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Temel ilkeler, Süzgeçler, Entegratörler, Türev alıcılar ve Enstrümentasyon (Cihaz) Kuvvetlendiricileri.

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... konulu sunumlar: "Temel ilkeler, Süzgeçler, Entegratörler, Türev alıcılar ve Enstrümentasyon (Cihaz) Kuvvetlendiricileri."— Sunum transkripti:

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2 Temel ilkeler, Süzgeçler, Entegratörler, Türev alıcılar ve Enstrümentasyon (Cihaz) Kuvvetlendiricileri

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5  Temel kuvvetlediriciler  Enstrümentasyon kuvvetlendiricileri – geliştirilmiş bir farksal kuvvetlendirici  Aktif süzgeçler  Entegratör (toplama) ve türev alma  Önemli ekler: ◦ Hassas doğrultucular ◦ Logaritmik kuvvetlendiriciler ◦ Negatif kapasitans kuvvetlendiricileri 4

6  Ideal işlemsel kuvetlendirici (operational amplifier – op-amp)  Eviren, evirmeyen ve enstrümentasyon (cihaz) kuvvetlendiricisi  Entegratör (toplama), türev (differensiyel)  Süzgeçler (Filters)  İşaret koşullandırma modül örneklari 5

7 6 Kazanç K veya G (Gain) = Çıkış Değeri / Giriş Değeri

8  Gerilim kontrollü gerilim kuvvetlendiricisi (basit op- amp)  Gerilim kontrollü akım kuvvetlendiricisi (operational transconductance amplifier – OTA  Akım kontrollü gerilim kuvvetlendiricisi (transimpedance)  Akım kontrollü akım kuvvetlendiricisi 7

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10  Özellikleri ◦ Sonsuz kazanç (A=∞) ◦ v 1 = v 2 ise V 0 = 0 ◦ Giriş empedansı sonsuz ◦ Çıkış empedansı sıfır ◦ Bant genişliği sonsuz ve faz kayması yok.  Temel Kurallar ◦ Op-amp’in çıkışı lineer bölgede ise, her iki girişteki gerilim aynıdır (birbirine eşittir) ◦ Op-amp’in girişinden hiç akım akmaz. 9

11 Devre Elemanı Olarak Op-amp 10

12 V0V0 ViVi A + - R1R1 RfRf I Sanal (Virtual) toprak Gerçek toprak I = V i / R 1 V 0 = - ( R f / R 1 ) * V i 11

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15 V 0 = - ( R f / R 1 ) * V 1 - ( R f / R 2 ) * V 2 I 2 = V 2 / R 2 I 1 = V 1 / R 1 14

16 ii v b ii oo oo  + +15V Time  i +  b /2 -10 (a)(b) 5 k  -15 V RbRb 20 k  RiRi 10 k  RfRf 100 k  Voltage, V (a) This circuit sums the input voltage  i plus one-half of the balancing voltage  b. Thus the output voltage  o can be set to zero even when  i has a nonzero dc component. (b) The three waveforms show  i, the input voltage; (  i +  b /2), the balanced- out voltage; and  o, the amplified output voltage. If  i were directly amplified, the op amp would saturate. 15

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18 Toplayıcı Kuvvetlendirici Örneği 17

19 18 R i 

20 19 V 0 = (1+ R f / R 1 ) * V i I f = V i / R 1 R i 

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22 Evirmeyen Kuvvetlendirici (temel) 21

23 Evirmeyen Kuvvetlendirici Örneği 22

24 23 V 2 = V 1 * R 4 / (R 3 +R 4 ) V 0 = (1+ R 2 / R 1 ) * V 2 V 0 = (R 4 /(R 3 +R 4 ))* (1+ R 2 / R 1 ) * V 1 R i = V 1 / I i = R 3 +R 4

25 24 Vo/Vi =

26 25 Vo/Vi =

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30 29 V 5 = V 4 * R 4 / (R 3 +R 4 ) I f = (V 5 - V 3 ) / R 3 = (V 0 – V 5 )/ R 4 V 0 = (V 4 - V 3 )*R 4 / R 3 + G C * (V 4 + V 3 )/2 CMRR = G d /G C G C =2*V 0 /(V 4 + V 3 ) G d =V 0 /(V 4 - V 3 ) = R 4 / R 3

31 30 R i3 =V 3 /I 3 = R 3 R i4 =V 4 /I 4 = R 3 + R 4 G d =V 0 /V d = R 4 / R 3 G C =V 0 /V CM = Deneysel olarak bulunur CMRR = G d /G C

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33  i R1 = i R2a = i R2b olduğundan farksal (diferensiyel) çıkış gerilimi:  Farksal giriş gerilimi ise:  Böylece farksal kazanç: R2R2 vava R2R2 R1R1 vbvb V S+ va'va' vb'vb' i R1 i R2a i R2b V S– 32

34  Giriş koşullandırma ortak işareti kuvvetlendirmez!  Bunun da nedeni şöyledir: ◦ v b – v a = i R1 R 1 olduğundan ◦ Böylece çıkış gerilimi de: R2R2 vava R2R2 R1R1 vbvb V S+ va'va' vb'vb' i R1 i R2a i R2b V S– 33

35 Toplam kazanç: R4R4 R3R3 vovo R3R3 R4R4 R2R2 vava R2R2 R1R1 vbvb V S+ Difference Amplifier Input Conditioner V S– 34

36 35 Devrenin kazancı K = K1*K2

37 36

38  V 0 =A(V 2 -V 1 )  V 0 = 0 şayet V 1 = V 2  V 0 = +V SAT şayet V 1 < V 2  V 0 = -V SAT şayet V 1 > V 2  Op-amp’in girişlerinden akım akmaz. 37

39 38 Şayet V - <0 ise çıkış +V SAT kalır V - = (V i *R 2 +V ref *R 1 )/(R 1 +R 2 ) Şayet V - >0 olursa çıkış –V SAT a gider -V SAT V SAT

40 39 V - = (V i *R 2 +V ref *R 1 )/(R 1 +R 2 ) V + = V 0 *R 3 /(R 3 +R 4 ) V - V + olursa çıkış –V SAT a gider

41 40 V H = -V ref *R 1 /R 2 – V SAT *R 3 /(R 3 +R 4 ) V L = -V ref *R 1 /R 2 + V SAT *R 3 /(R 3 +R 4 )

42 Hassas Yarım Dalga Doğrultucu Devresi Hassas Tam Dalga Doğrultucu Devresi

43 10 V (b) - 10 V oo ii 10 V  + (a) D3D3 R R o=o= ii  + D2D2 D1D1 D4D4 xR (1-x)R ii x (c) D v o ii  + R i = 2 k  R f = 1 k  R L = 3 k  42 (a) Full-wave precision rectifier. For  i > 0, the noninverting amplifier at the top is active, making  o > 0. For  i 0. Circuit gain may be adjusted with a single pot. (b) Input-output characteristics show saturation when  o > +13 V. (c) One-op-amp full-wave rectifier. For  i 0, the op amp disconnects and the passive resistor chain yields a gain of +0.5.

44 (a)(b) RfRf IcIc R f /9 oo RiRi ii  + 10 V -10 V v o ii 11  V 43 Figure 3.8 (a) A logarithmic amplifier makes use of the fact that a transistor's V BE is related to the logarithm of its collector current. With the switch thrown in the alternate position, the circuit gain is increased by 10. (b) Input-output characteristics show that the logarithmic relation is obtained for only one polarity;  1 and  10 gains are indicated.

45  Frekans tepkisi.pptx Frekans tepkisi.pptx 44

46  Low-pass  High-pass  Band-pass  Band-stop (notch)  All-pass (phase-shift) 45

47 HP BP LP k100 Frequency, Hz 10 fcfc 1 D I fcfc Figure 3.10 Bode plot (gain versus frequency) for various filters. Integrator (I); differentiator (D); low pass (LP), 1, 2, 3 section (pole); high pass (HP);bandpass (BP). Corner frequencies f c for high-pass, low-pass, and bandpass filters. 46

48 47 Sanal Toprak i i V 0 nun V i ye ilişkisini yaz Kazanç G nin ifadesini yaz Where  = R i C f

49 48 G/G L f/f c Slope =  /2 Fazf/f c -5  /4 -- G/G L nin f/f c ye göre değişimini çiz Faz değişimini f/f c göre çiz

50 49 Besleme akımı hangi eleman üzerinden akar?

51 50 G L = R f /R i ;  f = R f C f ; f c = 1/2  f Kazanç G’nin denklemini çıkarınız Önemli parametreler nelerdir?

52 51 Gain/G L f/f c Slope =  /2 Phasef/f c -5  /4 -- G L = R f /R i ;  f = R f C f ; f c = 1/2  f

53 Gain/G L f/f c eyim= R f olmadan R f ile R f DC besleme akımlarının akım yoludur 1Phasef/f c -3  /2 -5  /4 -- R f olmadan Alçak geç süzgeç Enteg.

54 53 vovo Electrode The piezoelectric sensor generates charge, which is transferred to the capacitor, C, by the charge amplifier. Feedback resistor R causes the capacitor voltage to decay to zero

55 54 Sanal Toprak I sC = I sR = 0

56 55 xixi Deflection vovo v o max v o max /e Time The charge amplifier responds to a step input with an output that decays to zero with a time constant  = R f C f

57 56 G H = -R f /R i ;  i = R i C i ; f c = 1/2  i Kazanç G nin denklemini çıkar Önemli parametreler nelerdir?

58 57 Gain/G H f/f c Eyim = - Phasef/f c -3  /4 -  /2 G/G H yi f/f c ye göre çiz Fazı f/f c ye göre çiz

59 58 Eyim = +1 f c = 1/2  Faz kayması = - 90 o Gain f/f c V 0 yu V i cinsinden ifade et Kazanç G nin ifadesini yaz

60 59 In an ideal differentiator, the high frequency response is limited by the open-loop gain of the op-amp yielding a very noisy output voltage.  f = R f C f ; f c = 1/2  f C f is used to limit the hf gain, hence the hf noise. The gain at hf: G H = - C i /C f

61 60 Slope = - Phasef/f c -3  /4 -  /2 Gain/G H f/f c Without C f With C f DifferentiatorHigh-pass filter

62 61 Mid-band gain =G MB = - R f /R i Time-constants:  I = R i C i ;  f = R f C f Critical frequencies: f L = 1/2  I ; f H = 1/2  f

63 62 Gain/G H f/f c Slope = -1 Slope = +1

64 +  ii oo CiCi +  RiRi ii oo +  RiRi RfRf (a) (b) (c) CfCf ii oo JH RfRf CfCf RfRf CiCi RiRi Active filters (a)A low-pass filter attenuates high frequencies (b)A high-pass filter attenuates low frequencies and blocks dc. (c) A band-pass filter attenuates both low and high frequencies. 63

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68  Fabricated using integrated- circuit technologies ◦ Inside an op-amp: 1)Transistors 2)Parasitic capacitors 3)Internal resistors  Op-amp chip configuration that we will use: Quad-channel (i.e. 4-in-1) ◦ 14 pins in the op-amp chip 67

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70 69 Level shifter Differential input preamplifier Inputs Output Output buffer Input Output Q k V+ V- -Vs +Vs Q1 Q2 Q3 10 k 3 k 1 k R2 Io Q4 200 Null 10 k R4

71 70 Gain [Gain-dB] 100K K [100] [80] [60] [40] [20] [0] [-20] Frequency (Hz) K1M100M Slopes

72 71 Phase 0 -  /2 -- -3  /2 Frequency (Hz) K 1M 100M

73 72 Gain 100K K1M100M Phase 0 -  /2 -- -3  /2 Frequency (Hz) Gain margin

74 73 Gain 100K K Frequency (Hz) K1M100M Uncompensated Compensated Open-loop gain Slopes

75 74 Gain 100K K Frequency (Hz) K1M Unity gain BW Slope = -1 Slope = -2

76 75 Gain Frequency K 10K 100K 10K 1M Open-loop gain Uncompensated Compensated Unity gain BW Amplifier BW Amplifier circuit gain Loop gain

77 76 Figure 3.13 Op-amp frequency characteristics early op amps (such as the 709) were uncompensated, had a gain greater than 1 when the phase shift was equal to –180º, and therefore oscillated unless compensation was added externally. A popular op amp, the 411, is compensated internally, so for a gain greater than 1, the phase shift is limited to –90º. When feedback resistors are added to build an amplifier circuit, the loop gain on this log-log plot is the difference between the op-amp gain and the amplifier-circuit gain.

78 77 Interim amplifier stage I max /C = dV 0 /dt = Slew rate (S r ) (V/  s) For sinusoidal signals with V 0 = V or *sin(  p t) where V or is the maximum rated output voltage and  p =2  f p is the maximum frequency of the signal; then the slew rate is S r = 2  f p V or yielding f p =Sr /(2  V or ) = full-power response Compensation capacitor

79 78 ViVi V0V0 A ViVi Time V0V0 V or -V or T1T1 Slope = S r = 2V or /T 1 V 0 =V or sin2  ft Maximum slope Limited by Slew rate = S r (dV 0 /dt) max = 2  f*S r Time V or -V or T

80 79 V0V0 A(V 2 – V 1 ) + V 1 V 2 RdRd R0R0 RfRf RLRL CLCL I0I0 IfIf ILIL Don’t let equivalent resistance seen from the output below a few k  Capacitive loads may impose slew-rate like limitations

81 80 VSVS -V S V CM V CM = common mode input voltage - V S – 0.7V < V CM < V S + 0.7V VOVO - V S + 1 V < V O < V S – 1 V 3 V < V S < 18 V V S = supply voltage

82 81 V0V0 Ideal Op-amp I b1 I b2 V b1 V b2 V1V1 V2V2 R 1eq R 2eq I b1 and I b2 are input bias currents; input offset current I io = I b1 - I b2 V b1 and V b2 are input bias voltages; input offset voltage V io = V b1 - V b2

83  Offset voltage ◦ Problem in amplifying low-level signals ◦ Can be nulled using a potentiometer ◦ Drift in time and with temperature ◦ Noise  Bias current ◦ Flows through feedback R; use small R (around 10 K, if R is too small then load current + f.b. current may exceed o/p current limit) ◦ Differential bias current (offset current) ◦ Drift ◦ Noise 82

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85 +  RdRd i RoRo RLRL CLCL ioio AdAd dd oo ii +  84 Figure 3.15 The amplifier input impedance is much higher than the op- amp input impedance R d. The amplifier output impedance is much smaller than the op-amp output impedance R o.

86 85 Unity-gain follower V 0 = AV d = A(V i – V 0 ) = AV i /(A+1) I i = V d /R d = V i /(A+1)R d R i = V i /I i = (A+1)R d  AR d -V d = V 0 = AV d + I 0 R 0 = -AV 0 + I 0 R 0 Yielding (A+1)V 0 = I 0 R 0 Hence, R 0 = V 0 /I 0 = R 0 /(A+1)  R 0 /A I 0 = C L dV 0 /dt

87 86 Frequency (Hz) Noise voltage (V.Hz -1/2 ) or Noise current (A.Hz -1/2 ) in rms 1/f noise White noise 1/f (flicker)noise Colored noise White noise Popcorn noise fLfL fHfH Bandwidth (BW)

88 87 V0V0 - + InIn InIn V1V1 V2V2 R1R1 R2R2 + - VnVn VdVd AV d Characteristic noise resistance R n = v n /i n


"Temel ilkeler, Süzgeçler, Entegratörler, Türev alıcılar ve Enstrümentasyon (Cihaz) Kuvvetlendiricileri." indir ppt

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