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MCH DAO NG M TN
I)S NGUYN L:
R1 R5 R6 R7 R9
R3 R11R10R8R2 R4 VR
+ - + -
+ -+ -
+
-
C1
C2
C3
C4
C5
C6
C7
C8
Ura
+9V
Trong cc thng s u vo:
R1 = R9 = 100KR2 = R3 = 10KR4 = R5 = R7 = R10 = 560 1.2KR8 = 1
5R11 = 2.2K VR = 2 10K
C1 = C2 = 10nF (103)C8 =(104)C3 = C4 = C5 = C6 = C7 = 10FR6 =
220320KT1,T2,T3=C828
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II.S LP RP :
R4R 3
R 2
R1 R 5 R6
C 1
2TT1 T3
R7R9
R10 R11R8
VR+
_
+ _
+ -
+ -
C2
C3
+ -C4 C6
C5
C7
C8
B1B3B2C1 C2 C3
E1 E2 E3
III . NGUYN L HOT NG:Mch gm ba khi chnh:
- Khi dao ng.
- Khi khuych i.
- Khi chng nhiu.
A)Khi dao ng.
_n T1: T1 lm nhim v dao ng nh c dng hi tip dng t ccCollector ca
T2 a v. Dng hi tip dng c a qua mch lc tn s
mc theo kiu cu Wien gm t C1, C2, v in tr R 2 ,R 3.
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_Ta c cng thc tnh tn s dao ng ca mch cu Wien:
Nu ly R 2 = R 3 = R v C2 = C3 = C th tn s dao ng s l:
_ Do mun thay i tn s lm vic ca mch, ta gi nguyn gi tr ca t,thay
i R hoc ngc li. (Thng thay i gi tr R d hn C).Khi sa dng xung: Gm t
C3 v bin tr VR, gy ra hi tip m xoay
chiu lm thay i bin ra ca tn hiu. Do vy, y l mch sa dngxung.
+Cc in tr R 1, R 2 , R 4 , R 5 dng phn p cho n T1.
r a
C 1
C 2
v a o
R 3
R 2
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+T C4 l t ni tng. T C5 dng n nh dng mt chiu.
B) Khi khuych i.
1)Khi khuych i chnh:
- Hai n T1 v T2 thuc loi C828 mc theo kiu EC do ch s khuych i ln
hn 1 v c hai n u lm vic ch A.- Cc in tr R1, R2, R4, R5 dng phn p
cho n T1.
- Cc in tr R6, R7, R8 dng phn p cho n T2.- T C3 v bin tr VR hi
tip ni tip in p (hi tip mxoay chiu) l mch sa dng xung.- T ha C4 l t
ni tng.- T C5 dng n nh dng mt chiu.2) Khi khuych i m :- C6 l t ghp
tng, t C8 lm nhim v chng nhiu, t C7ly tn hiu ra, n T3 mc theo kiu C
chung.- Khi ta cp cho mch mt in p 9V DCV do hin tng hitip dng gy t
kch lm cho T1 c dao ng, in p hnhsin t u vo s c khuych i qua T1 (lm
vic ch A) ly ti trn C nn qua T1 thu c 1 tn hiu hnh sin nhngngc pha
vi tn hiu u vo.- Tn hiu ny tip tc c ua n T2, qua T2 tn hiu lic
khuych i ln 1 ln na (T2 cng lm vic ch A)a tn hiu ra ng pha vi tn
hiu u vo.- T3 c tc dng nh dng c mc theo kiu C chung, do c h s
khuych i bng 1, tn hiu vo v ra la ngpha v ti ly ra trn E. T1 v T2
mc theo kiu E chung, n
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T1 lm nhim v dao ng a hi c hi tip dng, n T2ch yu lm nhim v
khuych i.- n T2 : L mch khuych i n EC c nhim v khuych i dng
xung
ca n T1.- Cc in tr R 6, R 7, R 8 dng phn p cho T2.
- T1,T2 mc theo kiu EC nn c h s khuch i ln hn 1
C)Khi chng nhiu v nhn dng xung u ra.
- T3 mc CC, do h s khuy ch i K1, cho thy khi 3 hon ton khngnh
hng g n bin v dng ca xung ra. Do trong qu trnh iu
khin mch, ta khng cn iu chnh cc in tr ca khi ny.- Cc in tr R 9,
R 10 dng phn p cho n T3.
- T C7 dng trit tiu dng mt chiu u ra.
IV)CCH IU CHNH 1.So snh s lp rp vi mch lp (cha cm VR).Cm ngun
9V,dngng h o thang o DC 10V o Uce ca c 3 n .
Uce ca ca transistor : T1 : 7.5 8.3V , T2 : 4 5-6V , T3 : 4 5V
_Sau chuyn thang o AC 10V ,o Ura c in p : 4V_Nu nh khng c in p 5V
coi nh mc mch sai,ta lp li .Quan st trnoxylo c dng xung vung2.Lp VR
vo,iu chnh VR c xung hnh sin nhng bin thpiu chnh VR cho xung ra ln
th c hin tng moC 2 dng mo : mo trn v mo di+Thay i R6 khi c mo c
ngha l thay i ch nh bin ca n. Nu tn hiu u ra c bin ln v vn mo,ta
tip tuc iu chnh R4+Tip tc quan st trn oxyl,thy bin ra vn cn hi thp
v b mo t thta iu chnh R5.Sau khi quan st trn oxyl thy dng xung mo
mmun tng bin th ta iu chnh R6 theo tng 100.
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_Nu Ura l ng thng th ta o li in p Uce s thy in p khc vi lc ban u
,c bit l n T2.Dng m hn tho vt n i,thay n khc.
V)THNG S O C Gi tr cc in tr :
R1 = R9 = 100KR2 = R3 = 10KR4 = R5 = R7 = 560 R10 =1KR8 = 1
R11 = 2.2KVR = 2 10KC1 = C2 = 10nF (103)C8(104)C3 = C4 = C5 = C6
= C7 = 10FR6 = 220 K
T1,T2,T3=C828Cc gi tr in p ca cc n:Uce(T1) = 8.2V, Uce(T2) = 6V,
Uce(T3) = 4.2VUra = 2.5V
VI)KT LUN _ Bin in p ra t yu cu, dng xung ra l dng chun hinh sin
vi bin nh nh bng 2.5V. _Bi thc tp mch dao ng m tn gip sinh vin hiu
r hn v hot ngca cu Wien, cc ch lm vic ca Transistor lng cc mc EC,
CC,cch iu chnh dng din ra trn cc cc Collector, Emittor v s dng
cc
thit b hn, dng c o nh ng h vn nng, Oscilloscope thnh tho hn.
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