Thevenin Equivalents Lab

In this experiment we studied a complex circuit which was transformed into a Thevenin equivalent circuit. Given the fact that the following circuit is linear,

we can simplify it to its Thevenin equivalent,

to find values of Resistance in R2, as well as the voltage and current flow through Load 2.

Problem:

We can start analyzing the circuit by obtaining Thevenin Resistance, Rth. In order to do so, the voltage sources must be turned off:

• Step 1A:

Use nodal analysis to obtain Vx or Vopen circuit by opening the circuit and disregarding RL2:

• Step 1B:

Next we use nodal analysis again to get the voltage Vy when short-circuiting the current:

• Step 1C:

If we look at the Thevenin equivalent circuit, we can determine:

1. Smallest permissible RL2 with voltage division technique
2. Short circuit current (Isc) through Ohm's Law (found above already)
3. Open-circuit voltage by inspection

Step 2: Lab equipment 
Stage 1: Build Thevenin equivalent circuit on breadboard

• Variable resistance boxes for Rth and RL2
• Power supply channel V1 set to Vth
• RL2 = RL2min and measure VLoad2
• Replace RL2 by open circuit and measure VLoad2

Thevenin Circuit Equivalent

Stage 2: Build original circuit on breadboard

• Variable reisistance box for RL2 and fixed resistors for Rc1, Rc2, Rc3, RL1
• Power supply channels V1 and V2 to implement Vs1 and Vs2
• RL2 = RL2min and measure VLoad2
• Replace RL2 by open circuit and measure VLoad2

Original circuit with fixed resistors

Step 3A: Measuring components before assembly of circuit

Component	Nominal Value	Measured Value	Power/Current Rating
Rth	66 ohms	65.9+/-0.1 ohms	1 W
RL2min	819.9 ohms	820 +/-0.1 ohms	1 W
Vth	8.64 V	8.66+/-0.1 V	10 mA

Step 4A: Thevenin Equivalent Circuit

RL2 = Infinite resistance (absence of black variable resistance box)

Step 5A: Performing the experiment

Config	Theoretical Value	Measured Value	Percent Error
RL2 = RL2min	Vload2 = 8V	7.77+/-.01	2.88%
RL2 = Infinite R	Vload2 = 8.64V	8.70+/-.01	0.69%

Step 3B: Measure components of original circuit before assembly

Component	Nominal Value	Measured Value	Power/Current Rating
Rc1	100 ohms	98+/-.1 ohms	1/8 W
Rc2	39 ohms	38+/-.1 ohms	1/8 W
Rc3	39 ohms	39+/-.1 ohms	1/8 W
RL1	680 ohms	667 +/-1 ohm	1 W
Vs1	9 V	9.10+/-.01 V	10 mA
Vs2	9 V	9.18+/-.1 V	11 mA

Step 4B: Building Original Circuit on breadboard

Step 5B: Performing the experiment

Config	Theoretical Value	Measured Value	Percent Error
RL2 = RL2min	Vload2 = 8V	6.90+/-.01	13.75%
RL2 = Infinite R	Vload2 = 8.64V	7.94+/-.01	8.10%

Transistor Switching Experiment

For this experiment we studied the behavior of a transistor. A good analogy for a transistor would be a faucet; in this case current is distributed from an emitter, and passes through a base that regulates the flow of current ending up in a collector.

First we set up the following circuit:

 

The switch for this circuit is modeled as a pair of wires that when they interact acts as an ON switch.

When doing this, the LED gets lit lightly.

The second part of this lab is to create a circuit where a finger tip will serve as a switch, there fore conducting current to the base of the transistor.

Two cables make contact with the finger tip skin and allow the flow of current.What the base of the transistor is doing here is amplifying changes in current coming from the skin.

For the final circuit layout, we constructed a circuit similar to the ones above. However, this time we put in a potentiometer that allowed us to regulate the flow of current throughout the transistor. The current from the base and the current going to the collector must be measured using ammeters.

R2 and R4 were removed from the original layout for the purpose of ease of measurement

The following currents were obtained:

Miliamps through A1	Miliamps through A2
0.25	44.3
0.30	47.2
0.35	50
0.40	52.5
0.45	54.5
0.55	58.3
0.60	60.5
0.65	61.7
0.70	63.2
0.75	64.65

If we take the ratio of the current going out of the emitter (A2) with respect to the current going to the base (A1), we could get BETA:

177.2
157.3333
142.8571
131.25
121.1111
106
100.8333
94.92308
90.28571
86.2

The average value for Beta would be 120.8

Graphical Representation of A1 vs A2

MatLab Laboratory

Simple Plot

Two Plots on a Graph

Solving simultaneous equations


 Assignment 1:

Given the circuit with V1=15V, V2=7V, R1=20ohms, R2=5ohms, R3=10ohms, find the current through R3 using FreeMat.

Using Mesh analysis,
At i1, 30i1 -10i2 = 15
At i2, -10i1 + 15i2 = 7

Solving the simultaneous equations we get;

 where i1 = 0.8429 and i2 = 1.0286.
At the upper node we have i1 = i2 + I, so

I = i1 -i2 = -0.1857A





Plotting exponentials and adding sinusoids
Assignment 1:
1. Circuit A has a time constant of 100ms, while circuit B has a time constant of 200ms. Because of the design of the circuit, the ouput is 2e^-t/tau. Plot the two outputs, and describe which circuit will have the lowest out put sooner.

Circuit A (time constant = 100ms) has the lowest output sooner.


2. The circuit was redesigned, so the output is now 2(1-e^-t/tau). Plot the output of both circuits.

Assignment 2:
1. Both theoretically and using MATHLAB, determine what the output would be when adding the following two sinusoids: 3sin(2t+10deg) and 5cos(2t-30deg). Does the MATHLAB output resemble your theoretical outpu - plot your theoretical output in MATHLAB, and compare.

Two sinusoids and the superposition of both (red)

Theoretical output of addition of both sinusoids

Theoretical graph of the superposition of both sinusoids

The theoretical and MATHLAB generated sinusoid superpositions differ by some phase angle.

2.  You now need to repeat this for a large number of frequencies. Write a script to implement this and demonstrate the output at 10Hz.

Superposition of the two sinusoids at 10Hz

Voltage Dividers

In this experiment we had to study how does the concept of voltage dividers work in a circuit. We will have three resistors connected in parallel which will serve as loads. Such circuit can be redrawn using Req (three parallel resistors):

Problem Statement: If R1=R2=R3=1000ohm and 1-3 resistors may be on at any time, determine Vs and Rs that will guarantee bus voltage to satisfy: 5.75<Vbus<6.25.

Step 1: Design calculations

Req, max (1 resistor) = 1000ohms
Req, min (3 resistors in parallel) ~ 333.333ohms

Vbus, max = 6.25V - Req, max
Vbus, min = 5.75V - Req, min

Vbus equations:

Vs = 6.53V               Rs = 45.45ohms

Ibus, max:
I = 6.25/1000 = 6.25mA

Ibus, min:
I = 5.75/333.333 = 17.25

Step 2: Laboratory equipment

•  Each load will be a fixed resistor
• Rs will be a variable resistance box
• Vs will be a unregulated power supply
• The switches will be short cables

Step 3: Measure the components

Color Code				Nominal Value (ohm)	Measured Value (ohm)	Wattage (W)
Br	Bl	Bl	Br	1000	978+/-001	1/8
Br	Bl	Bl	Br	1000	976+/-001	1/8
Br	Bl	Bl	Br	1000	979+/-001	1/8
Resistor Box				45.45 ohms	45+/-0.2 ohms	10
Power Supply				6.53V	6.50+/-0.01V	x

Is the maximum bus current within the capability of the power supply? Yes
Is the maximum current within the power capability of the resistor box? Yes

Step 4: Building circuit on the bread board

Mounted resistor in parallel (Req)

Rs

Step 5: Data collection

Config	Req (ohm)	Vbus (V)	Ibus (mA)	Pload (W)
1 Load	978+/-1	6.08+/-.05	6.04+/-.01	0.0378+/-.00044
2 Load	488.7+/-0.80	5.65+/-.01	11.27+/-.02	0.0653+/-.00020
3 Load	325.9+/-0.69	5.33+/-.02	15.72+/-.01	0.0871+/-.00050

Step 6: Calculations

a) Calculate the power delivered to the load for the three configurations and record the values in the table. Show the calculation for the 2 loads.

Pload2 = V^2/R = (5.65+/-.01)^2/488.7+/-.80 = 0.0653+/-.00020W


b) Calculate actual percentage in load voltage variation achieved (how much % up and how much % down, assuming 6V is nominal value). Why is this different from the +/-5% design goal?

 % up =100*|6 - 6.08+/-.05|/6 = 1.3+/-.83%

% down = 100*|6 - 5.33+/-.02|/6 = 11.2+/-.33%

The difference between these % in load voltage and the +/-5% design goal is probably due to some possible flaws in the circuit bread board and/or the fixed resistors not being exactly 1000 ohms (~976 ohms).


c) Given our source parameters (Vs and Rs the same), what would be the new load voltage variation if we added a fourth 1k ohm load?

With 4 - 1000 ohm resistors in parallel, Req will be 250, so;
Vbus = (Req)(Vs)/(Req + Rs) = (250)(6.53)/(250+45.45 = 5.53V


d) What would be the new source parameters required for the 3load case if we wanted to reduce the voltage load variation to +/-1%?

with 1% variation, we have 5.95 < Vbus < 6.05
so we use the Vbus formulas;

6.05 = 1000Vs/(1000+Rs)           and         5.95 = 333.333(Vs)/(333.333+Rs)

Solving the system of equations yields: Vs = 6.10V and Rs = 8.48 ohms.

Introduction To Biasing

For this lab we were given two LED lights, and each one had a specific voltage and current requirements. We set up the circuit; the two LEDs are connected in parallel across the voltage supply. We need to find the voltage and current necessary to be distributed between the two LEDs and regulate the flow with the use of resistors.

 Failure to deliver the correct voltage and current will result in a burned out LED, so that's why the circuit is connected in parallel. Our goal then is to make both LEDs light and take measurements of voltage and current.

The requirements for the LEDs are:
LED1: 5V and 22.75mA
LED2: 2V and 20mA

Circuit Diagram

Step 1: Laboratory equipment

• A regular classroom power supply will be set at 9V
• Although it was suggested to use a breadboard to build the circuit, we built the circuit using cables.
• The resistance on the LEDs (equivalent resistances = R=V/I) were:
  R LED1 = 219.78ohms
  R LED2 = 100ohms
• We have to find out the value of R1 and R2

 Step 2: Calculations

KCL is applied to both nodes on the picture above to find:
Ir1 - I LED1 = 0,  Ir1 = 22.7mA
Ir2 - I LED2 = 0,  Ir2 = 20mA

KVL is applied to both loops to get
Vr1 + 5 - 9 = 4V
Vr2 +2 -9 = 7V

We can apply Ohm's Law to find the needed resistance:
R1 = Vr1/Ir1 = 4/0.0227 = 175.82ohms
R2 = Vr2/Ir2 = 7/0.0200 = 350ohms

We also need to find the maximum power that can be consumed in order to avoid compromising lab equipment
Pr1 = Vr1*Ir1 = 0.09098W
Pr2 = Vr2*Ir2 = 0.14W

Next we had to pick R1 and R2 from restricted-value resistors in the classroom. We picked two fixed resistors; R1 = 220ohms and R2 = 470ohms

Step 3: Measurements

Color Code				Nominal (ohms)	Measured (ohms)	Wattage
Red	Red	Black	Black	270	217	1/8
Yellow	Purple	Black	Black	470	465	1/8

Config	I LED1 (mA)	V LED1 (V)	I LED2 (mA)	V LED2 (V)	I Supply (mA)
1	13.37+/-.01	6.27+/-.01	15.05+/-.01	2.14+/-.01	28.7+/-.1
2	13.40+/-.01	2.96+/-.01	x	x	13.36+/-.01
3	x	x	15.09+/-.01	2.14+/-.01	15.09+/-.01

Step 4: Calculations

A). Capacity of 9V alkaline battery is ~0.6Ahr; however, voltage drops significantly as we get to the 'end' of the battery life. Assuming 'useful' life of battery is 0.2Ahr. With both LEDs, how long can the circuit operate before battery voltage goes down too low?

when we have both LEDs;
0.2Ahr = (13.37+/-.01 + 15.05+/-.01)mA

hr = (28.42+/-.014mA)/200mA = 0.1421hrs or 8.53 minutes.


B). What is the percent error between achieved LED current and desired value for both LEDs in the same circuit?

LED1 = 41.23%
LED2 = 75.25%

The cause for high percent errors might have been because the ratings were for individual LED situations. Usage of the LEDs might have affected the current readings as well.


C). From data, determine circuit efficiency when both LEDs are in the circuit.
P LED1 = VI = (6.27+/-.01V)(13.37+/-.01mA) = 0.0838+/-.00015W
P LED2 = VI = (2.14+/-.01V)(15.05+/-.01mA) = 0.0322 +/-.00015W
Ptotal out = 0.116 +/-.00021W

Pin = VI = (9.26+/-.02V)(28.7+/-.1mA) = 0.266+/-.0011W

Efficiency, n = Pout/Pin = 43.6+/-.2%

C). If we repeated the design with 6V battery (changing biasing R1 and R2 but not changing LED parameters), would efficiency go up, down or same?
 With 6V,
       Vr1 = 1V                    Vr2 = 4V
       R1 = 43.96ohm          R2 = 200ohm
       Pr1 = 0.02275W        Pr2 = 0.08W                

n = Pout/Pin = (0.02275+0.08)/(0.1722) = 59.67%

If we had a 5V battery, Vr1 = 5volts battery - 5volts LED1 = 0. As we increase the voltage (i.e. 5.1V, 5.5V, the efficiency gets from 43% - 51.3%.

The most efficient battery would be around 5.9V - 6V.

Introduction to DC Circuits

The goal in this experiment is to measure the resistance of a hypothetical couple of cables that are represented as a variable resistor box. A fixed resistor will act as a resistor load in the circuit. The resistor box will be adjusted to get a value for the cable resistance and then it will be compared to the following assumptions:

P = 0.144W at 12V

Proper operation at V>11V

Voltage source = 12V with capacity of 0.8A-hr

In this experiment we set up a circuit consisting of a power supply set in DC mode. We were provided with a composition fixed resistor of 1000 ohms which represented the load resistance. The resistor was then connected in series to a resistor box (cable resistance).

Circuit Diagram

Circuit Layout

Modeled cable resistance

 The theoretical value of the load resistor according to the assumption values is:

R = V^2 / P = 12^2 / 0.144 = 1000 ohms

Once we had the circuit set up, we placed an ammerter in series with the cable resistance to measure the current. We also put a voltmeter across the resistance load to measure the voltage through the load resistor.

The power rating for the resistor box was: 1 Watt with 1% accuracy.
For the power supply we got  a maximum voltage supply of 12V at a current of 2A.

The colors of the fixed blue resistor were: Brown - Black - Red which was the color code for 10*10^2 = 1000 ohms. We then measured the actual resistance of the fixed resistor using a multimeter to get 0.988+/-0.001 Kohms. The last color code was Brown which was a 1% tolerance (not within the regular 5% tolerance). The wattage was found to be 1/8 W.

Once the circuit layout was approved by the professor, we turned on the power supply.

Using a voltmeter we found the voltage across the load (fixed resistor) to be:
Vload = 10.99 +/- 0.01V

The current running through the circuit was:
Ibatt = 11.22+/-0.01A

The total cable resistance (resistor box) was modified to be 105 ohms

Using the current measured we now can estimate the amount of time it takes for the battery to supply the load before being discharged.

0.8A-hr = (11.22+/-0.01A) t
t =  0.0713 +/- .000064hrs or about 4.28 minutes.

To find the power to the load, that is the power out we take the voltage through the load and the load resistance;

P out= V^2 / R = (10.99+/-0.01V)^2 / (988+/-1ohms) =  0.1223+/-0.00020 Watts

The power to the cable (power lost) is obtained by multiplying the current out of the power supply and the resistance from the resistor box;

Plost = (I^2)*(Rtotcable) = ( 11.22+/-0.01A)^2 * (105ohm) =  13220+/-17 Watts

The efficiency is found by computing:

n = [Pout/(Pout + Plost)] = 0.1223+/-0.00020 / (0.1223+/-0.00020 + 13220+/-17)     =0.000925+/-0.0000019%

Given that the power capability of the resistor box is  1%, we can say that we are not exceeding the power capability of the resistor box.

If the resistance of AWG #30 wire is 0.3451ohm/m, determine the maximum distance between the battery and the load.

(988+/-1 ohms)/(0.3451ohm/m) =  2863+/-2.9m.