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Electric Circuits ECE 2262 Department of Electrical and Computer Engineering

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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


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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:

𝑉2 = 𝑅. 𝑖𝑐(𝑡) + 𝑉𝑜(𝑡) => 𝑅𝐶𝑑𝑉𝑜(𝑡)

𝑑𝑡+ 𝑉𝑜(𝑡) = 𝑉2

This is a first-order linear ordinary differential equation (ODE), which has the following

solution:

𝑉𝑜(𝑡) = (𝑉1 − 𝑉2)𝑒−𝑡𝑅𝐶⁄ + 𝑉2

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


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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:


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𝑖𝐿 =(𝑉1 − 𝑉2)

𝑅𝑒−(𝑅

𝐿⁄ ) 𝑡 +𝑉2

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


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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.

Figure 6

Inverting differentiator

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.

http://en.wikipedia.org/wiki/Coil

http://en.wikipedia.org/wiki/Electrical_conductor

http://en.wikipedia.org/wiki/Magnetic_core

http://en.wikipedia.org/wiki/Ferromagnetic


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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 1.

4. 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


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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

Channel 1 Channel 2 TA signature

Volt/Div (Volts)

Time/Div (Seconds)

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


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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)


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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)


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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.


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.


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)


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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


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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)


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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)

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:

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

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