ELEC 2400 Electronic Circuits. Chapter 3: Op Amp and Circuits

Transcription

1 ELEC 2400 Electronic Circuits Chapter 3: Op Amp and Circuits Course Website: HKUST, Fall

2 Chapter 3: Operational Amp and Circuits 3.1 Amplifiers Types of Amplifiers Ideal Operational Amplifier Unity Gain Buffer Positive and Negative Feedback Characteristics of Ideal Operational Amplifier 3.2 Op Amp Circuits Non-Inverting Amplifier Inverting Amplifier Summing Amplifier and Synthesizer Difference Amplifier Instrumentation Amplifier Current Source Negative Impedance Converter oltage-to-current Converter Effect of Finite Op Amp Gain 3.3 Signal Processing Applications Analog-to-Digital Converter Digital-to-Analog Converter Integrator and Differentiator Oscillator Bistable Circuits: Schmitt Trigger, Hysteretic Comparator and Square Waveform Generator 3-2

3 3.1.1 Types of Amplifiers: Ideal oltage Amplifier An ideal voltage amplifier amplifies the input voltage signal in by a voltage gain A v to generate the output voltage signal out : out A v in It can be modeled by a voltage-controlled voltage source: (input port) + in A v in + out (output port) Examples of voltage amplifiers: 3-3

4 Practical oltage Amplifier A practical voltage amplifier has finite input impedance i instead of infinity; and non-zero output impedance o instead of zero. The excitation - the source voltage, may also have finite source resistance s ; and the response - the load, may be a load resistor L. s + o + s i i A v i L o source amplifier load The effective voltage gain due to loading effects ( s, i and o ) is out s i L + + s i o L A v 3-4

5 Example 3-1 Example 3-1: Compute the voltage gain of the voltage amplifier needed for the temperature sensor below, if we require to have o 1 when s 100m (at 100 o C). 0 C C 1 Temperature Sensor + i oltage Amplifier + o Meter Display s 0 at 0 o C +1m/ o C s 300Ω i 2kΩ o 400Ω L 10kΩ 3-5

6 Example 3-1 (cont.) Soln.: The temperature sensor can be modeled as follows: s 300Ω + i 2kΩ 400Ω A v i + o 10kΩ sensor amplifier meter out A v i k + 10k 2k 10k A v s k k 1 Av 2k 10k A 100m v k k / 21.6dB 20 log (gain) db

7 Current Amplifier A current amplifier can be modeled as a current-controlled current source, with a current gain of A i. The ideal current amplifier has zero input impedance ( i 0) and infinite output impedance ( o, in parallel with the dependent current source). I I o i A i I i AI i i I o Examples of current amplifiers: Current Source HiFi Power Amplifier Speaker 3-7

8 Practical Current Amplifier A practical current amplifier has non-zero (instead of zero) input impedance i and finite (non-infinite) output impedance o. I s s + i I i i AI i i o I o L + o source current amplifier load The effective current gain due to loading effects ( s, i and o ) is I I o s s o + + s i o L A i 3-8

9 Transresistance Amplifier A transresistance amplifier can be modeled as a current-controlled voltage source, and the gain m has the dimension of resistance. The ideal transresistance amplifier has zero input impedance ( i 0) and zero output impedance ( o 0). I o i m I i mii + o Examples of transresistance amplifiers: f I s I D s f Photodiode Amplifier Optical Sense Amplifier 3-9

10 Transconductance Amplifier A transconductance amplifier can be modeled as a voltagecontrolled current source, and the gain G m has the dimension of conductance. The ideal transconductance amplifier has infinite input impedance ( i ) and infinite output impedance ( o ). Io i G m + i G m I o i Examples of transconductance amplifiers: Universal Transconductance Amplifier Stereo Amplifier using acuum Tube 3-10

11 3.1.2 Ideal Operational Amplifier (Op Amp) Operational amplifiers, or commonly known as op amps, are probably the most versatile electronic components used in analog circuits. The first monolithic IC (integrated circuit) implementation of an op amp was invented in 1963 by Bob Widlar of Fairchild Semiconductor, labeled µa702. Op amps are so named because they can be configured to perform mathematical operations such as +,,,,, d/dx, log, exp, etc. They can amplify continuous signals and are the basic components of analog computer. 3-11

12 Op Amp as High Gain oltage Amplifier An op amp is a voltage amplifier with an extremely high voltage gain (for example, A v > 1000 /, or 60dB), while the ideal op amp has infinite gain. dd in + + A ss + A( ) o + + in + A A( ) o + Examples of op amps: 3-12

13 Ideal and Practical Op Amps The op amp can be described as a differential input single-ended output device. In the past, positive and negative power supplies were used, such as ±15 or ±5. Nowadays, the negative power supply is often just 0 (Ground, or GND). Some of the properties of an op amp are shown below: Ideal Practical Gain A 10 4 / 80dB Input Impedance i 10MΩ Output Impedance o 0 100Ω Input offset voltage os 0 ±2m Input offset current I + I 0nA ±5nA Bandwidth BW 100MHz 3-13

14 Circuit Model of Op Amp Ideal Op Amp Practical Op Amp dd dd + in + ( ) + o + in + i o A(+ ) o ss ss 3-14

15 Example 3-2 Example 3-2: Compute the output voltage o when in is (1) 0; (2) 100µ; (3) 1m; and (4) 10m in A o Soln.: 15 (1) o (2) o µ 1 (3) o m 10 (4) o m 100 However, 100 is even higher than the power supply voltage +15, and the output voltage will saturate at

16 Operating egions of Op Amp If the input voltage is too large, the output voltage will saturate at either the positive power rail (with + too positive) or the negative power rail (with + too negative). For example, consider the op amp shown below o + in o A m 0 1.5m negative saturation region linear region positive saturation region The input voltage in + can only range between 1.5m and +1.5m and is too narrow to be practical. Therefore, op amps are often used in a feedback configuration. 3-16

17 Op Amp as Comparator Due to the high gain of the op amp, it can be used as a comparator: when the input voltage in is higher than the reference voltage ref, the output voltage o is at dd ; and when in < ref, o ss. in, ref dd o dd in ref o 0 time ss time ss It should be noted that although feasible, it is not economical to use an op amp as a comparator. 3-17

18 3.1.3 Unity Gain Buffer The simplest feedback configuration is the unity gain buffer, by connecting the output voltage o back to the negative input terminal as shown below. The connection realizes negative feedback. in A o in + A(+ ) o The output voltage o can easily be computed: o A( in o ) o A in 1 + A in (A >> 1) 3-18

19 3.1.4 Positive Feedback vs Negative Feedback Qn. Ans. Qn. What is the difference between positive feedback and negative feedback? Negative feedback is stable and positive feedback is unstable (loosely speaking). Why? Negative Feedback Positive Feedback in A o in A o o A( in o ) o in A o 1 + A in 1 (A >> 1) o A( o in ) A 1 A A A 1 1 (A >> 1) 3-19

20 Problem with Positive Feedback For the positive feedback connection, let A1001. For in 2, then o Next, consider injection of noise at o (0) such that it changes slightly to ecompute o and o (1 st ) Go through the loop one more time and o (2 nd ) 1000! Clearly this is unrealistic, and o will stay at the negative power rail, say, at GND! in 2 A o A in A noise

21 Negative Feedback Provides Stability For the negative feedback connection, let A999. For in 2, then o Next, consider injection of noise at o (0) such that it changes slightly to ecompute o and o (1 st ) However, as o swings from to 2.997, it has to pass that is the solution to the circuit. Hence, when o swing to it will just stay there! And negative feedback gives stability in A in A o A + 1 noise

22 Unity Gain Buffer with Ideal Op Amp For the following op amp circuits, we will consider ideal op amps, that is, A, i, and o 0Ω. econsider the unity gain buffer. If A, then o A 1 + A 1 + in in in o in + N.B. When an ideal op amp is working as an amplifier in an op amp circuit, the positive input terminal + has the same potential as the negative input terminal. 3-22

23 Unity Gain Buffer can drive esistive Load A unity gain buffer is different from just a simple wire connected between the input and output of the buffer because the output of a unity gain buffer can drive resistive load. 2 1kΩ 1kΩ o 1 2 1kΩ 1kΩ 100Ω o 1 resistive divider divider cannot drive resistor The "internal" controlled voltage source enables the op amp to drive a large resistive load. o1 2 1kΩ 1kΩ I0A 100Ω o 1 o Ω o 3-23

24 Example 3-3 Example 3-3: A sensor outputs a sensed voltage of 5m and has an internal resistance of 25kΩ. (1) If the sensor is connected to a voltmeter with input resistance of 100kΩ, what is the measured voltage? (2) When a unity gain buffer is added between the sensor and the voltmeter, what is the measured voltage? Soln.: (1) measured 5m 100k 25k + 100k 4m sensed 5m sensor resistance 25kΩ input resistance of meter 100kΩ + measured The measured value is only 4m and is off (lowered) by 20%. sensor voltmeter 3-24

25 Example 3-3 (cont.) (2) sensed 5m sensor resistance 25kΩ I0A input resistance of meter 100kΩ + measured sensor unity gain buffer voltmeter Since i, I 0A; therefore, + measured 5m sensed + sensed 5m Using a unity gain buffer gives an accurate measured valued of the sensor output voltage. 3-25

26 3.1.5 Characteristics of Ideal Op Amp Circuits In order to analyze more complicated op amp circuits, we need to know two important characteristics of the ideal op amp. (i) + If the op amp operates in linear region (not in positive/negative saturation region) with negative feedback connection, the output is a finite voltage. Since o + A for an ideal op amp, A, then + A A( ) o + finite number

27 Ideal Op Amp Circuits (cont.) (i) I + I 0 The input currents of an op amp depend on the input resistance. For an ideal op amp, i. I + I in in i 0 + in I + I + i A(+ ) I ο o Therefore, the input currents of an ideal op amp are always zero. N.B. I o is not necessary equal to

28 Example 3-4 Example 3-4: Find o when the switch is (i) open; and (ii) closed. I o s SW I + Soln.: (i) When the switch is open and I + 0, so the voltage drop across that is connected to + is 0, giving s ( ) + and I 0 gives s o s s s o o s (a voltage follower circuit) 3-28

29 Example 3-4 (cont.) I o s SW I + (ii) When the switch is closed, + is shorted to ground, and Now I 0, + 0 ( ) s 0 0 o o s (a voltage inverter circuit) 3-29

30 3.2.1 Non-Inverting Amplifier To achieve a gain other than unity, we have two simple configurations: the non-inverting amplifier and the inverting amplifier. The non-inverting amplifier is shown below. in out 2 1 Now, and in out 2 in out in

31 Op Amp Gain and Closed Loop Gain It is important to distinguish two types of voltage gains. (1) The op amp gain A has the relation: A( ) out + In the previous case, A. (2) The non-inverting amplifier employs negative feedback, and the voltage gain relating the output voltage to the input voltage is the closed loop gain A cl : Acl out in

32 Example 3-5 Example 3-5: Let the op amp be ideal, (i) find o for s 2; and (ii) find o (t) for s (t) 2 sin(2π50t). s 2 o 4kΩ 1 2kΩ Soln.: (i) o s 4k k (ii) o(t) o(t) 6 sin(2π50t) o 6 t / ms

33 Example 3-6 Example 3-6: Find I 1 and I 2. o s 1 I 1 100Ω 400Ω I 2 2kΩ Soln.: The op amp circuit can be redrawn as shown in the right. + 1 I 1 10mA o Ω 100 I o mA 100Ω 2k 2k I 1 o I 2 2kΩ 3-33

34 3.2.2 Inverting Amplifier Another simple op amp circuit is the inverting amplifier: 2 in Now, virtual ground out As has the same potential as ground, but itself is not ground, hence, it is called the virtual ground. Next, in out in 1 2 out 2 A cl 1 Note that the polarity of out is opposite to in. 3-34

35 Example 3-7 Example 3-7: Let the op amp be ideal, (i) find o for s 1; and (ii) find o (t) for s (t) sin2π50t. 6kΩ s 1kΩ 3kΩ o Soln.: (i) o s 6k 3k 2 1k o (ii) o(t) 2 o(t) 2 sin(2π50t) t / ms

36 3.2.3 Summing Amplifier and Synthesizer By adding more inputs to the inverting amplifier, a summing amplifier can be realized. 2 f 1 s2 s1 o By noting that 0 we have s1 s2 0 o f f f o s1 1 2 s2 The output voltage o is the weighted sum of the inputs s1 and s

37 Example 3-8 Example 3-8: For the summing amplifier shown below, find o for s1 1 and (1) s2 2; and (2) s2 (t) 2sinωt. 2kΩ 6kΩ s2 s1 3kΩ o Soln.: (1) o 6k 6k 1 2 3k 2k (2) o(t) 6k 6k 1 2sin t 3k 2k ω 2 6 sinωt

38 Summing Amplifier as Audio Mixer By adding more inputs to the summing amplifier, and the gain of each input branch can be adjusted by a variable resistor, an audio mixer can be constructed. Other applications include karaoke, equalizer, synthesizer. n 2 f 1 sn s2 s1 o 3-38

39 Synthesizer A synthesizer is an electronic instrument that can produce various sounds by generating and combining different frequencies. It can produce thousands of different sounds and sound combinations by controlling different resistor ratios / 1, / 2, etc A a a A o 4 4 o

40 3.2.4 Difference Amplifier Besides addition, we may arrange to have a difference amplifier o 1 2 and o o o 2 ( 2 1 )

41 Example 3-9 Example 3-9: Find o of the following difference amplifier. 1 2kΩ 10kΩ 2 2kΩ 10kΩ o Soln.: o 10k (2 1) 2k

42 Disadvantages of Difference Amplifier The difference amplifier has two disadvantages: (1) To change the gain of the amplifier, two resistors (either all 1 or all 2 ) have to be changed. (2) An ideal voltage amplifier has infinite input resistance. Now, the input resistance is only 2 1, and it draws considerable current from the input ( 2 1 ). Moreover, if a larger gain is achieved by lowering 1 's, a larger current is drawn from ( 2 1 ). I in o I in 1 2 virtual short 3-42

43 3.2.5 Instrumentation Amplifier A difference amplifier can be realized by the instrumentation amplifier: Advantages: o + 1 I ( 2 ) 2 1 ( + 2 1) (2 1) (1) The input resistance is very high, and no current is drawn from the sources. (2) The gain can be adjusted by changing one resistor (). Disadvantage: The output is a differential voltage I 1 + O 3-43

44 IA with Single-Ended Output The previous instrumentation amplifier can be modified to have singled-ended output: o 2 2 I a 1 + I b I + I a b + o G 2 G o 2 1 2(2 1) 2 2 G ( 2 1 ) 1 2 G I a I b G o o G o (2 1) 1 G 3-44

45 IA with Single-Ended Output (cont.) Advantages: (1) The input resistance is very large (close to ), and draws negligible current from the source. (2) We only need to change G to adjust the voltage gain. (3) The output voltage o is now referenced to ground. 3-45

46 3.2.6 Current Source The following circuit provides a current source that is independent of the load resistance/impedance: I3 3 2 I2 s I Z L L 1 I 1 1 I 3 I s I L s I I s s

47 Op Amp Based Current Source (cont.) From the analysis, the current source I L is independent of the load Z L. It only depends on the voltage source s, and accurate resistors 1, 2 and 3 (but Z L is not connected to ground). Example 3-10: Find I L given s 2, 1 10kΩ, 2 1MΩ and 3 100kΩ. Soln.: I L 1M k 100k 2.02mA Note that I 3 s / 3 2/100k 20µA and is negligibly small. 3-47

48 3.2.7 Negative Impedance Converter A negative impedance converter realizes the negative value of a specify resistance or impedance Z. I i s i I 2 2 I 1 Ii i s 1 s 2 s s 1 I i

49 3.2.8 oltage-to-current Converter Based on the negative impedance converter, a voltage-to-current converter (-to-i converter) that is independent of the load can be realized. 1 I L s 2 1 Z L I L 1 IL IL s s Z L s Z L 1 ZL s 3-49

50 Example 3-11 Example 3-11: Find o and I s of the following circuit. 5 I s 1kΩ 1kΩ 1kΩ 1kΩ 1kΩ o 5 I s 0 10mA 5mA 1kΩ 5mA 1kΩ 5 5mA 1kΩ 1kΩ 1kΩ o 10 5mA 5 5mA 3-50

51 3.2.9 Effect of Finite Op Amp Gain For a practical op amp, the open-loop gain A is not infinite. It may be 10 4 / (80dB), 1000 / (60dB), etc. The circuit model of the inverting amplifier with a finite gain A is shown in the right figure. 2 2 in 1 A o in 1 + A o 3-51

52 Effect of Finite Op Amp Gain (cont.) For A, +, and o A. Now, in o in o A o o 2 A cl in A + 1 For A >> 1, then o in

53 Example 3-12 Example 3-12: Let 1 1kΩ and 2 10kΩ. Compute the closedloop gain A cl of the inverting amplifier for A 100, 1000, 10 4 and Soln.: A cl A ( ) A cl A ( 11) A cl A ( 11) A cl A ( 11) / / / /

54 3.3. Signal Processing Applications Signals such as voice and temperature vary continuously with time, and are known as analog signals. With the exploding development of digital computers, it is more efficient to convert an analog signal into its digital equivalent for processing (analogto-digital conversion, or A/D conversion), and then convert it back to analog signal as output (digital-to-analog conversion, D/A conversion). The circuit that performs A/D conversion is an analog-to-digital converter (ADC), and that performs D/A conversion is a digital-to-analog converter (DAC). analog input ADC digital output Digital Signal Processor digital input DAC analog output DSP 3-54

55 3.3.1 Analog-to-Digital Converter Digitizing an analog signal to a digital signal with 3-bit resolution. in (t) ref 7 ref /8 3 ref /4 5 ref /8 ref /2 3 ref /8 ref /4 ref / t/ms digital word 3-55

56 2-Bit Flash ADC Using comparators (op amps in open loop) and logic gates, a 2-bit flash analog-to-digital converter can be constructed. ref in 3 ref /4 CMP A ref /2 CMP B b 1 ref /4 CMP C Binary Encoder b

57 2-Bit Flash ADC (cont.) A flash ADC can make one conversion per clock cycle (instead of one conversion per several clock cycles). A 2-bit flash ADC needs 3 ( 2 2 1) comparators, and an n-bit flash ADC needs 2 n 1 comparators. Analog Signal A B C b 1 b 0 i < ¼ ref ¼ ref < i < ½ ref ½ ref < i < ¾ ref ¾ ref < i By considering "don't care" conditions, it can be shown that b 1 B (A+B)C b

58 3.3.2 Digital-to-Analog Converter A 4-bit DAC can be constructed using the summing amplifier as shown below. (b i 0 switch open, b i 1 switch closed) MSB b a ref LSB b 2 b 1 b A a A o o b b b b ref Let ref 2, For b 3 b 2 b 1 b , b b b b ref o CD player 3-58

59 3.3.3 Integrator and Differentiator An op amp can be connected to become an integrator: + (t) c1 C 1 1 in(t) I(t) o(t) in(t) 1 o(t) d I(t) c1(t) C1 dt 1 1 C d (t) o 1 1 t in( λ)dλ+ o(0) C 0 dt Apart from the integration constant, the output waveform is the integral of the input waveform. 3-59

60 Example 3-13 Example 3-13: Find o (t) given o (0) µ F ms 2ms s(t) 12.5kΩ I(t) o(t) Soln.: o(t) o(1ms) 1 t in( )d 2 0.1µ 12.5k λ λ+ 1 1m 5dλ m o(t) 0ms 2ms t 3-60

61 Differentiator An op amp can be connected to become a differentiator: + c1(t) 1 in(t) C1 I(t) o(t) (t) o 1 d I(t) c1(t) d in(t) C1 C1 dt dt o(t) C 1 1 d in(t) dt 3-61

62 Example 3-14 Example 3-14: Find and plot o (t). 10kΩ 0.05µ F ms 0.4ms s(t) I(t) Soln.: From t0 to t0.2ms, the input voltage has a constant slope of 2/0.2ms 10,000/s, and o(t) µ 10k 5 0.2m s(t) ms o(t) 5 o(t) 0.4ms t 0 t

63 3.3.4 Oscillator Using two integrators in cascade, an oscillator with natural (undamped) frequency defined by the 's and C's can be obtained. C 1 C 2 1 A 2 B C C A (t) (t) 1 2 C C 1 2 da (t) dt db (t) dt C (t) B (t) 3-63

64 Oscillator (cont.) Clearly, Define 2 ω o (t) (t) B C 2 d B (t) + ω 2 dt 1 CC o The solution gives an oscillator: 2 C B 2 (t) 1 C 1 da (t) 1C1 dt 2 d B (t) 2 dt 0 B (t) α cos( ω o t + β) Note that there is no input to the circuit, and the op amp circuit just oscillates by itself (due to positive feedback). 3-64

65 Comparator with Noisy Input Consider a comparator with a noisy input: dd dd in ref 0 dd CMP o multiple crossings dd If the comparator is very fast, it will response to noise near ref 0 immediately and results in multiple crossings before settling down to the required value. 3-65

66 3.3.5 Bistable Circuits A bistable circuit has two stable states. It can be generated by using positive feedback around an op amp. Consider swapping the + and of the non-inverting amplifier such that negative feedback is turned into positive feedback. As such, the output voltage will saturate at either the positive power rail ( dd ) or the negative power rail ( ss ). An alternative way to draw the circuit is shown to the right. in dd o ss o in

67 Schmitt Trigger The circuit is called a Schmitt trigger. For simplicity, let ss dd. 1 2 H 1 2 L in o dd in o dd Consider the case when in dd, giving o dd and dd H As long as in < H (high threshold), o remains at + dd. However, when in > H, o trips to dd, and dd Similarly, for in > L (low threshold), o remains at dd. L 3-67

68 Schmitt Trigger: Hysteresis The DC transfer characteristic can be constructed accordingly. It exhibits hysteresis that can be adjusted by 1 : o dd H dd L dd 0 in dd hysteresis window The quantity hys H L is defined as the hysteresis window. 3-68

69 Schmitt Trigger ejects Noise Let ref 0. By designing the hysteresis window appropriately, multiple crossings are avoided, although the output waveform is shifted as H L 0. in H 0 L dd CMP dd 2 expected output o actual output dd 1 dd 3-69

70 Square Waveform Generator By modifying the Schmitt trigger appropriately, that is, (1) making the hysteresis window hys large; and (2) using the output of the comparator to drive an C circuit; then we can construct a square waveform generator. If C time constant is much larger than the period, a triangular waveform can be generated. 1 2 dd o o 0 time CMP dd C C H C large C time constant 0 time L 3-70