Thevenin Equivalent

Introduction
           If we have a power system with multiple sources and loads, we sometimes get into a really complex circuit. In order to be able to easily calculate a complicated circuit, thevenin equivalents is used. This experiment serves the purpose of be able to make a complicated circuit and do some calculation on the thevenin equivalent.

Experimental

We first use a nodal analysis to calculate the open circuit voltage using the information given:
R_C1 = 100 ohm
R_C2 = R_C3 = 39ohm
R_L1 = 680ohm
V_S1 = V_S2 = 9V
V_load2 min = 8V

calculation to obtain V_x

Based on the calculation using nodal analysis, V_x is calculated to be 8.644V


Once V_x is obtained, another calculation is done for finding V_y:

calculation to obtain V_y

V_y of 5.112V is obtained

Once the V_y is obtained, the smallest permissible for R_L2 and short circuit current is calculated as follow:

Calculation for R_L2 and I

I = 9.77 mA

R_L2 = 819ohm

Once the calculation part is done, each components were measured and the circuit for the thevenin equivalent was built

Component	Nominal Value	Measured Value	Power or Current Rating
R_TH	65.9ohm	66.2 ± 0.1 ohm	0.3W
R_L2	819 ohm	819 ± 1 ohm	0.3W
V_TH	8.644V	8.66 ± 0.02 V	2A

The data taken is tabulated below:

Config	Theoretical	Measured	%error
R_L2 = R_L2 min	V_L2 = 8V	7.99 ± 0.01 V	0.125
R_L2 = infinity	V_L2 = 8.64V	8.66 ± 0.02V	0.231

Once the thevenin equivalent circuit is done, the actual complicated circuit will be built after the component was measured:

Component	Nominal Value	Measured Value	Power or Current Rating
RC1	100Ω	98 ± 1Ω	0.25W
RC2	39 Ω	38 ± 0.5 Ω	0.25W
RC3	39 Ω	38 ± 0.5 Ω	0.25W
RL1	680 Ω	680 ± 1 Ω	0.25W
VS1	9V	9.12 V	2A
VS2	9V	9.21V	2A

The breadboard for the circuit

Other setup like the resistor box that can't be put into the breadboard

The result was taken and tabulated as follow:

Config	Theoretical Value	Measured Value	Percent Error
RL2 = RL2, min	Vload2 = 8V	8.15 V	1.875
RL2 = ∞	Vload2 = 8.644V	8.81 V	1.920

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Once the data is obtained, we want to find maximum power through experiment. Theoretically, maximum power is obtained when R_L2 is equal to the thevenin resistance. In this case, we will try to see it by varying resistance of R_L2 to be 0.5 R_TH, R_TH , and 2 R_TH. The calculation for power is shown below:

Power = V^2 / R

The data is tabulated as follow:

Configuration	V­Load 2	PLoad 2
RL2 = 0.5RTH = 33.4 ± 0.1 Ω	2.93 ± 0.01 V	257 ± 2.52 mW
RL2 = R­TH = 66.4 ± 0.1 Ω	4.39 ± 0.01 V	290 ± 1.76 mW
RL2 = 2RTH = 131.9 ± 0.1 Ω	5.87 ± 0.01 V	261 ± 1.09 mW

Conclusion
            As we seen on the table above, the maximum power of 290 mW is obtained when the load resistance is equal to the thevenin resistance. The percent error of the actual circuit is larger than the thevenin circuit which can be expected. The reason for this is in the actual circuit has more component on it which will build up error from the discrepancy in the components value. This also contributes to the uncertainty in the power calculation.

Transistors Lab

Introduction
         In this lab, we try to explore the characteristic of a 2N3904 transistor. In this experiment, we will analyze the beta value of the transistor. We will use a variable resistor box in order to change the current at the base. We should see a linear relationship of the current in the base and emitter until it reach the saturation level. When the transistor is saturated, there will be no more gain in the emitter even though the current of the base keep increasing.

Experimental


The circuit that will be built is going to follow the schematic below:

Schematic of the circuit but change the pot to a variable resistor

Building the circuit is as follow:

The setup of the resistor and voltage source connected in series

The setup of the breadboard for this experiment

 The data taken is as follow:

Base Current (mA)	Emitter Current (mA)	R(kΩ)
0.247	38.5	15.05
0.240	38.4	15.5
0.308	40.9	12
0.284	40.1	13
0.257	38.5	14
0.227	37.3	16
0.330	41.3	11
0.369	42.8	10
0.219	37.4	17

After making some measurements, it is decided that the transistor will saturate if we use 8kohm resistor.

After making some calculation the saturation occur when the base is around 0.5mA.

By looking at the graph, however, we can figure out the gain of the transistor since the beta value is actually the slope which is 37.0.


Conclusion


This lab shows us that the base and the emitter current has a linear relationship. In the graph that is obtained, the R^2 value shows our precision as it gets closer to one. The R^2 value that we have for this appears to be 0.983, which is close to one. Some discrepancy that contributes to the R^2 value not being one is that the resistor box is not exact when we change it. These values contributes to the error we got.

FreeMat Lab

Introduction
       MatLab is one of the most popular application for numerical computations. MatLab is a computing language more powerful than traditional languages such as C, C++ or Fortran. In this exercise, we learn some of the commands used in the FreeMat program to solve equations for circuits. FreeMat used Matrix to do the computation.

Questions
            In this exercise, we want to find the current of R_3 of the following circuit:

V_1 = 15V

V_2 = 7V

R_1 = 20 Ω

R_2 = 5 Ω

R_3 = 10 Ω

The equation can be derived as follows:

After rearranging the equation 3 equations as follows are obtained:

Once the equation is rearranged, the FreeMat is used to do the calculation. The commands used are as follow:

The answers agree with the hint as the current through R_3 is predicted to be -0.186A and one of the current obtained is the -0.186A.

Introduction of Biasing

Introduction
 To establish the correct voltage across and current through component so it operates properly, biasing is required. Biasing is technique used for parallel circuit so that we don't burn the LED in the circuit. In this lab, 2 LEDs with different ratings are connected in paralel.

Prediction:
R_LED1 = 219.8Ω           R_LED2 = 100Ω
I_R1 = 22.75 mA              I_R2 = 20 mA
V_R1 = 4 V                      V_R2 = 7 V
R1 = 175.8 Ω                    R2 = 350 Ω
P_R1 = 91 mW                 P_R2 = 140 mW

The following is the set up of the lab

Based on the measurement that has been made,
R1 = 176.3 Ω                    R2 = 355.7 Ω

We then configure the circuit in 3 configuration:
1) Both LEDs in the circuit
2) LED2 is removed
3) LED1 is removed

The data is as followed:

Config	ILED1 (mA)	VLED1 (V)	ILED2 (mA)	VLED2 (V)	ISupply (mA)
1	19.60 ± 0.05	2.26	14.58 ± 0.01	6.77	26.4
2	19.77 ± 0.01	2.283	X	X	73.4
3	X	X	14.60 ± 0.01	6.83	64.1

Questions

1. The time operation of battery = 0.2 A-hr / 0.0264 A = 7.58hr
2. Based on the discussion, LED 2 (yellow LED) is not viable for error calculation, because the LED itself is not the same spec as it is from the site of purchase. The theoretical value for LED 1 is 20 mA while the actual data comes out to be 19.60mA. The percent error of this is calculated to be 2.0%. The cause of this error is probably because the ammeter itself has an internal resistance which will; reduce the value of the current in the actual circuit. We also ignore the resistance of the cable in this experiment which actually reduce the current too.
3. P battery = 26.4mA * 9 V = 237.6 mW. P LED = 19.60 * 2.26 + 14.58 * 6.77 = 143.0 mW. Efficiency = 143/237.6 *100% = 60%
4. If we repeated using 6V battery, the efficiency will go up because by reducing the voltage of the battery, the power supply will decrease while the power used stays the same which means the ratio increase and efficiency goes up. The efficiency will be the greatest when 5V supply is used since the P_out = 5 * 42.75 = 153.75 mW. P_in = 214 mW. Efficiency = 153.75/214 *100% = 71.9%

Conclusion

     According to the data taken, it is quite agree with the theoretical value. Some of the error in the experiment happens because the yellow LED is not the same as the specification of it. We also ignore the internal resistance from the cable which creates the percent error that we have.

Nodal Analysis

Introduction

In this lab, we are trying to build a reliable power system by using 2 sources and a couple of circuit breakers. we will be able to isolate the damaged equipment or reconfigure the system. We are using a resistor box for the load on the circuit. The nodal analysis is used in this experiment to calculate the voltage drop across the node. The voltage sources that we're going to use in this experiment are  9V voltage source and 12V voltage source.

Experimental

Component	Nominal Value	Measured Value	Power of Current Rating
RC1	100 Ω	99.1 Ω	0.25 W
RC2	220 Ω	218 Ω	0.25 W
RC3	220 Ω	220 Ω	0.25 W
RL1	1000 Ω	985 Ω	0.25 W
RL2	1000 Ω	986 Ω	0.25 W
VBat1	12 V	12.13 V	2 A
VBat2	9V	9.03 V	2 A

Once all the components has been measured, the circuit is set up as follow:

Once the circuit is made, several data is taken as follow:

Variable	Theoretical Value	Measured Value	Percent Error
IBat1	17.4mA	18.62 ± 0.005 mA	7.01%
IBat2	1.5mA	1.30 ± 0.10 mA	13.3%
V2	10.26 V	10.24 ± 0.01 V	0.194%
V3	8.67 V	8.37 ± 0.01 V	3.46 %
PBat1	208.8 mW	226 mW	8.23%
PBat2	13.5 mW	11.7 mW	13.3%

After the data is taken, the second part of the experiment is conducted where V_1 = V_2 = 9 V. Nodal equation is used and we found that the V_bat1 has to be 9.9 V and V_bat2 = 10.98 V. The setup of the second part is as follow:

Since the voltage in the previous part is unregulated voltage, when it is regulated, adjustment to the voltage source has to be made. The data for it is as follow:

Variable	Value
VBat1	10.3 ± 0.05V
VBat2	11.5 ± 0.1 V
V2	9.13 ± 0.05V
V3	7.83 ± 0.05V
IBat1	10.60 ± 0.005 mA
IBat2	8.10 ± 0.005 mA

Conclusion

      The result obtained has a fairly low percent error with the highest being 13%. This result can be found from several things. One of the reason of this error is the use of 5 resistor which each have a 15% uncertainty. Other reason is that we set the voltage source before we put it in the circuit, so when it is regulated by the circuit, the voltage decrease. We fail to take this into consideration, hence the percent error.