Ece 203 electrical circuit (3)

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Week 6

ECE 203 - ELECTRICAL CIRCUITS I PRELIMINARY MANUAL – 3 – Linear Resistive Circuits Background Knowledge: 1) Read the below notes on internal resistance. On your preliminary report state that you read the below notes;

Function Generator Internal Resistance and Capacitance: In general the function generator is thought as an ideal voltage source. A more accurate model would be as shown in figure1 where Vgen is an ideal voltage source.

Figure 1

In case of not connecting earth ground to the negative terminal of the generator; the components between the negative terminal and the earth ground, namely the 1MΩ resistor and the 45nF, may affect the operation of the circuit. Keeping in mind that the negative probes of the oscilloscope are connected to earth ground, this kind of effect is to be seen when the negative probes of the oscilloscope is not connected to negative terminal of the signal generator. If the earth ground is connected to somewhere else in the circuit, the part between the ground and the negative terminal of the generator, and the lower part of the signal generator model become in parallel. The 1MΩ resistor can be neglected when it is connected in parallel with relatively low load impedance (or resistance). The capacitor can also be neglected at low frequencies but it starts to become effective at high frequencies. If the negative terminal is connected to earth ground somehow, the capacitor will be shorted and it does not affect the circuit. However, in some cases we are not able to connect the negative terminal of the signal generator to earth ground. Therefore, in those cases we should be aware that the 45nF capacitor may affect the circuit at high frequencies.

When the negative terminal of the signal generator is connected to the ground, neither the 1MΩ resistor nor the 45nF capacitor affects the circuit. However, 50Ω resistor becomes in series with the load connected to the signal generator. Hence, if the load impedance (or resistance) is comparable with the 50Ω series resistance, an amplitude drop at the terminals of the signal generator is expected.

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Figure 2: AA TECH Function Generator that is used in LAB.

2) Consider the circuit given in figure3. The voltages V1(t) and V0(t) are measured by a multimeter having an internal resistance of 20 kΩ in voltage measurements.

Figure 3 a.

What would be the reading of the voltmeter when measuring V1(t)? What would be the reading of the voltmeter when measuring V0(t)? (Hint: keep in mind that voltmeter is connected in parallel)

b. The 1kΩ resistors R1 and R2 are replaced by 5kΩ resistors. Repeat part (a).

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Alternating Current: In alternating current (AC, also ac), the flow of electric charge periodically reverses direction.

In direct current (DC, also dc), the flow of electric charge is only in one direction. The abbreviations AC and DC are often used to mean simply alternating and direct, as when they modify current or voltage.

AC is the form in which electric power is delivered to businesses and residences. The usual waveform of an AC power circuit is a sine wave. In certain applications, different waveforms are used, such as triangular or square waves. Audio and radio signals carried on electrical wires are also examples of alternating current. In these applications, an important goal is often the recovery of information encoded (or modulated) onto the AC signal.[1]

3) Analyze the circuit given in figure4 and find V0(t). (Hint: use superposition principle.)

Figure 4

4) Read and study the below notes about RMS, Root Mean Square. On your preliminary report state that you studied the below notes; ROOT MEAN SQUARE (RMS):

There are two values to represent the amplitude of an AC voltage or current. One of them is the peak value and the other is RMS value. You need to understand the difference of these values. For instance, ammeters you are using in your laboratories are able to measure amplitude of voltage or current of signals. However, you need to know whether the amplitude value you are reading on ammeter display is peak value or RMS value.

RMS value is the amount of AC power that produces the same heating effect as an equivalent DC power. For example, a lamp connected to a 6V RMS AC supply will shine with the same brightness when connected to a steady 6V DC supply. However, the lamp will be dimmer if connected to a 6V peak AC supply because the RMS value of this is only 4.2V (it is equivalent to a steady 4.2V DC).

Little Brainstorming: Think about the electricity bills you receive every month. These bills are basedon the power you consumed every month.

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For DC sources, the power is For AC sources, the power is Normally calculating AC power would require some more complex calculations. However, using RMS values, AC power becomes easier to calculate. The RMS values are, in a sense, the equivalent to the values in a DC circuit.

Figure 5: A sinusoidal signal with peak amplitude 170 volts and RMS voltage of 115 volts .

The average current and the average voltage in a circuit supplied by an AC device are equal to zero. 

The RMS formula is:

The T here is the observed time period of the signal. So we take the square of the voltage function and take its integral in a time period of T. Then we divide it by T and finally take the square root to find the RMS voltage.

5) (please note that for below questions the DC offset is 0V) a. Consider an AC current rent of Remember

with period T, angular frequency

and peak cur-

Ip . Show that the average of I over time T is zero. = Iavg

b. Square the current I given in part a. Now obtain the average of squared current. What is the result? Is it still zero? c. Find the square root of the result you obtained in part b. d. Result of part c gives you IRMS. So, how do you represent IRMS in terms of

Ip?

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e. For triangular (ramp) signals, in order to obtain the RMS value, we divide peak value by matically.

. Prove it mathe-

f. For square signals, the RMS value is always the positive peak value. Prove it mathematically.

6) Now we will construct the circuit in figure6 on 5spice, a circuit simulation program, and do transient analysis.

Then we will observe the plots of V1(t) and V0(t).

Figure 6 Only follow the instructions and try to do them yourself. On your preliminary report, stating that you did the below tutorial would be fine.

5spice Analysis: First of all, go to the webpage: http://www.5spice.com/download.htm and download 5spice Analysis V2.0.

Click on “Click here to download�

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Click on “Save File”

Now run the 5SpiceAnaysis210.exe by double-clicking on it to setup the program.

Click on “Next”

Select “I accept the agreement” then click on “Next”

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Enter user name then click on “Next”

Select setup location then click on “Next”

Select start menu shortcuts folder then click on “Next”

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Now click on install

Wait for program install to finish

De-select “View the README file” then click on “Finish”

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5spice Analysis setup is finished. Now open the program.

From components section, hover your mouse over resistor symbol and click on resistor from the menu. Now there is a purple resistor symbol attached on your mouse. Place two resistors inside the yellow rectangle by left clicking then right click to de-select resistor.

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Now, from components section, hover your mouse over battery symbol and choose “Signal Source, voltage” from the menu. Just as before a purple element is attached on your mouse. Place one voltage source inside the yellow rectangle by left clicking and then right click to de-select the voltage source.

Double click on R1. A new menu “Component” is opened. From this menu change its value to 1,2K. Click “OK”. Do the same for the R2 and make its value 1,8K.

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Double click on Vs1. A new menu “Signal Source Vs1” is opened. From this menu, click on “TRAN/FFT analysis”. Choose “Sine wave” and change Amplitude to 2,5V and Frequency to 10Hz. Click “OK”.

Hover your mouse over battery symbol again and choose “Ground”. Place one ground inside yellow rectangle. Click on R1 and press R to rotate it once. Click and drag elements until they are like on the above figure.

Hover your mouse over pencil symbol and choose “Wire”. Now your cursor should look like a pencil. Left click on upper end of Vs1 and then left click on left-hand end of R1 to finish the wire. Now do the same thing to all connection points to finish wiring.

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Hover your mouse over “TP” symbol and choose “Test Point, differential voltage”. Place two test points. Make the wirings as in the figure above. You can rename these test points if you like.

Save your work from “File” menu.

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Click on “Analyze” on menu bar then click on “Spice Defaults…”.

“Analysis Dialog” menu is opened now. Click on “Analysis” tab. Then choose “Transient New” under “Select Analysis” section. Then, under “Time” section change “To” value to 0,4 seconds and “max time step” value to 5 microseconds (i.e. enter 5u).

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Click on “Graph/Table” tab. Under the “Plots” section from Plot1 combo-box choose TPdv1.

From Plot2 combo-box choose TPdv2. Then from the Axis menu choose left for both plots.

When finished click on “Apply and Run” button below.

After you pressed “Apply and Run” button the simulation starts. When it is finished the waveforms are shown in a graph under the newly added “Transient-1” tab. Here the blue waveform is voltage on R2 and red waveform is the voltage on R1.

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Right click on graph area and click on “Save Image to File…”. You can save your graph in “.bmp” extension format and can add it to your reports.

As always much more helpful and detailed tutorials can found through Google search. You are welcome to study those.

References: [1] Alternating Current, http://en.wikipedia.org/wiki/Alternating_current, accessed October 7th, 2014 [2]