BIN I ADBi:

Nguyn Trung Tp

BIN I AD & DA

C th ni s bin i qua li gia cc tn hiu t dng tng t sang dng s l cnthit v:

- H thng s x l tn hiu s m tn hiu trong t nhin l tn hiu tng t: cn thitc mch i tng t sang s.

- Kt qu t cc h thng s l cc i lng s: cn thit phi i thnh tn hiu tngt c th tc ng vo cc h thng vt l v th hin ra bn ngoi (th d ti to mthanh hay hnh nh) hay dng vo vic iu khin sau (th d dng in th tng t iu khin vn tc ng c)

Bin i s - tng t (digital to analog converter , DAC)

Mch bin i DAC dng mng in tr c trng lng khc nhau (Weightedresistor network)

(H 8.1)

Trong mch trn, nu thay OP-AMP bi mt in tr ti, ta c tn hiu ra l dng in.

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Nh vy OP-AMP gi vai tr bin dng in ra thnh in th ra, ng thi n l mtmch cng

Ta c v0 = -RF.I = -(23b3 + 22b2 + 2b1+b0)Vr.RF/23R

= -(2n-1 bn-1 + 2n-2 bn-2 + ........+ 2b1 + b0)Vr.RF /2n-1.R

Nu RF = R th:

v0 =-(2n-1 bn-1 + 2n-2 bn-2 + ........+ 2b1 + b0)Vr./2n-1.

Th d:

1/ Khi s nh phn l 0000 th v0 = 0

1111 th v0 = -15Vr / 8

2/ Vi Vr = 5V ; R = RF = 1k?

Ta c kt qu chuyn i nh sau:

Mch c mt s hn ch:

- S chnh xc ty thuc vo in tr v mc n nh ca ngun tham chiu Vr

- Vi s nh phn nhiu bit th cn cc in tr c gi tr rt ln, kh thc hin.

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Mch i DAC dng mng in tr hnh thang

(H 8.2)

Cho RF = 2R v ln lt

Cho b3 = 1 cc bit khc = 0, ta c: v0 = -8(Vr /24)

Cho b2 = 1 cc bit khc = 0, ta c: v0 = -4(Vr /24)

Cho b1 = 1 cc bit khc = 0, ta c: v0 = -2(Vr /24)

Cho b0 = 1 cc bit khc = 0, ta c: v0 = - (Vr /24)

Ta thy v0 t l vi gi tr B ca t hp bit B = (b3 b2 b1 b0 )2 v0 = -B(Vr /24)

Mch i DAC dng ngun dng c trng lng khc nhau

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(H 8.3)

c tnh k thut ca mch i DAC

Bit c ngha thp nht (LSB) v bit c ngha cao nht (MSB)

Qua cc mch bin i DAC k trn ta thy v tr khc nhau ca cc bit trong s nhphn cho gi tr bin i khc nhau, ni cch khc tr bin i ca mt bit ty thuc votrng lng ca bit .

Nu ta gi tr ton giai l VFS th bit LSB c gi tr l: LSB = VFS / (2n - 1)

v bit MSB = VFS .2n-1/ (2n - 1)

iu ny c th hin trong kt qu ca th d 2 trn.

(H 8.4) l c tuyn chuyn i ca mt s nh phn 3 bit

(a) (b)

(H 8.4)

(H 8.4a) l c tuyn l tng, tuy nhin, trong thc t ng trung bnh ca c tnhchuyn i i qua im 0 in th tng t ra c lm lch (1/2)LSB (H 8.4b). Nhvy in th tng t ra c xem nh thay i ngay gia hai m s nh phn vok nhau. Th d khi m s nh phn vo l 000 th in th tng t ra l 0 v in thtng t ra s ln nc k 000+(1/2)LSB ri nc k tip 001+(1/2)LSB.v.v....Tr tngt ra ng vi 001 gi tt l 1LSB v tr ton giai VFS = 7LSB tng ng vi s 111

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Sai s nguyn lng ha (quantization error)

Trong s bin i, ta thy ng vi mt gi tr nh phn vo, ta c mt khong in thtng t ra. Nh vy c mt sai s trong bin i gi l sai s nguyn lng ha v =(1/2)LSB

phn gii (resolution)

phn gii c hiu l gi tr thay i nh nht ca tn hiu tng t ra c th c khis nh phn vo thay i. phn gii cn c gi l tr bc (step size) v bng trnglng bit LSB.

S nh phn n bit c 2n gi tr v 2n - 1 bc

Hiu th tng t ra xc nh bi v0 = k.(B)2

Trong k chnh l phn gii v (B)2 l s nh phn

Ngi ta thng tnh phn trm phn gii:

%res = (k / VFS)100 %

Vi s nh phn n bit

%res = [1 / (2n - 1)]100 %

Cc nh sn xut thng dng s bit ca s nh phn c th c bin i ch phngii. S bit cng ln th phn gii cng cao (finer resolution)

tuyn tnh (linearity)

Khi in th tng t ra thay i u vi s nh phn vo ta ni mch bin i c tnhtuyn tnh

ng (accuracy)

ng (cn gi l chnh xc) tuyt i ca mt DAC l hiu s gia in th tngt ra v in th ra l thuyt tng ng vi m s nh phn vo. Hai s nh phn k nhauphi cho ra hai in th tng t khc nhau ng 1LSB, nu khng mch c th tuyntnh nhng khng ng (H 8.5)

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a/ Tuyn tnh b/ Tuyn tnh nhng khng ng

(H 8.5)

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BIN I ADBIN I AD & DABin i s - tng t (digital to analog converter , DAC)Mch bin i DAC dng mng in tr c trng lng khc nhau (Weighted resistor network)Mch i DAC dng mng in tr hnh thangMch i DAC dng ngun dng c trng lng khc nhauc tnh k thut ca mch i DACBit c ngha thp nht (LSB) v bit c ngha cao nht (MSB)Sai s nguyn lng ha (quantization error) phn gii (resolution) tuyn tnh (linearity) ng (accuracy)