EXPERIMENT 12

Op Amp Configurations - Integrator, Differentiator, Voltage Follower

In this experiment, we will look at two of the standard op amp configurations that give us the mathematical operations of differentiation and integration. Also we will see how we can add just about any kind of load to an amplifier and not change its performance by using a voltage follower. Before proceeding, please review section 6.2.3 of Gingrich in which he describes various mathematical operations that can be performed with op amps. You will see that it is quite easy to perform additions, subtractions, derivatives and integrals. That is why collections of op amp circuits have been used in the past to represent dynamic systems in what is called an analog computer. There is a nice discussion of analog computers and how they can be used to analyze a spring-mass system in the link below:

http://colosus.mathcs.rhodes.edu/~stuart/analog/analog.html

Also review sections 6.2.1 and 6.2.2 on non-inverting and inverting amplifiers, respectively. The first section introduces the voltage follower.

There are several basic circuit blocks that we will use in this experiment. Rather than drawing the entire circuit at once, it is more instructive to show them separately, so that each function can be identified. First there is the function generator, that we will represent as a Thevenin equivalent:

Next is the load resistor:

In this experiment, we will be connecting the load to the function generator, but not directly. We will first integrate or differentiate the source signal and then connect to the load. We will see that we can make the operation of the circuit more independent of the load by adding a voltage follower.

Next is the basic op amp configuration with enough components to be either an integrator or differentiator. (Note that a voltage divider has been included on the output of this amplifier. Recall that we used a potentiometer like this in the Project 2 circuit as a volume control.):

Finally, the voltage follower:

Part A The Voltage Follower

Begin by simulating the circuit consisting of the source, the basic op amp and the load. We will add the voltage follower next. Use a source voltage of 1 volt and a frequency of 1kHz. Do an AC Sweep from 1 Hz to 100 kHz. Obtain a plot of the output showing the voltages at the input to the op amp, at pin 6 of the op amp, and across the load. Note that you are plotting the input signal, the output of the basic op amp (pin 6), the output from the voltage divider and the voltage across the load resistor. In this case the latter two voltages are connected directly.

Now add the voltage follower circuit between the voltage divider output of the basic op amp and the load. Do the same AC Sweep and obtain a plot of the output. What similarities and differences do you observe between the two sweep outputs? It is said in Gingrich that the voltage follower (also called a buffer amplifier) is used to isolate a signal source from a load. From your results can you explain what he meant?

Voltage followers are not perfect. They are not able to work properly under all conditions. To see this, do a transient analysis of the circuit in steps of 10 us for a total time of 10 ms. You should see the circuit working more or less as you would expect. Now change the load resistor to 10 ohms and do the transient analysis again. What do you observe now? Can you explain it?

Part B Integrator

Now remove C1 from your circuit and replace it with a wire. Also, go back to the original 100 ohm load. Repeat the transient analysis of this circuit. You should observe that this circuit does work approximately as an integrator. What is there about the transient response that tells you this? If you have trouble answering this question, you might want to do all the tasks in part B and then come back to it.

Now repeat the AC Sweep. In addition to the voltages that you have been plotting, also plot the phase of the voltage across the load. What should the value of the phase be (approximately) if the circuit is to be working more or less like an integrator. (Hint - look at Gingrich) You should observe that this circuit will work better if you lower the characteristic RC frequency by increasing C2. Try C2=1uF. Repeat the simulation. Does it now work better? Why? Print a copy of the transient output and AC Sweep. Indicate on these plots why you think the integrator is now integrating.

A more direct way of demonstrating that integration can be accomplished with this circuit is to replace the source with a DC source and a switch. Note that the switch is set to close at time t=0.01 sec. Use a voltage of 0.1 volts to avoid saturation problems. Do the transient analysis for times from 0 to 50ms with a step of 10us. Rather than plotting the output voltage (voltage across the load resistor), plot the negative of the output voltage. You should see that this circuit does seem to integrate. Why? To show what our problem was before when it did not produce an output proportional to the integral, decrease C2 to 0.01uF and repeat the simulation. Decreasing the capacitance increases the RC corner frequency. See section 3.3.2 of Gingrich for the general discussion of approximate integrators. Print your output when you see integration occurring correctly.

Please note that the ideal integrator does not have a feedback resistor. We have been looking at the configuration with such a resistor because it is generally more practical.