AN INTRODUCTION TO THE EXPERIMENTS The following two experiments are designed to demonstrate the design and operation of the op-amp differentiator and integrator at various frequencies. These two experiments that you will perform can be summarized as follows: Experiment No. Purpose 1. Demonstrates the design and operation of an op-amp differentiator. 2. Demonstrates the design and operation of an op-amp integrator. EXPERIMENT NO. 1 Purpose The purpose of this experiment is to demonstrate the design and operation of an opamp differentiator, using a type 741 op-amp. Discussion Schematic Diagram of Circuit (Fig. 3-11) A differentiator is a circuit that produces an inverted output that approximates the rate of change of the input function. An ideal differentiator is shown in Figure 3-11. Notice how the placement of the capacitor and resistor differ from that of the integrator. The capacitor is now the input element. A differentiator produces an output that is proportional to the rate of change of the input voltage. Although a small-value resistor is normally used in series with the capacitor to limit the gain, it does not affect the basic operation and is not shown for the purpose of analysis. Design Basics Output voltage: Low-frequency response: When: f <fc, the circuit acts as a differentiator, f > fc, the circuit approaches that of an inverting amplifier with a voltage gain of -Rf/Rs.
Before setting up the lab, you must first design the circuit on microcap using the template on the next page. Once you have shown your instructor, he will instruct you on what to do next EXPERIMENT NO. 1 Step 1 Set your oscilloscope for the following settings: Channel 1: 0.5 volt/division Channel 2: 0.05 volt/division Time base: 0.5 msec/division DC coupling Step 2 Fig 3.1 Wire the circuit shown in the schematic diagram and apply power to the breadboard. Step 3 Adjust the peak-to-peak voltage of the triangle-wave input at 1 volt (0.5 volt peak) and the frequency at 400 Hz. At this point have the instructor observe a working circuit. Instructor signature As shown in the microcap exercise, you should see that the output signal is a square wave that is 180 o out-of-phase with the input. Step 4 Temporarily disconnect Ch 2. Switch Channel 2 probe to ground and position line on 0 volts. Switch back to DC mode and reconnect the probe to the output of the circuit as shown in the schematic diagram (Fig. 3-11). Measure the negative peak voltage of the square wave (with respect to ground), recording your result below: Step 5 Negative peak voltage = volts
Now measure the time period for which the square-wave voltage is negative (t1) t= usec Draw the waveforms to scale on the graph Figure 3-12Draw the waveforms in the graph Fig. 3-12 Fig. 3-12. Input and output waveforms for op-amp differentiator in Experiment No. 1. Step 6 For this negative output voltage, the equation for the peak output voltage of a triangledifferentiated square wave is given by: How does the value that you determined in Step 4 compare with the predicted value? When this was performed, I measured a peak voltage of 0.0825 volt, as compared to a calculated value of 0.0827 volt. Step 7 Adjust the input frequency so that there is 1 khz. Repeat Steps 4, 5, and 6. Measure the negative peak voltage of the square wave (with respect to ground), recording your result below:
Negative peak voltage = T = volts usec At this point have the instructor observe a working circuit. Instructor signature How does your experimental result compare with the calculated value? At what approximate frequency will this circuit cease to act as a differentiator (that is, approach the operation of an inverting amplifier)? Step 8 F = Hz Adjust the input frequency so that there is 30 khz. What does the output signal look like? Step 9 Instructor s signature Measure the peak-to-peak output voltage and determine the volt age gain. ACL=
EXPERIMENT NO. 2 Purpose The purpose of this experiment is to demonstrate the design and operation of an opamp integrator, using a type 741 op-amp. Discussion Schematic Diagram of Circuit (Fig. 3-14) By interchanging the position of the resistor and capacitor of the differentiator circuit of Figure 3-1, we now have an opamp integrator. As shown in Figure 3-8, the resistor R1, is the input element and the capacitor, C1 is the feedback element. The circuit is said to be the inverse of a differentiator circuit, which is consistent with the mathematical operations of differentiation and integration. In its integral form, the output voltage, as a function of time, is given by: The integrator is a circuit that produces an inverted output that approximates the area under the curve of the input function. An ideal integrator is shown in Figure 3-14. Notice that the feedback element is a capacitor that forms an RC circuit with the input resistor. Design Basics Output Voltage: Low-frequency response: When: f < f C, the circuit approaches that of an inverting amplifier with a voltage gain of R S /R 1. f > f C, the circuit acts as an integrator.
For minimum output offset due to input bias currents, Before setting up the lab, you must first design the circuit on microcap using the template on the next page. Once you have shown your instructor, he will instruct you on what to do next Step 1 Set your oscilloscope for the following settings: Channels 1 & 2: 0.5 volt/division Time Base: 20 usec/division DC coupling Step 2 Wire the circuit shown in the schematic diagram (Fig. 3-14), and apply power to the breadboard. Step 3 Adjust the peak-to-peak voltage of the input square wave at 1 volt (0.5 volt peak) and the frequency at 10kHz. At this point have the instructor observe a working circuit. Instructor signature You should see that the output signal is a triangle wave that is 180 o out-of-phase with the input. Step 4 Temporarily disconnect Ch 2. Switch Channel 2 probe to ground and position line on 0 volts. Switch back to DC mode and reconnect the probe to the output of the circuit as shown in the schematic diagram (Fig. 3-14). Measure the peak-to-peak triangle voltage, recording your result below: Step 5 Peak-to-peak voltage = volts Now measure the time period that the input square-wave takes to complete one-half cycle (t). T= usec
Step 6 For a square-wave input signal, the output peak-to-peak voltage of the triangle wave is: How does the value you determined in Step 4 compare with the value computed from the above formula? Step 7 Now adjust the input frequency so there is 4 KHz. Repeat steps 4,5,6. Measure the peak-to-peak triangle voltage, recording your result below: Peak-to-peak voltage = volts T= usec At this point have the instructor observe a working circuit. Instructor signature How does your experimental result compare with the calculated value? At what approximate frequency will this circuit cease to act as an integrator (that is, approach the operation of an inverting am plifier)? Step 8 F= Hz Adjust the frequency so that there is 100 Hz. What does the output signal look like? Measure the peak-to-peak triangle voltage, recording your result below: Peak-to-peak voltage = volts Now measure the time period that the input square-wave takes to complete one-half cycle (t). T= usec How does your ex perimental result compare with the calculated value? Step 9 Measure the peak-to-peak output voltage and determine the volt age gain. A CL =