EE 11L Lab 2

Experiment 2

Equivalent Sources and Superposition

Max Woo
11/17/14
Lab 1D, Professor Babaie

Objectives

The purpose of this experiment is to first verify the Principle of Superposition by testing it on a simple circuit network. Our second goal is to apply and verify the Thevenin and Norton Equivalence Theorems by comparing the output of a circuit and its Thevenin or Norton equivalent.

Theory

For the first part of the experiment, we investigate the Principle of Superposition, which states that the total voltage and current from all power sources across a circuit component (resistor, capacitor, or inductor) is equal to the sum of the voltages and currents induced by each power source individually. Thus, to verify this, we simply need to measure the voltage and current contribution for each power source and make sure they sum to the total voltage and current.

The second part of our experiment involves the two equivalence theorems for analyzing and reducing circuit networks. The Thevenin Equivalence Theorem states that any network of independent power sources and resistors can be modeled by a voltage source and resistor in series, whose output behaves exactly like the output of the original network. An example of this is shown below

Fig 1: A circuit diagram of a Thevenin equivalence circuit, with output at terminal A and B

As seen in Fig. 1, the Thevenin voltage and resistance are denoted by $V_{th}$ and $R_{th}$ respectively, and represent the values needed for the equivalence circuit to imitate the original. To calculate $V_{th}$, we use KVL and KCL to analyze the original circuit and find the voltage at the output terminals when no load is applied. To calculate $R_{th}$, we take the original circuit diagram and replace all current sources with an open circuit and all voltage sources with a wire, and then add an arbitrary power source to the output terminal and calculate the total resistance of the network. These two steps give us all the information needed to construct a Thevenin equivalence circuit.

The Norton Equivalence Theorem works very similarly to Thevenin’s Equivalence Theorem, stating that any network of independent power sources and resistors can be modeled by a current source and resistor in parallel. A diagram of one of these is shown below:

Fig 2: A circuit diagram of a Norton equivalence circuit, with output at terminal A and B

The values here are denoted by $I_{No}$ and $R_{No}$, and are calculated much the same way. For $I_{No}$, we add a wire short-circuiting the output terminals, and then calculate the current going through that wire. $R_{No}$ is actually equal to $R_{th}$ and thus calculated the same way, replacing all power sources and adding an arbitrary power source before calculating total resistance. That gives us all we need to construct a Norton equivalence circuit.

Throughout the experiment, we may sometimes need to calculate total resistances of resistor networks, for which we may need the equations derived in Lab 1:

$R_{series} = R_1 + R_2$
$R_{parallel} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$

Lastly, in preparation for our experiment we to make a current source because the myDAQ cannot produce one. To do so we build the circuit shown below:

Fig 3: Circuit diagram for the current source

The current output is given by the formulas:

$I_{out} = \frac{15 - V_{ref}}{R_3}$ (Eq. 1)
$V_{ref} = 15 \frac{R_2}{R_1 + R_2}$ (Eq. 2)

Thus, we can adjust the resistor values until we get the current output we need.

Part 1: Superposition

Procedure

We first build the circuit shown in Fig 4, adjusting the current source to get as close to 4 mA as possible:

Fig 4: Schematic for circuit used in part 1

The component we will be measuring is the $470 \ \Omega$ resistor. First we measure the voltage and current across the resistor due to both power sources. Then we replace the voltage source with a wire and measure the voltage and current again. Finally we put the voltage source back in but take out the current source (leaving an open circuit there), measuring the voltage and current across the resistor one last time.

Data

Data Analysis

To find the theoretical values for the each power source individually, we first replace the current source with an open circuit. Then we can reduce the resistors to one total resistance of approximately 1000 $\Omega$, which gives us a current of 5 mA through the 680 $\Omega$ resistor. Thus, for the 470 Ohm resistor we get a voltage drop of $5 - 680*0.005 = 1.6 V$ and a current of $1.6 / 470 = 3.40 mA$.

Then we replace the 5V voltage source with a wire, and put the current source back in. Now the total resistance is $217.474 \Omega$, giving us a voltage drop of $217.474*0.004 = 0.870 V$ and a current drop of $0.870/470 = 1.85 mA$.

The theoretical results are compared with the measured results in the table below

Discussion

In Table 2, the measured values all fell within the error bounds of the theoretical values, verifying our results. In addition, in Table 1, the measured value of the voltage and current for both power sources fell within the error bounds of the sum of the voltage and current for each power source, proving the Principle of Superposition.

Part 2: Equivalent Source Transform

For this part, we first build the circuit illustrated in Fig 5, with $R = 1000 \ \Omega$ and $R_L = 470 \ \Omega$

Fig 5: Diagram of circuit for part 2

We then measure the voltage and current across the load resistor $R_L$. After that, we need to construct a circuit that uses a current source instead of a voltage source, such that the voltage and current across the load resistor is the same. Note that Fig 3 is a Thevenin equivalent circuit. Therefore, we are trying to find a Norton equivalent of the Thevenin equivalent circuit, with the resistor R in parallel with the current source, as shown in Fig 6 below:

Fig 6: Diagram of Norton circuit needed for part 2

To start, we know the current through $R_L$ in the Thevenin circuit to be $I_L = V/(R+R_L)$. Thus the voltage is $V_L = IR_L = VR_L / (R+R_L)$. These values need to be the same in our Norton circuit.

Because they are the same, we can use the values of $I_L$ and $V_L$ derived in the Thevenin circuit to solve for the current source in the Norton circuit.
$I_R = V_L / R = \frac{VR_L}{R(R+R_L)}$.
$I_{source} = I_R + I_L = V/R$.

Thus, after we measure the actual values of the voltage source and resistor, we can calculate the appropriate value for the current source. Then we build the Norton circuit and check the voltage and current across the load resistor.

Data and Data Analysis

Table 3: Results along with error

Discussion

All of our measured values fell within the error bounds of our theoretical values, proving our results. In addition, we were able to derive and prove a relationship between the voltage source of a Thevenin circuit and current source of a equivalent Norton circuit, as shown by the low error values in Table 3. The relationship is:

$I_{No} = V_{th}/R$ (Eq. 3)

Part 3: Thevenin/Norton Equivalent

For this part, we construct the circuit shown in Fig 7 with the values $R_1 = 2.2 k\Omega, R_2 = 470 \Omega, R_3 = 1 k\Omega, R_4 = 10k\Omega, R_5 = 3.3k\Omega$.

Fig 7: Circuit diagram for part 3

We then measure the voltage and current across the terminals A and B to get the measured intended values for $V_{th}$ and $I_{No}$. We then replace the voltage sources with wires and measure the resistance across A and B to get the measured intended value for $R_{th}$.

Afterwards, we put the voltage sources back in and add a load resistor across A and B, measuring the resistance, voltage, and current across the load resistor.

Next we need to create Thevenin and Norton equivalence circuits. However, to do so, we first need to analyze the circuit to calculate $V_{th}$, $R_{th}$, and $I_{No}$. We start by defining the current loop $i_1$ as the loop on the left, and current loop $i_2$ as the loop in the middle. We then use Kirchoff’s Voltage Law to model the system:

$i_1R_1+(i_1-i_2)R_3=2$
$i_2(R_2+R_4+R_5)+(i_2-i_1)R_3=3$

Substituting in the actual resistor values gives us:

$2200i_1+1000(i_1-i_2)=2$
$i_2(470+10000+3300)+1000(i_2-i_1)=3$

And solving this system of equations gives us:

$i_1 = 0.000703355 A , i_2 = 0.000250735 A$

From this, we can calculate that:

$V_{th} = R_4 i_2 = 10000 i_2 = 2.50735 V$

Next, to find $R_{th}$, we replace all voltage sources with wires, and add an arbitrary voltage source in between the output terminals. We end up with a circuit shown in Fig 8

Fig 8: Setup for calculating $R_{th}$

We can easily calculate the total resistance of the network using formulas for parallel and series resistors defined in the Theory section:

$R_{th} = R_{total} = \frac{1}{\frac{1}{R_4}+\frac{1}{R_2+R_5+\frac{1}{\frac{1}{R_1}+\frac{1}{R_3}}}}$
$R_{th} = 3083.1748 \ \Omega$

Thus, we have both values needed for the Thevenin equivalence circuit. For the Norton’s equivalence circuit, we can refer to Part 2 of this experiment, which gave us a way of finding the Norton’s equivalence circuit of a Thevenin circuit. From the results of Part 2, we find that

$I_{No} = V_{th} / R_{th} = 0.000813 \ A = 0.813 \ mA$
$R_{No} = R_{th} = 3083.1748 \ \Omega$

With these values, we construct the corresponding Thevenin and Norton circuits, and then, using the same load resistor as we used for the circuit in Fig 7, we measure the voltage and current across the load resistor.

Data

Table 4: Actual measured values for resistors

Table 5: Calculated and measured values for thevenin and norton circuit values

Table 6: Voltage and current across the load resistor for all three circuits

Discussion

The voltage and current across the load resistor for both our Thevenin and Norton circuit fell within the error bounds of the measured values for our original circuit, verifying our calculations for the Thevenin and Norton source and resistor values.

Black Box Equivalence

Procedure

Note that the myDAQ power source is not ideal, and thus can be modeled more accurately as a Thevenin equivalent circuit. To determine the Thevenin equivalent voltage, we can simply measure the voltage across the terminals. To measure the Thevenin resistance, we can first find the Norton current by connecting the terminals of the myDAQ with a wire and measuring the current. Then, following the relationship derived in Part 2, we find the Thevenin resistance by dividing the Thevenin voltage by the Norton current.

We then verify our result by measuring the current and voltage across a load resistor.

Data

Discussion

For the myDAQ, we could not remove the voltage and current sources inside, so we could not directly measure the Thevenin resistance. Thus, in order to find the Thevenin equivalent circuit, we had to measure both the Thevenin voltage and the Norton current to get the Thevenin resistance. This technique is beneficial because it does not alter the myDAQ in any way, and thus this technique can be applied to any other black box.

Note that the Thevenin Equivalent circuit might still not have modeled the myDAQ completely, because the Thevenin equivalent circuit cannot factor in dependent voltage and current components like inductors or capacitors.

Conclusion

To conclude, in every single part of the experiment, our measured results fell within the error bounds of our theoretical values. Thus, we were able to verify the Principle of Superposition, and also the theory and methods behind the Thevenin and Norton equivalent circuits.