Mô hình hóa Full-Bridge

7/28/2019 M hnh ha Full-Bridge

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I. Tng quan v n1. Mc tiu

Vi hc phn n 2, chng em ra mc tiu t c nh sau:

Bit cch tnh ton, m hnh ha mt mch ngun hot ng theo nguynl Switching Mode Power Supply (SMPS). Bit cch tng hp b iu khin tng t cho mch hot ng theo

nguyn l trn.2. La chn cu hnh mch

Hnh 1: Cu hnh mch Full-Bridge

Vi nhng mc tiu t ra nh vy, chng em quyt nh la chn cu hnhmch Full-Bridge thc hin, vi nhng l do sau:

L cu hnh hot ng da trn nguyn l hot ng (SMPS). Tp hp nhiu n v kin thc nh m hnh ha, tng hp b iu

khin, thit k mch t,... C th thc nghim tch ly thm kinh nghim thc t.

Cc thng s vo ra yu cu ca mch: Vin = 300VDC; Vo = 24VDC; Iomax =5A; Vin = 1%Vin; Vo = 1%Vo; Io = 30%Io

3. Cc bc thc hin M hnh ha Tng hp biu khin tng t kiu b M phng Thit k mch in Thc nghim

II. Tnh ton v m hnh ha mch lc1. c im chung ca cc b bin i ngun DC-DC s dng nguyn l

SMPS

Da trn nguyn l bm xung p u ra phi c t ln san bng in p trn ti C th coi trong mt khong thi gian rt nh, vi chu k trch mu, in

p ra l khng i


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Gi thit ny cho php n gin ha ti a qu trnh phn tch cc s DC-DC

Cu trc vng iu khin in p ca mch c dng chung nh sau:

Hnh 2: Vng iu khin in p

2. M hnh tn hiu lnGi strng thi lm vic n nh, cc thng s ca mch nh Vin, Vo, Io, ts Duty D thay i khng ng k, n l t s vng dy gia cun s cp vcun th cp ca bin p xung. Theo nh lut Kirrkoff, ta lp c cc phngtrnh mch sau:

Trong thi gian T1 cp van Q1, Q2 dn,cp van Q3, Q4 khng dn, hoccp van Q1, Q2 khng dn, cp van Q3, Q4 dn:

Trong thi gian T2 cp van Q1, Q3 dn, cp van Q2, Q4khng dn, hoc

cp van Q1, Q3 khng dn, cp van Q2, Q4 dn:

Dng qua cun cm ni vng qua D1, D2.

Trong c chu kng mvan, ta c Gi thit Vin = const, Vo = const, dng qua cun cm thay i tuyn tnh. Ta c thdng in, in p ca mch:


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Hnh 3: Dng dng in, in p ca mch

(B) Xung PWM trn van

(C) in p cun dy th cp ca bin p xung

(D) Dng in trn cun cm L

Tnh gi tr LTrong c chu k,

, gi s t s duty-cycle D = 0.45 nn ta tnh c t s

vng dy ca bin p xung:n = Vo / (DVinmin) = 24/(0.45*270) = 0.197

Mt khc ta c VL = L.diL/dt

L.diL/dt = nDVinVo L = (nDVinVo).dt /diLChn tn s mch l fsw = 100kHz.Xt trong khong thi gian T1, ta c dt = Ton= D/fsw. bin thin dng trung bnh qua cun cm chnh l bin thindng ti IL = Io. Suy ra:

L (nVinminVo).D/(fsw.Io) = 90 uH Tnh gi tr t lc u ra C


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Hnh 4: Dng in p trn t CXt trong khong thi gian t thi im Tonn thi im (Ton + Toff/2), ta c:

Dng in cun cm gim t ILmax xung Io in p tin tang t Vo ln Vomax Nng lng tiu tn trn ti l WR= Vo.Io.Toff/2 = Vo.Io.(1-D)/(2.fsw)

Theo nh lut cn bng nng lng, trong khong thi gian Toff/2, nng lngphng ra trn cun cm s bng tng ca nng lng tin thu vo v nnglng trn ti tiu tn, tc l:

WL = WC + WR

0.5L.( Vo.Io.(1-D)/(2.fsw) C = = 15.6 uFChn tin theo cataloge ca nh sn xut: C = 47uF; ZC = 1.3Ohm

Thit k cun cmCng sut max ca mch:

Po = Vo . (Io+ 0.5Io) = 20 W

Dng in peak trn cun cm:

Ipk= Io+ 0.5Io = 4.5A

Nng lng tch t trong cun cm:

Energy = 0.5 L . = 0.91mJChn Bmax = 0.22T

H s quan h =

; chn = 1

Hng s Kec quyt nh bi t ha v iu kin vn hnh in:


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Ke = 0.145Po = 0.379Chn h s lp y Ku = 0.4

Hng s Kgc quyt nh bi hnh dng ca li t:

Kg = Mt khc Kg =

= 2.19e-6 Da vo hng s Kg , ta chn c li thp EE25 c cc thng snh sau:

Din tch mt ct tr gia: Ac = 0.4Din tch ca s: Wa = 0.65 di trung bnh 1 vng dy: MLT = 3.5cmH sm = 2500

Hng s Ap = Ac.Wa = 0.26Mt dng in:

J = = 796.5 A/

Dng in rms:

Irms = = 4.12ADin tch mt ct dy dn:

Aw = Irms/J = 0.5

(Chn dy 0.4)

S vng dy:

n = 0.6Wa /Aw = 75 vng

Khe hkhng kh:

lg = = 3mm

in trcun cm:

Rl =

= 0.0878Ohm

3. Thit k bin p xungCc thng su vo: in p vo: Vin = 300V; Dng in vo: Iin = 0.4A; p mch in p u vo: Vin = 10% = 30V; in p u ra: Vo = 24V; Dng in ra: Io = 5A; p mch dng in u ra: Io = 30% = 1.5A; Tn sng ct van: fsw = 100kHz

Chn cc h s: H s lp y: Ku = 0.25;


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H s quan h:f = 0.5% Cm ng t cc i ca vng dy: Bmax = 0.12T T s Duty-cycle max: Dmax = 0.48 Hiu sut: eff = 0.98

Cng sut u ra ca mch:Po = Io * (Vo + Vo) =

Cng sut u vo ca mch:

Pin = Po * 1.1 / eff =

Hng s Ke quyt nh bi tha v iu kin vn hnh in:

Ke = 0.145 * * * =Hng s Kg quyt nh bi hnh dng, kch thc ca li t:

Kg = Pin * Dmax * 1.35 / (f * Ke) =Da vo hng s Kg, ta chn li ... c cc thng snh sau:

Thit din tr gia ca li: Ae = Din tch ca s: Wa = di trung bnh ca 1 vng dy: MLT = Hng s Al =

S vng dy ca cun s cp:

n1 = (Vin0.5 * Vin) * Dmax *

/ (fsw * Ae * Bmax) = (vng)

Mt dng in yu cu:J = 2* Pin * * / (fsw * Ae * Bmax * Wa * Ku) = (A/)

Dng in trung bnh qua cun s cp:

Ip = Pin / ((Vin0.5 * Vin) * ) = (A)Thit din dy s cp:

Awp = Ip / J = ()in trdy s cp:

Rp = MLT * n1 * ro = (Ohm)S vng dy ca cun th cp:

n2 = = (vng)

n3 = n2 = (vng)

Dng trung bnh qua cun th cp:

Is = Io / = (A)Thit din dy th cp:

Aws = Is / J = ()


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in trca cun th cp:

Rs = MLT * n2 * ro = (Ohm)

in cm t ha ca bin p:

Lm = Al *

*

= (H)

Dng in t ha cc i:

Immax = (Io + 0.5 * Io) * (n1/n2) + Vo * (Io + 0.5 * Io) / Vin = (A)

4. M hnh tn hiu nh (tuyn tnh ha quanh im lm vic)p dng phng php phn tch tng t, trong c xt n cc gi trintrca tin v cun cm:

Trong thi gian cp van Q1, Q2 dn, cp van Q3, Q4 khng dn, hoccp van Q1, Q2 khng dn, cp van Q3, Q4 dn:

Trong thi gian cp van Q1, Q3dn, cp van Q2, Q4khng dn, hoc cp

van Q1, Q3 khng dn, cp van Q2, Q4 dn:

Trong c chu kng mvan, ta c

Gi scc i lng c bin thin rt nh quanh gi trn nh: = IL + iL;

= D + d;

= Vin + vin; =Vo + vo; = IC + iCThay vo hphng trnh mch, ta c


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V , v coi , ta c

Laplace ha: Thphng trnh 2 v 3 vo phng trnh 1, coi vin = 0, ta c hm truyn vo ph thucvo d:

Gvod = 2

[( ) 1]

( ) ( ( ) )

g

vod

nV R ESR CsG

LC ESR R s RC ESR L s R

a v dng chun: Gvod = Go .

Trong : Go = n*R*Vin / (R+Rl);

wzo = 1/(Rc * C);

wo = ;Q = (R+Rl) / ((C*(R*Rc+Rl*Rc+R*Rl)+L)*wo).

5. Hm truyn mch o:Mch o l mch phn p c hm truyn nh sau:

Go =


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Gi s in p ra Vo = 12V c chun ha v Vph = 3V, ta chn R1 =3.3kOhm, R2 = 1.1kOhm

6. Hm truyn PWMHm truyn PWM c tha v dng tng ng nh sau:

Gpwm = Vref= 3Vdc => Cpk= 12Vdc7. Hm truyn i tng

Hm truyn i tng c suy ra nh sau:Gt = Gvod . Go . Gpwm

III. Phn tch i tng v tng hp biu khin1.Phn tch i tng

Mun thit k b iu khin th i tng phi n nh v c th iu khin c.

Kho st th Bode ca i tng, ta nhn thy i tng n nh, c thiu khinc.

Hnh : th Bode ca i tng

Ngoi ra ta c th s dng tiu chun i s Routh v Hurwitz xc nh tnh n nhcng s cho kt qutng t.

2.Thit k biu khina)Tiu ch thit k

Da vo nhng kin thc trong mn hc L thuyt iu khin tng 1, chngem a ra mt s tiu ch tit knh sau:- Thit k biu khin trn min tn s, s dng th Bode;- Biu khin Gc c th l biu khin kiu b loi 1, 2, hoc 3, hay cc

biu khin kiu lead, lag, PID;


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- H kn phi n nh, khc cc nhiu tc ng, nh nhiu ti, nhiu inp u vo;

- H kn c cht lng tt khi h hc:+ d trbin :

10 dB

+ d tr pha: Cu trc vng phn hi in p u ra c dng nh sau:

b)Biu khin kiu b loi 2

Hnh : Cu trc biu khin kiu b loi 2

Hm truyn: Gc = =

Gc(s) PWM Gvod(s)

Go(s)

VoVref dVc


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Hnh : th bin pha ca biu khin kiu b loi 2

Chn tn s ct nm trong khongwc (wo*;*fsw) tha mn tiu ch thit k, b iu khin kiu b loi 2 cn tha mn h

phng trnh sau

|| || () () Gii hphng trnh trn, ta tm c cc tham s Kc, wz, wp ca mch.

Xut pht t hm truyn ca biu khin, ta tip tc tm c thng s cc linh kinR, C ca biu khin theo cng thc:

R1 = 10000 Ohm (chn trc);

C1 = ;C2 =

;R2 =

Thay tham s c th, ta tnh c cc gi trin tr, tin ca biu khin.

R1 = 10kOhm

C1 = 0.2nF

C2 = 2.16nFR2 = 31kOhm

Kt qu kho st th Bode ca vng htrn Matlab nh sau


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Hnh : th Bode ca vng h

Ta nhn thy biu khin tha mn nhng tiu ch thit ka ra ban u, nh v d tr pha ; .IV. Kt qu m phng

Kt qu m phng p ng in p u ra vi 2 khong thi gian: khng ti vng ti

Hnh : in p u ra

Mô hình hóa Full-Bridge

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