BI 1: MCH XN, MCH GHIM IN P I. Mc ch yu cu: Kho st hot ng ca mch
xn, mch ghim in p dng diode, cp sng vo v kho st dng sng ra sau khi
b xn hoc ghim. II. Tm tt l thuyt: 1. Mch xn: - Mch xn c nhim v ch
cho mt phn tn hiu ng vo a n ng ra. - Trong bi th nghim ny ta thc
hin mch xn dng diode mc theo kiu song song, c xn trn, xn di, v xn
hai mc. S mch xn mc kiu song song :R R
+D
+R t
+D
+R t
VinEc
Vout
VinEc
Vout
-
-
Ec 0
-
Mch xn mc trn
Mch xn mc di
- Coi diode D l l tng, VD = 0V - i vi mch xn trn, khi V in <
Ec th diode cha c phn cc thun, ton b V in c a ra Vout, khi Vin >
Ec th diode c phn cc thun, in p ra Vout = Ec, ta ni tn hiu b xn mc
Ec. khi tn hiu vo l sng sin th dng sng ra c dng:
vEc
Vin t Vout
- i vi mch xn di, khi V in > Ec th diode cha c phn cc thun,
ton b Vin c a ra Vout, khi Vin < Ec th diode c phn cc thun, in p
ra Vout = Ec, ta ni tn hiu b xn mc Ec. Khi tn hiu vo l sng sin th
dng sng ra c dng:
Trang :1
V
Vout
t - Ec Vin
+
Kt hp hai mch xn trn v di ta c mch xn hai mcR
+D 1 D 2
VinV1 V2
Vout -
R t
-
V V1 Vout
t - V2 Vin
2. Mch ghim: - Mch ghim in p l mch khi phc thnh phn mt chiu ca
tn hiu. N c dng n nh nn hoc nh ca tn hiu, mt mc xc nh no . - C hai
loi mch ghim c bn l mch ghim nh trn v mch ghim nh diC C
+D
+ VoutEc
+D
+ VoutEc
Vin -
Vin -
-
-
Mch ghim nh trn- Coi diode nh l tng, VD = 0V
Mch ghim nh di
Trang :2
- i vi mch ghim nh trn: Khi Vin tng t 0 n Vp(gi tr nh), th t C c
np n gi tr VC = Vp Ec, diode D dn, Vout = Ec. Khi Vin gim t Vp v 0,
in p t ln hai cc diode l V in VC Ec m VC = Vp Ec => VD = Vin Vp,
vy VD < 0 diode D tt, Vout = Vin VC, do Vout gim t Ec v (- Vp +
Ec). Khi Vin gim t 0 v - Vp, diode D vn phn cc ngc, t C vn cha x,
Vout = Vin VC gim t (- Vp + Ec) v (- 2Vp + Ec). Khi Vin tng t - Vp
n 0, th Vout tng t (- 2Vp + Ec) n (- Vp + Ec).C
+D
+ VoutEc
Vin -
-
Vp Ec
Vin
Ec - Vp + Ec
Vout
-
i vi mch ghim nh di(gii thch tng t), c dng sng vo ra:
Trang :3
C
+D
+ VoutEc
Vin -
-
Vp
Vin
- Ec
2Vp Ec Vp Ec Ec
Vout
Ta c th thay i Ec ghim cc mc mong mun. III. Dng c th nghim 1
FACET Base Unit. ( lp mch th nghim). 1 SEMICONDUCTOR FUNDAMENTAL
circuit board. 1 VOM. 1 Dao ng k. 1 My to sng. Cc dy ni v cc
connector. VI. Tin hnh th nghim 1. kho st mch xn s mch xn trn
Trang :4
V 0.6V
Dng my to sng, cp sng sin 20Vp-p, tn s 1kHz, iu chnh ngun dng
trn Base unit a V1 v 0V. Dng knh 2 ca dao ng k, o dng sng trn 2 u
in tr R2 th thu c sng sin b xn mc trn
t
0.6V l in p phn cc thun ca diode CR1. khi p vo vt qu 0.6V th
diode dn, ghim mc p ra 0.6V. S mch xn di
- Lm tng t cc bc nh trn. - Thu c dng sng ra 2 u R2 l sng sin b
xn mc diV
-0.6V
t
khi p vo nh hn - 0.6V th diode dn, ghim mc p ra - 0.6V. s mch xn
2 mc
- Quan st dng sng, ta thy dng sng ra c dng sng vung, 0.6Vp-p
Trang :5
V -0.6V t -0.6V
khi p vo nh hn - 0.6V th diode CR2 dn, ghim mc p ra - 0.6V, khi
p vo ln hn 0.6V th diode CR1 dn, ghim mc p ra 0.6V. - iu chnh V1 =
2V, V2 = 0V th mc xn trn tng ln 2.6 V - iu chnh V1 = 0V, V2 = -2V
th mc xn di gim xung 2.6 V 2. kho st mch ghim: S mch ghim nh
trn
-
V1 v 0V. Dng knh 2 ca dao ng k, o dng sng ra thu c in p ghim nh
trn mc 0,6VV 0.6V t
Dng my to sng, cp sng vung 10Vp-p, tn s 1kHz, iu chnh ngun dng
trn Base unit a
-9.4V
-
iu chnh V1 = 3V, th thu c in p ghim nh trn mc 3.6V
S mch ghim nh di
Trang :6
-
Dng my to sng, cp sng vung 10Vp-p, tn s 1kHz, iu chnh ngun dng
trn Base unit a V2 v 0V. Dng knh 2 ca dao ng k, o dng sng ra thu c
in p ghim nh di mc - 0,6V.9.4V
- 0.6V
t
-
iu chnh V1 = -2V, th thu c in p ghim nh trn mc -2.6V
BI 2: KHO ST MCH TCH PHN V MCH VI PHN DNG OP-AMP I. Mc ch yu cu:
Kho st mch tch phn v mch vi phn dng op-amp, kim tra dng sng vo ra
ca mch vi phn, tch phn. II. Tm tt l thuyt Mch tch phn v mch vi phn
l nhng mch lm thay i dng sng vo.
1. Mch tch phn - Mch tch phn th ng (ch bao gm cc phn t th ng):
in p ra l in p ly trn t, do t l vi XC. Nhng XC= 1/j2fC. Do khi f
tng th XC gim, Vo gim; khi f gim th XC tng, Vo tng. Mch tch phn lm
chc nng mch lc thng thp. - Gi s sng vo l sng vung c rng xung l PW:
Khi RC> PW: xung vo kt thc trc khi t np y, xung ra l xung tam
gic
Trang :7
Khi RC PW: thi gian np ca t xp x thi gian tn ti xung, xung ra l
xung rng ca Khi RC< PW: t np y trc khi xung vo kt thc, xung ra l
xung vung - Khi mc thm opamp vo mch nh sau ta c mch tch phn tch
cc:
- i vi tn hiu tn s thp hoc tn hiu mt chiu, X C0, t CF c th b
qua, mch hot ng nh l mt mch khuych i o, do lch pha gia ng ra v ng
vo l 1800. p ng tn s ca mch:
2.Mch vi phn: - Mch vi phn th ng (gm cc phn t th ng): in p ng ra
ly trn in tr R, V o t l nghch vi XC ca mch. M XC= 1/j2fC. Do khi f
tng, Vo tng; khi f gim, Vo gim. - Gi s sng vo l sng vung c rng xung
l PW: Khi RC< PW: t np y trc khi xung vo kt thc, xung ra l xung
nhn Khi RC PW: thi gian np ca t xp x vi thi gian tn ti xung vo,
xung ra l xung rng ca Khi RC> PW: xung vo kt thc trc khi t np y,
xung ra l xung vung
- Khi mc thm opamp vo mch nh sau ta c mch vi phn tch cc:
- i vi tn hiu tn s cao, X C. t CF c th b qua, mch hot ng nh l mt
mch khuych i o, do lch pha gia ng ra v ng vo l 1800.
Trang :8
p ng tn s ca mch:
III. Dng c th nghim 1 FACET Base Unit. 1 OPERATIONAL AMPLIFIER
APPLICATIONS circuit board. 1 VOM. 1 Dao ng k. 1 My to sng. Cc dy
ni v cc connector. VI. Tin hnh th nghim 1.Mch tch phn Mch th
nghim:
- Tn s ct fc ca mch (ti XC1=R3): fc=1/2R3C1= 159.5 Hz - t knh 1
ca my hin sng vo u vo ca mch, knh 2 vo u ra. - Cp ngun tn hiu sng
sin 1Vp-p tn s 20Hz. - T dng sng ra ta c: li : Gain=R3/R1= 10 . lch
pha gia in p ra v in p vo l 1800 . Mch lm vic nh mt mch khuch i
o.
Trang :9
- Tng tn s ca sng vo ln 2 kHz, v quan st tn hiu ra knh 2: li in
p gim i. Mch lm vic nh mt mch lc thng thp. lch pha 2700
V
Vin
Vout
tLch pha 270o
- i sng sin u vo thnh sng vung. Dng sng vo ra nh sau:
Trang :10
V Vin
t V Vout t
Mch lm vic trong vng tch phn 2. Mch vi phn Mch th nghim:
- Tn s ct ca mch ( ti XC1=R1): fc=1/2R1C1 = 15.9 kHz - t knh 1
ca my hin sng vo u vo ca mch, knh 2 vo u ra. - Cp ngun tn hiu sng
vung 1Vp-p tn s 100Hz. Dng sng vo raV
Vin
t V
Voutt
i sng vung thnh sng sin. Dng sng ng ra:
Trang :11
V
Vin
Vout t
Lch pha 90o
lch pha: 90o. Gain
