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EE 174 Fall 2019

EE 174 Fall 2019. Operational Amplifiers. Table of Contents. Introduction to Operational Amplifier (Op-Amp) Brief of History Op-Amp Internal Circuit Op-Amp: Practical, Ideal, symbol and Characteristics Table Op-Amp: CMRR, Golden Rules Op-Amp: Negative and Positive Feedback

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EE 174 Fall 2019

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  1. EE 174Fall 2019 Operational Amplifiers

  2. Table of Contents • Introduction to Operational Amplifier (Op-Amp) • Brief of History • Op-Amp Internal Circuit • Op-Amp: Practical, Ideal, symbol and Characteristics Table • Op-Amp: CMRR, Golden Rules • Op-Amp: Negative and Positive Feedback • Gain and Bandwidth (GBP) • Op-Amp: The Four Amplifier Types • Common Mode Rejection Ratio (CMRR) • Fundamentals of Op-Amps • Basic operation • Gain • Offset • CMRR • Applications

  3. Introduction to Op-Amp • Operational Amplifier (Op-Amp for short) which name comes from the early days of amplifier design, when the op amp was used in analog computersto perform various mathematical operations. • Op-Amp is an active circuit element consisting of transistors, resistors, diodes and capacitor. • Op-Amp is two-port networks in which the output voltage or current is directly proportional to either input voltage or current. Four different type of amplifiers exits: • Voltage amplifier (VCVS): Av = Vo / Vi • Current amplifier (ICIS): Ai = Io / Ii • Transconductance amplifier (VCIS): Gm = Io / Vi (Siemens) • Transresistance amplifier (ICVS): Rm = Vo / Ii • Op-Amp is ideal device which takes a relatively weak signal (sensor) as an input and produce a much stronger signal as an output without affecting its other properties. • Op-Amps are commonly used for both linear and nonlinear applications: Inverting/Non-inverting Amplifiers, Variable Gains Amplifiers, Summers, Integrators/Differentiators, Filters and Schmitt trigger, Comparators, A/D converters.

  4. Brief History of Op-Amp • Monolithic IC Op-Amp • First created in 1963 μA702 by Fairchild Semiconductor • μA741 created in 1968, became widely used due to its ease of use 8 pin, dual in-line package (DIP) • Further advancements include use of field effects transistors (FET), greater precision, faster response, and smaller packaging Vacuum Tube Op-Amps (1930’s-1940’s) Dual-supply voltage of +300/-300 V Output swing +/- 50 volts Open-loop voltage gain of 15,000 to 20,000, Slew rate of +/- 12 volts/µsecond Maximum output current of 1 mA George Philbrick Solid State Discrete Op-Amps (1960’s) Dual-supply voltage of +15/-15 V Output swing +/- 11 volts Open-loop voltage gain of 40,000, Slew rate of +/- 1.5 volts/µsecond Maximum output current of 2.2 mA

  5. Op-Amp Internal Circuit Compensation Cap • Op-Amp 3-stages: • Differential Amplifier (Blue) • Gain Stage (Magneta) • Output Stage (Cyan) • Current Mirrors/Sources (Red) • Voltage Level Shifter (Green)

  6. Op-Amp: Practical, Ideal, Symbol and Characteristics Table Non-Inverting Input Ouput Inverting Input

  7. Op-Amp: Common Mode Rejection Ratio (CMRR) Common-Mode Rejection Ratio (CMRR) describes the ability of an op-amp to amplify differential signals and reject common-mode signals. The CMRR is the ratio of the differential-mode gain to common-mode gain. CMRR = = , normally be expressed in decibels asCMRR = 20 log in dB Common Mode Gain (Acm) is defined as: Acm = < 1 typical for practical op-amp. Ideally, VOCM = 0, since V+ = V- Differential Mode Gain (Ad) is defined as: Ad = =  VO = AdVdwhere Vd = Example:Calculate the CMRR for the circuit with measurement values as follow: VCM = 1mV, VOCM = 0.1mV and V1 = - 0.2mV, V2 = 0.8mV, VO = 10V Solution: AC = = = 0.1, Ad = = = 104 CMRR = = = 105 or 20 log(105) = 100dB CMRR = 100,000 means that desired signal is amplified 100,000 times more than un-wanted noise signal.

  8. Op-Amp: Voltage Offset • Another practical concern for op-amp performance is voltage offset. • The primary cause of the Op-Amp offset is a slight mismatch of the base-emitter voltages of the differential input stage of the Op-Amp. • For an ideal op-amp with both inputs shorted to ground or when there is no difference between its inputs would produce output to zero. •  V+ = V– = 0  VO = Aol(V+ – V–) = 0. • However, for a pratical op-amp  An offset input can drive the output to either negative or positive saturated level even if the op-amp in question has zero common-mode gain (infinite CMRR). This deviation from zero is called offset. • The 741 OPAMP have offset voltage null capability by connecting 10kΩ potentiometer between pin 1 and pin 5. By varying the potentiometer, output offset voltage (with inputs grounded) can be reduced to zero volts. For the 741C the offset voltage adjustment range is ± 15 mV. • Effect of offset voltage on Op-Amp produces error at the output. • VO = – Vin+ (1 +) + • Desired output Error

  9. Op-Amp: Current Offset Ideally, no current flows into the input terminals of an op amp. In practice, there are always two DC input bias currents, IB+ and IB- . The input bias current IB can vary from 60 fA (1 electron every 3 μs) to tens of μA, depending on the device. The input bias current, IBIAS, is defined as: IBIAS = The input offset current, IOS, is defined as: IOS = IB+ − IB– The input offset voltage, VOS, is defined as: VOS = IOS Rin The output voltage error, VO(ERROR), is defined as: VO(ERROR) = Av IOS Rin In order to cancel the effects of the bias current in an inverting op-amp configuration, a simple bias compensation resistor, R3, (R3=R1||R2) can be used to introduce a voltage drop in the non-inverting input to match and thus compensate the drop in the parallel combination of R1 and R2 in the inverting input. This minimizes additional offset voltage error. VO = R2 (IB– – IB+) = R2 IOS = 0, If IB+ = IB– (well-matched) Neglecting VOS

  10. Op-Amp Offset Example • 1) Determine the output voltage for the open loop differential amplifier Fig #1. • vOS- = 5 µVdc, vOS+ = –7 µVdc • vOS- = –10 mVdc, vOS+ = 0Vdc • Specifications of the Op-Amp are given below: A = 200,000, Ri = 2 MΩ , RO = 75Ω, + VCC = + 15 V, - VEE = - 15 V, and output voltage swing = ± 14V. • Solution: • a) VO = A(vOS+– vOS-) = 200,000 (– 7 – 5) µV = 2.4 V • Remember that vo = 2.4 V dc with the assumption that the dc output voltage is zero when the input signals are zero. • b) VO = A(vOS+– vOS-) = 200,000 (0 – (– 10mV)) = 2000 V  Saturation VO = 14V • Thus the theoretical value of output voltage vo = 2000 V However, the OPAMP saturates at ± 14 V. • 2) Given circuit in Fig #2 with Vin = 0.25V, R1 = 1kΩ , R2 = 4 kΩ Vos+ = 5mV. Determine the offset voltage error. • Solution: • Use superposition to obtain: • VO = – Vin + (1 +) += – 4(0.25V) + 5(0.005V) = (– 1 + 0.025)V = – 0.975V Fig #1 Fig #2

  11. V V0 V+ V0 Vi + +   Op-Amp: Gain and Bandwidth Closed loop voltage gain (Acl) of the Op-Amp is defined as: Acl = • Open loop gain:  Gain is measured when no feedback is applied to the op amp. Open-loop gain (Aol) is very high. • Loop gain: The difference between the open-loop gain and the closed-loop gain. This is useful information because it gives you the amount of negative feedback that can apply to the amplifier system. Gain versus Bandwidth: Applying feedback will reduce the gain but increase the bandwidth. • Gain-bandwidth (GBW) product is defined as the op-amp gain A multiplied by the bandwidth (BW). • The product of the gain and bandwidth are constant. • Aclfcl = Aolfol • The GBW product is also equal to the unity gain frequency • fT = Acl fcl fcl = fT /Aclwhere fT is the unity-gain bandwidth. • Example: For closed loop gain 70 dB. What is closed-loop BW? • Gain: 20log(Acl)= 70dB •  log (Acl)= 70/20 = 3.5 or Acl = 103.5 = 3162 •  Closed-loop BW: fcl = fT /Acl = 1 MHz / 3162 = 316.3 Hz. A = 105, BW = 10Hz, GBW = 105x10Hz = 1MHz A = 1, BW = 1 MHz, GBW = 1x1MHz = 1MHz

  12. Op-Amp Slew Rate The slew-rate (SR) is the maximum possible rate of change of the op amp output voltage. Slew Rate (SR) = max inV/µs Square Wave: Applied a square wave input to the 741 Op-amp in Fig. 1, find the slew rate SR = ΔVO/Δt = 2.06V / 3.16 µs = 0.63V/ µs (Ref Lab #1). Sine Waves: The maximum rate of change for a sine wave occurs at the zero crossing and for a given vo = VP sin(2π f t), the slew rate SR is: Slew Rate (SR) = max = 2π fmax VP SR = 0.63V/ µs. Fig. 1 • Example: Given Slew rate SR = 1 V/μs; input voltage vin (t) = 1 sin(2π × 105t); • closed loop gain AV =10. • Sketch the theoretically output and the actual output in the same graph. • What is the maximum frequency that will not violating the given slew rate? • What is the maximum gain A that will not violating the given slew rate? • What is the required SR to prevent distortion output? • Solutions: • See graph Fig. 2 • fmax = SR / (A x 2π) = 1 x 106 / (10 x 2π) = 15.9 kHz • VPmax = SR / (f x 2π) = 1 x 106 / (105 x 2π) = 1.59 V  A = 1.59 • For voltage gain A = 10  vo(t) = 10 sin(2π × 105t)  SR = |= 2π x fx 10 = 2π × 105 x 10 = 6.28 V/μs Fig. 2

  13. Op-Amp: The Four Amplifier Types

  14. Op-Amp: Four Amplifier Types

  15. Op-Amp: Golden Rules vd= v+ –v– vo= Avd =A(v+ – v–) + Vd – i+ = i – = 0(Since Ri = ꚙ. This is always true for any feedback configuration). v+ = v–(In negative feedback, the output of the op amp will try to adjust its output so that the voltage difference between the + and − inputs is zero  V+ = V−). Rule 2. is not applied if VO in saturation. Note: The resistances used in an op-amp circuit must be much larger than Ro and much smaller than Ri in order for the ideal op-amp equations to be accurate.

  16. Op-Amp: Negative and Positive Feedback Negative feedback:Vout = A(V+ – V–) = A(Vin – Vout) If Vout > Vin  Vout↓ goes down toward Vin If Vout < Vin  Vout↑ goes up toward Vin This drives Vout = Vin or V+ = V-  Negative feedback Op-Amp can produce any voltage in the supply power range. Positive feedback:Vout = A(V+ – V–) = A(Vout – Vin) If Vout > Vin  Vout↑ goes way up to + saturation If Vout < Vin  Vout↓ goes way down to – saturation This drives Vout to either directions: + / – saturation  Positive feedback Op-Amp can only produce maximum and minimum voltages of the range. Positive feedback: Negative feedback:

  17. Op-Amp: Non-inverting Negative Feedback Negative feedback is used in almost all linear op-amp circuits because it stabilizes the gain and reduces distortion. A basic configuration is a noninverting amplifier with negative feedback the difference between Vi and Vf is very small, or Vi ≈Vf. The closed-loop gain for the noninverting amplifier can be derived as: Acl(NI) = 1 + The input resistance increases by: Ri(NI) = (1 + AolB)Ri where Ri is op-amp input resistance and feedback fraction B = The output resistance decreases by: Ro(NI) = Rowhere Ro is op-amp output resistance The low output resistance implies that the output voltage is independent of the load resistance (as long as the current limit is not exceeded). Example: Given a non-inverting op-amp with Rf = 24kW, R1 = 1kW, what are the input and output resistances and the gain of the noninverting amplifier? Assume the op amp hasAol = 100,000, Ri = 2 MW, and Ro = 75 W. Solution: The gain: Acl(NI) = 1 + = 1 + = 25 The feedback fraction: B = = The input resistance: Ri(NI) = (1 + AolB)Ri = (1 + (100,000)0.04)2MW = 8MW The output resistance: Ro(NI) = Ro = 75W = 0.019W

  18. Op-Amp: Inverting Negative Feedback Negative feedback forces the inverting input to be near ac ground for the inverting amplifier. For this reason, the input resistance of the inverting amplifier is defined as: Ri(I) = R1. The low input resistance is usually a disadvantage of this circuit. However, because the Ri(I) is equal to R1, it can easily be set by the user for those cases where a specific value is needed. The output resistance of the inverting amplifier is the essentially the same as the noninverting amplifier: Ro(NI) = Ro where Ro is op-amp output resistance The closed-loop gain for the inverting amplifier can be derived as: Acl(I) = – Example: Given a inverting op-amp with Rf = 24kW, R1 = 1kW, what are the input and output resistances and the gain of the noninverting amplifier? Assume the op amp hasAol = 100,000, Ri = 2 MW, and Ro = 75 W. Solution: The gain: Acl(I) = – = – = – 24 The feedback fraction: B = = 16 The input resistance: Ri(NI) = R1 = 1kW The output resistance: Ro(NI) = Ro = 75W = 0.018W Virtual Ground

  19. Op-Amp: Saturation • The op-amp output voltage is limited by the supply voltages.  In practice the limits are about 1.5 to 2 V below the value of the supply voltages. • The op amp has three distinct regions of operation: • Linear region: −VEE < Vo < VCC  Vo/Vi = A = constant • Positive saturation: Vo > VCC Vo = VCC • Negative saturation: Vo < −VEE  Vo = –VEE • Note that the saturation voltage, in general, is not symmetric. • For an amplifier with a given gain, A, the above range of Vo translate into a certain range for Vi • − VEE < Vo < VCC or − VEE < AVi < VCC •  − VEE / A < Vi < VCC / A • Any amplifier will enter its saturation region if Vi is raised above certain limit. The figure shows how the amplifier output clips when amplifier is not in the linear region. • Example: For A= 105, –VDD = –12V, VCC = +15V, find range of input Vi to prevent saturation. • Solution: –12V / 105 < Vi < 15V / 105 or – 0.12mV< Vi < 0.15mV

  20. Op-Amp: Two Basic Op-Amp Configurations Negative feedback is when the occurence of an event causes something to happen that counteracts the original event. In negative-feedback configuration, op-amp always “wants” both inputs (inverting and non-inverting) to be the same value. The Non-Inverting Op Amp The Inverting Op Amp Negative feedback: Vi = V+ = V– = Vo Closed loop gain: Acl = = = 1 + When Rf = 0 and R1= ∞, the non-inverting amplifier becomes a voltage follower or unity gain buffer (gain Acl = 1). This configuration offers very high input impedance and its very low output impedance. Negative feedback: V+ = V– =  I1 = =  = Closed loop gain: Acl = = V+ Vi Vo Negative feedback: Vi = V+ = V–= Vo Closed loop gain: Acl = = 1 V –

  21. Op-Amp: Top Op-Amp Application Configurations Differential Amplifier Non-inverting Summing Amplifier Inverting Summing Amplifier Vo =) Vo = Intrumentation Converter current – voltage Negative resistance two op-amps to buffer the input signals and a third to cancel out the common-mode noise.

  22. Op-Amp: Op-Amp Circuits V = 2V, R = 1kΩ The Howland Current Pump Circuit IR = Vi/2R IO = IS + IR V = ½ Vo I1 + I2 + IL = = 0 = 0 0  constant Ballanced Differential Amplifier Op-Amp #1 is non-inverting: Vo1= (1 + RF1/R1) Vi Op-amp #2 is inverting: Vo1 = RF2/R1Vi Ballanced amplifier when: Vo1 = Vo2 #1 #2

  23. Op-Amp: Top Op-Amp Application Configurations Integrator (Low pass filter): Bode plot is shown below. Gain: = – = – When ZC = 1/(2πfC) = R Gain = 1 Integrator (Low pass filter with Gain): Bode plot is shown below. DC Gain: = – AC gain: = – Corner Frequency: f0= Example: For LPF with gain, find suitable components to achieve a −3-dB frequency of 1 kHz with a dc gain of 20 dB and an input resistance of at least 10 kΩ At what frequency does gain drop to 0 dB?

  24. Op-Amp: Op-Amp Circuits (a) Find i, if and vo. i = 0 A (b) For the ideal op amp shown below, what should be the value of resistor Rf to obtain a gain of 5? Solution: Want Vo = 5Vi For non-inverting: For Input V+ = and Output VO = 5Vi  Gain == Gain = 1 +  Rf = 19.5kΩ Another way: i = = Vo = (Rf + )i = (Rf + ) = 5Vi Rf = 19.5kΩ V+

  25. Op-Amp: Op-Amp Circuits • (a) Given an op-amp circuit below. If the power supplies for the op-amp are ± VCC = ± 12V. Determine: • Overall gain A = VOUT / VIN, VOUT and IOUT . • If op-amp in saturation, what is voltages at V+ and V-. • Solution: • (a) Vx= • i = • Vz= • Vy= • IOUT = 2 mA + 2 mA = 4 mA • VOUT = = 16V  A = VOUT / VIN = 16V/ 4V = 4 • However, VOUT = 16V > Power supply 12V • Saturation  VOUT = 12V Clamp • For VOUT = 12V  IOUT = 12V / ( + //2 • i = 3 mA / 2 = 1.5 mA • Vz= • Vy= V- = 1kΩ x 1.5 mA = 1.5V • V+ = Vx = 2V • V+ ≠ V- when op-amp in saturation mode.

  26. Op-Amp: Op-Amp Circuits (a) Given gain A1 = RF/RS = 100, 741 Op-Amp GBW product = A0 ω0 = K = 2π × 106 . Determine the overall 3-dB bandwidth of the cascade amplifier below: Solutions: The 3-dB bandwidth for each amplifier is: ω1 = = 2π × 104 rad/s BW of the cascade = 2π × 104 & cascade gain AT = A1A1= 102 × 102 = 104.  Gain AT = 104 with BW = 104 Hz. To get the same gain AT = 104 with a single-stage amplifier having the same K BW: ω2 = = 2π × 102 rad/s  Gain AT = 104 with BW = 102 Hz. (b) Voltage follower application: Solutions: Given Rs = 1 kΩ, RL = 100Ω For source and load connected directly: VL = 0.1 Vin  Huge attenuation of the signal source. For source and load connected via voltage follower (buffer amplifier):  VL = Vp = Vin

  27. Op-Amp: Op-Amp Circuits (1) Find Vx , Vo (3) Find Vy , Vo Vx = 3.2V Vo = -11.2V Vy =3/2Vo Vo = -11.4V (2) Find Vx , Vo (4) Find Vo Vx = 0.5Vo Vo = 0.8Vi v+= v-= 6V

  28. Basic Comparators • An ideal comparator compares two input voltages (inverting and noninverting input) and produces an output indicating the relationship between them. • If V+ > V– Vout is HIGH • If V+< V– Vout is LOW • The inputs can be two signals (V1,V2) or an input signal (Vin) and a fixed dc reference voltage (Vref). The output that usually swings from rail to rail. • A typical comparator has low offset, high gain, and high common-mode rejection. • Comparators are designed to work as open-loop systems, to drive logic circuits, and to work at high speed, even when overdriven. • The voltage at which a comparator changes from one level to another is called the crossover (or threshold) voltage. • The output logic voltages can be symmetrical or asymmetrical depends on the digital logic requirements. Symmetrical Assymmetrica

  29. Basic Comparators Non-Inverting zero-crossing (Vref= 0V) comparator Inverting zero-crossing (Vref= 0V) comparator Non-Inverting non-zero reference (Vref≠ 0V) comparator Inverting non-zero reference (Vref ≠ 0V) comparator

  30. Basic Schmitt Comparator • Schmitt trigger uses positive feedback: • When op-amp output VO rises then (V+ − V−) will increase. This causes output to positive saturation  VO = +Vsat. • If VO = +Vsat = +15V, then Vf = 5V. For any Vi < 5V, then (V+ − V−) > 0  output stays at +Vsat VO = +15V. • If Vi > 5V, then (V+ − V−) < 0  output switch to its minimum value or negative saturation VO = −Vsat = −15V. • If VO = −Vsat = −15V, then Vf = −5V  output will only switch back to VO = +Vsat = +15V when Vi < −5V. • Notes: • Negative feedback stabilizes the output to make V+ ≃ V−. • Positive feedback adjusts the output to maximize |V+ − V−|. Output will switch between its maximum and minimum values, e.g. ±Vsat. • Switching will happen when V+ = V−.

  31. The Comparators w/o and w/Hysteresis – Single power supply Noise or signal variation at the comparison threshold will cause multiple transitions. Hysteresis sets an upper and lower threshold to eliminate the multiple transitions caused by noise. Calculate VH and VL: 100kΩ // 576kΩ= 85kΩ VH = = 2.7V VL = = 2.3V

  32. Design of Hysteresis Comparator Equations (1) and (2) can be used to select the resistors needed to set the hysteresis threshold voltages VH and VL. Rh VL Rx VH – VL (1) Ry VL Rx VCC– VH (2) = = • Example: The design requirements are as follows: • Supply Voltage: +5 V • Input: 0V to 5V • VL (Lower Threshold) = 2.3V • VH (Upper Threshold) = 2.7V • VH– VL= 0.4V • Vth ± 0.2V = 2.5V ± 0.2V • Equations (1) and (2) can be used to select the resistors needed to set the hysteresis threshold voltages VH and VL. • One value (RX) must be arbitrarily selected. In this example, Rxwas set to 100kΩ to minimize current draw. • Rhwas calculated to be 575kΩ, so the closest standard value 576kΩ was used

  33. When To Use Op-amp Comparator Circuit??? Although op amps are not designed to be used as comparators, there are, nevertheless, many applications where the use of an op amp as a comparator is a proper engineering decision. WHY USE AN OP AMP AS A COMPARATOR? • Convenience • Economy (e.g. use spare op-amps in quad package and cheaper than comparator) • Low bias current (IB) • Low offset voltage (VOS) Note: This is not a good design practice. WHY NOT USE AN OP AMP AS A COMPARATOR? • Speed • Inconvenient input structures (Comparator is designed for large differential input voltages) • Inconvenient logic structures (Compartor TTL, CMOS) • Stability/hysteresis (stability)

  34. References: • https://www.analog.com/media/en/training-seminars/tutorials/MT-038.pdf • http://www.ume.gatech.edu/mechatronics_course/OpAmp_F08.ppt • http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/operational-amplifier-models/ • http://www.ti.com/lit/an/slaa068a/slaa068a.pdf • http://www.radio-electronics.com/info/circuits/opamp_basics/operational-amplifier-slew-rate.php • http://www.planetanalog.com/author.asp?section_id=483&doc_id=562347 • http://www-ferp.ucsd.edu/najmabadi/CLASS/ECE60L/02-S/NOTES/opamp.pdf • http://www-inst.eecs.berkeley.edu/~ee105/fa14/lectures/Lecture06-Non-ideal%20Op%20Amps%20(Offset-Slew%20rate).pdf • http://nptel.ac.in/courses/117107094//lecturers/lecture_6/lecture6_page1.htm • http://www.ti.com/ww/en/bobpease/assets/AN-31.pdf • http://www.cs.tut.fi/kurssit/TLT-8016/Chapter2.pdf • http://www.electronics-tutorials.ws/opamp/op-amp-comparator.html • http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/schmitt.html • http://lpsa.swarthmore.edu/Bode/BodeExamples.html

  35. References: http://www.allaboutcircuits.com/worksheets/inverting-and-noninverting-opamp-voltage-amplifier-circuits/ Fundamentals of Electrical Engineering, Giorgio Rizzoni, McGraw-Hill Higher Education http://chrisgammell.com/how-does-an-op-amp-work-part-1/ http://electronicdesign.com/analog/whats-all-noise-gain-stuff-anyhow http://howtomechatronics.com/how-it-works/electrical-engineering/schmitt-trigger/ http://www.analog.com/media/en/training-seminars/design-handbooks/Basic-Linear-Design/Chapter1.pdf http://www.ti.com/lit/an/sloa011/sloa011.pdf http://www.analog.com/media/en/technical-documentation/application-notes/AN-849.pdf http://www.ti.com/lit/ug/tidu020a/tidu020a.pdf https://en.wikipedia.org/wiki/Instrumentation_amplifier https://www.tubecad.com/2012/07/blog0238.htm https://www.allaboutcircuits.com/technical-articles/the-howland-current-pump/

  36. Top Fundamental Op Amp Circuits Voltage Follower / Unity Gain Buffer Inverting Amplifier Non-inverting Amplifier Inverting Summing Amplifie Non-inverting Summing Amplifier Differential Amplifier Vo Vo Vo = Vo = https://www.arrow.com/en/research-and-events/articles/fundamentals-of-op-amp-circuits

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