PSpice Tutorial

The purpose of this lab is to learn and use PSpice to analyze circuits as well as finding the Thevinin and Norton equivalents using the DC sweep and graphing functions.

Procedure:
Part 1:
To start, we built this simple circuit in PSpice and was able to find the current and voltage across each element.

Part 2:
After the circuit was built, we inserted a dummy Independent Current Supply of 0A to find the Thevinin Resistance.

We can see that the Thevenin Voltage is 10V.
The slope (R_th) is calculated to be approx. 3.33 Ω.
I_n = V_th/R_th = 3 A

Afterwards, the dummy current supply was replaced with a varying RL going from 1 Ω to 10 Ω in increments of 0.1 Ω.
From the maximum power peak in the graph, we see that R_th = 3.3 Ω

Conclusion:
R_th was verified both ways and showing that the lab was a success.

Max Power Lab

The purpose of this experiment is to confirm the Maximum Power Transfer Theorem in a series circuit as well as determine the Thevinin Resistance using the theorem.

Procedure:

We start by building the simple circuit connected in series.

To record 15 different values for the potentiometer, we incremented the pot and recorded V_0 for about every 600 Ω.

The second part of the experiment was building the circuit below.

Power supply 4.5 V= 4.61 V
Power supply 9.0 V = 9.07 V

Once operational, we used logger pro to record current and voltage data at V_0 when changing the pot.

Analysis:
Part 1:
For the first circuit, we obtain these set of values.

Measured V0 (V)	Measured Rx (Ω)	Calc. P0 (W)
0	13.5	0.0000E+00
0.16	330	7.7576E-05
0.34	691	1.6729E-04
1.36	2280	8.1123E-04
1.53	2780	8.4205E-04
1.7	3270	8.8379E-04
1.87	3800	9.2024E-04
2.04	4440	9.3730E-04
2.21	5120	9.5393E-04
2.38	5940	9.5360E-04
2.54	6850	9.4184E-04
2.87	9200	8.9532E-04

Graph of measured V_0 vs R_x.



 From the plot of Power Vs. R_x we see that the highest point is around 5500 Ω.



The theoretical R_x for max power is 5600  Ω.
Error = 1.79%

Part 2:
From logger pro, we were able to obtain voltage, current, and power.

From the graph of power however, we could not determine the maximum point of power because of the noise due to the current.

Conclusion:
Part 1 was a success for the most part verifying that the Maximum Power Transfer Theorem; however, we could not confirm it for Part 2 since it was a failure because we were unable to locate the point where max power occured. The problem was because of the fact that the recording device for logger pro was too accurate when measuring small currents, and it recorded unwanted data/noise where as the DMM was much better at recording data. 

Thevenin Equivalents

For this experiment, the objective is to reduce this network of resistors and power supplies below to its Thevenin equivalent, so we can find the smallest resistance of Load #2 that can be connected to the system.

Prelab:

R_C1	R_C2	R_C3	R_L1	V_S1	V_S2	V_L2,Min
100 Ω	39 Ω	39 Ω	680 Ω	9 V	9 V	8 V

To find V_Th, we remove R_L2 and solve for the V_oc.

Since Voc = V_x, we used nodal analysis which gives us the equation:
(V_x-V_S2)/R_C2 + (V_x - V_S1)/R_C1 + V_x/R_L1 = 0
V_x = 8.6 V

As finding R_Th from reducing resistors is harder than it seems, we rely on the formula R_Th = V_oc/I_sc.

After we create a short circuit from R_L2, we can find I_sc.
Using nodal analysis we get the equation:
(V_y-V_S2)/R_C2 + (V_y - V_S1)/R_C1 + V_y/R_L1 + V_y/R_C3= 0
V_y = 5.1 VI_sc = V_y/R_C3 = 0.1307 mA
R_Th = 65.7 Ω

Given that R_L2,Min = 8V, we use the voltage divider to find R_L2
R_L2 = V_L2*R_Th/(V_Th - V_L2) = 876 Ω 
Short Circuit R_L2 current I_sc = 0.1307 mA
Open Circuit R_L2 voltage V_oc = V_Th = 8.6 V

Procedure:

Build circuit with these components

Component	Nominal Value	Measured Value	Power or Current Rating
R_Th	66 Ω	65.8 Ω	1 W
R_L2,min	876 Ω	873 Ω	1 W
V_Th	9 V	9.21 V	2 A

To perform the experiment, we measure the voltage across Load 2.

Config	Theoretical Value	Measured Value	Percent Error
R_L2 = R_L2,min	V_Load2 = 8 V	8.5V	6.25 %
R_L2 = ∞ Ω	V_Load2 = 8.6 V	9.17 V	6.63 %

Next we disassemble and build the second circuit by replacing the Thevenin Equivalents with the original circuit.

Component	Nominal Value	Measured Value	Power or Current Rating
R_C1	100 Ω	99.1 Ω	1/8 W
R_C2	39 Ω	38.1 Ω	1/8 W
R_C3	39 Ω	38.6 V	1/8 W
R_L1	680 Ω	667 Ω	1/8 W
V_S1	9 V	9.26 V	2 A
V_S2	9 V	9.21 V	2 A

Now remove R_L2 so it becomes infinite resistance.

We perform the experiment again by measuring V_Load2.

Config	Theoretical Value	Measured Value	Percent Error
R_L2 = R_L2,min	V_Load2 = 8 V	8.20 V	2.5 %
R_L2 = ∞ Ω	V_Load2 = 8.6 V	8.84 V	2.79 %

Next put R_L2 back, but change configuration to find power for a set of resistances

Analysis:
The power going into R_L2 = R_Th is V^2/R = 0.280 W
Calculating the first row is 2.97^2/(0.5*R_Th) = 0.267 W

Config	V_Load 2	P_Load 2
R_L2 = 0.5R_Th	2.97 V	0.267 W
R_L2 = R_Th	8.21 V	1.021 W
R_L2 = 2R_Th	5.89 V	0.264 W

Conclusion:
Seeing how we can vary different resistors, we can easily find the voltage in connection with a Thevenin Equivalent. It becomes very useful when just adding a new element to an existing large circuit.