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1 CHƯƠNG 2: KĐ THUẬT TOÁN Operational Amplifiers Giảng viên: TS. Nguyễn Phương Huy Bộ môn Kỹ thuật Điện tử Khoa Điện tử Trường ĐH Kỹ thuật công nghiệp 06/24/22

C2_Khuech Dai Thuat Toan

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Khuech Dai Thuat Toan

Text of C2_Khuech Dai Thuat Toan

  • *CHNG 2: K THUT TONOperational AmplifiersGing vin: TS. Nguyn Phng HuyB mn K thut in tKhoa in tTrng H K thut cng nghip*

  • Ni dung chnhKhuch i thut ton l tngCu trc oCu trc khng oB khuch i vi saiMch tch phn v mch vi phnnh hng ca h s khuch i vng h gii hn v bng thngMch khuch i thut ton tn hiu ln

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  • Khuch i thut ton l tngCc u vo/ra ca khuch i thut ton**K hiu mch ca khuch i thut tonS th hin ca khuch i thut ton khi kt ni vi ngun cp mt chiu

  • Khuch i thut ton l tng*Tr khng vo v cng lnTr khng ra bng 0H s khuch i ch chung bng 0 hay tng ng, s loi b ch chung l v hnH s khuch i vng h A v cng lnDi tn s lm vic v cng lnMch tng ng ca khuch i thut ton l tngChc nng v c tnh ca khuych i thut ton l tng

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  • Khuch i thut ton l tng**Tn hiu vo vi sai l s sai lch gia hai tn hiu u voTn hiu vo ch chung l gi tr trung bnh ca hai tn hiu vo Biu din ngun tn hiu v theo cc thnh phn tn hiu vo vi sai v tn hiu vo ch chungTn hiu vi sai v tn hiu ch chung

  • Cu trc oGii thiuMt cu trc dng o ca mch khuch i thut ton*

  • Cu trc oH s khuch i vng knH s khuch i vng kn: *

  • Cu trc oH s khuch i vng knPhn tch mch dng o. Cc s c khoanh trn cho thy cc bc phn tchH s khuch i vng kn: *

  • Cu trc oTr khng vo v ra trnh suy gim tn hiu ta chn R1 ln ?R2 ?Tr khng ra ca b khuch i vng kn l bng 0Tr khng vo *

  • Cu trc oTr khng vo v raGii php tng tr khng vo *

  • Cu trc oS dng IC KTD khuch i dngMch K dng vi tr khng vo bng 0, v tr khng ra bng v cng*

  • Cu trc oS dng IC KTD khuch i dngMch K dng vi tr khng vo bng 0, v tr khng ra bng v cng*

  • Cu trc oB cng c trng s*

  • Cu trc oB cng c trng sB cng c kh nng thc hin php cng hai h s khc du*

  • Cu trc khng oGii thiu*

  • Cu trc khng oH s khuch i vng kn*

  • Cu trc khng oH s khuch i vng knPhn tch cu trc khng o. Th t cc bc phn tch c th hin bi cc s khoanh trn.*

  • Cu trc khng oc im ca mch khuch i khng o*H s khuch i ca b khuch i khng o l dng Tr khng u vo bng v cngTr khng u ra ca b khuch i bng 0.

  • Cu trc khng oMch lp in p*

  • Cu trc khng oBi tp v da. S dng nguyn l xp chng tm in p ra ca mch?b. B in tr 1 k i v thay vo l mt ngun tn hiu vo v3Tm v3 theo v1 , v2 , v3 ?

  • B khuch i vi saiGii thiuT s loi b tn hiu ch chung (CMRR):

  • B khuch i vi saiMch khuch vi sai dng khuch i thut ton n

  • B khuch i vi saiMch khuch vi sai dng khuch i thut ton nXc nh h s khuch i ch chung

  • B khuch i vi saiMch khuch vi sai dng khuch i thut ton nHn ch!

  • B khuch i vi saiB khuch i vi sai ci tinVn cn hn ch!

  • B khuch i vi saiB khuch i vi sai ci tinMch ti u?

  • B khuch i vi saiB khuch i vi sai ci tin

  • B khuch i vi saiB khuch i vi sai ci tinXc nh h s khuch i ch chung

  • Mch tch phn v mch vi phnCu hnh o vi tr khng

  • Mch tch phn v mch vi phnCu hnh o vi tr khngTm biu thc hm truyn Vo(s)/Vi(s). Hy tm h s khuch i mt chiuTm tn s 3 dB. Thit k mch khuch i vi h s khuch i mt chiu bng 40 dB, tn s 3 dB l 1 kHz, v tr khng vo l 1 k.

  • * (a) Xc nh hm truyn t.Gii:

    Mch tch phn v mch vi phn

  • *Mch tch phn v mch vi phn

  • *Mch tch phn o

  • *Mch tch phn o

  • *Fig2.42 The Miller integrator with a large resistance RF connectedin parallel with C in order to provide negative feedback and hence finite gain at dcMch tch phn o

  • *Mch vi phn

  • *Mch vi phn

  • *2.6.1 Offset Voltagenh hng ca ngun mt chiuin p lch khng

    To rest of circuit

  • *Phn cc u vo v dng lch khngnh hng ca ngun mt chiu

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  • *nh hng ca ngun mt chiu

  • *nh hng ca ngun mt chiu

  • nh hng ca h s khuch i vng h gii hn v bng thngS ph thuc tn s ca h s khuch i vng hH s khuch i vng h ca mt mch KTT c b trongViDi thng h s khuch i: Vy:

  • nh hng ca h s khuch i vng h gii hn v bng thngp ng tn s ca b khuch i vng knH s khuch i vng kn ca b khuch i o H s khuch i mt chiu=?

  • nh hng ca h s khuch i vng h gii hn v bng thngp ng tn s ca b khuch i vng knH s khuch i mt chiu=? H s khuch i vng kn ca b khuch i khng o

  • Mch khuch i thut ton tn hiu lnS bo ha in p u raL+ v L- nh hn khong 1V so vi in p ngun dng v mVD : mt b khuch i thut ton hot ng vi ngun cung cp l 15 V th n s b bo ha khi in p u ra t khong +13V mc dng v -13V mc m

  • Mch khuch i thut ton tn hiu lnGii hn dng in u raDng in u ra b gii hn mc cc i nht nh. V d vi b khch i 741, dng in u ra cc i thng thng l 20mA. Phi m bo rng dng in u ra khng c vt qu 20mA ( bao gm c dng in trong mch hi tip cng nh dng in cp ti in tr ti)

  • *Mch khuch i thut ton tn hiu lnCho mch khuch i khng o nh hnh di. B KTT c thit k vi in p bo ha u ra l 13V, Dng in bo ha u ra l 20mA. (a) Vi vP =1V v RL=1k, xc nh tn hiu u ra b khuch i

  • *Mch khuch i thut ton tn hiu ln(a) Vi VP =1,5V v RL=1k, xc nh tn hiu u ra b khuch i

  • *(d) Find Vp=1V, what is the lowest value of RL for which an undistorted sine-wave output is obtained(c) Find RL=1k, what is the maximum value of VP for which an undistorted sine-wave output is obtainedMch khuch i thut ton tn hiu ln

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