Operational Amplifier Circuits - MIT OpenCourseWare
Operational Amplifier CircuitsReview:Ideal Op-amp in an open loop configurationIp Vp ViVnRo RiVoAViInAn ideal op-amp is characterized with infinite open–loop gainA The other relevant conditions for an ideal op-amp are:1. Ip In 02. Ri 3. Ro 0Ideal op-amp in a negative feedback configurationWhen an op-amp is arranged with a negative feedback the ideal rules are:1. Ip In 0 : input current constraint2. Vn Vp : input voltage constraintThese rules are related to the requirement/assumption for large open-loop gainA , and they form the basis for op-amp circuit analysis.The voltage Vn tracks the voltage Vp and the “control” of Vn is accomplished via thefeedback network.Chaniotakis and Cory. 6.071 Spring 2006Page 1
Operational Amplifier Circuits as Computational DevicesSo far we have explored the use of op amps to multiply a signal by a constant. For theinverting amplifier the multiplication constant is the gain RR12 and for the non invertingamplifier the multiplication constant is the gain 1 RR12 . Op amps may also perform othermathematical operations ranging from addition and subtraction to integration,differentiation and exponentiation.1 We will next explore these fundamental“operational” circuits.Summing AmplifierA basic summing amplifier circuit with three input signals is shown on Figure 1.Vin1Vin2Vin3R1I1R2I2R3I3IFRFN1VoutFigure 1. Summing amplifierCurrent conservation at node N1 givesI1 I 2 I 3 I F(1.1)By relating the currents I1, I2 and I3 to their corresponding voltage and resistance byOhm’s law and noting that the voltage at node N1 is zero (ideal op-amp rule) Equation(1.1) becomesVin1 Vin 2 Vin 3V outR1 R 2 R3RF(1.2)1The term operational amplifier was first used by John Ragazzini et. al in a paper published in 1947. Therelevant historical quotation from the paper is:“As an amplifier so connected can perform the mathematical operations of arithmetic and calculus on thevoltages applied to its inputs, it is hereafter termed an ‘operational amplifier’.”John Ragazzini, Robert Randall and Frederick Russell, “ Analysis of Problems in Dynamics by ElectronicsCircuits,” Proceedings of IRE, Vol. 35, May 1947Chaniotakis and Cory. 6.071 Spring 2006Page 2
And so Vout isRFRF RF Vout Vin1 Vin 2 Vin 3 R2R2 R1 (1.3)The output voltage Vout is a sum of the input voltages with weighting factors given by thevalues of the resistors. If the input resistors are equal R1 R2 R3 R, Equation (1.3)becomesVout RF(Vin1 Vin 2 Vin3 )R(1.4)The output voltage is thus the sum of the input voltages with a multiplication constantRFgiven by. The value of the multiplication constant may be varied over a wide rangeRand for the special case when RF R the output voltage is the sum of the inputsVout (Vin1 Vin 2 Vin 3 )(1.5)The input resistance seen by each source connected to the summing amplifier is thecorresponding series resistance connected to the source. Therefore, the sources do notinteract with each other.Chaniotakis and Cory. 6.071 Spring 2006Page 3
Difference AmplifierThis fundamental op amp circuit, shown on Figure 2, amplifies the difference betweenthe input signals. The subtracting feature is evident from the circuit configuration whichshows that one input signal is applied to the inverting terminal and the other to the noninverting terminal.Vin1Vin2R1I1IFR2N1R3VoutI2R4Figure 2. Difference AmplifierBefore we proceed with the analysis of the difference amplifier let’s think about theoverall behavior of the circuit. Our goal is to obtain the difference of the two inputsignals (Vin 2 - Vin1 ) . Our system is linear and so we may apply superposition in order tofind the resulting output. We are almost there once we notice that the contribution of thesignal Vin2 to the output is R4 R2 Vout 2 Vin 2 1 R1 R3 R 4 (1.6)and the contribution of signal Vin1 is R2 Vout1 -Vin1 R1 (1.7)R2 R4 R2 Vout Vout 2 - Vout1 Vin 2 1 - Vin1R1 R1 R3 R 4 (1.8)And the output voltage isChaniotakis and Cory. 6.071 Spring 2006Page 4
Note that in order to have a subtracting circuit which gives Vout 0 for equal inputs, theweight of each signal must be the same. Therefore R4 R2 R2 1 R1 R1 R3 R 4 (1.9)R4 R2 R3 R1(1.10)R2(Vin 2 - Vin1 )R1(1.11)which holds only ifThe output voltage is nowVout which is a difference amplifier with a differential gain of R2/R1 and with zero gain forthe common mode signal. It is often practical to select resistors such as R4 R2 andR3 R1.The fundamental problem of this circuit is that the input resistance seen by the twosources is not balanced. The input resistance between the input terminals A and B, theVdifferential input resistance, Rid (see Figure 3) is Rid inIA V inBIR1 V -R3 V R2VoutR4Figure 3. Differential amplifierSince V V- , Vin R1 I R3 I and thus Rid 2 R1 . The desire to have large inputresistance for the differential amplifier is the main drawback for this circuit. This problemis addressed by the instrumentation amplifier discussed next.Instrumentation AmplifierFigure 4 shows our modified differential amplifier called the instrumentation amplifier(IA). Op amps U1 and U2 act as voltage followers for the signals Vin1 and Vin2 which seethe infinite input resistance of op amps U1 and U2. Assuming ideal op amps, the voltageChaniotakis and Cory. 6.071 Spring 2006Page 5
at the inverting terminals of op amps U1 and U2 are equal to their corresponding inputvoltages. The resulting current flowing through resistor R1 isV V(1.12)I1 in1 in 2R1Since no current flows into the terminals of the op amp, the current flowing throughresistor R2 is also given by Equation igure 4. Instrumentation Amplifier circuitSince our system is linear the voltage at the output of op-amp U1 and op-amp U2 isgiven by superposition asR2 R2 V01 1 Vin2 Vin1 R1 R1 (1.13)R2 R2 V02 1 Vin1 Vin2 R1 R1 (1.14)Next we see that op amp U3 is arranged in the difference amplifier configurationexamined in the previous section (see Equation (1.11)). The output of the differenceamplifier isVout The differential gain,R4 2R2 1 ( Vin2 Vin1 )R3 R1 (1.15)R4 2R2 1 , may be varied by changing only one resistor: R1.R3 R1 Chaniotakis and Cory. 6.071 Spring 2006Page 6
Current to voltage convertersA variety of transducers produce electrical current in response to an environmentalcondition. Photodiodes and photomultipliers are such transducers which respond toelectromagnetic radiation at various frequencies ranging from the infrared to visible to γrays.A current to voltage converter is an op amp circuit which accepts an input current andgives an output voltage that is proportional to the input current. The basic current tovoltage converter is shown on Figure 5. This circuit arrangement is also called thetransresistance amplifier.RN1VoutI inFigure 5. Current to voltage converterIin represents the current generated by a certain transducer. If we assume that the op ampis ideal, KCL at node N1 gives V 0 I1 out 0 Vout RI1 R (1.16)The “gain” of this amplifier is given by R. This gain is also called the sensitivity of theconverter. Note that if high sensitivity is required for example 1V/µV then the resistanceR should be 1 MΩ. For higher sensitivities unrealistically large resistances are required.A current to voltage converter with high sensitivity may be constructed by employing theT feedback network topology shown on Figure 6.In this case the relationship between Vout and I1 is R2 R2 Vout 1 I1R1 R Chaniotakis and Cory. 6.071 Spring 2006(1.17)Page 7
RN1R1R2VoutI inFigure 6. Current to voltage converter with T networkChaniotakis and Cory. 6.071 Spring 2006Page 8
Voltage to Current converterA voltage to current (V-I) converter accepts as an input a voltage Vin and gives an outputcurrent of a certain value.In general the relationship between the input voltage and the output current isI out SVin(1.18)Where S is the sensitivity or gain of the V-I converter.Figure 7 shows a voltage to current converter using an op-amp and a transistor. The opamp forces its positive and negative inputs to be equal; hence, the voltage at the negativeinput of the op-amp is equal to Vin. The current through the load resistor, RL, thetransistor and R is consequently equal to Vin/R. We put a transistor at the output of theop-amp since the transistor is a high current gain stage (often a typical op-amp has afairly small output current limit).VccRLV inRFigure 7. Voltage to current converterChaniotakis and Cory. 6.071 Spring 2006Page 9
Amplifiers with reactive elementsWe have seen that op amps can be used with negative feedback to make simple linearsignal processors. Examples include amplifiers, buffers, adders, subtractors, and for eachof these the DC behavior described the apparent behavior over all frequencies. This ofcourse is a simplification to treat the op amp ideally, as through it does not contain anyreactive elements. Providing we keep the operating conditions out of the slew rate limitthen this is a reasonable model. Here we wish to extend this picture of op amp operationto include circuits that are designed to be frequency dependent. This will enable theconstruction of active filters, integrators, differentiators and oscillators.The feedback network of an op-amp circuit may contain, besides the resistors consideredso far, other passive elements. Capacitors and inductors as well as solid state devicessuch as diodes, BJTs and MOSFETs may be part of the feedback network.In the general case the resistors that make up the feedback path may be replaced bygeneralized elements with impedance Z1 and Z2 as shown on Figure 8 for an invertingamplifier.Z2Z1VoutVinFigure 8. Inverting amplifier with general impedance blocks in the feedback path.For an ideal op-amp, the transfer function relating Vout to Vin is given byZ (ω )Vout 2VinZ1 (ω )(1.19)Now, the gain of the amplifier is a function of signal frequency (ω) and so the analysis isto be performed in the frequency domain. This frequency dependent feedback results insome very powerful and useful building blocks.Chaniotakis and Cory. 6.071 Spring 2006Page 10
The Integrator: Active Low Pass FilterThe fundamental integrator circuit (Figure 9) is constructed by placing a capacitor C, inthe feedback loop of an inverting amplifier.CICRV inV outIRRFigure 9. The integrator circuitAssuming an ideal op-amp, current conservation at the indicated node givesI R IC(1.20)VindV C outRdtRearranging Equation (1.20) and integrating from 0 to t, we obtain dVout Vin (τ )dτRCt Vout (t ) 1Vin (τ ) dτ Vout (0)RC 0(1.21)The output voltage is thus the integral of the input. The voltage Vout (0) is the constant ofintegration and corresponds to the capacitor voltage at time t 0.The frequency domain analysis is obtained by expressing the impedance of the feedbackcomponents in the complex plane. The transfer function may thus be written as1VoutZjjωC C VinZRRω RCChaniotakis and Cory. 6.071 Spring 2006(1.22)Page 11
The above expression indicates that there is a 90o phase difference between the input andthe output signals. This 90o phase shift occurs at all frequencies. The gain of the amplifierV1is also a function of frequency. For dc signals withgiven by the modulus out Vinω RCω 0 the gain is infinite and it falls at a rate of 20dB per decade of frequency change. Theinfinite gain for dc signals represents a practical problem for the circuit configuration ofFigure 27. Since the equivalent circuit of a capacitor for ω 0 is an open circuit, thefeedback path is open. This lack of feedback results in a drift (cumulative summing) ofthe output voltage due to the presence of small dc offset voltages at the input. Thisproblem may be overcome by connecting a resistor, RF, in parallel with the feedbackcapacitor C as shown on Figure 10.CICRFRV inV outIRRFigure 10. Active Low Pass filterThe feedback path consists of the capacitor C in parallel with the resistor RF. Theequivalent impedance of the feedback path isRFR ΖRFjωCZF F C 11 jω RF CRF ΖC R FjωCThe transfer function(1.23)Z (ω )Voutbecomes FVinZ1 (ω )Z (ω )VoutRR11 F F FVinZ1 (ω )R1 1 jω RF CR1 1 jω(1.24)ωΗChaniotakis and Cory. 6.071 Spring 2006Page 12
WhereωH 1RF C(1.25)Voutversus frequency. At frequencies much lessVinRthan ωH (ω ωH) the voltage gain becomes equal to F , while at frequencies higherRthan ωH (ω ωH) the gain decreases at a rate of 20dB per decade.Figure 11 shows the logarithmic plot ofFigure 11. Bode plot of active low pass filter with a gain of 5.So we have seen that the integration is achieved by charging the feedback capacitor.For an integrator to be useful it must be allowed to be reset to zero. Since the output isstored in the charge of the feedback capacitor, all we need to do is to short out thecapacitor in order to reset the integrator.Integrators are very sensitive to DC drift, small offsets lead to a steady accumulation ofcharge in the capacitor until the op amp output saturates. We can avoid this by providinganother feedback path for DC. The circuit incorporates a shorting path across thecapacitor as shown on Figure 12.Chaniotakis and Cory. 6.071 Spring 2006Page 13
CICRFRV inV outIRRFigure 12. Integrator with reset buttonChaniotakis and Cory. 6.071 Spring 2006Page 14
The Differentiator: Active High Pass FilterA differentiator circuit may be obtained by replacing the capacitor with an inductor forFigure 9. In practice this is rarely done since inductors are expensive, bulky andinefficient devices. Figure 13 shows a fundamental differentiator circuit constructed witha capacitor and a resistor.RCV outV inRFigure 13. The differentiator circuitFor an ideal op-amp, the current flowing through the capacitor, Ccurrent flowing through the resistor,dVin, is equal to thedtVout, and thusRVout RCdVindt(1.26)The output is thus proportional to the derivative of the input.As the integrator is sensitive to DC drifts, the differentiator is sensitive to high frequencynoise. The differentiator thus is a great way to search for transients, but will add noise.The integrator will decrease noise. Both of these arguments assume the common situationof the noise being at higher frequency than the signal.Chaniotakis and Cory. 6.071 Spring 2006Page 15
Active Band Reject FilterThe integrator and differentiator demonstrate that op amp circuits can be designed to befrequency dependent. This permits the design of active filters, a filter that has gain. Wesaw before that we could design passive filters based on LC circuits, active filters are nomore complicated. Simple selective filters can be made through a frequency dependentimpedance in the feedback loop.Consider the band reject circuit shown on Figure 14.CLRFIFR1V inV outIRRFigure 14. Active band reject filterWe can understand how this circuit works without any detailed calculations. All we needto do is look at the feedback loops. There are two paths in the feedback loop: a frequencyindependent path with resistance RF, and a frequency dependent path with impedancegiven by1 Zω j ω L ωC (1.27)Let’s look at the behavior of the circuit as a function of frequency.For DC signals ( ω 0 ) the capacitor acts as an open circuit and the equivalent circuit isshown on Figure 15.Chaniotakis and Cory. 6.071 Spring 2006Page 16
RFIFR1V inV outIRRFigure 15Similarly at high frequencies ( ω ) the inductor acts as an open circuit and theequivalent circuit is the same as the one shown on Figure 15.Therefore the voltage transfer characteristics at DC and at high frequency are the samewith a gain given byG VoutR FVinR1(1.28)The other frequency of interest if the resonance frequency, which occurs when Zω asgiven by Equation (1.27) is equal to zero. The resonance frequency is1LCand the circuit reduces to the one shown on Figure 16.ωo (1.29)IFR1V inV outIRRFigure 16which gives Vout 0 at ω ωo .Chaniotakis and Cory. 6.071 Spring 2006Page 17
So this is a filter that passes and amplifies every frequency except the resonancefrequency.For the full analysis of this active filter we may write down the complete expression forthe impedance of the feedback loop which is1 j ωL Rj (ω 2 LC 1) RFωC F Z F Zω // RF 1 ω RF C j (ω 2 LC 1) RF j ω L ωC (1.30)And thus the general transfer function of the filter isj (ω 2 LC 1) RFVoutZF VinR1 R1 ω RF C j (ω 2 LC 1) Chaniotakis and Cory. 6.071 Spring 2006(1.31)Page 18
Diodes and transistors in op-amp circuits.Diodes and transistors may also be used in op-amp circuits. The nonlinear behavior ofthese devices result in very interesting and useful non-linear op-amp circuits.Logarithmic AmplifierIf we are interested in processing a signal that has a very wide dynamic range we takeadvantage of the exponential i-v characteristics of the diode and design an amplifierwhose output is proportional to the logarithm of the input.In practice we may have the voltage signal that corresponds to a certain chemical activitysuch as the activity, or concentration, of hydrogen ions in a solution which represents thepH of the solution. In this case the voltage is exponentially related to the concentration(pH).Vi Vo k ln( pH )(1.32)If we use this signal as the input to an inverting amplifier we may linearize the signal byusing a diode in the feedback path of the amplifier.Recall the i-v relationship for a diodeI Io eqV / kT 1 IoeqV / kT(1.33)Consider the circuit shown on Figure 17.IR1IRViVnVoVpRFigure 17. Logarithmic AmplifierKCL at the indicated node givesChaniotakis and Cory. 6.071 Spring 2006Page 19
Vi Vn Ioe q (Vn Vo ) / kTR1(1.34)Vi Ioe q ( Vo ) / kTR1(1.35)And since Vn Vp 0 we obtainAnd solving for Vo gives the desired relationship.Vo kTVilnqIo R1Vo kTkTln(Vi ) ln( IoR1)qqN a(1.36)bVo a ln(Vi ) bSimilarly and antilogarithmic amplifier may be constructed by placing the diode inseries with the signal source as shown on Figure 18.R2IFViVnIVoVpRFigure 18. Antilogarithmic amplifierHere you may show thatVo IoR 2e qVi / kTChaniotakis and Cory. 6.071 Spring 2006(1.37)Page 20
Superdiode. Precision half wave RectifierThe diode rectifier circuit and its associated voltage transfer characteristic curve areshown on Figure 19(a) and (b).VoutDVin Vd R11VoutVdVin(b)(a)Figure 19. Diode rectifier circuit (a) and voltage transfer curve (b)The offset voltage Vd is about 0.7 Volts and this offset value is unacceptable in manypractical applications. The operational amplifier and the diode in the circuit of Figure 20form an ideal diode, a superdiode, and thus they eliminate the offset voltage Vd from thevoltage transfer curve forming an ideal half wave rectifier.SuperdiodeVoutV-I1V1V VinDId Vd -I2RVoutVinFigure 20. Precision half wave rectifier circuit and its voltage transfer curve.Let’s analyze the circuit by considering the two cases of interest: Vin 0 and Vin 0.For Vin 0 the current I2 and id will be less than zero (point in a opposite direction to theone indicated). However, negative current can not go through the diode and thus thediode is reverse biased and the feedback loop is broken. Therefore the current I2 is zeroand so the output voltage is also zero, Vout 0. Since the feedback loop is open thevoltage V1 at the output of the op-amp will saturate at the negative supply voltage.Chaniotakis and Cory. 6.071 Spring 2006Page 21
For Vin 0, Vout Vin and the current I2 Id and the diode is forward biased. Thefeedback loop is closed through the diode. Note that there is still a voltage drop Vd acrossthe diode and so the op-amp output voltage V1 is adjusted so that V1 Vd Vin.Chaniotakis and Cory. 6.071 Spring 2006Page 22
ProblemsP1. Resistors R1 and R2 of the circuit on Figure P1 represent two strain gages placedacross each other on a beam in ord
Ideal Op-amp in an open loop configuration Ro Ri _ Vp Vn Vi _ AVi Vo Ip In An ideal op-amp is characterized with infinite open–loop gain A The other relevant conditions for an ideal op-amp are: 1. Ip In 0 2. Ri 3. Ro 0 Ideal op-amp in a negative feedback configuration When an op-
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E80 Lecture 4.2: Basic Electrical Measurements Agenda: Operational Amplifier o Recap: Non-inverting amplifier and unity gain buffer o Inverting amplifier (multiplication) o Summing amplifier (add and subtract) o Differentiator and integrator o Difference amplifier o Instrumentation amplifier o Transimpedance amplifier o Active filters 2
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