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KULIAH MINGGU ke 2 Elektronika Dasar 1 OPERASIONAL AMPLIFIER OP-AMP Jurusan Teknik Elektro 2007.

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Presentasi berjudul: "KULIAH MINGGU ke 2 Elektronika Dasar 1 OPERASIONAL AMPLIFIER OP-AMP Jurusan Teknik Elektro 2007."— Transcript presentasi:

1 KULIAH MINGGU ke 2 Elektronika Dasar 1 OPERASIONAL AMPLIFIER OP-AMP Jurusan Teknik Elektro 2007

2 2 Op-amp : suatu IC analog + _ Input 1 Input 2 output + V CC - V EE SIMBOL

3 SIFAT IDEAL 3 Ideally, 1.No current can enter terminals V + or V -. Called infinite input impedance. 2.V out =A(V + - V - ) with A → ∞ 3.In a circuit V + is forced equal to V - 4.An opamp needs two voltages to power it V cc and -V ee. A V OUT = (AV - AV ) = A (V - V )

4 4 Operational Amplifier (Op Amp) An operational amplifier (Op Amp) is an integrated circuit of a complete amplifier circuit. Op amps have an extremely high gain (A=10 5 typically). Op amps also have a high input impedance (R=4 MΩ, typically) and a low output impedance (in order of 100 Ω, typically). (in order of 100 Ω, typically). - + V i1 V out A B V i2

5 5 Characters of Operational Amplifiers  high open loop gain  high input impedance  low output impedance  low input offset voltage  low temperature coefficient of input offset voltage  low input bias current  wide bandwidth  large common mode rejection ratio (CMRR) Offset null Not used

6 6 Voltage Output from an Amplifier The linear range of an amplifier is finite, and limited by the supply voltage and the characteristics of the amplifier. A Linear region Non-linear region V out V in Daerah Linier ini sangat Kecil If an amplifier is driven beyond the linear range (overdriven), serious errors can result if the gain is treated as a constant. Kalau A = 10 6 dan V CC = 12 Volt maka daerah linier = 24 μV V in = V 2 -V 1

7 OPAMP: COMPARATOR (bekerja di daerah jenuh) 7 V out =A(V in – V ref ) If V in >V ref, V out = +∞ but practically hits +ve power supply = V cc If V in

8 8 OPAMP: ANALYSIS The key to op amp analysis is simple 1.No current can enter op amp input terminals. => Because of infinite input impedance 2.The +ve and –ve (non-inverting and inverting) inputs are forced to be at the same potential. => Because of infinite open loop gain 3.Use the ideal op amp property in all your analysis

9 9 Inverting Amplifier (bekerja di daerah linier) Point B is grounded, point A is called Virtual Grounded. Voltage across R 1 is V in, and across R F is V out. The output node voltage determined by Kirchhoff's Current Law (KCL). Circuit voltage gain determined by the ratio of R 1 and R F. - + V in V out R1R1 RFRF A B

10 10 PENGUAT INVERTING (bekerja di daerah linier) Kondisi fisik 1 R2R2 R1R input output R3R3

11 11 OPAMP: INVERTING AMPLIFIER 1.V - = V + 2.As V + = 0, V - = 0 (VG) 3.As no current can enter V - and from Kirchoff’s Ist law, I 1 =I I 1 = (V IN - V - )/R 1 = V IN /R 1 5. I 2 = (0 - V OUT )/R 2 = -V OUT /R 2 => V OUT = -I 2 R 2 6. From 3 and 6, V OUT = -I 2 R 2 = -I 1 R 2 = -V IN R 2 /R 1 (NEG) 7. Therefore V OUT = (-R 2 /R 1 )V IN

12 12 Analysis of Inverting Amplifier Ideal transfer characteristics: - + V in V out R1R1 RFRF A B R i+i+i+i+ iFiFiFiF i1i1i1i1 i-i-i-i- - or V IN KCL at A:

13 13 OPAMP: NON – INVERTING AMPLIFIER (bekerja di daerah linier) 1.V - = V + 2.As V + = V IN, V - = V IN 3.As no current can enter V - and from Kirchoff’s Ist law, I 1 =I I 1 = V x /R 1 = V IN /R 1 5. I 2 = (V OUT - V IN )/R 2  V OUT = V IN + I 2 R 2 6. V OUT = I 1 R 1 + I 2 R 2 = (R 1 +R 2 )I 1 = (R 1 +R 2 )V IN /R 1 7. Therefore V OUT = (1 + R 2 /R 1 )V IN (tak berlawanan) VxVx

14 14 Op-amp circuit is a voltage divider. Noninverting Amplifier - + V in V out R1R1 RFRF A B Circuit voltage gain determined by the ratio of R 1 and R F. Point V A equals to V in.

15 OPAMP : VOLTAGE FOLLOWER (BUFER) (bekerja di daerah linier) 15 V + = V IN. V - = V + Thus V out = V - = V + = V IN !!!! So what’s the point ? The point is, due to the infinite input impedance of an op amp, no current at all can be drawn from the circuit before V IN. Thus this part is effectively isolated. Very useful for interfacing to high impedance sensors such as microelectrode, microphone… i = 0

16 16 Differential Amplifier Point B is grounded, so does point A (very small). Voltage across R 1 is V 1, and across R 2 is V 2. Normally: R 1 = R 2, and R F = R 3. Commonly used as a single op-amp instrumentation amplifier. RFRF - + V1V1 V out R1R1 A B R3R3 V2V2 R2R2

17 17 Analysis of an Instrumentation Amplifier Design a single op-amp instrumentation amplifier. R 1 = R 2, R F = R 3 Determine the instrumentation gain. - + V1V1 V out R1R1 RFRF A B R3R3 V2V2 R2R2

18 SUMMING AMPLIFIER 18 V OUT = -R f (V 1 /R 1 + V 2 /R 2 + … + V n /R n ) IfIf Recall inverting amplifier and I f = I 1 + I 2 + … + I n Summing amplifier is a good example of analog circuits serving as analog computing amplifiers (analog computers)! Note: analog circuits can add, subtract, multiply/divide (using logarithmic components, differentiat and integrate – in real time and continuously.

19 19 For the following circuit, calculate the input resistance. R1 Rf R2 V in V out

20 20 INSTRUMENTATION AMPLIFIER

21 21 INSTRUMENTATION AMPLIFIER Inverting amplifier Non-inverting amplifier very high input impedance - So, you can connect to sensors Differential amplifier -> it rejects common-mode interference -> so you can reject noise Gain in the multiple stages: i.e. High Gain – so, you can amplify small signals

22 INSTRUMENTATION AMPLIFIER: STAGE 1 22 I1I1 Recall virtual ground of opamps I 1 = (V 1 – V 2 )/R 1 Recall no current can enter opamps and Kirchoff’s current law I 2 = I 3 = I 1 Recall Kirchoff’s voltage law V OUT = (R 1 + 2R 2 )(V 1 – V 2 )/R 1 = (V 1 – V 2 )(1+2R 2 /R 1 ) I2I2 I3I3 I1I1

23 INSTRUMENTATION AMPLIFIER: STAGE 2 23 I1I1 Recall virtual ground of opamps and voltage divider V - = V + = V B R 4 /(R 3 + R 4 ) Recall no current can enter opamps (V A – V - )/R 3 = (V - – V OUT )/R 4 Solving, V OUT = – (V A – V B )R 4 /R 3 I2I2 I3I3 VAVA VBVB

24 24 INSTRUMENTATION AMPLIFIER: COMPLETE V OUT = – (V 1 – V 2 )(1 + 2R 2 /R 1 )(R 4 /R 3 )

25 25 As force is applied on the sensor, the value of the variable resistor changes which results in a specific voltage output. Gain = Vout/Vin = 1 Resistor Values for the Inverting OP-Amp can be changed to modify gain of converter or to amplify the signal of interest. Analog Signal Conditioner (Current to Voltage Converter, LM-324) +5 VDC Sensor – Variable Resistor 4.7k 10k 5k V out 0-5 VDC I1I1 I2I2 I3I3

26 Single-Ended Input Ref:080114HKNOperational Amplifier26 + terminal : Source – terminal : Ground 0 o phase change + terminal : Ground – terminal : Source 180 o phase change

27 Double-Ended Input Ref:080114HKNOperational Amplifier27 Differential input 0 o phase shift change between V o and V d Qu: What V o should be if, (A) (B) Ans: (A or B) ?

28 Distortion Ref:080114HKNOperational Amplifier28 The output voltage never excess the DC voltage supply of the Op-Amp

29 Ref:080114HKNOperational Amplifier29 Common-Mode Operation Same voltage source is applied at both terminals Ideally, two input are equally amplified Output voltage is ideally zero due to differential voltage is zero Practically, a small output signal can still be measured Note for differential circuits: Opposite inputs : highly amplified Common inputs : slightly amplified  Common-Mode Rejection

30 Common-Mode Rejection Ratio (CMRR) Ref:080114HKNOperational Amplifier30 Differential voltage input : Common voltage input : Output voltage : G d : Differential gain G c : Common mode gain Common-mode rejection ratio: Note: When G d >> G c or CMRR   V o = G d V d

31 CMRR Example Ref:080114HKNOperational Amplifier31 What is the CMRR? Solution : NB: This method is Not work! Why? (1) (2)

32 Op-Amp Properties Ref:080114HKNOperational Amplifier32 (1)Infinite Open Loop gain -The gain without feedback -Equal to differential gain -Zero common-mode gain -Pratically, G d = 20,000 to 200,000 (2) Infinite Input impedance -Input current i i ~0A -T-  in high-grade op-amp -m-A input current in low-grade op-amp (3) Zero Output Impedance -act as perfect internal voltage source -No internal resistance -Output impedance in series with load -Reducing output voltage to the load -Practically, R out ~ 

33 Frequency-Gain Relation Ref:080114HKNOperational Amplifier33 Ideally, signals are amplified from DC to the highest AC frequency Practically, bandwidth is limited 741 family op-amp have an limit bandwidth of few KHz. Unity Gain frequency f 1 : the gain at unity Cutoff frequency f c : the gain drop by 3dB from dc gain G d GB Product : f 1 = G d f c 20log(0.707)=3dB

34 GB Product Ref:080114HKNOperational Amplifier34 Example: Determine the cutoff frequency of an op-amp having a unit gain frequency f 1 = 10 MHz and voltage differential gain G d = 20V/mV Sol: Since f 1 = 10 MHz By using GB production equation f 1 = G d f c f c = f 1 / G d = 10 MHz / 20 V/mV = 10  10 6 / 20  10 3 = 500 Hz 10MHz ? Hz

35 Ideal Vs Practical Op-Amp Ref:080114HKNOperational Amplifier35 IdealPractical Open Loop gain A  10 5 Bandwidth BW  Hz Input Impedance Z in  >1M  Output Impedance Z out 0   Output Voltage V out Depends only on V d = (V +  V  ) Differential mode signal Depends slightly on average input V c = (V + +V  )/2 Common-Mode signal CMRR  dB

36 Ideal Op-Amp Applications Analysis Method : Two ideal Op-Amp Properties: (1)The voltage between V + and V  is zero V + = V  (2)The current into both V + and V  termainals is zero For ideal Op-Amp circuit: (1)Write the kirchhoff node equation at the noninverting terminal V + (2)Write the kirchhoff node eqaution at the inverting terminal V  (3)Set V + = V  and solve for the desired closed-loop gain Ref:080114HKNOperational Amplifier36

37 Ref:080114HKNOperational Amplifier37 Noninverting Amplifier (1)Kirchhoff node equation at V + yields, (2)Kirchhoff node equation at V  yields, (3)Setting V + = V – yields or

38 Ref:080114HKNOperational Amplifier38 Noninverting amplifierNoninverting input with voltage divider Voltage follower Less than unity gain

39 Inverting Amplifier (1)Kirchhoff node equation at V + yields, (2)Kirchhoff node equation at V  yields, (3)Setting V + = V – yields Ref:080114HKNOperational Amplifier39 Notice: The closed-loop gain V o /V in is dependent upon the ratio of two resistors, and is independent of the open-loop gain. This is caused by the use of feedback output voltage to subtract from the input voltage.

40 Multiple Inputs (1)Kirchhoff node equation at V + yields, (2)Kirchhoff node equation at V  yields, (3)Setting V + = V – yields Ref:080114HKNOperational Amplifier40

41 Inverting Integrator Ref:080114HKNOperational Amplifier41 Now replace resistors R a and R f by complex components Z a and Z f, respectively, therefore Supposing (i)The feedback component is a capacitor C, i.e., (ii)The input component is a resistor R, Z a = R Therefore, the closed-loop gain (V o /V in ) become: where What happens if Z a = 1/j  C whereas, Z f = R? Inverting differentiator

42 Op-Amp Integrator Ref:080114HKNOperational Amplifier42 Example: (a)Determine the rate of change of the output voltage. (b)Draw the output waveform. Solution: (a) Rate of change of the output voltage (b) In 100  s, the voltage decrease

43 Op-Amp Differentiator Ref:080114HKNOperational Amplifier43

44 Non-ideal case (Inverting Amplifier) Ref:080114HKNOperational Amplifier44 3 categories are considering  Close-Loop Voltage Gain  Input impedance  Output impedance  Equivalent Circuit

45 Close-Loop Gain Ref:080114HKNOperational Amplifier45 Applied KCL at V– terminal, By using the open loop gain,   The Close-Loop Gain, A v

46 Close-Loop Gain Ref:080114HKNOperational Amplifier46 When the open loop gain is very large, the above equation become, Note : The close-loop gain now reduce to the same form as an ideal case

47 Input Impedance Ref:080114HKNOperational Amplifier47 Input Impedance can be regarded as, where R is the equivalent impedance of the red box circuit, that is However, with the below circuit,

48 Input Impedance Ref:080114HKNOperational Amplifier48 Finally, we find the input impedance as,  Since,, R in become, Again with Note: The op-amp can provide an impedance isolated from input to output

49 Output Impedance Ref:080114HKNOperational Amplifier49 Only source-free output impedance would be considered, i.e. V i is assumed to be 0 Firstly, with figure (a), By using KCL, i o = i 1 + i 2 By substitute the equation from Fig. (a),  R  and A comparably large,

50 50 KOMPARATOR Rangkaian komparator digunakan untuk membandingkan tegangan masukan dan tegangan referensi. Tegangan keluaran hanya ada dua kondisi yaitu tegangan tinggi atau rendah (negatif). Kondisi ini ditentukan oleh besarnya tegangan masukan apakah lebih tinggi terhadap tegangan referensi atau lebih rendah. Persoalan dalam komparator sederhana adalah stabilitas. Bila tegangan masukan bervariasi sekitar tegangan referensi maka tegangan keluaran akan berubah-ubah tidak stabil. Hal tersebut dapat dihilangkan dengan rangkaian schmitt.

51 51 KOMPARATOR SEDERHANA

52 52 KURVA HUBUNGAN TEGANGAN Vi < Vr  Vo = +Vsat Vi > Vr  Vo = -Vsat Vr>0

53 53 KURVA HUBUNGAN TEGANGAN Vi < Vr  Vo = +Vsat Vi > Vr  Vo = -Vsat Vi Vo Vr=0

54 54 KURVA HUBUNGAN TEGANGAN Vi < Vr  Vo = +Vsat Vi > Vr  Vo = -Vsat Vi Vo Vr<0

55 55 STABILITAS KOMPARATOR SEDERHANA V cc /2 V in R V cc R Teganga n masukan Tegangan keluaran V CC /2

56 56 Positive Feedback RANGKAIAN SCHMITT Positive Feedback Rangkaian ini disebut komparator Schmitt trigger. Rangkaian resistor membuat positive feedback. R V out R V out /2 V in

57 57 CARA KERJA Schmitt trigger Anggap tegangan masukan kecil, tegangan keluaran menjadi tinggi. Bila V out is 4 V, maka masukan non-inverting V + adalah 2 Volt. Kondisi keluaran tetap selama V in kurang dari 2 Volt. Bila V in diperbesar sehingga lebih besar dari 2 V, maka V out akan nol, dan V + akan nol juga. Kondisi output ini akan tetap, selama V in lebih besar 2 V. R V out R V out /2 V in

58 58 TAK STABIL

59 59 STABIL STABIL histerisis

60 60 STABIL

61 61 STABIL

62 62 RANGKAIAN dan OUTPUT

63 63 KETERANGAN SCHMITT Schmitt trigger adalah sebuah aplikasi comparator yang mengubah tagangan keluaran menjadi negatif bila mtegangan masukan lebih besar tegangan referensi. Schmitt trigger adalah sebuah aplikasi comparator yang mengubah tagangan keluaran menjadi negatif bila mtegangan masukan lebih besar tegangan referensi.comparator Kemudian menggunakan negative feedback untuk mencegah agar tegangan keluaran tdk kembali ke kondisi semula saat tegangan kembali kurang dari tegangan referensi, sampai nanti masukan lebih kecil dari yang ditentukan. Kemudian menggunakan negative feedback untuk mencegah agar tegangan keluaran tdk kembali ke kondisi semula saat tegangan kembali kurang dari tegangan referensi, sampai nanti masukan lebih kecil dari yang ditentukan.negative feedbacknegative feedback

64 64 APLIKASIAPLIKASIAPLIKASIAPLIKASI

65 65 PERHITUNGAN Kerja Schmitt trigger merupakan proses komparasi dengan threshold ganda. Persamaan arus di titik A: Kerja Schmitt trigger merupakan proses komparasi dengan threshold ganda. Persamaan arus di titik A:Schmitt triggerSchmitt trigger Karena hanya 2 pers, maka harus ada satu R yang ditentukan dulu. Karena hanya 2 pers, maka harus ada satu R yang ditentukan dulu. Ingat : V out = V CC saat V in diatas batas atas (V 2 ) V out = -V EE saat V in dibawah batas bawah (V 2 ’).

66 66 u

67 LM741 67

68 NOMOR KAKI 68

69 69

70 70

71 71

72 72

73 73

74 74

75 KESIMPULAN Op-amp dapat digunakan sebagai : 1.Penguat INVERTING 2.Penguat NON INVERTING 3.BUFER 4.Penguat PENJUMLAH 5.Penguat INSTRUMENTASI 6.Pengubah ARUS KE TEGANGAN atau sebaliknya 7.KOMPARATOR 75

76 PR Buktikan rumus untuk menghitung R 1, R 2 dan R 3 pada komparator Schmitt (slide 41), bila batas atas dan batas bawah diketahui. Rencanakanlah sebuah komparator Schmitt dengan menggunakan sebuah op-amp, yang menggunakan single supply 5 Volt. Batas tegangan yang dideteksi adalah diatas 3 Volt memberikan tegangan output tinggi, dan dibawah 1 Volt menghasilkan tegangan output rendah. Tentukan nilai R yang diperlukan. 76


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