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 E d = 0 Thus I = E i /R i 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 V DC Max rms Input voltage: 5 V rms Max Peak Input voltage: 5 V Peak Max Peak-Peak Input voltage: 5 V Peak-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 = 5V DC 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

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 50 microamps times 63.6k ohms = 3.18V Ave On the Peak-Peak setting it has a 31.8K ohm resistor to Common

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 V O is shown as a ramp voltage Analysis I in 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 (E d = 0V)

Data Acquisition ET 228 Op –Amp Applications Integrator Circuits Analysis I in or input current with respect to time {i(t)} Voltage across a capacitor is and when i(t) replaces I in the voltage equation and realize the output of an inverting amplifier is a negative voltage we have YIELDS For t greater than the start time of the input voltage step t 0 Note: If E i is a sine wave the output will be a negative cosine wave.

Data Acquisition ET 228 Op –Amp Applications Integrator Circuits Analysis When t = 0, V O = 0V (initial equilibrium voltage) When t = 1/10 of R i C f, V O = x e in V When t = 1/2 of R i C f, V O = x e in V When t = 0.9 of R i C f, V O = x e in V When t = R i C f, V O = - e in 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 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: E S = 0V, I S = 0A I in = e in /R in, I f = I in

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 (V F )causes the integrator’s output to stop at “-2” times the input voltage »V F is set at ½ the output voltage »How would you set V F at a different fraction of the output, thus changing the steady-state gain? »V C must climb to 3 times V in and the circuits time constant is 3 times that of a simple series RC circuit. Thus τ = 3R i C 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 E i changes and It stops when ΔV C = 3 ΔE i

Data Acquisition ET 228 Op –Amp Applications Servoamplifier Why does V C need to be 3 time E in -- Why does τ equal 3 time R i C I in shrinks to zero as V F approaches E in and becomes zero when they are equal When the voltages are stable, V F = ½ V O and the output voltage of the integrator = - V O = -2 V F = -2 V in By inspection then V C must = 3 times (note polarity shown in figure 5-16) Also, τ is 3 times R i C since the Cap must charge to 3 times the input voltage to reach stability Other cases If V F = V O, then V C would equal 2E i and τ would = 2R i C If V F = 1/4V O, then what would V C 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 V O equals the negative of the derivative of E i (see text book on page 145 in figure 5-17 for the equation) OR

Data Acquisition ET 228 Op –Amp Applications Differentiators Operation Notes: E d = 0V, I d = 0A I in = {Δ E in C in R f }/Δt, I f = I in

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 X C 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