max)(max) 50%
L
S
P
P
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1.4.3 Mch khuch i ch B
1.4.3.1 Mch khuch i y ko i xng b (ngc)Hnh (1.39) l mch khuch i ch B gm 2 transistor khc loi. Khi khng c tn
hiu vo uv = 0, c2 transistor u trong trng thi ngt v in p ra ur = 0. Gi sin p
UBE transitor dn l 0.6V th in p ra s duy tr l 0 khi in p vo trong khong -0.6V
u v 0.6V.
Hnh(1.39): Mch khuch i y ko b chB c bn.
Khi uvdng v ln hn 0.6V th Q1 s dn v hot ng nh mt mch lp Emitter s
cung cp dng in iLdng trn ti. Q2trong trng hp ny s ngt do tip gip B-E phn
cc ngc. Khi uv m v ln hn 0.6V th Q2 dn v hot ng nh mt mch lp Emitter
cung cp dng iL m trn ti.
Mch ny c gi l mch y-ko b. Hai transistor Q1 v Q2 s thay nhau dn trong
mi na chu k.
Q2 ngt trong khi Q1 dn v ngc li. Khng c khuch i p nhng h s khuch i
dng in ln khuch i cng sut. Mi transistor c phn cc ch B v ch dntrong na chu k. Tuy nhin c mt s suy bin nhc bit khi tn hiu nh(mo im 0).
Mo im 0
Hnh (1.39) l c tuyn truyn t in p ca mch ny. Khi mt trong hai transistor dn,
khuch i in p (hay dc ca c tuyn) vc bn bng 1 (do tnh cht ca mch lp
Emitter). Trong di in p gn 0V, c2 transistor u ngt v in p ra bng 0 v to ra
mo im 0nh trn hnh(1.39) vi tn hiu vo dng sin. Mo im 0 ny c thc khc
phc bng cch phn cc tnh vi gi tr nh cho c Q1 v Q2 v mch khuch i kiu ny
hot ng trong ch AB. Mch ny sc trnh by trong phn tip theo.
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Hnh 1.39: c tuyn truyn t v dng sng mch khuch i ch B
Hiu sut
Xt trng hp l tng ca mch trn, khi in p UBE transistor dn l 0V, ngha
l mi transistor s dn hon ton trong mt na chu k tn hiu sin. Mch ny l mch chB v dng tn hiu ra trn ti s lp li dng tn hiu vo.
im cng tc tnh Q c dng ICQ = 0 hay c hai transisor u ngt. Do , cng sut tiu
th trn mi transistor khi khng c tn hiu vo bng 0.
in p ra cho mch ch B ny l:
sino p
u U t
Trong in p ra ln nht Up = VCC.
Cng sut tiu th tc thi trn transistor Q1 l :
1 1 1.
Q CE C P u i
Dng collectorc tnh theo cng thc :
1
sin ; 0
0; 2
p
C L
Ut t
i R
t
in p uCE1c vit theo cng thc:
1 sinCE CC pu V U t
Do , tng cng sut tiu th tc thi trn transistor Q1 l:
1
sin sinpQ CC pL
UP V U t t
R
vi 0 t
V PQ1 = 0 vi 2t
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Do cng sut tiu th trung bnh trn Q1 l:
1
2
4
CC p p
Q
L L
V U UP
R R
Cng sut tiu th trung bnh trn Q2ng bng cng sut tiu th trung bnh trn Q1 do
tnh cht i xng.
th hm cng sut tiu th ca mi Transitor theo gi trin p ra ln nht Up. Cng
sut tiu th ln nht l:
1
2
2(max) CC
Q
L
VP
R khi
2CC
p
VU
Cng sut trung bnh trn ti l:
21
2p
L
L
UP
R
Do mi ngun cp dng in trong mi na chu k, dng cp trung bnh l /( )p LU R . Do
, cng sut trung bnh ca mi ngun cp l:
( )pS S CC
L
UP P V
R
Cng sut trung bnh do c 2 ngun cp l:
2. ( )p
S CC
L
UP V
R
Hiu sut ca mch l:
21
2
42. ( )
p
pL L
p CCSCC
L
U
UP R
U VPV
R
Hiu sut ln nht khi tn hiu ra ln nht vi Up=VCC. Ta c:
max 78.5%
4
Ta thy hiu sut chuyn i ca mch ch B ln hn nhiu so vi mch khuch i
ch A tiu chun (ti in tr).
Thc t, hiu sut ca mch ch B thp hn gi tr trn do cc suy hao khc ca mch
v in p ra nh (Up) phi nhhn VCCtrnh a transistor vo min bo ho. Khi bin
in p ra tng th mo tn hiu ra cng tng. mo mc chp nhn c, in p ra
nh Upthng nhhn VCCvi Volt.
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Theo kt qu trn ta thy cng sut tiu th ln nht trn transistor khi 2p CCU V .
Ti gi tr ny, hiu sut ca mch khuch i ch B l:
2.50%
4 4CC
p
CC CC
VU
V V
khc phc mo im 0 trong mch khuch i ch B, dng mch y ko ch
AB nh hnh (1.40).
Hnh1.40: Mch khuch i y ko i xng b ch AB
in trR l in trphn cc dn dng n 2 diode D1 v D2 to in p khng i gia
Q1 v Q2. Hai transistor Q1 v Q2 c cng VBE nn I1 = I2. Khi in p ri trn diode bng
VBE, ta c ti im A v im C c in p bng VBE, im B c in p l 2VBE.
Hnh b. l c tuyn truyn t. Khi uv = 0 (khng c tn hiu vo) th ur= VBE nn khng
c mo im 0.
1.4.3.2 Mch khuch i y ko dng bin pDng mch c bn ca mch y ko dng bin p nh sau
Hnh(1.41): Mch khuch i y ko ghp bin p dng n cng loi
Trong s mch hnh (1.41), bin p o pha u vo c nhim v to 2 in p vo
ngc pha nhau a ti 2 transistor Q1 v Q2.
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Khi khng c tn hiu vo, c Q1 v Q2u ngt, khng c dng in trong mch, in p
ra trn ti bng 0.
Khi c tn hiu vo, trong na chu kdng ca tn hiu, Q1 dn. Dng i1chy qua bin
p u ra to cm ng cp cho ti nn trn ti c na sng dng. Trong na chu k ny, tn
hiu a vo Q2 m nn Q2 ngt.
n na chu k tip theo, tn hiu a vo Q2dng nn Q2 dn. Dng i2 chy qua binp u ra to cm ng cung cp cho ti nn trn ti c na sng dng. Trong khi , tn hiu
a vo Q1 m nn Q1 ngt.
Do i1v i2 chy ngc chiu nhau trong bin p ra nn in th cm ng bn cun th cp
cng ngc pha nhau, chng kt hp vi nhau to c chu k ca tn hiu trn ti.
Tuy nhin khi bt u mt chu k, transistor khng dn in ngay m phi chkhi bin
vt qua in thngng VBE. Do tng khuch i y ko ghp bin p ch B cng b
mo khi tn hiu vo nh(mo im 0). khc phc dng mch chAB nh hnh (1.42).
Hnh (1.42): Mch khuch i y ko ch AB ghp bin p
Trong mch chAB ny, hai in tr R1 v R2dng cp dng in tnh cho Q1 v
Q2 ch AB sao cho gi tr dng tnh:
max)20
1
10
1( cCQ II
Icmaxl gi tr dng IC ln nht.
in trRE l in trn nh nhit (thng vi ) do transistor cng sut lm vic vidng ln nn thng c nhit cao lm thay i cc thng s ca mch.
Trong cc mch khuch i y ko ghp bin p th hiu sut ca tng c tnh nh sau:
= b.a.B
Trong b.a l hiu sut ca my bin p (khong t 80%90%), Bl hiu sut khuch
i ca tng khuch i ch B khi khng ghp bin p. Theo kt qu tnh ton phn trn ta
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c B (max)= 78.54%. Vy c hiu sut ln nht ca tng khuch i ch B ghp bin p
l:
max = b.a.B(max) = 60% 70%
1.4.3.3 Mch khuch i kt cui n vi 1 ngun cung cp s dng ch 1 ngun cung cp nh s mch h
nh (1.43) th ti s phi c ni
vi mt t in c gi tr cao (khong vi trm mF). Trong trng hp ny, in p trn t s
l hng s trong sut chu k hot ng, ging nh mt ngun cung cp th 2. Nu 2
Transistor ging nhau, ti im chung A c in p Vcc/2 v t s duy tr in p ny.
Q2
Q1
+Vcc
uv
Bin po phau vo
C
RL
UCo
+A
Hnh1.43 : Mch khuch i kt cui n vi 1 ngun cp
Nh vy, hot ng ca mch s ging nh trng hp 2 ngun cung cp. Khi Q1 dn,
in p cung cp cho mch s l hiu ca Vcc v in p trn t, tc l bng Vcc/2. Cn khi
Q2 dn, ch c ngun cung cp bi t l hot ng, cng bng Vcc/2.
TC ng vai tr nh ngun 1 chiu vi gi trin p l: UCo = Vcc/2
iu kin chn gi tr tC nh sau:
1 1L
d d L
R hay CC R
Trong d l tn s gii hn di ca b khuch i.
V d vi tn s gii hn difd=100Hz, in trti RL = 8 th gi tr ca t C
khong vi nghn F.
1.4.4 Mch khuch i ch C
Mch khuch i chC transistor c phn cc trong min ngt, im lm vic cn thp
hn im ngt. Ti mt sim khi tn hiu ln vt qu ngng ngt (trong na chu
kdng ca tn hiu) th mi xut hin tn hiu ra. Do trong ch ny transistor ch dn
trong khong nhhn na chu k. Vi tn hiu vo hnh sin, tn hiu ra s l cc xung vi
rng nhhn na chu knh hnh (1.44). Tn hiu ra ca mch khuch i ch C l nhng
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xung hp. Mo trong trng hp ny l rt ln nn khng s dng tng khuch i n hoc
tng y ko.
Mch khuch i ch C c khnng cung cp cng sut ln vi hiu sut ln hn 78.5%,
tuy nhin ch C to mo ln trong tn hiu ra. Cc mch khuch i ch C ch yu c
ng dng trong khuch i tn s cao dng ti cng hng RLC thng dng trong cc my
pht ca Tivi hoc i. y l mt lnh vc cn c xem xt ring (trong phn K thut siucao tn) nn trong phn ny khng phn tch cc mch khuch i cng sut ch C
Hnh 1.44 : Dng sng ra ca mch khuch i ch C.
1.5 Mch khuch i c hi tip1.5.1 Cc khi nim c bn:Hi tip l vic a mt phn tn hiu u ra (in p hoc dng in) trvu vo thng
qua mng hi tip.
Hi tip ng vai tr rt quan trng trong k thut mch tng t, cho php ci thin cc tnh
cht ca b khuch i, nng cao cht lng ca b khuch i.
Hnh 1.45: S khi b khuch i c hi tip
K: h s khuch i ; Kht: h s hi tip; Xv: Tn hiu vo; Xh: Tn hiu hiu; Xr: Tn hiu ra;
Xht: tn hiu hi tip
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Ngi ta phn loi 2 loi hi tip c bn:
Hi tip m: tn hiu hi tip ngc pha vi tn hiu vo lm yu tn hiu vo
Hi tip dng: tn hiu hi tip ng pha vi tn hiu vo lm mnh tn hiu vo
( thng dng trong cc mch to dao ng)
Mch in ca b khuch i c hi tip c chia lm loi
Hi tip ni tip in p: Tn hiu hi tip a vu vo ni tip vi ngun tn hiu
ban u v t l vi in p u ra
Hi tip song song in p: Tn hiu hi tip a vu vo song song vi ngun tn
hiu ban u v t l vi in p ra
Hi tip ni tip dng in: Tn hiu hi tip vu vo ni tip vi ngun tn hiu
ban u v t l vi dng in ra
Hi tip song song dng in: tn hiu hi tip vu vo song song vi ngun tn
hiu ban u v t l vi dng in ra
a. Hi tip ni tip - in p b. Hi tip song song- in p
c. Hi tip ni tip dng in c. Hi tip song song dng in
Hnh 1.46: Phn loi mch hi tip
1.5.2 Cc phng trnh c bn
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Hnh 1.47: S khi ton phn ca b khuch i c hi tip
Tt c 4 loi mch hi tip trn u c th quy vs khi tng qut trn.
Ts ta c mi quan h sau:
Xr= K.Xh ; Xv= KnXn
Xh= Xv Xht ; Xht = Kht.Xr
Ta c h s khuch i ca mch c hi tip:htv
r
KK
K
X
XK
.1'
H s khuch i ton phn:n
n
rtp
KKX
XK .'
t: Kv = K.Kht l h s khuch i vng
g = 1+ Kv= 1 + K.Kht: l su hi tip
Khi |g| > 1 => |K| < |K| : tng ng vi hi tip mKhi |g| < 1 => |K| > |K| : tng ng vi hi tip dng
Trng hp c bit: nu K >>1 thht
hthtv
r
KK
K
KK
K
X
XK
11
1
.1'
Vy nu mch c hi tip c h s khuch i rt ln th hm truyn t ca n ch ph thuc
vo mng hi tip nn sthay i cc tham s ca cc phn t tch cc v tp tn ca n
khng nh hng n cc tnh cht ca b khuch i c hi tip. V vy, mun xy dng cc
b khuch i chnh xc, phi dng linh kin (thng l in tr) chnh xc trong khu hi
tip.
1.5.3 nh hng ca hi tip m n cc thng sca mch1.5.3.1 Hi tip m lm gim h skhuch iTa c h s khuch i ca mch c hi tip:
g
K
KK
KK
ht
.1
'
Hi tip m g > 1 nn K < K hi tip m lm gim h s khuch i.
1.5.3.2 n nh h skhuch iCc thng s ca cc phn t tch cc (BJT hay FET) thay i rt nhiu theo nhit v vi
mi linh kin cng loi cng khc nhau. Do khi nhit thay i hoc khi thay i cc
linh kin th h s khuch i ca mch sthay i.
Xt mch c hi tip:htv
r
KK
K
X
XK
.1'
Ly vi phn 2 v
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50
22
'
).1().1(
.).1.(
htht
htht
KK
dK
KK
dKKKKKdKdK
=>ht
KK
KdK
K
dK
.1
/'
'
Vy h s khuch i c hi tip sn nh gp (1+K.Kht) ln khi khng c hi tip. Khi thc
hin hi tip m su th h s khuch i ca mch ch cn ph thuc vo mch hi tip.
Thng thng mch hi tip thng c xy dng bng in trnn h s Kht rt bn vng.
1.5.3.3 Gim moGm:
Mo tn sdo khuch i khng ng u cc tn s khc nhau
Mo phi tuyn do c tuyn khng tuyn tnh ca transistor l pht sinh cc hi chng
ln tn hiu c khuch i lm mo tn hiu u ra.
Vy u ra ngoi thnh phn tn hiu vo c khuch i cn c mt thnh phn nhiu do
mo gi l D.
Tn hiu u ra: Xr= K.Xv +D
Khi c hi tip m, nu gi Xvkhng i th tn hiu ra s gim v K < K ng thi mo cng
gim theo do t l vi K. Khi c hi tip m, mch khuch i K vn cho thnh phn nhiu D
nhng u ra ca mch ton phn nhiu ch cn l D.
D= D Kht.K.D => D(1+ K.Kht) = D =>
htKK
DD
.1'
Vy nhiu cng gim i (1+K.Kht) ln khi c hi tip m.
1.5.3.4 Gia tng di tn hot ngKhi c hi tip m di tn ca mch sc mrng (1+K.Kht) ln so vi khng c hi tip
Ko
Ko
ft fcf
K
ft fc
Hnh 1.48. Di tn ca mch
Trong : Ko = Ko/(1+K.Kht): H s khuch i gim (1+K.Kht) ln
Bng thng: B = (1+K.Kht).B: Bng thng c mrng (1+K.Kht) ln
1.5.3.5 in tru voin tru vo Rv = Uv /Iv
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Hnh 1.49. S tng ng u vo mch khuch i
Chia lm 2 trng hp: hi tip song song v hi tip ni tip
a. Hi tip song song
Thay ngun hi tip bng ngun dng in p dng cng dng in
Rh
Iv Ih
I
IhtRht
Xr
Kht
K
Hnh 1.50. S tng ng tnh Zv hi tip song song
in trra hi tip bng in trra ca ngun hi tip.
I l dng in chy trn in trra cu ngun hi tip : I = Uv / Rht ( do Uv gy nn)
Khi khng hi tip Iht = 0 Iv = Ih + I
+ Rv = Uv / Iv = Uv / (Ih + I)
+ Yv = 1/Rv = Ih/ Uv + I/Uv = 1/Rh + 1/ Rht
i vi ngun dng in in trra rt ln ( l tng = ), i vi ngun in p in trra
rt nh ( l tng = 0)
Vy c 1/Rrht
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Vyg
RR v
v '
Vy b khuch i hi tip song song c in trvo gim g ln so vi b khuch i khng
hi tip
b. Hi tip ni tip
Thay ngun hi tip bng ngun in p p dng cng in p
Hnh 1.51. S tng ng tnh Zv hi tip ni tip
Rrht: in trra ca ngun hi tip ( 0)
U: l in p ra trn in trra ca ngun hi tip do Iv gy nn
U = Iv. Rrht
Khi khng hi tip: Uht = 0
Uv = Uh + U
Rv = Uv /Iv = Uh/Iv + U/Iv = Rh + Rrht Rh
Khi c hi tip :
Uht = Uh + U + Uht = Uh + U + Xr.Kht = Uh + U+ Uh.K.Kht = g.Uh + U
Rv = Uv / Iv g.Rv
Kt lun: B khuch i hi tip ni tip c Rv tng ln g ln so vi khng hi tip.
1.5.3.6 in tru ra Cch tnh in trra
Ur= Urh - IrRr
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+ Urh: L in p ra khi hmch ti
r
r
Ir
rr
I
U
I
UR
(Du - biu thUr v Ir ngc pha nhau)
Xt trng hp gii hn: Hmch v ngn mch
r
r
rng
rh
rhrng
rngrhr
IU
IU
IIUUR
a. Hi tip in p
Thay ngun tn hiu ra bng ngun in p
Rrk
Urh
Xv Xh
Xht
Kht
Rvht
K
RtUr
Hnh 1.52. S tng ng tnh Zr hi tip in p
Khi cha c hi tip: Rr= Rrk// Rvht , do Rrkrt nh nn c th ly Rr Rrk
Khi c hi tip:
+ Urh = Xh . Kh = Xv . Kh = Xv .hth
h
KK
K
.1
+ Irng =rk
hh
rk
rh
R
KX
R
U .
Khi ngn mch Ur= 0 khng c hi tip Xht = 0 Xv = Xh
Vy Irng =rk
hv
R
KX .
in trra khi c hi tip:
g
R
g
R
RK
X
KKK
X
I
UR rrk
rk
hv
hth
hv
rng
rhr
.
).1(.
'
Vy khi hi tip m in p, in trra gim g ln so vi khng hi tip
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b. Hi tip m dng in:
Thay ngun tn hiu ra b khuch i l ngun dng in.
rng
rh
rI
UR
+ Irng: dng in ra khi ngn mch. H s khuch i khi gi l h s khuch i ngnmch.
Irng = Xh.Kng ( K ngn mch) = Xv .htng
ng
KK
K
.1
Urh = Irng.Rrk = Xh.Kng.Rrk
Trong Rrk: in trra ca b khuch i.
Khi hmch s khng c dng qua in trvo hi tip ngha l mt hi tip, ta c
Xht = 0 Xv = Xh Urh = Xv.Kng.Rrk.
in trra khi c hi tip: Rr =
rkng
h
rkngv
rng
rh Rg
g
KX
RKX
I
U.
.
..
in trra khi khng hi tip: Rr= Rrk+ Rvht Rrk( do Rrk>> Rvht)
Vy khi c hi tip m dng in th in trra stng g ln so vi khng hi tip.
Tng kt cc loi hi tip:
Cc loi hi tip:
Hi tip dng
K > K
Hi tip m
K < K
Hi tip in p
Xht = K.Ura
Hi tip dng in
Xht = K.Ira
Hi tip ni tip
Uv = Uh + Uht
Hi tip song song
Iv = Ih + Iht
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Thng s k thut
Cc thng s k thut Hi tip
m dng
in ni
tip
Hi tip
m in p
ni tip
Hi tip m
in p
song song
Hi tip m
dng in
song song
Tng tr vo: Zv ZV.g ZV.g Zv /g Zv /g
Tng tr ra: Zr Zr.g Zr /g Zr/g Zr.g
khuch i in
p: KU
Ku/g Ku/g Ku/g Ku/g
rng bng thng:
B
B.g B.g B.g B.g
Trong g =1K.Kht
Cc mch khuch i hi tip m lm tng tng tr u vo thng dng cho tng tin
khuch i, khng lm gim bin ca tn hiu hu ch, cc mch hi tip m lm gim
tng tr u ra thng dng cho cc tng cui(cng sut), tng kh nng cp dng cho ti.
Ngoi cc thng s thng k trn, mch hi tip cn c tc dng gim bin nhiu,
gim mo phi tuyn v mo tn s.
1.5.4 Phng php phn tch b khuch i c hi tip
1.5.4.1 Phngphp chung Bc 1: Phn tch loi hi tip
+ Nu l hi tip in p: Xra = Ura
+ Nu l hi tip dng in: Xra = Ira
+ Nu l hi tip song song cng dng in: Xv = Iv; Xh = Ih; Xht = Iht
+ Nu l hi tip ni tip cng in p: Xv = Uv ; Xh = Uh ; Xht = Uht
Bc 2: Lp cc phng trnh
+ Xra = K.Xh K = .
+ Xht = Kht.Xra Kht =
+ Xv = Kn.Xn Kn =.
Bc 3: Vs khi v p dng cc cng thc tnh K, Ktph
V d 1: Cho mch hi tip nh hnh v
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Hnh 1.53. Mch khuch i hi tip m in p ni tip
Bc 1 : Xc nh loi hi tip
+ Thnh phn hi tip: in trRE ni tu ra vu vo
+ Uht = Ur y l hi tip in p.
+ Do phn t hi tip RE, ngun tn hiu u vo ca b khuch i (UBE) mc ni tip vi
nhau nn y l hi tip ni tip
+ Uht = URe = Ura = IE.RE
Khi Uhttng UBE = Uh = Uv Uht s gim IE s gim Ur= Uht s gim hi tip
m (chng li s bin i ban u)
Vy y l mch hi tip m ni tip in p
Ta c: Xv = Uv
Xht = Uht
Xh = Uh = UBE
Bc 2 : Lp phng trnh
Ur = f(Uh) Ur= f(UBE)
Ur= IE.RE = IB (1+)RE = EBE
BE RR
U)1(
Ch : RBEl in trxoay chiu ca transistor ( UBEo/RBEo) ;B
BEBE
I
UR
+ H s khuch i: Xr= f (Xh) K =
BE
E
BE
r
h
r
R
R
U
U
U
U
1
+ H s hi tip
Xht = f (Xr) , Uht = Ur Kht = 1r
ht
r
ht
U
U
X
X
1kHz
En
nR
bi
er
B1R B2R ER+
-
Vi
2Cp
ERB2R
B1R
Q7
1Cp
Vcc
C
Uv UrUrUv
BE
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+ Xv = f (Xn) Uv = f(Un) =nv
nv
RR
UR
.
( thao t s phn p)
Rv = RB1 // RB2 //RvT ( RvT: in trvo ca transistor )
Bc 3: vs khi
Hnh 1.54. S khi mch hi tip
11
.1'
hthtKKK
KK (nu hi tip m su g >> 1)
nv
vn
ht
ntf
RR
RK
KK
KKK
.1
.' (nu hi tip m su)
1.5.4.2 Phng php chuyn i s khiPhng php: Ts khi thc t vit phng trnh quan h ca tt ccc i lng
trong mch. Sau lp s khi theo cc phng trnh trn v chuyn i s
khi thnh s khi tiu chun bit ca b khuch i c hi tip ri p dng
cng thc tnh cc thng s khuch i.
Mt vi php chuyn i s khi
Hnh 1.55. Mt s php chuyn i s khi
A BX1 X2X3 C = A.B
X1 X3
A
B
X1 X4C =
A/(1+A.B)X1 X4
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1.6 Mch khuch i thut ton v mt sng dng c bn1.6.1 C bn vb khuch i thut tonVit tt OP-amp ( operational amplifier). Khuch i thut ton v c bn l mch s dng
cho tnh ton : cng, tr, bin i cng tr Ngy nay, KDTT c s dng nh mt b
khuch i a chc nng v l cc IC tiu biu trong cc mch tng t.
B khuch i thut ton ging khuch i 1 chiu thng thng l c th khuch i U, P
hoc I. Tuy nhin, n c im khc khuch i thng thng l cc b khuch i thng
thng thng c kt cu khc nhau tu theo yu cu s dng, cn KTT v c bn c kt
cu bn trong ging nhau c to nn t tng khuch i vi sai. Vic s dng ca n khc
nhau tu theo cc phn t mc mch ngoi.
m bo tnh cht trn b KTT cn c nhng c im sau:
H s khuch i vi sai ln ( Kd ln)
H s khuch i ng pha nh ( KC nh)
Trkhng vo rt ln
Trkhng ra rt nh
p ng tn s khng i
B KTT l vi mch tch hp c h s khuch i rt ln. Chng thng c hai u vo tn
hiu, mt u ra, hai u vo cp ngun, v cc chn b in p lch, b tn s (thng
thng b KTT l IC c 8 chn dng DIP).S mch v hnh dng thc t caIC KTT
741.
Hnh 1.56: Cu to bn trong v hnh dng thc t ca IC 741
Chn 7, 4: c p ngu n cung c p+/-VChn 2: u vo oChn 3: u vo khng oChn 1, 5: iu chnh lch 0Chn 6: u ra
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Nh vy, KTT thc cht l mt mch bao gm mt b khuch i vi sai tng vo, cc b
khuch i m v cui cng l b khuch i cng sut. Cc mch khuch i ny c th l
transistor lng cc (BJT) hoc transistor trng (FET), v vy cc thng s ca cc b
KTT cng khc nhau t nhiu.
K hiu khuch i thut ton (Hnh 1-57)
Hnh 1-57: K hiu khuch i thut ton UP, IP: in p v dng vo ca thun
UN, IN: in p v dng vo ca o
Ud: in p vi sai (in p sai lch gia 2 u vo thut ton)
Ud = UP - UN
Quan h gia u vo v u ra ca KTT
Kd>0 : H s khuch i vi sai ca KTT ( 105 106 )
Nu UN = 0 Ur = Kd .UP> 0 in p ra ng pha vi in p vo ca P
l ca thun
Nu UP = 0 Ur = - Kd.UN< 0 in p ra ngc pha vi in p vo
ca N l ca o.
Ngoi ra cng mt s cc khc:
Cc cc hi tip m in p gia cc tng vi sai
Cc cc ngun cung cp gia cc tng vi sai Cc cc iu chnh im khng
1.6.2 Cc tham sc bn ca b khuch i thut ton1.6.2.1 H s khuch i hiu ( h s khuch i vi sai) Kd
NP
r
d
r
dUU
U
U
UK
Uo = Kd. Ud= K ( UP UN)
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Kd 104 rt ln , Urmax Ec= (10 20)V
Ud = Ur /Ud rt nh c mV.
1.6.2.2 c tuyn truyn t in p ca khuch i thut ton
Hnh 1-58: c tuyn truyn t ca KTT
Do Kd ln nn phn khuch i c dc ln. H s khuch i Kd s suy gim khi ln min
tn s cao do s ph thuc cc tham s transistor v in dung k sinh trong s .
1.6.2.3 Dng vo tnh v in p lch khngKhuch i mt chiu c hin tng tri im khng do s ph thuc vo nhit ca cc
tham s transistor. cn bng ban u cho OA thng cho mt trong cc u vo ca n
mt in p ph thch hp hoc mt in tr iu chnh dng thin p u vo.
Dng vo tnh: 2NP
t
III
vi UP = UN = 0
Dng vo lch khng : Io = IP IN vi UP = UN= 0 (thng Io = 0.1 It)
in p lch khng: U o = UP UN khi Ur= 0
Trong thc t khi Ud= 0 th Urvn khc 0 do in p lch khng u vo gy ra. in p
lch khng l in p cn t vo u vo OA Ur= 0
Cc gi tr I0 v Uou ph thuc nhit khi nhit thay i s dn n hin
tng tri dng lchkhng v in p lch khng.
Vi KTT l tng th cc gi tr Io v Uo bng 0.
1.6.2.4 T snn tn hiu ng phaT s nn tn hiu ng pha CMRR(common mode rejection ratio): Nu t vo u vo o
v u vo khng o cc in p bng nhau th theo l thuyt Ur phi bng 0. Nhng trn
thc t li khng nh vy, lc ny s c:
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Ur = Kc.Ucm
KC: H s khuch i ng pha ( vi khuch i thut ton l tng KC = 0)
Ucm: in p ng pha
T s nn ng pha : G = Kd / Kc = (103 105) hoc theo n v dB vi:
G[dB]= 20logC
d
KK khong (76 100)dB
H s nn ng pha s nh gi kh nng lm vic ca b KTT thc t so vi b KTT l
tng. Vi KTT l tng G.
Urph(V)
Uc(mV)Ucmax+
Ucmax-
Hnh 1-59: c tuyn truyn t in p ng pha
Khi in p ng pha tng th in p ra cng tng. Nu UC> UCmax th in p ra tng rtmnh nn yu cu nhit bin i trong khong cho php Uc < UCmax
1.6.3 Cc s c bn ca b khuch i thut ton n nh ch cng tc ca KTT s dng mch hi tip m.
to dao ng dng hi tip dng.
1.6.3.1 B khuch i oS mch khuch i o c bn nh hnh 1-60
Hnh 1-60: Mch khuch i o
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Ca P c ni t, ca N c ni vi in tr u vo Z iv in tr hi tip Zf .
Tn hiu vo c a vo ca o. Do OA c Rv nn dng vo cc ca 0 ( IPIN 0)
C: Kd= Ur /Ud= Ur/ (UP UN)
M UP = 0 UN = - Ur /Kd
M Kd
nn UN
0Vi KTT l tng coi UP = UN
Phng trnh dng in ti nt N:
i
f
v
ruv
i
f
r
f
Nr
i
N
Z
Z
U
UKU
Z
ZU
Z
UU
Z
UUv
.0
Vy h s khuch i p:i
f
uZ
ZK
Trkhng vo: Zi = Vi /ii = ZiNhn xt:
Khi tn hiu vo ca N thng qua Zi v hi tip l Zf th h s khuch i ch ph
thuc vo t s (Zf/Zi) m khng ph thuc vo h s khuch i thut ton (Kd)
Ku< 0 ngha l in p ra ngc pha in p vo, y c gi l mch khuch i
o.
Tng khuch i o c tr khng vo nh ( = Zi). Nu tng tr khng vo s lm
gim h s khuch i.
Nu cho Zi = Zht th Ku = -1 to tng o lp li in p
Nu cho Zi =0 th dng in vo Iv = -Ur / Z f Ur= -IV.Zf in p ra t l vi
dng vo b bin i dng sang p.
1.6.3.2 Mch khuch i khng oS mch khuch i khng o nh hnh 1-61.
Hnh 1-61: Mch khuch i khng o
in p vo (Vi) c t vo ca thun. Mch hi tip in p t vo ca o.
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Do UP = UN Vi = V1 = V2
V if ii
f
fZ
VVi 20
ifi
iZ
V
Z
VV
Z
Vi 2202
Vy h s khuch i p:
i
f
i
VZ
Z
V
VA 10
Nhn xt:
Zf, Zi c th c bt k dng no.
V0 v Vi cng c th c bt k dng no.
Khi Zf, Zi l in tr thun th u ra V0 s c cng pha vi u vo Vi (nn mch
c gi l mch khuch i khng o v uvo (+) c gi l u vo khng
o).
Zf cng ng vai tr hi tip m. tng khuch i Av, ta c th tng Zf hoc
gim Zi.
Mch khuch i c tn hiu 1 chiu khi Zf v Zi l in tr thun. Mch cng gi
nguyn tnh cht khng o v c cng cng thc vi trng hp ca tn hiu xoay
chiu.
Khi Zf=0 ta c Av=1 V0=Vi hoc Zi= ta cng c Av=1 v V0=Vi (hnh 1.62 ). Lc
ny mch c gi l mch voltage follower thng c dng lm mch m
(buffer) v c tng tr vo ln v tng tr ra nh nh mch cc thu chung BJT
Hnh 1.62: Mch m in p
1.6.3.3 Mch khuch i tngS mch nh hnh 1.63
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R1V1
V2
Vn
R2
Rn
i1
i2
in
V2=0
V1
Rf
if
V0
Hnh 1.63: Mch khuchi tng
Cc dng in chy qua cc in tr l:
n
n
nR
Vi
R
Vi
R
Vi ;....;;
2
22
1
11
Tng cc dng in ny chy qua Rfv to thnh V0 nn ta c:
n
j
jj
n
nf
vkVR
V
R
V
R
VRV
10
2
2
1
10 .....
Trong :n
f
n
ff
R
Rk
R
Rk
R
Rk ;....;
2
2
1
1
Nu: Rf=R1=R2=...=Rn th ta c:
n
j
jvV
10
Tn hiu u ra bng tng cc tn hiu uvo nhng ngc pha. Ta ch l Vil 1 in th
bt k c th l mt chiu hoc xoay chiu
1.6.3.4 Mch khuch i hiuDng mch c bn hnh 1.64.
Ri
V1
V2Vm
Rf
V0Vn
Ri
Rf
Hnh 1.64: Mch khuch i hiu
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Ta c:
if
f
nmRR
RVVV
1
Dng in vo t V2 qua Ri s qua Rfnn:
f
m
i
m
RVV
RVV 02
Thay tr s ca Vm vo biu thc trn ta tm c:
210 VVR
RV
i
f
Nu Rf=Ri ta c: V0=(V1-V2)
1.6.3.5 Mch tch phnS mch nh hnh 1.65.
R1V1
V2
Vn
R2
Rn
i
C
i
V0
Hnh 1.65: Mch tch phn
Dng in u vo:
n
n
R
V
R
V
R
VI ...
2
2
1
1
Dng ny np vo t C v to ra V0
t
n
n
t
dtR
V
R
V
R
V
CdtI
CV
0 2
2
1
1
0
....1
.1
0
Hay
t
n
n
dtVCR
VCR
VCR
V0
2
2
1
1
0 ..1
....1
.1
t
n
j jjdtVkV
010
..
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Vi kJ l h s khuch i ca cc u vo: kj = 1/Rj.C
c bit khi R1= R2 = . = Rn = R th:
t
n
j jdtV
RCV
0
10..
1
Vy in p ra s bng tch phn ca in p vo chia cho hng s thi gian = RC
Bin c th c nh ngha nh l thi gian cn thit cho in p Vr t ti bin bng
vi in p vo, bt u t iu kin 0 v vi in p vo l hng s.
Nu xt n iu kin ban u UC0l in p trn t C ti t = 0 th ta c:
v0 = 01
1C
t
o
n
j
iUdtv
RC
Xt c tuyn bin tn si vi khuch i thun hoc khuch i o th h s khuch i khng ph thuc vo tn
s tn hiu vo () vi KTT l tng.Nhng trong thc t th n s suy gim tn s cao
do s ph thuc ca Kd vo tn s c dng nh hnh 1.66.
Hnh 1.66: Suy gim h s khuch i khi tn s tng.
i vi mch tch phn ta c |Ku | = ()
Xt tn hiu vo dng sin vi bin Uvm ta c: Uv = Uvm. sint v in p ra ban u (trn t
C) = 0 ta c:
RCRCU
UK
RC
UU
tURC
dttURC
dttUvRC
U
oo
vm
rmu
vmrm
vmvmr
1;
1
cos.1
.sin.1
)(1
Ku[dB] = 20log|Ku|=20logo 20 log = b-a.x, y l phng trnh ng thng vi :
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b = 20logo
a = 20
x = log
Ta c c tuyn bin tn s ca mch tch phn nh hnh (1.67)
Hnh 1.67: c tuyn bin tn s ca mch tch phn
Khi x = log = 0 |Ku| dB = b = 20 logo
|Ku| dB = 0 khi log = logo = o
Khi |Ku|dB = 0 Ku = 1 gi o l tn s n v.
c tnh tn s ca ca mch tch phn gim khi tn s tng , y l mch lc thng thp.
nh hng ca in p lch khng
Xt vi b KTT thc, ta c th tm c in p lch khng, xut hin nh l in p DC ti
u vo v khi c tch phn s xut hin ti u ra nh l mt in p tng tuyn
tnh.Tng t, mt phn ca dng thin p cng c tch phn, to nn s thay i ca in
p ra.Hai nguyn nhn gy li trn thc t s a b KTT n trng thi bo ho. y chnh l
mt hn ch ca mch. Vn ny s c khc phc bi vic ni thm 1 in tr gia u
vo khng o v t, b nh hng ca dng thin p; ng thi thm in tr mc song
song vi t C trung ho nhhng ca in p lch nh sau.
Hnh 1.68: Mch tch phn thc t
ng dng ca mch tch phn c dng rng ri trong cc my tnh tng t gii cc
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phng trnh vi phn v trong s m phng cc h vt l. Trong cc mch nh vy t c
np vi tr s ban u cn thit v in p c duy tr vi h mch ph cho ti khi vic
gii c bt u vi t = 0.
1.6.3.6 Mch vi phn
Hnh 1.69: S mch vi phn
Tn hiu vi np vo t C bng dng in iic tr s:dt
dvCi ii
y cng chnh l dng in chy qua in tr R, vy:
dt
dvRCV i0
Du tr biu th s ngc pha ca Urv vi phn ca Uv
Xt quan h theo tn s: cho tn hiu vo dng sin: Uv = Uvm.sin t
=>
RC
CR
U
UKUCRU
tURCdt
tdUCRtUr
o
ovm
rm
uvmrm
vmvm
1;..||...
cos..sin
...)(
Theo n v dB: Ku[dB] = 20log - 20logo= ax b phng trnh ng thng vi o l
tn s n v.
Mch vi phn c tnh cht ca mch lc thng cao do Kutng khi tn s tng.
|Ku|dB
log w
-20log wo
log wo
Hnh 1.70: c tuyn bin tn s mch vi phn
Rf
C
if
Vi
V0
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Vn tp m
Tp m lun xut hin trong cc mch in t s tr nn mnh hn trong cc mch ly vi
phn. Tp m c xu hng thay i t ngt, bt ng to ra cc xung nhn. V tn hiu ra
ca b vi phn thc t t l vi tc bin thin ca tn hiu vo nn nhng thay i t
ngt lm tng tp m ti u ra. c bit khi tn s cao th tp m cng c khuchi nhiu do Kutng khi tn s tng. C th gim c vn tp m nu s dng tn
s thp.
khc phc mt phn no, mc thm mt in tr ni tip vi t C u vo nh hnh
1.71. Lc ny mch ch c c tnh vi phn tt khi tn s ca tn hiu nh hn 1/2Ri.C.
Hnh 1.71: Mch vi phn thc t
1.6.3.7 Mch so snhMch so snh s so snh bin in p a vo vi mt in p chun (Uch). Thng thng
in p chun c nh trc c nh.
Mch so snh mc 0 khi in p chun c a vo ca o, in p vo a ti ca thun
(Hnh 1.72)
Hnh 1.72: Mch so snh mc 0 in p vo ca khng o
Hot ng:
Khi Ei< Vref th Ud = Ei - Vref< 0 Vo = - Urmax mc bo ho m
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Khi Ei> Vref th Ud= Ei - Vref> 0 Vo = + Urmax mc bo ho dng
Hiu in p Ud s