ECE:3410 Electronic Circuits Reiew of Op-Amps Sections of Chapters 9 & 14 A. Kruger Op-Amp Reiew-1
Real-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 Reiew-
Operational Amplifiers Inputs Output Schematic symbol Real 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 Reiew-3
Ideal Op-Amp Equialent Circuit Inerting input Ideal oltage-controlled, oltage source Noninerting input o = A od 1 Control oltage A od = Open-loop differential gain Ideally A od = Common-mode input signal Common-mode rejection A. Kruger Op-Amp Reiew-4
Parameters of Ideal Op Amps +Vcc Effectie input resistance = Ideally A od = o Effectie output resistance = 0 A od ( ) 1 No common-mode amplification -Vcc o 1 A A od o od ( 0 1 ) Because A od, the implication is that for finite output oltage, the differential input oltage is ery small (0) A. Kruger Op-Amp Reiew-5
Op-Amp Basics Inerting Amplifier Op-amp inerting amplifier Because A od, 1 = 0 +Vcc -Vcc Thus, 1 is at irtual ground Virtual ground => terminal is at ground potential, but not connected to ground Closed-loop/negatie feedback A. Kruger Op-Amp Reiew-6
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 + 0 = 0 R 1 R I R 1 O R = 0 O = R R 1 I A. Kruger Op-Amp Reiew-7
A. Kruger Op-Amp Reiew-8 Op-Amp Basics 1 1 R i I 1 R R A I o Input resistance 1 1 R i R I i Set gain with two external resistors Set input impedance/resistance with external resistor i 1 R I i Inerting Amplifier
Op-Amp Basics Summing Inerting Amplifier i 1 = 0 I1 R 1 i = 0 I R i 3 = 0 I3 R 3 i 4 = 0 o R F +Vcc KCL at _ i 1 + i + i 3 + i 4 = 0 -Vcc 0 I1 R 1 + 0 I R + 0 I3 R 3 + 0 o R F = 0 Virtual Ground Ground I1 R 1 I R I3 R 3 0 R F = 0 Special case: R 1 = R = R 3 o = R F R 1 I1 + I + I3 Summing inerting amplifier A. Kruger Op-Amp Reiew-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 = 18 6 + 1 18 1 = 4.5 V The second amplifier is also a summing inerter, and the output oltage is V O = ( 4.5) 36 6 3 36 1 = 18 V A. Kruger Op-Amp Reiew-10
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 Z1 A Z Z 1 A. Kruger Op-Amp Reiew-11
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 C1 + 0 O + 0 = 0 R C 1 d 0 I dt O R = 0 C 1 d I dt O R = 0 O = R C 1 d I (t) dt A. Kruger Op-Amp Reiew-1
Op-Amp Basics Integrator (Special Type of Inerting Amplifier) +Vcc -Vcc O O KCL: sum of current flowing out of node = 0 Voltage here is zero (irtual ground concept) 0 I R 1 + i C + 0 = 0 I d C +C = 0 R 1 dt d 0 I O +C = 0 R 1 dt This circuit is also called a Miller integrator, and R 1 C is the time-constant. d I O C = 0 R 1 dt d O dt = 1 R 1 C I O = 1 R 1 C I t dt + V C 0 t Voltage at t = 0 A. Kruger Op-Amp Reiew-13
Op-Amp Basics Follower +Vcc -Vcc I O + = - => irtual short Why? A o I 1 Impedance transformer R L loads R S resulting in ery large error +Vcc -Vcc o I R L RL R S 1 1100 0.01 Input resistance ery high => no loading of source o I 1 A. Kruger Op-Amp Reiew-14
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 R 1 + V n V O R f = 0 V in R 1 + V in V O R f = 0 V O = 1 + R f R 1 V in A = 1 + R f R 1 A. Kruger Op-Amp Reiew-15
+Vcc -Vcc Op-Amp Basics Difference Amplifier 0 I Inerting amplifier R O1 I1 R1 1 = => irtual short Superposition - Linear circuit 0 I1 Non-inerting amplifier R 4 b I R3 R4 O R 1 O b R1 R R 1 R R 4 3 1 R R1 I 4 3 - Analyze with i = 0 - Analyze with i1 = 0 - Add results O O O1 O Superposition R R 1 R R 4 3 1 R / R1 I R / R1 I 1 R 4 R3 R R1 4 3 O R R 1 I1 I A. Kruger Op-Amp Reiew-16
Op-Amp Basics Differential Input Resistance Note differential source: generator is not connected to ground. This is a purely differential input signal. 1 = => irtual short i d +Vcc KVL I + ir 1 + ir 1 = 0 I = i(r 1 ) -Vcc I i = R id = R 1 Differential input resistance R id = I i A. Kruger Op-Amp Reiew-17
General Strategy for Soling (Ideal) Op-Amp Circuits Find o Step 1: Find N Step : Find P Step 3: Set N = P + +Vcc -Vcc Techniques + KVL KCL Superposition Voltage diision Step 1: Find N KVL N 4 0 N 4 o o 30 3 Step : Find P Voltage Diision P o o 0 30 5 3 Step : Set N = P o o 4 5 o 10 V A. Kruger Op-Amp Reiew-18
Sidebar pn Junction Diode Sect. 1..5 i D D nvt I S e 1 V T = kt/e 6 mv at T = 300 K i I D 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 Reiew-19
Op-Amp Applications Current-to-Voltage Conerter Current source e.g., photodiode Ideal current-controlled oltage source K = transresistance (transimpeadance) gain Application i 1 + i = 0 1 = =0 = > irtual ground In most cases R S large, so i 1 essentially i s i s + i = 0 i s + 0 o R F = 0 o = R F i s Transresistance/transimpedance gain = -R F A. Kruger Op-Amp Reiew-0
Op-Amp Applications Current-to-Voltage Conerter Another iew is that with R s large, all of i s through R F (nothing flows into the op-amp. Then: o = R F i s Photodiode amplifiers are used in CD/DVD players TV Remote controls Optical fiber communications A. Kruger Op-Amp Reiew-1
Op-Amp Applications Voltage-to-Current Conerter Simple oltage-to-current conerter i i 1 I R 1 1 = =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 R 1 A. Kruger Op-Amp Reiew-
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 R 1 A. Kruger Op-Amp Reiew-3
Op-Amp Applications Precision Half-Wae Rectifier 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 Reiew-4
Op-Amp Applications Precision Half-Wae Rectifier 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, O1 tends to go negatie, which tends to produce negatie load and diode currents => O = 0 A. Kruger Op-Amp Reiew-5
Op-Amp Applications Simple logarithmic amplifier Log Amplifier KCL: sum of current flowing out of node = 0 1 = =0 = > irtual ground 0 I R 1 + i D = 0 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. i D I s e D V T = I s e O V T I R 1 + I s e O V T 0 O V T ln I I s R 1 Output oltage is proportional to natural log of input oltage A. Kruger Op-Amp Reiew-6
Simple antilog or exponential amplifier Op-Amp Applications Antilog or Exponential Log Amplifier KCL: sum of current flowing into node = 0 1 = =0 = > irtual ground i D + O 0 R = 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 R 0 O R I s e I V T Output oltage is an exponential function of the input oltage A. Kruger Op-Amp Reiew-7
Real Op-Amps Effectie input resistance Effectie output resistance 0 o => R L will load/cause oltage drop A ) od( 1 A cm cm A od Common-mode signal is amplified o 1 A A od o od ( 0 1 ) A. Kruger Op-Amp Reiew-8
Real Op-Amps Voltage transfer characteristic Real amplifiers need power supplies. + Supply rail Slope A od Saturation effect - Supply rail A. Kruger Op-Amp Reiew-9
Real 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 1 I R 1 1 i I R o i i 1 The output oltage is A o od I I A o od R1 10 k R 100 k i 1 I A R 1 o od i o Aod R o A o I R R 1 1 1 A od 1 1 R R 1 Most op-amps hae A od ~ 10 5 A. Kruger Op-Amp Reiew-30
Real 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 Reiew-31
Real 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 Reiew-3
Real 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 Reiew-33
Real 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: 100-00 na JFET: ~0.5 na Input offset current I OS I P I N GP 10-0 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 Reiew-34
Real Op-Amps- Compensating for I B Using superposition, one can easily show that V O R 1 R R R 1 1 I N R P I P Setting R P R 1 R leads to R VO 1 R1 R I OS R 1 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 14.5. of 4 th edition of Neaman A. Kruger Op-Amp Reiew-35
Real 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 Reiew-36
The op-amps below hae offset oltages of 10 mv, but are otherwise ideal. What is the worst-case output oltage with I = 0? Example Step 1, add offset oltages using the standard model (note polarities) With respect to its offset oltage, the first amplifier is a noninerting amplifier with gain 11 so that the worst-case o1 is o1 = 110 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 =5 110 + 6 10 = 610 mv A. Kruger Op-Amp Reiew-37
Difference and Common-Mode Signals 1 1 Id Difference signal / 1 Common-mode signal A. Kruger Op-Amp Reiew-38
Difference and Common-Mode Signals 1 Ideally, only difference signal is amplified o A ) d ( 1 A cm cm Common-mode signal cm / 1 Ideally, commonmode gain is 0 A. Kruger Op-Amp Reiew-39
Difference and Common-Mode Signals o A ( ) d Id A cm cm Id cm 1 / 1 Icm 1 id Icm id A. Kruger Op-Amp Reiew-40
Op-Amp Basics Difference Amplifier Inerting amplifier 0 I R O1 I1 R1 1 = => irtual short Superposition - Linear circuit 0 I1 Non-inerting amplifier R 4 b I R3 R4 O R 1 O b R1 R R 1 R R 4 3 1 R R1 I 4 3 - Analyze with i = 0 - Analyze with i1 = 0 - Add results O O O1 O Superposition R R 1 R R 4 3 1 R / R1 I R / R1 I 1 R 4 R3 R R1 4 3 O R R 1 I1 I A. Kruger Op-Amp Reiew-41
Op-Amp Applications Difference Amplifier Difference amplifier What happens when R R R 4 3 R1 Answer: output is not 0, when I1 = I, and the common-mode signal is amplified cm I1 I Common-mode signal O R R 1 R R 4 3 1 R / R1 I R / R1 I 1 R O 4 3 4 R3 R R1 R R 1 I I1 A cm O cm CMRR A A Common-mode gain d cm Ad CMRR(dB) 0log 10 A Common-mode rejection ratio cm Good differential amplifiers hae CMRR s 80 100 db Thus, output = 0 when I1 = I A. Kruger Op-Amp Reiew-4
Effect of Imbalance Imbalance factor One can show that 1 R R1 CMRR(dB) 0log10 4 0.05 5% 0.011% Resistors Resistors What CMRR can we achiee R /R 1 = 10, using 1% resistors? Answer: 50 db What tolerance do we need for an 80 db CMRR? Answer: 0.03% A. Kruger Op-Amp Reiew-43
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 Reiew-44
Section 9.5.4 O1 = 1 + R R 1 I1 R R 1 I Our preious difference amplifier O = R 4 R 3 O O1 O = 1 + R R 1 I R R 1 I1 Combining these expressions gie O = R 4 R 3 1 + R R 1 I I1 Make sure you can derie this equation A. Kruger Op-Amp Reiew-45
Instrumentation Amplifiers Sense Output Monolithic IA Ground Reference Sense Output Single gainsetting resistor Single gain-setting resistor Reference A. Kruger Op-Amp Reiew-46
INA16 Single gain-setting resistor $ A. Kruger Op-Amp Reiew-47
Biasing Input Stage Input stage must be biased Input stage must be biased Input stage must be biased A. Kruger Op-Amp Reiew-48
Using the REF Pin Sense is connected internally Voltage at REF pin is the reference ground A. Kruger Op-Amp Reiew-49
INA11 Downside: GBP = 600 khz $7 A. Kruger Op-Amp Reiew-50
Op-Amp Powering Often not shown, but decoupling capacitors close to power supply pins are highly recommended 0.1 μf is a good place to start. Dual power supply: for many years: ±15 V, supply currents measured in ma Output oltage swings to 1- V of the supply rails Most modern op-amps can run on much lower power supplies: ±1 V to ±18 V, supply current as low as 1 µa, 0.5 pa input bias currents, Many modern op-amps are rail-to-rail on input and/output A. Kruger Op-Amp Reiew-51
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 Reiew-5
Generating Vcc/ Resistor proide dc oltage at Vcc/ Voltage is at Vcc/ Capacitor proides ac short (at what frequency?) A. Kruger Op-Amp Reiew-53
Generating Vcc/ Resistors 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 Reiew-54
Single Supply Operation TLE46 Precision Rail Splitter Vcc Vcc/ $0.70 in bulk A. Kruger Op-Amp Reiew-55
Single Supply Inerting Amplifier A. Kruger Op-Amp Reiew-56
Single Supply Non Inerting Amplifier Can you spot the bug in this circuit? Answer: no bias current for V + input. A. Kruger Op-Amp Reiew-57
Single Supply Inerting Summer A. Kruger Op-Amp Reiew-58
Frequency Response (Reiew form Lecture 1) Oscilloscope Signal Generator + LTI System (t) A I cos ωt 0 log 10 A O A I (db) t (t) A O cos ωt + θ t LTI delays and scales signal 10 0 10 1 10 10 3 ω or f θ rad or degrees 0 10 0 10 1 10 10 3 ω or f 45 90 A. Kruger Op-Amp Reiew-59
o (t) V Step response 0.9 0 = 18 0 0.1 0 = 4 t (ms) t r 1.8 ms 0 log 10 O I 10 db (db) 3 db Frequency Response 10 1 10 10 3 10 4 10 5 f in Hz 3-dB Bandwidth =190 Hz For an STC the following two expressions are ery useful. t r.τ BW 3dB 0.35 t r (Hz) A. Kruger Op-Amp Reiew-60
Single Time Constant (STC) Circuits R 1 C 1 τ = RC t t r =.τ = e t τ B 3dB = 0.35 t r = 1 πτ Example of an RC STC Bandwidth is generally taken as 3-dB bandwidth A 0 A (db) 3 db 3-dB Bandwidth 1 10 100 1,000 f or ω f b f b = break frequency, pole frequency, 3-dB frequency, bandwidth A. Kruger Op-Amp Reiew-61
Dominant Pole Concept R 1 R C 1 C τ R 1 C 1 t r.τ t e t τ B 3dB 0.35 t r 1 πτ Most circuits will hae multiple poles, and the mathematics can become complex. If, for example R R 1 and C C 1, their effect will be small. A (db) f 1 f This situation appears in many circuits, and we can approximate the system with an STC. This leads to the concept of a dominant pole. If f 1 f, then we call f 1 the dominant pole. 1 10 100 1,000 f b f or ω A. Kruger Op-Amp Reiew-6
Op-Amp Frequency Response - + O A = A = O I I We often express gain in decibel (db). Voltage gain is A (db) = 0 log O I Real amplifiers hae a finite range of frequencies oer which they will amplify signals A (db) A (db) A 0 3 db 3-dB Bandwidth f or ω 1 10 100 1,000 Frequeny of I Note the logarithmic spacing on the frequency axis 1 10 100 1,000 f b f or ω A. Kruger Op-Amp Reiew-63
A f ) 1 A j f ( 0 A ) 1 A0 A( s) 1 s b f b A j ( 0 b Frequency Response DC openloop gain Dominant pole is at the break frequency A db A 0 Slope is -0 db/decade Transition Frequency is where open-loop gain = 0 db ( 1) Dominant pole Slope is -45 o decade A. Kruger Op-Amp Reiew-64
Sidebar: Octaes and Decades The number of octaes between two frequencies f 1 and f refer to the number of times f 1 was doubled to get to f. f 1 = khz and f = 8 khz There are two octaes between - and 8 khz, since we hae to double khz twice to get to 8 khz. How many octaes are there between f 1 = khz and f = 10 khz? n OCT = log f f 1 = log 10 = log 5 = log 10 5 log 10 =.3 Electrical engineers often use the number of decades as a measure of the spacing between frequencies f 1 and f. n DEC = log 10 f f 1 The number of decades between - and 10 khz is log 10 10 = 0.699 A. Kruger Op-Amp Reiew-65
A f ) 1 Dominant-Pole Model and GBP A j f ( 0 A0 A( s) 1 s f b b A ) 1 A j ( 0 b Dominant-pole amplifiers hae a constant gain-bandwidth product (GBP) GBP A 0 f B f t Most op-amps are specifically manufactured to hae a (single) dominantpole, since it makes them easier to use. The amplifiers behae as STCs with gain. A. Kruger Op-Amp Reiew-66
Small-Signal Transient Response o ( t) V (1 e t ) A0 A( s) 1 s A V s) s 1 b s 0 o( o b ( t) V (1 e t ) Laplace transform of input (step) 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 Reiew-67
Large Signal Transient Response Not an exponentially-output, but close to linear. Amplifier is said to be slewing, and the slope is called the slew rate or SR. SR has units of V s. In datasheets it is often V μs A. Kruger Op-Amp Reiew-68
Slew-Rate Limited Response A. Kruger Op-Amp Reiew-69
Slew-Rate Limited Response Full slewing A. Kruger Op-Amp Reiew-70
A. Kruger Op-Amp Reiew-71