55:041 Electronic Circuits

Transcription

1 55:04 Electronic Circuits Lecture -5 eiew of Op-Amps Sections of Chapters 9 & 4 A. Kruger Op-Amp eiew-

2 eal-world Op-Amp In earlier courses, op-amp were often considered ideal Infinite input resistance Infinite open-loop gain Infinite bandwidth Noiseless Zero output resistance Zero amplification for common-mode signals Modern op-amps hae remarkable specifications, and in many cases approximate the ideal op-amp quite well Howeer, there are also many case where a clear understanding of the limitation of real op-amps are ery important We will start with a quick-paced reiew of op-amps A. Kruger Op-Amp eiew-

3 Operational Amplifiers Inputs Output Schematic symbol eal amplifiers need power supplies. Note the dual power supplies There are single-supply opamps on the market Integrated circuit (IC) op-amp A. Kruger Op-Amp eiew-3

4 Ideal Op-Amp Equialent Circuit Inerting input Ideal oltage-controlled, oltage source Noninerting input o A oo Control oltage A od Open-loop differential gain Ideally A od Common-mode input signal Common-mode rejection A. Kruger Op-Amp eiew-4

5 Parameters of Ideal Op Amps +Vcc Effectie input resistance Ideally A od o Effectie output resistance 0 A od ( ) No common-mode amplification -Vcc o A A od o od ( 0 ) Because A od, the implication is that for finite output oltage, the differential input oltage is ery small ( 0) A. Kruger Op-Amp eiew-5

6 Op-Amp Basics Inerting Amplifier Op-amp inerting amplifier Because A od, 0 +Vcc -Vcc Thus, is at irtual ground Virtual ground > terminal is at ground potential, but not connected to ground Closed-loop/negatie feedback A. Kruger Op-Amp eiew-6

7 Op-Amp Basics Inerting Amplifier +Vcc -Vcc KCL: sum of current flowing out of node 0 Voltage here is zero (irtual ground concept) 0 I + 0 O I O 0 O I A. Kruger Op-Amp eiew-7

8 Op-Amp Basics Inerting Amplifier i I i A o I I I Input resistance i i i Set gain with two external resistors Set input impedance/resistance with external resistor A. Kruger Op-Amp eiew-8

9 Op-Amp Basics Inerting Amplifier Example. What is the alue of V o in the circuit below? A Note the sign Solution. The first amplifier is a summing inerter and the oltage at node A is V A V The second amplifier is also a summing inerter, and the output oltage is V A ( 4.5) V A. Kruger Op-Amp eiew-9

10 Soling (Ideal) Op-Amp Circuits Find o Find N Find P Set N P + + Techniques KVL KCL Superposition Voltage diision Find N KVL N N 4 3 Find P Voltage Diision P o 5 o o 3 Set N P o o 4 5 o 0 V A. Kruger Op-Amp eiew-0

11 Op-Amp Basics Inerting Amplifier +Vcc -Vcc Similar to inerting amplifier seen before, except resistances are replaced with impedances: resistor(s), capacitor(s), indictor(s), or combinations of these elements Using a similar analysis of for the resistoronly-case, one can show that: Z O I Z A Z Z A. Kruger Op-Amp eiew-

12 Op-Amp Basics Differentiator (Special Type of Inerting Amplifier) +Vcc +Vcc KCL: sum of current flowing out of node 0 -Vcc -Vcc Voltage here is zero (irtual ground concept) i C + 0 O C d 0 I dd O 0 C d I dd O 0 O C d I (t) dd A. Kruger Op-Amp eiew-

13 Op-Amp Basics Integrator (Special Type of Inerting Amplifier) Voltage here is zero (irtual ground concept) +Vcc -Vcc O 0 I + i C + 0 O 0 I d C +C 0 dd d 0 I O +C 0 dd KCL: sum of current flowing out of node 0 This circuit is also called a Miller integrator, and C is the time-constant. d I O C 0 dd d O dd C I O C I t dd + V C 0 t Voltage at t 0 A. Kruger Op-Amp eiew-3

14 Op-Amp Basics Follower +Vcc -Vcc + I O + - > irtual short Why? A o I Impedance transformer L loads S resulting in ery large error +Vcc -Vcc o I L L + S Input resistance ery high > no loading of source o I A. Kruger Op-Amp eiew-4

15 Noninerting Amplifier V p V n > irtual short KCL: sum of current flowing out of node 0 No current flows into inerting input Voltage at noninerting input (V n, or V - ) is V in, because the op-amp maintains a irtual short between its inputs V n + V n V O f 0 V ii + V ii V O f 0 V O + f V ii A + f A. Kruger Op-Amp eiew-5

16 Op-Amp Basics Difference Amplifier Inerting amplifier I 0 O I > irtual short Superposition - Linear circuit 0 I Non-inerting amplifier 4 b I O + b O ( + ) I Analyze with i 0 - Analyze with i 0 - Add results + Superposition O O O O 4 3 ( + / ) ( ) I / I ( ) 4 3 O I I A. Kruger Op-Amp eiew-6

17 Op-Amp Basics Differential Input esistance Note differential source: generator is not connected to ground. This is a purely differential input signal. > irtual short i d KVL I + i + i 0 I i( ) I i ii Differential input resistance ii I i A. Kruger Op-Amp eiew-7

18 Op-Amp Basics Summing Inerting Amplifier i 0 II i 0 II i 3 0 II 3 i 4 0 o F KCL at _ i + i + i 3 + i II + 0 II + 0 II o F 0 Virtual Ground Ground II II II 3 0 F 0 Special case: 3 o F II + II + II Summing inerting amplifier A. Kruger Op-Amp eiew-8

19 Sidebar pn Junction Diode Sect...5 i D D nvt I S e V T kt/e 6 mv at T 300 K i D I S e V D T The cut-in oltage depends on the type of pn junction. For Si diodes it is about 0.7 V. For LEDs it is higher. The cut-in oltage is temperature-dependent. A. Kruger Op-Amp eiew-9

20 Op-Amp Applications Current-to-Voltage Conerter Current source e.g., photodiode Ideal current-controlled oltage source K transresistance (transimpeadance) gain Application i + i 0 0 > irtual ground In most cases S large, so i essentially i s i s + i 0 i s + 0 o F 0 o F i s Transresistance/transimpedance gain - F A. Kruger Op-Amp eiew-0

21 Another iew is that with s large, all of i s through F (nothing flows into the op-amp. Then: Op-Amp Applications Current-to-Voltage Conerter o F i s Photodiode amplifiers are used in CD/DVD players TV emote controls Optical fiber communications A. Kruger Op-Amp eiew-

22 Op-Amp Applications Voltage-to-Current Conerter Simple oltage-to-current conerter i i I 0 > irtual ground Problem: output current does not flow to ground (floating load) Solution: there are other topologies that where one end of the load is grounded. Battery s internal resistance changes as it is charged, but circuit forces a constant current though is regardless of the changing resistance i charge I A. Kruger Op-Amp eiew-

23 Op-Amp Applications Voltage-to-Current Conerter Current though LED does not depend on color of LED, temperature, aging, etc. The circuit forces a constant current through the LED. i LED I A. Kruger Op-Amp eiew-3

24 Op-Amp Applications Precision Half-Wae ectifier Diode does not conduct significant current (i.e., turn on) if the forward oltage across it is less that ~ 0.7 V (Si) Signal I Load I Load Voltage ~ 0.7 V for Si diode A. Kruger Op-Amp eiew-4

25 Op-Amp Applications Precision Half-Wae ectifier For I > 0, the circuit behaes as a oltage follower. The output oltage O I, the load current is positie. A positie diode current flows such that i L i D The feedback loop is closed through the forward biased diode. The output of the op-amp adjusts itself to absorb the oltage drop of the diode. For I < 0, O tends to go negatie, which tends to produce negatie load and diode currents > O 0 A. Kruger Op-Amp eiew-5

26 Op-Amp Applications Simple logarithmic amplifier Log Amplifier KCL: sum of current flowing out of node 0 0 > irtual ground Problem: V T and I S are functions of temperatures, and I S aries between diodes Solution: Special circuits hae been deeloped to account for this. Special logarithmic amplifiers are aailable. 0 I + i D 0 i D I s e D V T I s e O V T I + I s e O V T 0 O V T ln I I s Output oltage is proportional to natural log of input oltage A. Kruger Op-Amp eiew-6

27 Simple antilog or exponential amplifier Op-Amp Applications Antilog or Exponential Log Amplifier KCL: sum of current flowing into node 0 0 > irtual ground i D + O 0 0 i D I s e D V T I s e O V T Problem: V T and I S are functions of temperatures, and I S aries between diodes Solution: Special circuits hae been deeloped to account for this. I s e O V T + O 0 O I s e I V T Output oltage is an exponential function of the input oltage A. Kruger Op-Amp eiew-7

28 eal Op-Amps Effectie input resistance Effectie output resistance 0 A ) + o > L will load/cause oltage drop od ( A cm cm A od Common-mode signal is amplified o A A od o od ( 0 ) A. Kruger Op-Amp eiew-8

29 eal Op-Amps Voltage transfer characteristic eal amplifiers need power supplies. + Supply rail Slope A od Saturation effect - Supply rail A. Kruger Op-Amp eiew-9

30 eal Op-Amps - Effect of Finite Gain Consider an op-amp that is ideal (infinite input impedance, zero output impedance, ) except that it has finite differential-mode gain. i I i I o i i The output oltage is A o od I I A o od 0 kω 00 kω i I + A o od i o Aod o A o I + A od + Most op-amps hae A od ~ 0 5 A. Kruger Op-Amp eiew-30

31 eal Op-Amps - Input Bias Currents Op-amps need bias currents at their inputs. With FET input op-amps this current is ery small, but must still come from somewhere OK OK What happens when the input oltage source is remoed? Where will I P come from? Be careful A. Kruger Op-Amp eiew-3

32 eal Op-Amps - Errors Caused by I B This current causes a small oltage to deelop, that is then amplified This current is integrated by C and a oltage deelops at the output. This will saturate the output Question: Where does this current come from? Answer: From the output, o A. Kruger Op-Amp eiew-3

33 eal Op-Amps Input Bias Models I P I P I N I N I B I P + I N I OS I P I N A. Kruger Op-Amp eiew-33

34 eal Op-Amps - Input Bias Currents Depending on the type (pnp/npn), etc.) the current flow could be in different direction Input bias current I B I P + I N GP: na JFET: ~0.5 na Input offset current I OS I P I N GP 0-0 na JFET: ~0.05 na Note, if I OS << I B, then I P I N I B is a good approximation A. Kruger Op-Amp eiew-34

35 eal Op-Amps- Compensating for I B Using superposition, one can easily show that E O + [( ) I I ] N P P Setting P leads to [( ) ] EO + I OS A oltage will deelop here, which can be used to cancel the oltage resulting from I N This is typically an order of magnitude smaller E O can be further reduced by making resistors smaller See section 4.5. of 4 th edition of Neaman A. Kruger Op-Amp eiew-35

36 eal Op-Amps - Input Offset Voltage If we connect the two inputs of an ideal op-amp together, the output should be zero. In practice, howeer there is a small output oltage. This is modeled by adding a small oltage source of the ideal op-amp. Manufacturers proide V OS in their data sheets. V OS ranges from few mv down to few microolt On some op-amps one can trim effects of V OS away. A. Kruger Op-Amp eiew-36

37 The op-amps below hae offset oltages of 0 mv, but are otherwise ideal. What is the worst-case output oltage with I 0? Example Step, add offset oltages using the standard model (note polarities) With respect to its offset oltage, the first amplifier is a noninerting amplifier with gain so that the worst-case o is o 0 mv This is then amplified by the second amplifier by a factor 5. With respect to its offset oltage, the gain of the second amplifier is 6, so that the worstcase o (using superposition) is o mv A. Kruger Op-Amp eiew-37

38 Difference and Common-Mode Signals Id Difference signal ( + )/ Common-mode signal A. Kruger Op-Amp eiew-38

39 Difference and Common-Mode Signals Ideally, only difference signal is amplified A ) + o d ( A cm cm Common-mode signal ( )/ cm + Ideally, commonmode gain is 0 A. Kruger Op-Amp eiew-39

40 Difference and Common-Mode Signals A ( ) + o d Id A cm cm Id ( )/ cm + Icm id Icm + id A. Kruger Op-Amp eiew-40

41 Op-Amp Basics Difference Amplifier Inerting amplifier I 0 O I > irtual short Superposition - Linear circuit 0 I Non-inerting amplifier 4 b I O + b O ( + ) I Analyze with i 0 - Analyze with i 0 - Add results + Superposition O O O O 4 3 ( + / ) ( ) I / I ( ) 4 3 O I I A. Kruger Op-Amp eiew-4

42 Op-Amp Applications Difference Amplifier Difference amplifier What happens when 4 3 O 4 3 ( + / ) ( / ) I I O 4 3 ( ) I I Answer: output is not 0, when I I, and the common-mode signal is amplified A cm ( ) + cm I I Common-mode signal O cm CM A A Common-mode gain d cm CM(dB) 0log 0 Common-mode rejection ratio A A d cm Good differential amplifiers hae CM s db Thus, output 0 when I I A. Kruger Op-Amp eiew-4

43 Effect of Imbalance Imbalance factor One can show that + CM(dB) 0log0 4ε ε % ε 0.0 % esistors esistors What CMM can we achiee / 0, using % resistors? Answer: 50 db What tolerance do we need for an 80 db CMM? Answer: 0.03% A. Kruger Op-Amp eiew-43

44 New stuff Instrumentation Amplifier Our preious difference amplifier N Inerting input Very high input resistance O P Non-inerting input Section 9.45 Very high quality difference amplifier A. Kruger Op-Amp eiew-44

45 Section 9.45 OO + II II Our preious difference amplifier O 4 3 OO OO OO + II II Combining these expressions gie O II II Make sure you can derie this equation A. Kruger Op-Amp eiew-45

46 Instrumentation Amplifiers Sense Output Monolithic IA Ground eference Sense Output Single gainsetting resistor Single gain-setting resistor eference A. Kruger Op-Amp eiew-46

47 INA6 Single gain-setting resistor $ A. Kruger Op-Amp eiew-47

48 Biasing Input Stage Input stage must be biased Input stage must be biased Input stage must be biased A. Kruger Op-Amp eiew-48

49 Using the EF Pin Sense is connected internally Voltage at EF pin is the reference ground A. Kruger Op-Amp eiew-49

50 INA Downside: GBP 600 khz $7 A. Kruger Op-Amp eiew-50

51 Op-Amp Powering Often not shown, but decoupling capacitors close to power supply pins are highly recommended. Dual power supply: for many years: ±5 V, supply currents measured in ma Output oltage swings to - V of the supply rails Most modern op-amps can run on much lower power supplies: ± V to ±8 V, supply current as low as µa, 0.5 pa input bias currents, Many modern op-amps are rail-to-rail on input and/output A. Kruger Op-Amp eiew-5

52 Using a Single Power Supply Must supply a half supply here Half supply does not hae to be exactly 0.5 of supply oltage. A. Kruger Op-Amp eiew-5

53 Generating Vcc/ esistor proide dc oltage at Vcc/ Voltage is at Vcc/ Capacitor proides ac short (at what frequency?) A. Kruger Op-Amp eiew-53

54 Generating Vcc/ esistors proide dc oltage at Vcc/ Much lower output resistance and lower cutoff frequency Output oltage is at Vcc/ with no input signal Capacitor proides ac short (at what frequency?) A. Kruger Op-Amp eiew-54

55 Single Supply Operation TLE46 Precision ail Splitter Vcc Vcc/ $0.70 in bulk A. Kruger Op-Amp eiew-55

56 Single Supply Inerting Amplifier A. Kruger Op-Amp eiew-56

57 Single Supply Non Inerting Amplifier Can you spot the bug in this circuit? Answer: no bias current for V + input. A. Kruger Op-Amp eiew-57

58 Single Supply Inerting Summer A. Kruger Op-Amp eiew-58

59 Op-Amp Frequency esponse - + O A A O I I We often express gain in decibel (db). Voltage gain is A (db) 0 log O I eal amplifiers hae a finite range of frequencies oer which they will amplify signals A (db) A (db) A 0 3 db 3-dB Bandwidth 0 00,000 f or ω 0 00,000 f or ω Frequeny of I Note the logarithmic spacing on the frequency axis A. Kruger Op-Amp eiew-59

60 A f ) + A j ( 0 A ω) + A ( f ) A j ( 0 A s) + s ω ( 0 b f b ( ω ω ) b Frequency esponse DC openloop gain Dominant pole A db A 0 Slope is -0 db/decade Transition Frequency is where open-loop gain 0 db ( ) Dominant pole Slope is -45 o decade A. Kruger Op-Amp eiew-60

61 A f ) + Dominant-Pole Model and GBP A j ( 0 A ( f ) A s) + s ω ( 0 A ω) + A j ( 0 f b b ( ω ω ) b Dominant-pole amplifier hae a constant gain-bandwidth product (GPB) GPB A 0 f B f t Many op-amps are specifically manufactured to hae a (single) dominatpoles A. Kruger Op-Amp eiew-6

62 Small-Signal Transient esponse o ( t) V ( e t τ ) A0 A( s) + s ω A V s s ) + ωb s 0 o( o b ( t) V ( e t τ ) Laplace transform of input This result also follows from Circuits course Exponentially-rising ramp one would expect from single (dominant) pole response t r 0.35 BW BW in Hz A. Kruger Op-Amp eiew-6

63 Large Signal Transient esponse Not an exponentially-output, but close to linear. Amplifier is said to be slewing, and the slope is called the slew rate or S A. Kruger Op-Amp eiew-63

64 Slew-ate Limited esponse A. Kruger Op-Amp eiew-64

65 Slew-ate Limited esponse Full slewing A. Kruger Op-Amp eiew-65

66 A. Kruger Op-Amp eiew-66