faculty.psau.edu.sa · Web viewv c t = V 0 × e -t τ (1) Where, V 0 is the value of the capacitor...
Transcript of faculty.psau.edu.sa · Web viewv c t = V 0 × e -t τ (1) Where, V 0 is the value of the capacitor...
Experiment 6: Transient Response of First Order Circuits
Objectives:
The objective of this experiment is to observe the response of the first order R-C and
R-L circuits. The experiment demonstrates a method for measuring the time constant.
Equipments:
1. Function generator (Signal generator).
2. Digital multi-meters.
3. Resistors. 1k Ω , 500 Ω.
4. Variable Capacitor (0.1 µF).
5. Variable inductor: 100 mH.
Theoretical Background:1.1 Measurement of the Natural Response of First Order Circuits.
The natural response of an R-C circuit, shown in Figure 1, is given by:
vc (t )=V 0 ×e−tτ (1)
Where, V 0is the value of the capacitor voltage vc at t=0s, τ = RC is the time constant
of the circuit. The time constant gives the rate at which the voltage decays to zero. In
circuits, this decay response is due to Ohmic losses (discharging).
C 10.1u F
R 1
500 ohm
0
V s
Figure 1: RC circuit.
The time constant can be measured by one of the following graphical methods.
1. The Tangential Line. A line tangential at a certain point of the response curve
is drawn. The line intersects the time axis exactly in one time constant from
the point of tangent. Fig. 2 shows the graph of the response from eq. (1). Note 1
that the response begins at t= 0.05 s. With reference to Fig. 2, a line is drawn
tangential to the curve at t= 0.05 s. The line is extended until it intersects the
time axis. This occurs at t= 0.25 s. Therefore, the time constant of the response
is τ = 0.25 - 0.05 = 0.2 s.
This method is suitable if a hard copy of the response graph is available.
2. The 63.22% Decay. The second method is suitable for measurements on the
oscilloscope. From eq. (1), every time interval equal to one time constant the
response decays by 63.22%. Equivalently, at the end of every interval equal to
one time constant, the response is at the 100-63.22= 36.78 % of its value at the
beginning of the interval. This method of measurement is demonstrated in Fig.
2. The response begins its decay at t=0.05 s, at that point the value of the curve
is 1. The response reaches 36.78 % of its initial value at t= 0.25 s. Thus the
time constant is t= 0.25-0.05=0.2 s.
The results of 1 and 2 are in agreement.
Figure 2: Measurement of time constant.
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1.2 The Step Response of First Order Circuits.
The step response of the first order circuit in Fig. 1 is given by:
vc (t )=V ss ×(1−e−tτ ) (2)
Where V ssis the constant steady state value of the response. The step response of the
circuit is characterized by the same time constant as the natural response. The time
constant affects the step response in a similar manner as the natural response.
Therefore, the methods discussed previously apply in this case, as well.
On the oscilloscope the time constant of the step response can be measured by
measuring the time the output requires to reach 63.22% of its final value.
The natural and step responses of a first order R-L circuit, shown in Fig. 3, are the
same as in eq. (1) and (2) given for the circuit current iL (t). The circuit time constant
is τ=L/ R.
I
L 1
100m H
1
2
R 1
100V 1
TD = 0
TF = 0P W = 5 mP E R = 1 0 m
V 1 = 6
TR = 0
V 2 = 0
R 2
10
Figure 3: RL circuit.
Procedure:1. Connect the circuit shown in Fig. 4 with the values indicated in the figure.
2. Set the function generator to 5 V p−p, 1 kHz, square wave.
C 10.1u F
R 1
500 ohm
0
V s
Figure 4: First order RC circuit.
3. Connect Ch 1 of the oscilloscope with the input signal and Ch 2 to the
capacitor. Set appropriate grounding for the two channels and set the
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oscilloscope to dual mode. Sketch the wave forms together with an appropriate
scale.
4. Obtain vc (t ) according to the following table:
Charging Discharging
Time
vc (t )
Time
vc ( t )
Practic
al
Theoretica
l
Percentage
Error
Practica
l
Theoretica
l
Percentage
Error
5. Plot the source voltage v(t) and capacitor voltage vc ( t ) on the same graph.
6. Measure the time constant of the circuit (τ ¿ .
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7. Set the function generator to 2kHz and 5 kHz and notice the differences.
8. Use the variable resistance box and fill the following table:
R (Ohm)Time constant (τ ¿
Theoretical Practical
500
1k
4k
9. Disconnect the circuit and connect the circuit given in Fig. 5.
V 1
TD = 0
TF = 0P W = . 5 mP E R = 1 m
V 1 = -2 . 5
TR = 0
V 2 = 2 . 5
L 1
100m H1 2
R 1
1k
0
Figure 5: First order RL circuit.
10. Set the function generator to 5 V p−p, 1 kHz, square wave.
11. Connect Ch 1 of the oscilloscope with the input signal and Ch 2 to the resistor
R1. Set appropriate grounding for the two channels and set the oscilloscope to
dual mode. Sketch the waveforms together with an appropriate scale.
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12. Move channel 2 to L1 and repeat step 11.
13. Measure the time constant of the circuit (τ ¿ .
Questions:
1. Calculate the time constant of the circuit in fig. 4 mathematically and compare it
to the measured value from the waveforms you got in the Lab. Then, find the
percentage error.
2. Calculate the time constant of the circuit in fig. 5 mathematically and compare it
to the measured value from the waveforms you got in the Lab. Then, find the
percentage error.
3. Explain the response observed in steps 3 and 11. Indicate which interval
represents the natural and which interval represents the step response of the
circuit.
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