Name: William Tyler Dodson
EEE 202 Lab 3: Superposition, Thevenin’s and Norton’s Models
Data Sheet
Pre-Lab Work (Hand Calculations and/or LTSpice Simulations)
Part 1
a) Build circuit 1 in LTspice and run a DC simulation
Include a screenshot of your LTSpice circuit here (Have a look at an example screenshot at the end
of this document):
Simulated Results
Node Voltage at N1
3V
=_________________
Node Voltage at N2
662.21mV
=_________________
Node Voltage at N3
0
=_________________
1
b) Build circuit 2 in LTspice and run a DC simulation:
Include a screenshot of your LTSpice circuit here (Have a look at an example screenshot at the end
of this document):
Simulated Results
Node Voltage at N1
0
=_________________
Node Voltage at N2
0.55V
=_________________
Node Voltage at N3
5V
=_________________
c) Build circuit 3 in LTspice and run a DC simulation:
2
Include a screenshot of your LTSpice circuit here (Have a look at an example screenshot at the end
of this document):
Simulated Results
Node Voltage at N1
3V
=_________________
Node Voltage at N2
1.214V
=_________________
Node Voltage at N3
5V
=_________________
Do the node voltages from the first two circuits sum to match the voltages of the third circuit?
Note any additional observations. What circuit analysis technique does this demonstrate?
Yes, the voltage for N2 from both part a and part b sum to be the same value as in part c. This is to
be expected due to the superposition technique. Splitting a 2 supply system into 2 circuits each
with one supply, and adding the results.
3
Part 2:
Use the circuit shown to calculate the following:
a) Calculate the voltage across and power dissipated of the load resistor (show your work)
VL = 158.8mV
PL = 76.42µW
b) Draw the Thevenin equivalent circuit and recalculate the numerical value of the voltage
and power of the load resistor. Show your work. You can draw by hand or by any software
such as LTSpice. No need to simulate this circuit.
4
VL = 158.72V P = 76.33µW
L
Do the values match your answers from Part a? (explain)
Slightly off, due to rounding errors.
c) Use the Thevenin circuit to create an equation for P as a function of R
L L (start by finding an
equation for V as a function of R , R and/or V ):
L L th th
Could you do this with the original circuit? Explain.
Yes, but it would require much more work. Such as finding Vth and Rth as part of the equation.
Not just finding Vl .
d) Draw the Norton equivalent circuit and recalculate the numerical value of the voltage and
power of the load resistor (as you did in part b above). Show your work. You can draw by
hand or by any software such as LTSpice. No need to simulate this circuit.
5
VL = 158.72mV P = 76.33W
L
Do the values match your answers from Part a and Part b? (explain)
Yes, they match because I used the same exact numbers for the conversion to a Norton equivalent.
This circuit is identical to the Thevenin equivalent.
Lab Work (Hardware Kits):
Step 1 (Part 1): Superposition
1. Build circuit 1 in hardware and using multimeter measure the voltage at each node with respect
to ground.
Include a photo of your hardware-built circuit here (Have a look at an example photo at the end of
this document):
6
Measured Results
Node Voltage at
2.98V
N1=_________________
Node Voltage at
662mV
N2=_________________
Node Voltage at
0
N3=_________________
2. Build circuit 2 in hardware and using multimeter measure the voltage at each node with respect
to ground.
Include a photo of your hardware-built circuit here (Have a look at an example photo at the end of
this document):
7
Measured Results
Node Voltage at
0V
N1=_________________
Node Voltage at
555mV
N2=_________________
Node Voltage at
4.99V
N3=_________________
3. Build circuit 3 in hardware and using multimeter measure the voltage at each node with respect
to ground:
Include a photo of your hardware-built circuit here (Have a look at an example photo at the end of
this document):
8
Measured Results
Node Voltage at
2.98V
N1=_________________
Node Voltage at
1.22V
N2=_________________
Node Voltage at
5.00
N3=_________________
Do the node voltages in the first two circuits sum to the node voltages of the third circuit?
Yes, the 2 superposition circuits do sum to the third original circuit.
Step 2: Thevenin’s and Norton’s Theorems
Part 2a LTSPICE
9
Build this circuit using LTSpice and run a DC Sweep analysis:
Include a screenshot of your LTSpice circuit here (Have a look at an example screenshot at the end
of this document):
1. On LTSpice software, plot the load voltage (V ) vs. the supply voltage (V1)
L
Include a screenshot of your plot here:
10
2. Build the Thevenin equivalent circuit in LTSpice and run a DC Sweep analysis.
Include a screenshot of your LTSpice circuit here (Have a look at an example screenshot at the
end of this document):
11
Plot the load voltage vs. the Open Circuit Voltage, V (i.e., vs. the source voltage V ).
OC th
Include a screenshot of your plot here:
Does your plot EXACTLY match the plot from Part2a #1 (the original circuit in the previous
question)? Highlight your answer:
- Yes
- No
3. Build the Norton equivalent circuit in LTSpice and run a DC Sweep analysis.
Include a screenshot of your LTSpice circuit here (Have a look at an example screenshot at the
end of this document):
12
Plot the load voltage vs. the Short Circuit Current, I (i.e., vs. the source current I ).
SC N
Include a screenshot of your plot here:
Does your plot EXACTLY match the plot from Part2a #1 (the original circuit)? Highlight your
answer:
13
- Yes
- No
Part 2b Hardware
Build this circuit in hardware (use a DC power supply, breadboard, resistors):
Include a photo of your hardware-built circuit here (Have a look at an example photo at the end of
this document):
14
1. Use a multimeter to measure the voltage across the load resistor (V ).
L
VL = 162.2mV
Calculate the load power dissipation (P ). (show your work)
L
PL = 79.79µW
2. Build the Thevenin equivalent circuit. (if you can’t find an equivalent R , build one out of
TH
parallel and series combinations). Use a multimeter to measure V and calculate the load power
L
PL and the source power P of the source in this Thevenin circuit.
S
First start by drawing your Thevenin equivalent circuit here (you can draw by hand or by any
software such as LTSpice):
15
Second, if you don’t have R in your inventory, identify the resistors you will need to create
TH
R :
TH
Since my Rth is 2248.12Ω. I will have to round it because I do not have any fraction of an ohm
resistors. I will go with 2248 for simplicity. I will use a 2.2kΩ, a 47Ω, and a 1Ω in series to make
2248Ω
Third, build the circuit on hardware.
Include a photo of your hardware-built circuit here (Have a look at an example photo at the
end of this document):
Fourth, measure VL
VL = 157.8mV
Fifth, calculate P mathematically. Your equation for P should depend on the variables you
L L
measured only (V and R , and not I )
L L L
Equation:
75.46µW
Value:75.46µW
PL = 75.46µW
Sixth, calculate P similarly.
S
Equation:
16
Value:
PS =596.4µW
3. Consider the same Thevenin circuit as in Part 2b #2. Now assume that the source V cannot
TH
supply more than 50% of the power P you calculated before. If you were asked to change the
S
value of R to reduce the power supplied by the source V to be below the maximum power it
L th
can supply, what range of values of RL can you choose? What is the corresponding range of the
load power, P in this case?
L
Derivation of the range of R . Hint: an equation relating source power P to R might be
L S L
helpful.
Type or print on a piece of paper and include its photo:
RL range = 0Ω<Rl<29.93Ω
Derivation of the range of P . Hint: You can depend on your result in Part 2-c in the pre-work
L
at the beginning of this Data Sheet document.
Type or print on a piece of paper and include its photo:
PL range = 0W<Pl<85.5µW
Could you have done this equation as easily without the Thevenin equivalent circuit? Explain.
No way, With out the Thevenin there would be many more sub calculations. Total resistance,
voltage to the load resistor, just to name a few.
What do you think would happen to the source if your load’s resistance was outside this range?
With it below this resistance the maximum power will continue to flow. Too far above the value
will leave the power at next to nothing or practically nothing.
17
Step 3 (Part 3): Non-Ideal Voltage Sources
Note in this expirement:
- If you don’t have a Carbon Zinc Battery, use the waveform W1 (set to 5V DC) on your
Analog Discovery 2 kit.
- If you don’t have an Alkaline Battery, use the waveform W2 (set to 5V DC) on your Analog
Discovery 2 kit.
First find the open circuit voltage of each supply (no resistor connected):
Supply Open Circuit Voltage (Vin)
Carbon Zinc Battery 9.67V
Alkaline Battery 9.54V
Next, build the circuit as shown with the three different supplies for V1 and using two different
resistors and record your results.
Note:
The V1 supply consists of Vin and
Rin if it’s a cell battery. sRin as 0 for ideal
Supply Vout for R1 = 1k
Vout for R1 =
1M
Ideal supply + R1 Calculate Vout
9V 9V
Carbon Zinc Battery Measure Vout
9.28V 9.30V
Alkaline Battery Measure Vout
9.32V 9.33V
a) Is a cell battery a reliable constant voltage source for all load resistances? Why do you think the
measured output voltages didn’t match the calculated ones? Explain.
Not perfect but a decent one. The output will be determined by the resistance of the circuit and
will never be perfect due to many different factors. Like the internal resistance of the battery.
With more of a load on it, it will adjust the voltage output to keep the power output the same for
all circuits.
b) Are all battery chemistry types the same? Explain.
No, not all battery chemistry is the same. Take the carbon zinc battery, and the Alkaline battery
used for this lab.
c) If the internal resistance of a battery is high, what does that do to the output voltage if the load
18
resistance is low? For your experiment, which battery has higher internal resistance and how do
you know?
Causes the battery to heat up and the output voltage to drop. This is because the internal
resistance is in series with the load, which is a simple calculation to determine the divided voltage,
and power absorbed by the internal resistor.
Step 4 (Part 4): Thevenin Model for Non-Ideal Sources
Alkaline Battery Modeling
Do you expect R for the Alkaline battery to be high or low? Why?
TH
Low, because the voltage with a the different load resistors is relatively unchanged. This means
the internal resistance is negligible compared to the load resistors.
VTH (V)
Open Circuit voltage
measured without
connecting RL
RL (K Ohms)
Measured using
Ohmmeter to be precise
IL (mA)
Measured using Ammeter
(don’t forget to connect its
wires in the correct socket)
RTH (K Ohms)
Calculated
9.33 982Ω 9.35mA 15.86Ω
9.4 1.01MΩ 9.3µA 752.68Ω
8.97 98.8Ω 89.9mA 0.977Ω
Average Value: 256.5Ω
Alkaline Battery Thevenin Model:
VTH = 9.24V R = 256.5 Ohms
TH
Carbon Zinc Battery Modeling
Do you expect R for the Carbon Zinc battery to be high or low? Why?
TH
Higer, the voltage with changes more when the load resistor changes. This means it is more
significant that in the alkaline battery.
VTH (V)
Open Circuit voltage
measured without
connecting RL
RL (K Ohms)
Measured using your
Ohmmeter to be precise
IL (mA)
Measured using Ammeter
(don’t forget to connect its
wires in the correct socket)
RTH (K Ohms)
Calculated
9.54 982Ω 9.4mA 32.89Ω
9.64 1.01MΩ 9.5µ 4.7kΩ
7.98 98.8Ω 80.4mA 0.453Ω
Average Value: 1589Ω
Carbon Zinc Thevenin Model:
VTH = 9.05V R = 1589 Ohms
TH
19
Which battery chemistry would be better for small load resistances and why? Was the Thevenin
Model useful in understanding the non-idealities of the batteries? Why do you think the calculated
RTH may have changed as the load resistance changed? Comment on your findings.
I think the Alkaline is better for most applications, but especially ones with smaller load
resistance. This means that more power and voltage will be dissipated at the load side, and not in
the battery. If the power dissipates in the battery, it will make the battery heat up.
The Rth changes because with battery chemistry, the output to current ratio is not necessarily
linear. This explains the very high values of Rth for high resistance, and low values for low
resistance. The 1kΩ values are much closer to the actual values of the internal resistance for these
two batteries.
20
LTSpice Screenshot Example (Note your name, date and time, readable circuit, and labelled nodes
– screenshots might vary based on the operating system you are using) – This applies to
screenshots of circuits only. Screenshots of graphs/plots/anything else can include only the date
and time.
Hardware Image Example. Note your name:
1- on a piece of paper; OR
2- typed electronically (must be typed on the breadboard WITHOUT any “text background”.
Breadboard must show up in the background of your name.)
21
22
