KULIAH MINGGU ke 2 Elektronika Dasar

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

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

3. VOUT = (AV - AV ) = A (V - V ) + - - + 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

4. - Vi1 Vout A Vi2 + B 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) .

5. Offset null Not used 7 6 5 8 Offset null 1 2 3 4 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)

6. Vout A Non-linear region Vin Linear region Voltage Output from an Amplifier The linearrange of an amplifier is finite, and limited by the supply voltage and the characteristics of the amplifier. If anamplifieris 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. Vcc VREF -Vee VIN 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

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) 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'sCurrent Law (KCL). Circuit voltage gain determined by the ratio of R1 and RF. RF R1 - Vin A Vout + B

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

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. RF iF - i- R1 i1 i+ - Vin A Vout + B R or Analysis of Inverting Amplifier KCL at A: VIN 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. RF R1 - A + Vout Vin B Circuit voltage gain determined by the ratio of R1 and RF. Noninverting Amplifier Point VA equals to Vin. Op-amp circuit is a voltage divider.

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

18. SUMMING AMPLIFIER Recall inverting amplifier and If = I1 + I2 + … + In If 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. +5 VDC 4.7k 10k Sensor – Variable Resistor 10k Vout 0-5 VDC 5k Analog Signal Conditioner (Current to Voltage Converter, LM-324) 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 = 1 Resistor Values for the Inverting OP-Amp can be changed to modify gain of converter or to amplify the signal of interest.

26. Single-Ended Input • + terminal : Source • – terminal : Ground • 0o phase change • + terminal : Ground • – terminal : Source • 180o phase change Operational Amplifier

27. 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) Operational Amplifier

28. Distortion The output voltage never excess the DC voltage supply of the Op-Amp 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 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 Operational Amplifier

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

32. 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 ~ 20-100  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 = Gdfc Operational Amplifier

34. ? Hz 10MHz 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 = Gdfc • fc = f1 / Gd = 10 MHz / 20 V/mV • = 10  106 / 20  103 • = 500 Hz Operational Amplifier

35. Ideal Vs Practical Op-Amp 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 Operational Amplifier

37. Noninverting Amplifier • Kirchhoff node equation at V+ yields, • Kirchhoff node equation at Vyields, • Setting V+ = V– yields • or Operational Amplifier

38. Noninverting amplifier Noninverting input with voltage divider Less than unity gain Voltage follower Operational Amplifier

39. Inverting Amplifier • Kirchhoff node equation at V+ yields, • Kirchhoff node equation at Vyields, • Setting V+ = V– yields Notice: The closed-loop gainVo/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. Operational Amplifier

40. Multiple Inputs • Kirchhoff node equation at V+ yields, • Kirchhoff node equation at Vyields, • Setting V+ = V– yields Operational Amplifier

41. 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 Operational Amplifier

42. 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 Operational Amplifier

43. Op-Amp Differentiator Operational Amplifier

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

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

46. 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 Operational Amplifier

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

48. 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 Operational Amplifier

49. 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, Operational Amplifier

50. KOMPARATOR • Rangkaian komparator digunakan untuk membandingkan tegangan masukan dan tegangan referensi. • Tegangan keluaran hanya ada dua kondisi yaitu tegangantinggi 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.