I just need help to go the (( PROPAGATION OF ERROR )) for part 1 section (B) AND Part 2 for section (B). I HIGHLIGHT THEM WITH ( YELLOW COLOR )
majeed1990
ENGR206Lab5FORWEB.docLab Report 7/22/2014
Experiment 5: Thévenin Equivalent Circuits
Experiment 5: Thévenin Equivalent Circuits
Objective:
In this lab we will explore the theory of Thévenin Equivalent Circuits. We will take many voltage and current measurements to verify the theory of Thévenin Equivalent Circuits. We will use several different circuit/source configurations to gather our measurements. Additionally, we will use the theory of Thévenin circuits to predict the maximum power that can be drawn from a circuit with a load resistor.
Apparatus:
· E3631A Triple Output Power Supply --Figure 1
· Agilent 34401A DMM (Digital Multimeter) --Figure 2
· Resistors (1kΩ-10kΩ)
· Wiring Package (Alligator Clips, Test leads….)
· Breadboard
· Decade Box
Procedures, Data, Results, and Analysis:
1) Thévenin equivalent, one source. The purpose of this part is to find the Thévenin equivalent of a circuit with one source by measuring find Voc and Isc.
a) Construct the circuit of Figure 7. Choose R1 = R2 = R3 = 10kΩ and set the value of Vs = 6v using the 10V supply.
b) Measure the open-circuit voltage, Voc, and the short-circuit current, Isc, and thereby find Vth and Rth. Note: Do not attempt to measure Rth directly with the ohmmeter! Explain in your report why this is a bad idea.
· Voc = 2.991 v Isc = 0.2016 mA
Vth = Voc = 2.991 v
Rth = Voc/Isc = 2.991/0.2016 = 14.84 kΩ
Trying to measure Rth directly with an ohmmeter is a bad idea because it will be measuring the resistance of the power supply (Vs) which will not give an accurate measurement. Also, if the supply is disconnected for the purpose of measuring Rth, this will still give a false reading because it will simply read the resistance of R2 and R3 in series, in this case.
c) Using PSpice, create a simulation of the circuit. Attach a current source (IDC on PSpice’s list of sources) to the terminals of the circuit, in a manner similar to what we did in Figure 4, and do a DC sweep of the current from at least -3 mA to 3 mA. Consult the tutorial for instructions on doing such a sweep. Plot the I-V response, as we did in Figure 2b, and include it in your laboratory report. From this plot, read off values of Voc and Isc, and compare these to your theoretical calculations and the experimental measurements.
|
|
Voc (v) |
Isc (mA) |
Rth (kΩ) |
Match? |
|
Theoretical calculations: |
3 |
0.2 |
15 |
Yes |
|
Plot values: |
3 |
0.2 |
14.84 |
Yes |
|
Experimental measurements: |
2.991 |
0.216 |
14.84 |
Yes |
d) Now we’ll check the values of Vth and Ith by injecting a current into the circuit and measuring the voltage, as in Figure 2a. We’ll use the +25V voltage source and measure the voltage it takes to inject a given amount of current into the circuit, as we did in Experiment #2. So, leaving in place the 10V supply in place for Vs, connect the
+25V supply to the terminals of the circuit read the current flowing into the circuit from the power supply’s display as you adjust the voltage. Measure the voltage it takes to inject 2 ma of current into the circuit. Repeat with 0 ma. Do these points lie along the I-V curve you plotted in part 1b?
(Figure 2a)
|
Injected current |
Voltage Required |
|
1.537 mA |
25.75 v |
|
0 A |
3.05 v |
We were not able to achieve 2mA of injected current. The highest amount that we got was 1.537 mA.
2) Thévenin equivalent, two sources. The purpose of this part is to find the Thévenin equivalent of a circuit with two voltage sources and verify that you have the right values of Vth and Rth by constructing the equivalent circuit with those values.
a) Construct the circuit of Figure 8 using the values V1 = -5v (note the minus sign!), V2
= 10v, R1 = 5.1kΩ, R2 = 2.2kΩ, R3 = 3.3kΩ, R4 = 6.8kΩ.
b) Measure the open-circuit voltage, Voc, and the short-circuit current, Isc, the same way you did in part 1b and thereby find Vth and Rth. Plot a theoretical I-V response curve.
· Voc = 5.480 v Isc = 1.151 mA
Vth = Voc = 5.480 v
Rth = Voc/Isc = 4.761 kΩ
c)
Connect a 1kΩ load resistor across the output terminals of the circuit and measure the resulting voltage across the terminals. Also measure the current through the load resistor. Repeat the measurement of voltage and current using a 10kΩ load resistor. Plot these two points on your I-V graph in part a). Do they fit the characteristic? Discuss.
· The lines of the I-V Response for the load resistors fit the characteristics quiet well. The reciprocal of their slope corresponds to Rth, which is the value of the load resistors.
d) Construct the Thévenin equivalent of the circuit of Figure 8. You can leave the circuit of Figure 8 in place if you like (for reference) and create the second circuit another part of your circuit board using the 6V supply for Vth and the decade resistor box for Rth. Find Voc and Isc for this circuit. Do they agree with those for the original circuit?
· Vth = 5.480 v Rth = 4.761 kΩ
Voc = 5.483 v Isc = 1.151 mA
Yes, these values agree with the original circuit. Voc is the same as Vth, and Isc is the same value as measured before in the actual circuit.
e) Repeat step 2c for the Thévenin equivalent circuit. How closely do the resulting I-V values correspond to those of the original circuit?
|
Load Resistor Rl (kΩ) |
Voc (v) |
Isc (mA) |
|
1 |
0.947 |
0.951 |
|
10 |
3.699 |
0.373 |
The resulting Voc and Isc are very close to the original values measured for the real circuit of figure 8 with the load resistors.
Section 3:
Thévenin equivalent, voltage and current sources: The purpose of this part is to find the Thévenin equivalent of a circuit which contains both voltage and current sources.
a. Construct the circuit of Figure 3a using the values = −10V, = 1kΩ, = 2.2kΩ and
=1kΩ. Configure the +6V source as the current source, as you have done in previous laboratory experiments.
· We configured the circuit in figure 3a, and set our power supply to the prescribed settings. (Figure 3a is pictured below.)
b. Measure the open-circuit voltage, , and the short-circuit current, , the same way you did in part 1b and thereby find and . Plot the theoretical I-V response curve.
·
c. Now, hook up the +25V source as a current source to the output terminals of your circuit, as shown by the red source in Figure 3b. Plot the voltage across the terminals for current values 0, 2, 4, 6, 8 and 10 mA on the graph you did in part b). Do the results agree with those of part b)?
Part C: Short Circuit Current and Open Circuit Voltage.
|
(mA) |
(V) |
(kΩ) |
|
0 |
3.02 |
----------N/A---------- |
|
2 |
2.00 |
1.0 |
|
4 |
4.00 |
1.0 |
|
6 |
6.00 |
1.0 |
|
8 |
8.00 |
1.0 |
|
10 |
10.00 |
1.0 |
Our results--for the value of --were in close agreement with part b.
( )
Section 4:
Maximum power: In this part of the laboratory, we will use the Thévenin equivalent to predict the value of a load resistor that draws the maximum power from a circuit, and then verify this via simulation and experiment.
a. Consider the circuit of Figure 9 with =6.0V, = 1kΩ, = 3.3kΩ, =2.2kΩ and = 1kΩ. By any method you choose (i.e. theoretical analysis or PSpice simulation), find and, and hence, the value of a load resistor that will extract maximum power from the circuit.
· By theoretical analysis (use of voltage divider formula) we determined the values of , and .
b. Construct the circuit on your breadboard. Measure the open-circuit voltage, , and the short-circuit current, , and thereby find VTH and RTH. Do the values agree with those in part a?
· We constructed the circuit of figure 9 (Pictured below)
· =1.885mA
c. Attach the decade resistor box across the terminals of your circuit to act as a load resistor,
. Measure the voltage across the load (or the current through the load, if you wish) as you vary the resistance in 100Ω increments over a range of about 50% of to 200% of
. Plot the power consumed by the experimental circuit at each value of . From the graph, find the value of that corresponds to the maximum power point.
Values of , and respective Power calculations:
|
R_L |
V_L |
P_L |
|
0.75 |
0.932 |
1.158165333 |
|
0.85 |
1.01 |
1.200117647 |
|
0.95 |
1.081 |
1.230064211 |
|
1.05 |
1.148 |
1.255146667 |
|
1.15 |
1.208 |
1.268925217 |
|
1.25 |
1.265 |
1.28018 |
|
1.35 |
1.357 |
1.364036296 |
|
1.45 |
1.365 |
1.284982759 |
|
1.55 |
1.411 |
1.284465161 |
|
1.65 |
1.453 |
1.279520606 |
|
1.75 |
1.493 |
1.273742286 |
|
1.85 |
1.53 |
1.265351351 |
|
1.95 |
1.565 |
1.256012821 |
|
2.05 |
1.598 |
1.245660488 |
|
2.15 |
1.629 |
1.234251628 |
|
2.25 |
1.659 |
1.223236 |
|
2.35 |
1.687 |
1.211050638 |
|
2.45 |
1.713 |
1.197701633 |
|
2.55 |
1.738 |
1.184566275 |
|
2.65 |
1.762 |
1.171563774 |
|
2.75 |
1.785 |
1.158627273 |
|
2.85 |
1.801 |
1.138105614 |
Plot of Values of and respective powers:
d. Using PSpice, create a simulation of the circuit with a load resistor, RL, across the terminals with a value ‘{Rload}’, as shown in Figure 10 (note the squiggly brackets).
· (Pictured Below)
e. Do a parametric sweep of RL from about100Ω to10kΩ in100Ω increments. Consult the tutorial for instructions on doing a parametric sweep. Now plot the power consumed by RL. You can do this as follows: After you have run the simulation, in the probe window: T
Expression’ box at the bottom: ‘I(RL)*I(RL)*Rload’. That plots the power ( 2 P IR = )! From the graph, determine the value of RL that corresponds to the maximum power consumed by the load. Does this value agree with the theoretically predicted and experimentally measured values? Explain reasons for possible discrepancies.
From the plot the maximum power occurred when R_load was 1.45kohms—nearly R_TH. This is in close agreement with our theoretical and experimental values.
Conclusion:
In this lab we did extensive work in studying PSpice simulations. We made voltage and current measurements of complex circuits to convert them into their simpler equivalents. We found equivalent circuits for one, two, and three source configurations. Additionally we used experimentation combined with PSpice simulation to learn how to draw the maximum power with a load resistor.
Figure 1
Figure 2
I (mA)
V (v)
-0.15
-0.2
-0.25
y = 0.0674x - 0.2016
-0.05
-0.1
3.5
3
2.5
2
1.5
1
0.5
0
0
I-V Response
I (mA)
I (mA)
V (v)
-1
-1.2
-1.4
y = 0.21x - 1.151
-0.6
-0.8
6
5
4
3
2
1
0.2
0
-0.2 0
-0.4
I-V Response
V (v)
-1.4
-1.151
-1.2
-0.953
-1
-0.6
-0.8
-0.375
-0.4
-0.2
6
5
2
1
0
0
0
4
3
0
0
I-V Response
Load Resistor Rl (kΩ) 1
10
Voc (v) 0.9376
3.692
Isc (mA) 0.953
0.375
Rth (kΩ) 0.9838
9.845
Voltage (V)
-1
-1.5
-2
y = 1.0145x - 1.895
Current (mA)
-0.5
2
Section 3 Part b i-V Response
0.5
0
0 0.5 1 1.5
3
2.5
2
1.5
R_Load (kOhms)
1
0.5
0
1.2
1.15
1.1
1.25
Power (mW)
1.4
1.35
1.3
Power vs. R_Load Value
