KULIAH MINGGU ke 2 Elektronika Dasar

Presentasi berjudul: "KULIAH MINGGU ke 2 Elektronika Dasar"— Transcript presentasi:

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

2 Op-amp : suatu IC analog
+ VCC + _ Input 1 Input 2 output SIMBOL - VEE

3 SIFAT IDEAL Ideally, No current can enter terminals V+ or V-. Called infinite input impedance. Vout=A(V+ - V-) with A → ∞ In a circuit V+ is forced equal to V- An opamp needs two voltages to power it Vcc and -Vee. A VOUT = (AV - AV ) = A (V - V ) + -

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=105 typically). Op amps also have a high input impedance (R=4 MΩ , typically) and a low output impedance (in order of 100 Ω , typically) . - + Vi1 Vout A B Vi2

5 Characters of Operational Amplifiers
1 2 3 4 8 7 6 5 Offset null Not used 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)

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 Vout Vin If an amplifier is driven beyond the linear range (overdriven), serious errors can result if the gain is treated as a constant. Vin= V2-V1 Kalau A = 106 dan VCC = 12 Volt maka daerah linier = 24 μV Daerah Linier ini sangat Kecil

7 OPAMP: COMPARATOR (bekerja di daerah jenuh)
Vout=A(Vin – Vref) If Vin>Vref, Vout = +∞ but practically hits +ve power supply = Vcc If Vin<Vref, Vout = -∞ but practically hits –ve power supply = -Vee A A (gain) very high Application: detection of a complex signal in ECG Vout Vcc -Vee VIN VREF

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

9 Inverting Amplifier (bekerja di daerah linier)
RF Point B is grounded, point A is called Virtual Grounded. Voltage across R1 is Vin, and across RF is Vout. The output node voltage determined by Kirchhoff's Current Law (KCL). Circuit voltage gain determined by the ratio of R1 and RF. R1 - Vin A Vout + B

10 PENGUAT INVERTING (bekerja di daerah linier)
Kondisi fisik 1 R2 R1 2 3 4 8 7 6 5 input output R3

11 OPAMP: INVERTING AMPLIFIER
V- = V+ As V+ = 0, V- = 0 (VG) As no current can enter V- and from Kirchoff’s Ist law, I1=I2. 4. I1 = (VIN - V-)/R1 = VIN/R1 5. I2 = (0 - VOUT)/R2 = -VOUT/R2 => VOUT = -I2R2 6. From 3 and 6, VOUT = -I2R2 = -I1R2 = -VINR2/R1 (NEG) 7. Therefore VOUT = (-R2/R1)VIN

12 Analysis of Inverting Amplifier
- + Vin Vout R1 RF A B R KCL at A: iF i1 i- - VIN i+ or Ideal transfer characteristics:

13 OPAMP: NON – INVERTING AMPLIFIER (bekerja di daerah linier)
V- = V+ As V+ = VIN, V- = VIN As no current can enter V- and from Kirchoff’s Ist law, I1=I2. Vx 4. I1 = Vx/R1=VIN/R1 5. I2 = (VOUT - VIN)/R2  VOUT = VIN + I2R2 6. VOUT = I1R1 + I2R2 = (R1+R2)I1 = (R1+R2)VIN/R1 7. Therefore VOUT = (1 + R2/R1)VIN (tak berlawanan)

14 Noninverting Amplifier
Point VA equals to Vin . - + Vin Vout R1 RF A B Op-amp circuit is a voltage divider. Circuit voltage gain determined by the ratio of R1 and RF.

15 OPAMP : VOLTAGE FOLLOWER (BUFER) (bekerja di daerah linier)
V+ = VIN. V- = V+ Thus Vout = V- = V+ = VIN !!!! i = 0 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 VIN. Thus this part is effectively isolated. Very useful for interfacing to high impedance sensors such as microelectrode, microphone…

16 Differential Amplifier
RF - + V1 Vout R1 A B R3 V2 R2 Point B is grounded, so does point A (very small). Voltage across R1 is V1, and across R2 is V2. Normally: R1 = R2, and RF = R3. Commonly used as a single op-amp instrumentation amplifier.

17 Analysis of an Instrumentation Amplifier
Design a single op-amp instrumentation amplifier. R1 = R2, RF = R3 Determine the instrumentation gain. - + V1 Vout R1 RF A B R3 V2 R2

18 SUMMING AMPLIFIER Recall inverting amplifier and If = I1 + I2 + … + In
VOUT = -Rf (V1/R1 + V2/R2 + … + Vn/Rn) 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 For the following circuit, calculate the input resistance.
Rf R1 Vin Vout R2

20 INSTRUMENTATION AMPLIFIER

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

22 INSTRUMENTATION AMPLIFIER: STAGE 1
Recall virtual ground of opamps I1 = (V1 – V2)/R1 Recall no current can enter opamps and Kirchoff’s current law I2 = I3 = I1 Recall Kirchoff’s voltage law VOUT = (R1 + 2R2)(V1 – V2)/R1 = (V1 – V2)(1+2R2/R1) I2 I1 I1 I3

23 INSTRUMENTATION AMPLIFIER: STAGE 2
Recall virtual ground of opamps and voltage divider V- = V+ = VBR4/(R3 + R4) Recall no current can enter opamps (VA – V-)/R3 = (V- – VOUT)/R4 Solving, VOUT = – (VA – VB)R4/R3 VA I2 VB I1 I3

24 INSTRUMENTATION AMPLIFIER: COMPLETE
VOUT = – (V1 – V2)(1 + 2R2/R1)(R4/R3)

25 Analog Signal Conditioner (Current to Voltage Converter, LM-324)
+5 VDC Sensor – Variable Resistor 4.7k 10k 5k Vout 0-5 VDC I2 I1 I3 As force is applied on the sensor, the value of the variable resistor changes which results in a specific voltage output. Gain = Vout/Vin = Resistor Values for the Inverting OP-Amp can be changed to modify gain of converter or to amplify the signal of interest.

26 Operational Amplifier
Single-Ended Input + terminal : Source – terminal : Ground 0o phase change + terminal : Ground – terminal : Source 180o phase change Ref:080114HKN Operational Amplifier

27 Operational Amplifier
Double-Ended Input Differential input 0o phase shift change between Vo and Vd Qu: What Vo should be if, Ans: (A or B) ? (A) (B) Ref:080114HKN Operational Amplifier

28 Operational Amplifier
Distortion The output voltage never excess the DC voltage supply of the Op-Amp Ref:080114HKN Operational Amplifier

29 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 Ref:080114HKN Operational Amplifier

30 Common-Mode Rejection Ratio (CMRR)
Differential voltage input : Common voltage input : Common-mode rejection ratio: Output voltage : Note: When Gd >> Gc or CMRR  Vo = GdVd Gd : Differential gain Gc : Common mode gain Ref:080114HKN Operational Amplifier

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

32 Operational Amplifier
Op-Amp Properties Infinite Open Loop gain The gain without feedback Equal to differential gain Zero common-mode gain Pratically, Gd = 20,000 to 200,000 (2) Infinite Input impedance Input current ii ~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, Rout ~  Ref:080114HKN Operational Amplifier

33 Frequency-Gain Relation
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. 20log(0.707)=3dB Unity Gain frequency f1: the gain at unity Cutoff frequency fc: the gain drop by 3dB from dc gain Gd GB Product : f1 = Gd fc Ref:080114HKN Operational Amplifier

34 Operational Amplifier
GB Product Example: Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV Sol: Since f1 = 10 MHz By using GB production equation f1 = Gd fc fc = f1 / Gd = 10 MHz / 20 V/mV = 10  106 / 20  103 = 500 Hz 10MHz ? Hz Ref:080114HKN Operational Amplifier

35 Ideal Vs Practical Op-Amp
Open Loop gain A  105 Bandwidth BW 10-100Hz Input Impedance Zin >1M Output Impedance Zout 0   Output Voltage Vout Depends only on Vd = (V+V) Differential mode signal Depends slightly on average input Vc = (V++V)/2 Common-Mode signal CMRR 10-100dB Ref:080114HKN Operational Amplifier

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

37 Noninverting Amplifier
Kirchhoff node equation at V+ yields, Kirchhoff node equation at V yields, Setting V+ = V– yields or Ref:080114HKN Operational Amplifier

38 Operational Amplifier
Noninverting amplifier Noninverting input with voltage divider Less than unity gain Voltage follower Ref:080114HKN Operational Amplifier

39 Operational Amplifier
Inverting Amplifier Kirchhoff node equation at V+ yields, Kirchhoff node equation at V yields, Setting V+ = V– yields Notice: The closed-loop gain Vo/Vin 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. Ref:080114HKN Operational Amplifier

40 Operational Amplifier
Multiple Inputs Kirchhoff node equation at V+ yields, Kirchhoff node equation at V yields, Setting V+ = V– yields Ref:080114HKN Operational Amplifier

41 Operational Amplifier
Inverting Integrator Now replace resistors Ra and Rf by complex components Za and Zf, respectively, therefore Supposing The feedback component is a capacitor C, i.e., The input component is a resistor R, Za = R Therefore, the closed-loop gain (Vo/Vin) become: where What happens if Za = 1/jC whereas, Zf = R? Inverting differentiator Ref:080114HKN Operational Amplifier

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

43 Op-Amp Differentiator
Ref:080114HKN Operational Amplifier

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

45 Operational Amplifier
Close-Loop Gain Applied KCL at V– terminal, By using the open loop gain,   The Close-Loop Gain, Av Ref:080114HKN Operational Amplifier

46 Operational Amplifier
Close-Loop Gain 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 Ref:080114HKN Operational Amplifier

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

48 Operational Amplifier
Input Impedance Finally, we find the input impedance as,  Since, , Rin become, Again with Note: The op-amp can provide an impedance isolated from input to output Ref:080114HKN Operational Amplifier

49 Operational Amplifier
Output Impedance Only source-free output impedance would be considered, i.e. Vi is assumed to be 0 Firstly, with figure (a), By using KCL, io = i1+ i2 By substitute the equation from Fig. (a), R and A comparably large, Ref:080114HKN Operational Amplifier

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 KOMPARATOR SEDERHANA

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

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

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

55 STABILITAS KOMPARATOR SEDERHANA
Tegangan masukan VCC/2 Vin Vcc R Vcc/2 R Tegangan keluaran

56 RANGKAIAN SCHMITT Positive Feedback
Vin Vout Vout /2 R R Rangkaian ini disebut komparator Schmitt trigger. Rangkaian resistor membuat positive feedback.

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

58 TAK STABIL

59 STABIL histerisis

60 STABIL

61 STABIL

62 RANGKAIAN dan OUTPUT

63 KETERANGAN SCHMITT Schmitt trigger adalah sebuah aplikasi comparator yang mengubah tagangan keluaran menjadi negatif bila mtegangan masukan lebih besar tegangan referensi. 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.

64 APLIKASI

65 PERHITUNGAN Kerja Schmitt trigger merupakan proses komparasi dengan threshold ganda. Persamaan arus di titik A: Karena hanya 2 pers, maka harus ada satu R yang ditentukan dulu. Ingat : Vout = VCC saat Vin diatas batas atas (V2) Vout = -VEE saat Vin dibawah batas bawah (V2’).

66 u

67 LM741

68 NOMOR KAKI

69

70

71

72

73

74

75 KESIMPULAN Penguat INVERTING Penguat NON INVERTING BUFER
Op-amp dapat digunakan sebagai : Penguat INVERTING Penguat NON INVERTING BUFER Penguat PENJUMLAH Penguat INSTRUMENTASI Pengubah ARUS KE TEGANGAN atau sebaliknya KOMPARATOR

76 PR Buktikan rumus untuk menghitung R1, R2 dan R3 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.