PSPICE

Statement of purpose
Learn the pspice program to design and simulate circuits.
 
Procedure
Part 1. Build and simulate a simple circuit.

Part 2. Find the Thevenin equivalent of a circuit.

We set up a current supply with no current, and we steady increase this current and model it with a graph.

Data

The data obtained shows that the Thevenin voltage is 10V and the slope or Thevenin resistance is 3.33Ohms. Which means the Norton current is 3A.

We then replace the current supply with a variable resistor and we obtain this graph with the results. The vertex of this graph is where maximum power transfer occurs, exactly the value we obtained for the Thevenin resistance.

Conclusion 
Pspice provides a good tool to model circuits that would otherwise required physical labor and plenty of mathematical calculations.

Maximum Power Transfer

Statement of purpose
To affirm that the maximum power transfer to a load resistance occurs when the load resistance is the same as the Thevenin resistance of the circuit.

Part A
 
Procedure

Measure the resistance of the fixed resistor.

Resistance = 5.61Ohms

Data 1

Measured values

Measured V0 (V)	Measured Rx (Kohms)	Calculated P0 (W)
2.80+/-.01	14.64+/-.01	0.000536+/-.0000028
2.68+/-.02	13.73+/-.01	0.000523+/-.0000055
2.51+/-.01	12.55+/-.02	0.000502+/-.0000029
2.43+/-.01	12.08+/-.01	0.000488+/-.0000028
2.32+/-.02	11.50+/-.01	0.000468+/-.0000058
2.17+/-.01	10.77+/-.01	0.000437+/-.0000029
1.98+/-.01	10.00+/-.01	0.000392+/-.0000028
1.80+/-.02	9.34+/-.01	0.000347+/-.0000055
1.63+/-.02	8.78+/-.02	0.000303+/-.0000053
1.41+/-.01	8.17+/-.02	0.000244+/-.0000025
1.28+/-.01	7.82+/-.01	0.000210+/-.0000023
0.84+/-.01	6.89+/-.01	0.000103+/-0000017
.02+/-.01	5.65+/-.02	0.00000007+/-.00000
.00+/-.001	5.62+/-.02	~0

Theretical values

using voltage divider,

Theoretical V0 (V)	Theoretical Power (W)
3.25	0.000721
3.20	0.000746
3.11	0.000771
3.07	0.000780
3.03	0.000798
2.96	0.000814
2.88	0.000829
2.81	0.000845
2.75	0.000861
2.67	0.000873
2.62	0.000878
2.48	0.000893
2.26	0.000904
2.25	0.000901

Part B

Procedure

Data 2

	Nominal (kohms)	Measured (kohms)
R1	1	0.966
R2	10	9.85
R3	10	9.86
R4	1	0.976
R5	1	0.979

	Nominal (V)	Measured (V)
Voltage source 1	4.5	4.54+/-.01
Voltage source 2	9	9.05+/-.01

LoggerPro graph

 

Conclusion
The maximum amount of power transfer occurs when the resistance of the load is the same as the Thevenin equivalent of the the network.

Thevenin Equivalent

Statement of Purpose

To represent a circuit network by a resistor and a voltage source in series.

Procedure

Using mesh currents we find that the current through resistor RL1 (680 Ohms) is 12.71mA which means the Thevenin voltage is 8.64V. 

We also solve for resistance by turning off all independent source voltages which reduces the circuit to a simple parallel and series resistance network. We find the Thevenin resistance to be 65.95 Ohms.

Data

We adjust values for our variable resistor and measure voltages across that resistor for values.

We then compare those voltage values to the values obtained by measuring the Thevenin voltage in the original circuit.

Conclusion
The values obtained from both circuits are reasonably close, any difference may be caused by the tolerance of the resistors.

Transistor Switching

Statement of purpose
To control current flow with a transistor switch.

Procedure

Using a switch to turn on a LED with a transistor.

Finger-tip transistor switching

Data
Too see a precise understanding of how a transistor works. We adjust the resistance of the potentiometer to control the voltage that is sent to the base.

This is the current emitted by the transistor.

Conclusion

Controlling electricity by a transistor is "expensive", so it is noteworthy that to know that a transistor takes at least .25A at the base just to switch the flow of electricity.

FreeMat

Statement of Purpose
Learn the basic functions with graphing functions and solving system of linear equations by matrix operations.

Procedure

Graph two sine waves with different phase.

Set up a matrix and solve a circuit by using inverse matrix and Ohm's Law

 Graph two sine waves and the superposition of both waves.

Change the angular frequency of the function.

 

 This the resulting graph. Notice the change in the x-axis.

Conclusion

Freemat is very useful for calculating arithmetic problems, solving linear equations, basic matrices operations, and graphing functions.

Voltage DIviders

Objective: To learn the function of an unregulated power supply and the Thevenin Equivalent circuit.

Procedure:

Calculate the maximum and minimum values possible for Req.

 Ibus,max = 15mA

Ibus, min = 6mA

Config         REQ                 Vbus                Ibus                 Pload

1                  1kOhm             6.33V              5.88mA           37.22mW       

2                  800Ohm           6.17V              11.24mA         69.35mW      

3                  330Ohm           6.05V              15.21mA         68.87mW       

Questions

•  Power for each configuration
• P1 = (6.33*0.00588) = 0.0372 W
• P2 = (6.17*0.01124) = 0.0694 W
• P3 = (6.05*0.01521) = 0.0689 W

•  Load voltage variation: 1-6.33/6 = +5.5% 
•  Load voltage variation: 1-6.05/6 = +0.8%
• The difference is caused by the unregulated power supply.
• The voltage would not change because it is connected in parallel.

Introduction to Biasing

Objective: To determine the voltage and current necessary for electrical loads by using resistors to solve the problem of biasing.

Equipment:

• resistors (Ohms) : 100, 150, 220, 330, 470 //one each
• LED: 5v, 2v                                                         //one each
• 9v power supply                                              //one

Procedure:
Calculate the equivalent resistance of each LED using Ohm's Law with its rated voltage and current.

Given that the the current flowing through the LED is 22.75mA and that the rated voltage is 5v, using Ohm's Law gives,

V=IRled1

5=(.00275)Rled1

Rled1 = 219.78 Ohms

 The same calculation holds for led2

Rled2 = 100.00 Ohms

Given that this is a parallel circuit, the voltage across each branch is 9V. Calculate and determine the resistance needed to drop the voltage for each LED.

Look and measure some resistors needed for the circuit

Color Code          Nominal Value          Measured Value          Wattage

2200                      220 Ohm                    213 Ohm                         1/8 W

3600                      360 Ohm                    352 Ohm                         1/8 W

4700                      470 Ohm                    459 Ohm                         1/8 W

Build the circuit and record data for the following configuration

Configuration 1: Both LED in the circuit

Configuration 2: Remove LED2 from the circuit

Configuration 3: Remove LED3 from the circuit

Configuration         I-LED1          V-LED1          I-LED2          V-LED2          I-supply

1                                 10mA             5.9V               14mA            2.11mA          24mA

2                                 5.9mA            5.9V                   X                     X                 10mA

3                                      X                    X                  14mA            2.11mA         14mA

Bonus 

In this case, both LEDs need to be in the circuit otherwise there will be too much current following through a single LED.

Questions

• a.   If this Fig. 2 circuit used a 9V battery supplying 0.2A-hr, it lasts 0.2/(0.02275+0.02) = 4.68 hr 
• b.   The percent error for the LED1 current (0.014-0.02275)/0.02275=38.5%
• b.   The percent error for the LED2 current (0.010-0.020)/0.020*100 = 50.0%
• c.    Determine circuit efficiency 
• Pin=IV=9*0.04275=0.38475 W
• Pout= 5.9*0.010 + 2.11*0.024=0.059 + 0.051 = 0.110 W
• Efficiency = 0.11/0.38475 * 100 = 27.48% 

• d.   Efficiency would increase because less power would be used since V^2/R. Theoretically, a 5V power supply would be best because only one LED will need a resistor.

Introduction to DC Circuits

When we discuss electronic circuits, most often we refer to either direct current (DC) or alternating current (AC). The properties of a circuit can be expressed by Ohm's law.

V = IR

The most obvious of concerns when engineering simple circuits come from selecting the amount of current or voltage for the required source. Simple circuit diagrams explains perfectly well how to build circuits and describe the voltage, current, and resistance of each element. 

The equipment we wish to model has a load of .144W when supplied 12V. The battery will have a voltage of 12V with a capacity of 0.8Ahr.

 Obviously, the equipment that we will be powering will have a resistance as well, which we emulate by selecting a resistor with the same resistance of our equipment. Using Ohm's Law we find that the resistance of the load is 1000 Ohms.

A simple circuit diagram may very well be too "simple" by not explicitly stating the material, diameter, or even length of the wire. With considerable distances wires have considerable resistances. Thus, resistance should be as much of a concern as voltage and current when engineering circuits.

R = pL/A

Since we do not possess such long wires, we can emulate one by using a resistance box. A simple calculation using the expression above could give us an idea of what kind of current will run through the wire created by the voltage source.

We find the measurements to be:

Voltage of load: 11V

Current of battery: 12.4mA

Resistance of cable: 89 Ohms

With the specifications of our equipment, the time it would take for our battery to discharge is 64.5 hrs.

With the Joules Heating Law,

P = V^2/R = I^2R

we can find the power lost by the cable and consumed by the load which we calculated to be:

lost by the cable =  0.0137W

consumed by the load = 0.153W

efficiency of the circuit = 91.78%

Our resistance box has a power rating of 1W which can perfectly handle the power lost by the cable. Had we be handling AWG #30 wire which has a resistance of 0.3451 Ohms/meter, we could have a wire that is about 260 meters.

In the case of the robosub, using AWG #28 wire (resistance of .0764 Ohm/foot) and nominal rating is 0 -0.4 V for low and 2.6-5V for high. So, if we leave 2.6V for the load and 2.4V for the resistance of the cable, then we have sufficient voltage for our circuit. The current that is a 60 foot tether is 20mA. We can use Ohm's law to solve the problem for the maximum length that this wire can be to keep in proper communication with the sub.

2.4V = (20mA)(0.0764 Ohm/foot)(# of feet)

# of feet = 5/[(.02)(.0764)]

# of feet = 1570.68 ft

When sending 48V at 10A through a 60 ft tether the calculation for a proper guage for the wire required can also be solved using Ohm's laws,

48V = (10A)(60ft)(Resistance)

Resistance = 48/[(10)(60)]

Resistance = 0.08

We look this number up on a chart and find that the guage most appropriate is between #28 and #30 guage wire.