faculty.psau.edu.sa · Web viewv c t = V 0 × e -t τ (1) Where, V 0 is the value of the capacitor...

9
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: v c ( t ) =V 0 ×e t τ (1) Where, V 0 is the value of the capacitor voltage v c at t=0s, τ = RC is the time constant of the circuit. The time constant gives the rate at which the voltage decays to 1

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...

Page 1: faculty.psau.edu.sa · Web viewv c t = V 0 × e -t τ (1) Where, V 0 is the value of the capacitor voltage v c at t=0s, τ = RC is the time constant of the circuit. The time constant

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

Page 2: faculty.psau.edu.sa · Web viewv c t = V 0 × e -t τ (1) Where, V 0 is the value of the capacitor voltage v c at t=0s, τ = RC is the time constant of the circuit. The time constant

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.

2

Page 3: faculty.psau.edu.sa · Web viewv c t = V 0 × e -t τ (1) Where, V 0 is the value of the capacitor voltage v c at t=0s, τ = RC is the time constant of the circuit. The time constant

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

3

Page 4: faculty.psau.edu.sa · Web viewv c t = V 0 × e -t τ (1) Where, V 0 is the value of the capacitor voltage v c at t=0s, τ = RC is the time constant of the circuit. The time constant

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 ¿ .

4

Page 5: faculty.psau.edu.sa · Web viewv c t = V 0 × e -t τ (1) Where, V 0 is the value of the capacitor voltage v c at t=0s, τ = RC is the time constant of the circuit. The time constant

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.

5

Page 6: faculty.psau.edu.sa · Web viewv c t = V 0 × e -t τ (1) Where, V 0 is the value of the capacitor voltage v c at t=0s, τ = RC is the time constant of the circuit. The time constant

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.

6