Lab 3 Transient Response of RC & RL Circuits

Transcription

1 Lab 3 Transient Response of RC & RL Circuits Last Name: First Name: Student Number: Lab Section: Monday Tuesday Wednesday Thursday Friday TA Signature: Note: The Pre-Lab section must be completed prior to the lab session. Lab objectives o Transient response of RC circuits o Transient response of RL circuits o Simulated inductor circuit using Op-Amps Equipment o Oscilloscope o Signal generator o ADALM1000 (M1K) o Resistor, capacitor, inductor, op-amp Pre Lab Mark: /2 Lab Mark: /8 Total: /10 Page 1 of 13 V.4, Fall 2018

2 Pre Lab RC Circuits Assume that the switch S has been at the position shown in Figure 1 for a long time and it toggles at time t=0. Answer the following questions. 1. As you will see in your upcoming course notes, Vo can be derived by writing a KVL after the switch has toggled. The output, Vo, can be calculated in terms of V1 and V2 as a function of time for t 0 as follows: V 2 = R. i c (t) + V o (t) => RC dv o(t) + V dt o (t) = V 2 This is a first-order linear ordinary differential equation (ODE), which has the following solution: V o (t) = (V 1 V 2 )e t RC + V 2 This equation is called the transient response of the capacitor voltage. How much is Vo at t=0+ (that is the moment right after the switch is toggled) and t=? Explain if these values are what you expected. 2. An RC/RL circuit which has only one C or L and consists of DC sources and switches has a general solution for the voltages and currents: x(t) = A + Be t τ τ is the time constant of the circuit which is equal to RC in this case as can be seen in the comparison of the last two equations. How much is Vo at t=5τ? Normally t=5τ is the time needed to allow the circuit reach its steady state. Figure 1 Page 2 of 13 V.4, Fall 2018

3 3. How does it affect the circuit time constant if the internal resistor of the voltage source V2 was nonzero (Rs 0)? 4. Oscilloscopes are able to display periodic waveforms. In order to monitor the transient response of the circuit using an oscilloscope, we need to toggle the switch S every T seconds. One solution is to replace the voltage sources and the switch in Figure 1 with a voltage source generating the square waveform shown in Figure 2 connected to the terminals A and B in the circuit of Figure 1. In order to allow the circuit reach its steady state at each half cycle of the square wave, how much should the period of the square wave, 2T, be at least? Explain how Vo changes if the square wave s period is selected too short. Figure 2 RL Circuits The voltage source Vs in the circuit of Figure 3 generates the square wave shown in Figure 2. Answer the following questions. 1. Based on the general solution described in the previous section, an equation for il can be derived in terms of V1 and V2 as a function of time between t=0 and t=t. This equation is called the transient response of the inductor current and it has the following form: Page 3 of 13 V.4, Fall 2018

4 i L = (V 1 V 2 ) e (R L ) t + V 2 R R How much is il at t=0 + and t=? Explain whether the results are acceptable. 2. Assuming T=5τ, so the period is long enough such that the circuit reaches its steady state. How much is the inductor current at t=t? 3. What is the time constant of the circuit? How does it affect the circuit time constant if the internal resistor of the voltage source V2 was nonzero (Rs 0)? 4. Oscilloscopes are basically voltmeters with the capability of displaying the measured voltages. However, in the circuits including inductors usually inductors currents are of more interest to monitor. In such cases, one may pick a voltage in the circuit which is proportional to the inductor s current to display on the oscilloscope. Is there any element in the circuit of Figure 3 whose voltage is proportional to the inductor current? Explain. Figure 3 Page 4 of 13 V.4, Fall 2018

5 Simulated inductor An inductor is usually constructed as a coil of conducting material, typically copper wire, wrapped around a core either of air or of magnetic material. Figure 4 shows some examples of coil inductors. Building an inductor has always been a challenge in electrical engineering. Especially when a large value of inductance is needed, the size and the weight of the inductors are undesired. Figure 5, for instance, shows a case that the inductor is significantly larger than the rest of the circuit components. Figure 4 Figure 5 In low voltage applications, it is very efficient to use a circuit that is equivalent to an inductor. Figure 6.a shows a circuit consisting of two resistors, one capacitor, and one Op-Amp that is equivalent to an inductor of L=R1 R2 C in series with the resistor R1. Selecting a small value for R1, the equivalent circuit will be close to an ideal inductor. The limit of this model is that one of the terminals of the inductor must be grounded. Inverting differentiator Figure 6 Figure 7 shows a simple differentiator circuit. Answer the following questions. 1. Derive an equation for Vout in terms of Vin as a function of time. Page 5 of 13 V.4, Fall 2018

6 2. What is the waveform of Vout if Vin is a triangular waveform? Figure 7 3. Why is this circuit called inverting differentiator? Refer to equation you derived for the question It is often desired to have a differentiator rather than an inverting differentiator circuit. Draw your suggested circuit which if added to the output of the circuit of Figure 7, corrects the output voltage sign. Lab Work (In-Lab) RC circuit 1. First try to cancel the dc value of the generated signal coming out of channel A of M1K using an adder circuit you learned in lab 2 so you can apply a 2 Vp-p square waveform as Vs in Fig. 8. For this purpose you should set the voltage amplitude to 4 V and then subtract 2 V from it using an op-amp circuit. Then Build the circuit of Figure 8. Use R=1 kω, and C=100 nf (Ceramic). Using M1K, apply a 500 Hz, 2 Vp-p square waveform as Vs to the circuit. Use channel B of M1K to display the output voltage Vo and channel A to display the square waveform Vs simultaneously. Draw both of the waveforms together in Figure 9, as you see on the oscilloscope. Do not forget to write the volt/div and time/div used for both channels. Have a TA Page 6 of 13 V.4, Fall 2018

7 sign in the box after showing the results. Write down the peak to peak voltage of Vo. Calculate the time constant of the circuit. Vo(p-p)=..Volts Time constant of the circuit = =..Seconds Figure 8 Figure 9 Volt/Div (Volts) Time/Div (Seconds) Channel 1 Channel 2 TA signature Now change the frequency of the square waveform to 2 khz and do not make any changes to its amplitude. How much is the peak to peak voltage of Vo? Explain why it is not the same as the previous case. Vo(p-p)=.Volts Page 7 of 13 V.4, Fall 2018

8 Explain here: Change the frequency of the square waveform to 10 khz and draw both the square waveform Vs and the output voltage Vo together in Figure 10. Ask a TA sign in the box. What type of waveform does Vo look like? Answer here: It looks like a... waveform. How much is the peak to peak voltage of Vo? Vo(p-p)= Volts. Check the one which appropriately represents the function of this circuit when f=10 khz? Differentiator Integrator Figure 10 Volt/Div (Volts) Time/Div (Seconds) Channel 1 Channel 2 TA signature Page 8 of 13 V.4, Fall 2018

9 RL circuit 2. Build the circuit of Figure 11. Use R=100 Ω, and L=10 mh. This time using a function generator, apply a 500 Hz, 2 Vp-p square waveform as Vs to the circuit. Use channel 1 of the oscilloscope to display the output voltage Vo and channel 2 to display the square waveform Vs simultaneously. Draw both of the waveforms together in Figure 12, as you see on the oscilloscope. Do not forget to write the volt/div and time/div used for both channels. Have a TA sign in the box after showing the results. Write down the peak to peak voltage of Vo. Calculate and write the peak to peak current of the inductor using Vo(p-p) measured. Also, calculate the time constant of the circuit including the equation you used. See the voltage at AB terminal. Why is it different from the square waveform you set in the function generator? Explain in the space provided. Figure 11 Vo(p-p)=.Volts il(p-p)=...=.. Amperes Time constant of the circuit =. =..Seconds Figure 12 Volt/Div (Volts) Time/Div (Seconds) Channel 1 Channel 2 TA signature Page 9 of 13 V.4, Fall 2018

10 Explain here: Now, change the frequency of the square waveform to 3 khz and do not make any change to its amplitude. How much is the peak to peak voltage of Vo? Calculate and write the peak to peak current of the inductor using Vo(p-p) measured. Vo(p-p)=.Volts i L(p-p)=...=.. Amperes Change the frequency of the square waveform to 10 khz and draw both of the square waveform Vs and the output voltage Vo together in Figure 13. Ask a TA sign in the box. How much is the peak to peak voltage of Vo? Calculate and write the peak to peak current of the inductor using Vo(p-p) measured. Vo(p-p)=.Volts i L(p-p)=...=.. Amperes Check the one which appropriately represents the function of this circuit when f=10 khz? Differentiator Integrator Figure 13 Volt/Div (Volts) Time/Div (Seconds) Channel 1 Channel 2 TA signature Page 10 of 13 V.4, Fall 2018

11 Simulated inductor 3. Using Multisim, simulate the circuits of Figure 14 together on the same Multisim page. Use the values for the elements of your simulation the same as those shown in the Figure 14. Calculate the value of the capacitance C in this circuit so that two circuits of Figure 14 be equivalent. Refer to the Pre Lab section for more information. There are two identical voltage sources VS in this circuit. Apply a 2 Vp-p, 500 Hz square waveform to both the sources. Pick Vo1 and Vo2 signals by channels 1 and 2 of the oscilloscope respectively and display them together. You should see the two signals very similar. This implies that these two circuits are equivalent since their responses are the same for similar inputs. Show your results to a TA. C=..μF. Equation used:.. What is the main source of inaccuracy in this modeling? Write your answer here: Figure 14 (R1=1 Ω, R2= 1k Ω, R=100 Ω and L=10mH) TA signature Page 11 of 13 V.4, Fall 2018

12 Inverting differentiator 4. Build the circuit of Figure 15 with the specified values for the components. Use a ceramic capacitor for your circuit. Using a function generator, apply a 2 Vp-p, 50 Hz, triangular waveform as the voltage source Vin. Display the triangular waveform and the output voltage Vout using the oscilloscope and display both of them at the same time. Adjust the volt/div for both channels as well as the time/div in an appropriate way to clearly have two complete cycles on your screen. Draw the two signals together on the screen of Figure 16. Also show your results to a TA. Do not forget to write your volt/div for both channels and the time/div you applied. Figure 15 (R= 10k Ω and C=1µF) Figure 16 Volt/Div (Volts) Time/Div (Seconds) Channel 1 Channel 2 TA signature Page 12 of 13 V.4, Fall 2018

13 How much is the peak to peak voltage of Vout? Vout(p-p)= Volts. Now, change the function generator to generate a 2 Vp-p, 33 Hz, sine waveform and draw both Vin and Vout on the screen of Figure 17 using appropriate volt/div and time/div selections to have two complete periods of the signals on the screen. Again, write the volt/div for each channel as well as the time/div used. Show the result to a TA. Figure 17 Volt/Div (Volts) Time/Div (Seconds) Channel 1 Channel 2 TA signature How much is the peak to peak voltage of Vout? Vout(p-p)= Volts. Based on your observations, explain what is the relationship between Vout and Vin in this circuit. Describe how you observed both the inverting and the differentiating properties of this circuit. Write your answer here:... Page 13 of 13 V.4, Fall 2018