Lesson 3: Operational Amplifier Circuits In Analog Control

8/31/2016Lesson 3: OperationalAmplifier Circuits inAnalog ControlET 438aAutomatic Control Systems Technologylesson3et438a.pptx1After this presentation you will be able to: List the characteristics of an ideal Operational Amplifier (OPAMP) circuit.Identify and utilize fundamental OP AMP circuits to amplifyand signals.Use OP AMP circuits to reduce inter-stage loading effects insensor circuits.Average sensor signals using OP AMP circuits.LEARNING OBJECTIVESlesson3et438a.pptx21

8/31/2016Zout 0Zero output resistanceNo offset voltage,Vo 0 with Vd 0IinInfiniteZinInfinite voltage gain,AvBandwidth InfiniteGain constant for all fInstant recoveryfrom saturationIin 0 due toinfinite ZinIdeal OP AMP Characteristicslesson3et438a.pptx3OP AMPs are voltage amplifiers designed originally for use inanalog computersOP AMPs are direct coupled (dc) amplifiers that amplify both ac anddc signals simultaneously. Requires bipolar supplies. V Vin V1Vo Vin V2 -V0V1 0, Vo 0 V0V2 0, Vo 0-VSchematic symbol for non-ideal OP AMPTwo inputs:V1 inverting inputV2 non inverting inputNon-Ideal OP AMPslesson3et438a.pptx42

8/31/2016Inverting Voltage AmplifierAV R fRin R Vo Vi f Rin Vo Limited by saturationLarge Av causes Vo VNon-Inverting Voltage Amplifier Rf A V 1 Rin Rf Vo Vi 1 Rin Rin Rin of OP AMPAv has minimum value of 1Fundamental OP AMP Circuitslesson3et438a.pptx5Voltage follower (Impedance buffer) circuit used to reduce circuitloading. (Has a high Zin and low Zout) VCircuit causesminimumloading onpreviousstageZinZo-VCharacteristicsPractical Circuit (LM741)Av 1Zin 1 MWZo 10 WIdealAv 1Zin infiniteZo 0Voltage Follower Circuitlesson3et438a.pptx63

8/31/2016Voltage divider formula only valid for infinite load resistanceFind Vo under load 12 Vdc10kNo load Vo5k RLVo5k5 kW Vo 12 V 5 kW 10 kW V0 4.0 VWith load resistorR L 5 kW 5 kW 5 kW 2.5 kW2.5 kW VoL 12 V 2.4 V2.5kW 10 kW ANSExample 3-1 Buffered Voltage DividerCircuitlesson3et438a.pptxAdd impedance buffer 12 Vdc10k5kRinRo5k RL7Find Vo with load and OP AMP bufferAssume LM741 with Ri 1 MW andRo 10 W AV 1 so Vin VoWith load resistorR L 1 MW 5 kW 1 MW R eqVo1 MW 5 kW 4975 W1 MW 5 kW4975 W Vin 12 V 3.987 V4975W 10kW ANSVin Vo 3.987 VR eq Example 3-1 Buffered VoltageDivider Circuit (1)lesson3et438a.pptx84

8/31/2016Inverting Summing AmplifierFind total output usingsuperpositionIdealcircuitGain v1 RfR1Gain v2 RfR2Gain v3Total output RfR3 vvv v 0 Rf 1 2 3 R1 R2 R3 Output is inverted sum of v1, v2, and v3Electronic Addition and Subtractionlesson3et438a.pptx9Improved circuit (non-ideal OP AMP)Practical OP AMP chips requirebias currents to operate.Unequal R values at each inputcause voltage differences thatproduce output errors.Biascompensation RRc R1 R2 R3 RfElectronic Addition and Subtractionlesson3et438a.pptx105

8/31/2016Non-inverting Summing AmpAssuming R1 R2 R3 R v v 2 v3 v 0 1 f 1 R 3 For any number, n, inputs. R v v 2 . v n n inputv 0 1 f 1 averageR n Assuming R1 R2 R3 .RnElectronic Addition and Subtractionlesson3et438a.pptx11For circuit shown n 3R1 R2 R3 56kRf 9k R 1kV1 0.5 VdcV2 0.37 VdcV3 0.8 Vdc Find Vo R v v 2 v3 v 0 1 f 1 R 3 9k 0.5 0.37 0.8 v 0 1 5.56673 1k ANSExample 3-2 Non-inverting Averagerlesson3et438a.pptx126

8/31/2016For the circuit shown find Vo with Vin 2 Vdc10k5kExample 3-3:Inverting Amplifierlesson3et438a.pptx132.5kLet Rf 2.5 kW and findAv and Vo for Vi 2.0 Vdc5kReduces input voltageOutput is smaller than input. Circuit divides input by 2(0.5 ½)Example 3-3: Solution (2)lesson3et438a.pptx147

8/31/2016Find Vo and Av given values of R and Vi -1 Vdc. Assume nonideal OP AMP with power supply values of 15 Vdc50k100k 15RinNo sign change-15Note: Rin of non-inverting OPAMP is infinite (Ideally). Circuitwill not load previous stagesignificantlyExample 3-4 Non-Inverting Amp (1)lesson3et438a.pptx50k100k 1515Vin rises to 6 Vdc. What isVo? Assume a non-ideal OPAMP with given power supplyvalues of 15 Vdc-15This value can’t be achieved since the OP AMP saturatesbetween 13-15 Vdc. Power supplies limit output. Ac signalsdistorted (clipping)Example 3-4 Solution (2)lesson3et438a.pptx168

8/31/2016Find Vo given v1 0.2 Vdc v2 0.1 Vdcv3 -0.1 Vdc25k15k100k 1510k-15GainsExample 3-5 Inverting SummingAmplifierlesson3et438a.pptxLetting R1, R2 and R3be potentiometersproduces an audiomixer17R1V1R2V2R315VVo V3R4-15VWhen R1 R2 R3Output voltage is theaverage of the inputvaluesSumming Amplifier Applicationslesson3et438a.pptx189

8/31/201650k50k50kLM34 - temperature sensors. Gain 10 mV/FT1 50 F T2 45 F T3 40 FAverage the temperature using a gain of -1 and -5. Find thevalue of Rf and Vo for each gain value.Example 3-6: Averaging Sensor Signalslesson3et438a.pptx19To average let R1 R2 R3 50 kWSumming equationSince R1 R2 R3 vv vV0 R f 1 2 3 R1 R 2 R 3 v R v vV0 R f 1 2 3 f v1 v 2 v3 RRR11 1 R1 Find relationship for average with 3 inputs and gain of -1 v v 2 v3 Vave 1 1 3 R 1 Vo Vave f v1 v 2 v3 v1 v 2 v3 3 R1 Example 3-6 Solution (1)lesson3et438a.pptx2010

8/31/2016Complete Algebra tofind value of Rf R 1 f 3 R1 RRf 13Make equation more general by letting n be the number of inputsand Av be the desired gain factor. v R vv V0 R f 1 2 . n f v1 v 2 . v n R1 R1 R1 R1 v v 2 . v n Vave A v 1 n Example 3-6 Solution (2)lesson3et438a.pptx21Equate the OP AMP output and the average formula R A Vo Vave f v1 v 2 . v n v v1 v 2 . v n n R1 R A f v n R1 A RRf v 1nUse this formulaFind sensor output voltages using temperature and gain valueT1 50 F V1 10 mV/F 50 F 0.5 VT2 45 F V2 10 mV/F 45 F 0.45 V Example 3-6 Solution (3)T3 40 F V3 10 mV/F 40 F 0.40 Vlesson3et438a.pptx2211

8/31/2016Find Rf and Vo for a gain of -1 and R1 50 kW, n 3Rf A v R1 1 50,000 W 16,670 Wn3ANS R 16.67k f v1 v 2 v3 0.50 0.45 0.40 0.45 V 50k R1 Use averageformula tocheck output 0.50 0.45 0.40 Vave 1 0.453 ANSChecksExample 3-6 Solution (4)lesson3et438a.pptx23Find Rf and Vo for a gain of -5 and R1 50 kW, n 3A v R 1 5 50,000 WANS 83,330 Wn3 R 83.33k Vo f v1 v 2 v 3 0.50 0.45 0.40 2.25 V 50k R1 Rf ANS 0.50 0.45 0.40 Vave 5 2.253 Note: both values of Rf are not standard values. Use potentiometerand closest standard value then calibrate circuit to get desired outputPractical Rf82 k5kExample 3-6 Solution (5)lesson3et438a.pptx2412

8/31/2016Differential Voltage Amplifier CircuitInput/output Formula R R1 R 4 R V2 2 V1Vo 2 R1 R 4 R 3 R1 To simplify letR1 R3 R2 R4 R Vo 2 (V2 V1) R1 Amplifiers the difference between - terminalsDifferential Voltage Amplifierlesson3et438a.pptx25Can use voltage differential amp to generate an errorsignalMeasuredvalueer mBlock diagrammererrorsetpointe r-mDifferential Amplifier Applicationslesson3et438a.pptx2613

8/31/2016End Lesson 3: Operational AmplifierCircuits in Analog ControlET 438aAutomatic Control Systems Technologylesson3et438a.pptx2714

f 0.45 3 .500.450.40 Vave 1 Use average formula to check output ANS ANS Checks Example 3-6 Solution (5) lesson3et438a.pptx 24 Find R f and V o for a gain of -5 and R 1 50 kW, n 3 2.25 3 .500.450.40 V 5 .500.450.40 2.25 V 50k 83.33k v v v R R V 83,330 3 550,000 n A R R ave 1 2 3 1 f o v 1 f W W ANS ANS Note: both values of R f