Sep.

30, ‘20 ETRO 210 Electronic Technology I Fall 2020

Lab Assignment #2
Integrating and Differentiating Circuits
Due: Oct. 5, ‘20

Materials Needed
One 10kΩ resistor
Capacitors: One 0.01 µF, one 1000 pF
Inductor: 100 mH

OBJECTIVES
After performing this experiment, you will be able to:
1. Explain how an RC or RL series circuit can integrate or differentiate a signal.
2. Compare the waveforms for RC and RL circuits driven by a square wave generator.
3. Determine the effect of a frequency change for pulsed RC and RL circuits.

SUMMARY OF THEORY
In mathematics, the word integrate means to sum. If we kept a running sum of the area under a
horizontal straight line, the area would increase linearly. An example is the speed of a car. Let's
say the car is traveling a constant 40 miles per hour. In 1/2 hour the car has traveled 20 miles. In
1 hour the car has traveled 40 miles, and so forth. The car's rate is illustrated in Figure 2-1(a).
Each of the three areas shown under the rate curve represents 20 miles. The area increases
linearly with time and is shown in Figure 2-1(b). This graph represents the integral of the rate
curve.

Figure 2-1

A similar situation exists when a capacitor starts to charge. If the applied voltage is a constant,
the voltage on the capacitor rises exponentially. However, if we examine the beginning of this
exponential rise, it appears to rise in a linear fashion. As long as the voltage change across the
capacitor is small compared to the final voltage, the output will represent integration. An
integrator is any circuit in which the output is proportional to the integral of the input signal. If
the RC time constant of the circuit is long compared to the period of the input waveform, then the
waveform across the capacitor is integrated.
The opposite of integration is differentiation. Differentiation means rate of change. If the RC
time constant of the circuit is short compared to the period of the input waveform, then the
waveform across the resistor is differentiated. A pulse waveform that is differentiated produces
spikes at the leading and trailing edges as shown in Figure 2-2. Differentiator circuits can be
used to detect the leading or trailing edge of a pulse. Diodes can be used to remove either the
positive or negative spike.

Input
Pulse

Output

Figure 2-2

An RL circuit can also be used as an integrator or differentiator. As in the RC circuit, the time
constant for the RL integrating circuit must be long compared to the period of the input
waveform, and the time constant for the differentiator circuit must be short compared to the input
waveform. The RL circuit will have similar waveforms to the RC circuit expect that the output
signal is taken across the inductor for the differentiating circuit and across the resistor for the
integrating circuit.

PROCEDURE
1. Measure the value of a 100 mH inductor, a 0.01 µF and a 1000 pF capacitor, and 10 kΩ
resistor. Record their values in Table 1. If it is not possible to measure in inductor or
capacitors, use listed values.

Table 1

Listed Measured
Value Value

L1 100 mH
C1 0.01µF
C2 1000 pF
R1 10 kΩ

2. Construct the circuit shown in Figure 1. Set the generator for a 1.0 Vpp square wave with load
at a frequency of 1.0 kHz. You should observe that the capacitor fully charges and discharges
at this frequency because the RC time constant is short compared to the period. On Plot 1,
 sketch the waveforms for the generator, the capacitor, and the resistor. Label voltage and
time on your sketch. To look at the voltage across the resistor, use the difference function
technique.

R1
VS
10kΩ
1kHz
V1 C1
1V 0.01µF VR

VC
Figure 1
Plot 1

3. Compare the RC time constant for the circuit. Include the generator’s Thevenin impedance as
part of the resistance in the computation. Enter the computed time constant in Table 2.

Table 2

Computed Measured

RC time
constant

4. Measure the RC time constant using the following procedure:

(a) With the generator disconnected from the circuit, set the output square wave on the
oscilloscope to cover 5 vertical divisions (0 to 100 %).
(b) Connect the generator to the circuit. Adjust the SEC/DIV and trigger control to
stretch the capacitor-charging waveform across the scope face to obtain best
resolution.
(c) Count the number of horizontal division from the start of the rise to the point where
the waveform crosses 3.15 vertical divisions (63% of the final level). Multiply the
number of horizontal divisions that you counted by the setting of the SEC/DIV
control.
Alternatively, if you have cursor measurements on your oscilloscope, you may find
they allow you to make a more precise measurement. Enter the measured RC time
constant in Table 2.

5. Observe the capacitor waveform while you increase the generator frequency to 10 kHz. On
Plot 2, sketch the waveforms for the generator, the capacitor, and the resistor at 10 kHz.
Label the voltage and time on your sketch.
 VS

VR

VC

Plot 2

6. Temporarily, change the generator from a square wave to a triangle waveform. Describe the
waveform across the capacitor.

7. Change back to a square wave at 10 kHz. Replace the capacitor with a 1000 pF capacitor.
Using the difference channel, observe the waveform across the resistor. If the output were
taken across the resistor, what would this circuit be called?

8. Replace the 1000 pF capacitor with a 100 mH inductor. Using 10 kHz square wave, look at
the signal across the generator, the inductor, and the resistor. On Plot 3, sketch the
waveforms for each. Label the voltage and time on your sketch.

VS

VR

VL

Plot 3

QUESTIONS (17 pts)

1. Suggest how you might find the value of an unknown capacitor using the RC time
constant. (5 pts)

2. (a) Compute the percent difference between the measured and computed RC time
constant. (2 pts)
(b) List some factors that affect the accuracy of the measured result. (2 pts)

3. What account for the change in the capacitor voltage waveform as the frequency was
raised in step 5? (3 pts)
4. What is the time constant (τ) of an RC integrator at 1.3 ms when the input is a square
wave with an amplitude of 1.0 V and the instantaneous voltage at the point is 0.75 V? (5
pts)

 log e 
* log e e a = a ln (e ) = a 10  = a
 log10 e 