The Integrating and Differentiating Op-Amp Circuits - Experiment 8 | ECE 225, Lab Reports of Electrical Circuit Analysis

Download The Integrating and Differentiating Op-Amp Circuits - Experiment 8 | ECE 225 and more Lab Reports Electrical Circuit Analysis in PDF only on Docsity! Boise State University Department of Electrical and Computer Engineering ECE225L – Circuit Analysis and Design Lab Experiment #8: The Integrating and Differentiating Op-Amp Circuits 1 Objectives The objectives of this laboratory experiment are: • To demonstrate the input-output relationships of the integrating and differentiating amplifier configurations using operational amplifiers. • To simulate a first-order linear differential equation using an integrating op-amp circuit. 2 Theory The dynamic characteristics of capacitors and inductors produce signal processing functions that cannot be obtained using resistors. The op-amp circuit of Figure 1(a) is similar to the inverting amplifier except for the capacitor in the feedback path. To determine the input-output relationship of this circuit, we use the ideal op-amp equations as well as Kirchhoff’s voltage and current laws. Begin by writing Kirchhoff’s current law (KCL) at the inverting node N, iR(t) + iC(t) = iN (t) (1) and substitute the i-v characteristics of the resistor and capacitor, vi(t)− vN (t) R + C dvC dt = iN (t) (2) Then use the ideal op-amp equations iN (t) ∼= 0 (3) vN (t) ∼= vP (t) = 0 (4) together with the fact that vC(t) = vo(t)− vN (t) ∼= vo(t)− vP (t) = vo(t) (5) and rewrite KCL as vi(t) R + C dvo dt = 0 (6) To solve for the output voltage vo(t), multiply this equation by dt, solve for the differential dvo, and integrate to get ∫ dvo = − 1 RC ∫ vi(t) dt (7) Assuming the output voltage is known at time t = 0, the integration limits are ∫ vo(t) vo(0) dvo = − 1 RC ∫ t 0 vi(t) dt (8) 1 R (b)(a) o v i v − + − + + − o v i v C C i R i N i P N P N N i C i R i C R − + − + + − Figure 1: Op-Amp (a) Integrator and (b) Differentiator Circuits which yields vo(t) = − 1 RC ∫ t 0 vi(t) dt + vo(0) (9) From Equation (5), the initial condition vo(0) is actually the voltage on the capacitor at time t = 0. When this voltage is initially zero at time t = 0, the circuit input-output relationship reduces to vo(t) = − 1 RC ∫ t 0 vi(t) dt (10) The output voltage is proportional to the integral of the input voltage when the initial capacitor voltage is discharged. This circuit is an inverting integrator since the proportionality constant (1/RC) is negative. This constant has the units of sec−1 so that both sides of Equation (10) have units of volts. Interchanging the resistor and the capacitor in Figure 1(a) produces the op-amp differentiator in Figure 1(b). To derive the input-output characteristic of this circuit, write KCL as before at the inverting node N: iC(t) + iR(t) = iN (t) (11) Then, use the ideal op-amp equations iN (t) ∼= 0 (12) vN (t) ∼= vP (t) = 0 (13) together with the fact that vC(t) = vi(t)− vN (t) ∼= vi(t)− vP (t) = vi(t) (14) and rewrite KCL as C dvi dt + vo(t) R = 0 (15) Solving this equation for vo(t) produces the circuit input-output relationship vo(t) = −RC dvi dt (16) 2 4 Procedure + − − + − v(t) + − R 6 + 1 Vdc 10k 10k 10k 10k 100k 100n Figure 4: Integrating Op-Amp Test Circuit Simulation of a First-Order Linear Differential Equation 1. Build the circuit of Figure 4 on your protoboard. Use precision resistors and try to match the capacitor C as close to 100 nF as possible by measurement. 2. Apply a 10-Hz, 1-V peak-to-peak square wave with 0.5-V offset and 50% duty cycle to the input of the integrating op-amp. Make sure the function generator is set on HIGH Z. Observe the input square waveform v1(t) on Channel 1 and the output waveform v2(t) on Channel 2 of the scope. Does the output resemble the theoretical solution given by Equation (18)? (You may need to add a large resistor R6 with a value ranging between 1 MΩ and 20 MΩ. Vary R6 and record its value when you obtain the expected output. Save a picture of both waveforms for the report questions.) 3. Using the cursors, measure the response values of the output waveform at time values of τ , 2τ , 3τ , and 4τ . 5 Report Questions 1. Solve the differential equation in PLA-1 for v(t). What is the time constant τ of the circuit? What is the final steady state value vss of v(t)? 2. Find the theoretical values of v(t) at t = τ, 2τ, 3τ, 4τ . Compare these values to the ones recorded in the lab. 3. One purpose of the resistor R6 is to discharge the capacitor C before applying the input Vdc to the circuit so that v(0) = 0 V. How did the resistor R6 affect the steady state output waveform vss of the integrating op-amp? What value worked the best? 4. Draw a tangent line to the output waveform at time t = 0 and verify that its crosses the steady state output at time t = τ . 5 Boise State University Department of Electrical and Computer Engineering ECE225L – Circuit Analysis and Design Lab Experiment #8: The Integrating and Differentiating Op-Amp Circuits Date: Data Sheet Recorded by: Equipment List Equipment Description BSU Tag Number or Serial Number HP/Agilent 54810A Infinium Oscilloscope HP/Agilent 33120A Function/Arbitrary Waveform Generator HP/Agilent E3631A Triple Output Power Supply Simulation of a First-Order Linear Differential Equation: τ 2τ 3τ 4τ t (ms) v(t) (V)