Data Acquisition ET 228 Op –Amp Applications • Subjects Covered • • • • • • Overview of OP Amp Applications High Resistance Voltmeters Phase Shifter Circuit Integrators Diferentiators Example use – Servo amplifiers Data Acquisition ET 228 Op –Amp Applications • Overview of Op-Amp Applications • Chapter five covers a varied set of OP-Amp applications • Measuring short circuit current • Measure output from photo detectors • Equalize audio tones of different amplitudes • Control high currents • Allow matching of semiconductor characteristics • High resistance DC/AC voltmeter • A phase shift circuit • Integrators and differentiators • Servo Amplifiers Data Acquisition ET 228 Op –Amp Applications • High Resistance DC Voltmeter • See Figure 5.1 on page 119 • Voltage to be measured is input to the + pin • Ed = 0 • Thus I = Ei /Ri • Limits per design • Input voltage range: -1.0 to 1 V • Do Example problem 5-1 on page 120 • Two simple steps to change the voltmeter into one with an input range of -10.0V to 10V • _____________ • _____________ • Do Example problem 5-2 on page 120 Data Acquisition ET 228 Op –Amp Applications • Universal High Resistance Voltmeter • See Figure 5.2 on page 121 • Measures DC, rms, Peek, and Peak-Peak voltages • Sector switch for the Ranges • Bridge circuit around meter – current only flows through it in one direction • Limits per design – Full Scale reading per selected Scale • • • • Max DC Input voltage: 5 VDC Max rms Input voltage: 5 Vrms Max Peak Input voltage: 5 VPeak Max Peak-Peak Input voltage: 5 VPeak-Peak Data Acquisition ET 228 Op –Amp Applications • Universal High Resistance Voltmeter • Design considerations • Meter has full scale rating of 50 micro-amps it only reads average voltage • On the DC setting it has a 100K ohm resistor to Common • 50 micro-amps X 100k ohms = 5VDC • On the rms scale it has a 90 k ohm resistor and an average voltage on the input of 4.5V will yield a full scale deflection Vrms 0.707 xVPeak 5Vrms 0.707 x7.07VPeak 7.07VPeak from bridge rectifier 0.636 xVPeak 0.636 x7.07VPeak 4.50VPeak Data Acquisition ET 228 Op –Amp Applications • Universal High Resistance Voltmeter • Design considerations • On the Peak setting it has a 63.6K ohm resistor to Common VPeak 1 1 xV Ave 5VPeak x3.18V Ave 0.636 0.636 • 50 microamps times 63.6k ohms = 3.18VAve • On the Peak-Peak setting it has a 31.8K ohm resistor to Common VPeak Peak 1 1 xV Ave 5VPeak Peak x1.59V Ave 0.318 0.318 Data Acquisition ET 228 Op –Amp Applications • Phase Shifter Circuit • See figure 5-13 on page 137 • • • • Low pass passive filter on Op-Amp + input Voltage on Op-Amp – input = Voltage on + input (VCap) Voltage across other Input resistor = Ei – VCap = I Voltage out has a phase shift equal to twice the phase shift on the input Cap • Output amplitude =Input amplitude • Design procedure • Phase angle shift only depends upon Ri, Ci, and frequency of the input signal 𝜽 𝜽 = 𝟐 𝒂𝒓𝒄𝒕𝒂𝒏 𝟐𝝅𝒇𝑹𝒊 𝑪𝒊 OR 𝑹𝒊 = 𝒕𝒂𝒏(𝟐) 𝟐𝝅𝒇𝑪𝒊 Data Acquisition ET 228 Op –Amp Applications • Phase Shifter Circuit • Design procedure • Steps • Determine the desired phase shift and frequency of input signal • Set Ci to an appropriate value for the input frequency • Determine the value of Ri using one of the equations above • Example problems 5-13 and 5-14 • Proof of concepts • 𝑽𝑪𝒂𝒑 = −𝒋𝑿𝑪 𝒙 𝑹𝒊 −𝒋𝑿𝒄 𝑬𝒊 𝑬𝒊 = 𝑨 𝒔𝒊𝒏𝟐𝝅𝒇𝒕 • Voltage on pin 2 = 𝑽𝑪𝒂𝒑 • Voltage across other input resistor =𝑬𝒊 - 𝑽𝑪𝒂𝒑 Data Acquisition ET 228 Op –Amp Applications • Phase Shifter Circuit • Proof of concepts Note; phase shift across the resistor from the input = the phase shift from the LP filter. There will be the same phase shift across the feedback resistor THUS double the phase shift on the Cap 𝑬𝒊 = 𝑨𝒔𝒊𝒏𝟐𝝅𝒇𝒕 = 𝑨@𝟎𝒐 𝑰𝒇 = 𝑬𝒊 −𝑽𝑪 𝑹𝒇 𝑽𝑪 = −𝒋𝑿𝑪 𝑬 𝑹𝒊 −𝒋𝑿𝑪 𝒊 𝑽𝑪 = 𝑬𝒊 − 𝑰𝒇 𝑹 • With f=1kHz and Ri = 15.9 k ohms and A=1V 𝑬𝒊 = 𝑨𝒔𝒊𝒏 𝟔. 𝟐𝟖𝒌𝒕 = 𝟏@𝟎𝒐 −𝑗15.9𝑘 𝑽𝑪 = 𝟏@𝟎𝒐 = 𝟎. 𝟕𝟎𝟕@ − 𝟒𝟓𝒐 15.9𝑘 − 𝑗15.9𝑘 Data Acquisition ET 228 Op –Amp Applications • Phase Shifter Circuit • Proof of concepts With f=1kHz and Ri = 15.9 k ohms and A=1V 𝑰𝒇 = 𝑬𝒊 −𝑽𝑪 𝑹𝒇 = 𝟏−(𝟎.𝟓−𝒋𝟎.𝟓) 𝟏𝟎𝟎𝒌 𝒐𝒉𝒎𝒔 = 𝟎.𝟓+𝒋𝟎.𝟓 𝟏𝟎𝟎𝒌 𝒐𝒉𝒎𝒔 = 𝟕. 𝟎𝟕𝝁𝑨@𝟒𝟓𝟎 𝑽𝑹−𝟏𝟎𝟎𝒌 = 𝑽𝑹𝒇 = 𝑰𝒇 𝒙𝟏𝟎𝟎𝒌 𝒐𝒉𝒎𝒔 = 𝟎. 𝟕𝟎𝟕@𝟒𝟓𝒐 𝑽𝑶 = 𝑽𝑪 − 𝑽𝑹𝒇 = 𝟎. 𝟕𝟎𝟕@ − 𝟒𝟓𝒐 − 𝟎. 𝟕𝟎𝟕@𝟒𝟓𝒐 𝑽𝑶 = 𝟎. 𝟓 − 𝒋𝟎. 𝟓 − 𝟎. 𝟓 + 𝒋𝟎. 𝟓 = −𝒋𝟏. 𝟎 = 𝟏@ − 𝟗𝟎𝒐 Data Acquisition ET 228 Op –Amp Applications • Integrators • Reference Circuit: • Fig 5-15 on page 141 • Inverting Op-Amp circuit with the feedback resistor replaced with a capacitor • The case of the input voltage being a step function is shown • VO is shown as a ramp voltage • Analysis • Iin or input current with respect to time {i(t)} • The voltage difference between the two Op-Amp input terminals is assumed to be 0V DC (Ed = 0V) Data Acquisition ET 228 Op –Amp Applications • Integrator Circuits • Analysis • Iin or input current with respect to time {i(t)} ein i (t ) Ri 1 v idt C • Voltage across a capacitor is and when i(t) replaces I in the voltage c equation and realize the output of an inverting amplifier is a negative voltage we have 1 vO Ri C f eindt YIELDS eint vo Ri C f For t greater than the start time of the input voltage step Note: If Ei is a sine wave the output will be a negative cosine wave. t 0 Data Acquisition ET 228 Op –Amp Applications • Integrator Circuits • Analysis • • • • • eint vo Ri C f When t = 0, VO = 0V (initial equilibrium voltage) When t = 1/10 of RiCf, VO = - 0.1 x ein V When t = 1/2 of RiCf, VO = - 0.5 x ein V When t = 0.9 of RiCf, VO = - 0.9 x ein V When t = RiCf, VO = - ein V • Conclusion • For input step functions to an integrator the output is a ramp style output voltage • Goes from the output before the input and ramps to the saturation voltage with the opposite polarity Data Acquisition ET 228 Op –Amp Applications • Integrator Circuits • Conclusion eint vo Ri C f • For other input functions over a range of time find the indefinite integral answer • Solve using the value of time for the end of the range • Solve using the value of time for the start of the range • Subtract the second answer from the first Notes: ES = 0V, IS = 0A Iin = ein /Rin, If = Iin Data Acquisition ET 228 Op –Amp Applications • Servoamplifier • Uses an Integrator to delay the full effect of an input voltage change to the output voltage • See Figure 5-16 on page 143 • The first Op-amp is the integrator • The second Op-amp is an inverter with a gain of “-1” • The positive feedback voltage (VF)causes the integrator’s output to stop at “-2” times the input voltage » VF is set at ½ the output voltage » How would you set VF at a different fraction of the output, thus changing the steady-state gain? » VC must climb to 3 times Vin and the circuits time constant is 3 times that of a simple series RC circuit. Thus τ = 3RiC • Work Example problem 5-15 Starting on bottom of page 143 Data Acquisition ET 228 Op –Amp Applications • Servoamplifier • How long of a delay • Stability in 5 τ • In addition noise during the transition is zeroed out • The time delay is caused by the time required for the Cap to charge to a new value required by the change in the input voltage • The voltage on the cap in figure 15-6 changes per the following: t starts at 0 when Ei changes and It stops when ΔVC = 3 ΔEi Ei t VC Ri C Data Acquisition ET 228 Op –Amp Applications • Servoamplifier • Why does VC need to be 3 time Ein -- Why does τ equal 3 time RiC • Iin shrinks to zero as VF approaches Ein and becomes zero when they are equal • When the voltages are stable, VF = ½ VO and the output voltage of the integrator = - VO = -2 VF = -2 Vin • By inspection then VC must = 3 times (note polarity shown in figure 5-16) • Also, τ is 3 times RiC since the Cap must charge to 3 times the input voltage to reach stability • Other cases • If VF = VO , then VC would equal 2Ei and τ would = 2RiC • If VF = 1/4VO , then what would VC equal?? Data Acquisition ET 228 Op –Amp Applications • Differentiators • Looks like an Integrator with the Cap and resistor swapped • Operation • Performs the mathematical operation of differentiation • VO equals the negative of the derivative of Ei (see text book on page 145 in figure 5-17 for the equation) dei VO R f Ci dt OR VO Ei R f Cin t Data Acquisition ET 228 Op –Amp Applications • Differentiators • Operation Notes: Ed = 0V, Id = 0A Iin = {Δ Ein CinRf}/Δt, If = Iin Data Acquisition ET 228 Op –Amp Applications • Differentiators • Problems • Unstable – may oscillate • However it is a plus factor when making a multivibrator by adding positive feedback • Gain increases with frequency • XC decreases with increased frequency, thus the gain will increase with increased frequency • Solution • Add a parallel Cap to the Feedback Resister and a series resistor to the input Cap • Design procedure • See bottom of page 145 Data Acquisition ET 228 Op –Amp Applications • Differentiators • Design procedure • Work example problem 5-18 • Device recommendations • Op Amps with high slew rates