Thevenin Equivalents

           We were given the engineering problem of finding the smallest equivalent load resistance that can be used in a system of multiple sources and loads. Using Thevenin Equivalent, we wanted to simplify everything into a circuit with only VTh, RTh, and Rload 2. Therefore, using nodal analysis, we calculated Vth = 8.64V and I load 2 = 0.131 A, and found RTh with RTh = VTh/ I load 2 = 66 Ω. With a min. Voltage of 8V across load 2, we used Ohm's Law to find RL2 = 825 Ω.

 

 

              We first built the thevenin equivalent circuit on a breadboard. After measuring values, we treated the RL2 to have infiinite resistance and measured the voltage. The values were as followed:

Config                     Theoretical Value        Measured Value         Percent Error

RL2 = RL2, min          VLoad 2 = 8V                 VLoad 2 = 7.77V            2.9%                

RL2 = ∞Ω                VLoad 2 = 0V                 VLoad 2 = 0V                  0%

 

 

 

       

          Afterwards, we built the original circuit, consisting two power sources and more Rs, and calculated:

Config                     Theoretical Value        Measured Value         Percent Error

RL2 = RL2, min          VLoad 2 = 8V                 VLoad 2 = 8.04V            0.5%              

 

Values were fairly close with both circuits. Percent error may have been the result of using 9.11V for power sources instead of an ideal 8V, and using R's slightly different from nominal values.

Pspice Introduction

          This lab offered an introduction to using the program Pspice in order to create circuits, thus obtaining values of current and voltage electronically.The program is quite user friendly, where elements and wires could easily be placed, rotated, or adjusted as commanded. Values of elements can be changed by clicking on the specified elements, and having to add ground makes it practical to real world appliances. In the end, the circuit could be run as a simulation to obtain values of current and voltages wherever specified, as shown above.

         For the circuit below,we had to find the voltage drop across R1 and the total power being produced by VS. Using V=IR, the voltage drop through R1 was 8.66 Voltage. For VS, taking the difference of the two V values on both sides gives the V drop. Using P= VI, (P = (30)(5.118)), the power for VS was around 153.5V.

Nodal Analysis

 

 In this lab we constructed a circuit designed to be 'reliable,' where damage to one part of the circuit may not destroy the entire circuit, leaving other parts functional. The idea is to have two loads supplied by two different batteries, so that if one load fails,  circuit breakers (not included in this circuit) would separate the damaged circuit and the other battery can continue to support the functional parts of the circuit.

       Given resistor and battery voltage values, the currents can be found using nodal analysis by finding the voltages of the nodes and choosing a ground. The 3 upper resistors act as cable resistance (100, 220, and 220 ohm  Rs from left to right) and the 2 vertical 1000 ohm Rs are the loads. With Bat1(nearer to 100 R) = 12V and Bat2 = 9V, the values for I bat 1and2 and Voltages of the nodes were calculated and measured.

Variable    Theoretical Value     Measured Value         Percent Error 

Ibat1         17.5  mA                 17.52  mA                 6.3%

Ibat2         1.5    mA                 1.62    mA                 8.0%

V2            10.3  V                    10.23  V                    0.7%

V3            8.67  V                    8.70    V                    0.3%

(V2 and V3 being the nodes above first and second 1000 ohm Rs, respectively).

Surprisingly, there were no large percent errors when dealing with a small current such as 1.5 mA. It was to be expected that with a 12V Ibat1 would have lots of current since it has a small 100 ohm R.

Voltage Dividers

An unregulated power supply was tested with three resistors (1 kOhm each) in parallel to the circuit. In this lab, we wanted to maintain 4.75-5.25V (5% variation about 5V) while varying the # of Resistances in parallel. Simplification of calculations was made by adding R's and solving Req, and using the Req and max. and min. of the voltage of the load (Vbus) gave the voltage & resistance of the battery source (Vs and Rs).

Vs = 5.54 V          Rs = 54.9 Ohm

Using V=IR, the calculated min. and max. Ibus can be found, which were 5.25mA and 14.25A respectively. For the lab, the power supply had only intervals of 2V, so 6V was used instead of 5V, and a Resistor box measured around 55.4 for the Rs. Measured values of circuits with different # of resisters (loads) were as followed:

Config     Req (Ohm)     Vbus (V)     Ibus (mA)     Pload (W)

1 Load    1000               5.72              5.81 -min      0.033

2 Load    500                 5.43              10.97            0.060

3 Load    333                 5.18              15.6 -max     0.081

All Vbus were within the 5% variation except for the 5.72V (circuit with 1 load). Naturally, the use of 6V would increase all the values of Vbus. The higher Vs (6V, not 5V) also explains the higher min. and max. Ibus, as more current flows as Vs increases and R stays constant. This lab demonstrates that with less Req, there is more current to flow, and using P=VI, shows that more power is applied to the load as well.



Introduction to Biasing

  

The goal here was to have a 9V Alkaline battery light up two LEDs with the following requirements:
                       LED1 rated 5V & 22.75mA            LED2 rated 2V & 20mA.
 LEDs were placed 'in parallel' to regulated their individual currents, with a resistor 'in series' to each LED to control their voltages. Using Kirchoff's current and voltage law we determined the ideal current, voltage, and resistors necessary, and used resistors closest to the calculated values.
                      Calculated Rs:   R1 = 176ohm     R2 = 350ohm
                      Actual Rs:         R1 = 150ohm     R2 = 360ohm
Using an ammeter and voltmeter we measured the current, voltage, and resistance of the circuit with different configuration as followed:
                      Config                    ILED1       VLED1       ILED2        VLED2      ISupply
                      Both LEDs             14.5mA   6.86V       21.0mA    1.65V      35.4mA
                      No LED2 branch    14.5mA   6.81V       X              X             14.5mA
                      No LED1 branch    X             X             20.8mA    1.65V       20.8mA
The bulbs lit up and did not explode because of too much power or current. There was a 26.4% error between the achieved LED current and the desired value, which was most likely the result of using different R values than calculated.

Simple DC Circuit

The objective of this lab was to approximate the max. distance between a battery and load (the 1000 ohm resistor) that would still allow the circuit’s proper functioning, as a 12V battery would work accurately if the voltage is greater than 11V. Rather than use AWG #30 wires, we used a resistor box to act as the resistance of the cables. The box reached its peak at 87 ohm by the time the battery lowered to 11V, and finding the power used by the load (0.1232 W) and the power lost to the cable (0.011 W), the efficiency (= Pout /(Pout + Plost ) was 92%. Assuming the resistance of an AWG #30 wire to be 0.3451 ohm/m, the max. distance rounded to about 252m. With the cable looping back to the battery, the displacement from the battery to the load would actually be 126m, which is still a long distance. Don’t worry anymore about connecting to your neighbor’s battery supply during days of short circuits!