LAB 6 - THEVENIN EQUIVALENTS

Goal: To determine the smallest equivalent load resistance that can be successfully used given the minimum acceptable voltage. The maximum power delivered to the load would also be calculated.

Step 1A: determining open circuit voltage in the following circuit using nodal analysis given Rc1=100 ohms, Rc2=Rc3=39 ohms, RL1=680 ohms, and Vs1=Vs2=9 V.

Step 1B: determining Vy, Isc, and Rth from in the following circuit

Step 1C: determining smallest permissible RL2 using voltage divider, Isc using Ohm's Law, and Voc by inspection

Step 2A: Build the following circuit and collect some measurements. Although Vth was determined to be 8.64V, 9V was used instead. I was determined to be 0.0097 A using Ohm's law. Using this value of I, the theoretical voltage across load 2 can be easily calculated using the Ohm's Law.

Step 2B:

Step 3: Calculate Pmax and verify by experiment

Theoretical Pmax = (Vth^2)/4Rth = 0.3068 W

Conclusion:  The maximum load for load 2 was calculated to be 825 ohms with about 8 V delivered to the load. The amount of power decreases when we set R = 0.5 Rth and 2 Rth. The values agree with the fact that the maximum power happens when RL2 = Rth.

Extra Credit Selfie

LAB 5 - TRANSISTOR SWITCHING

Goal: to measure β value for the transistor 2N3904


Experiment:


1) Build a circuit according to the following scheme on a breadboard. Instead of using push button to turn on, we just closed the circuit to turn on the LED. LED turned bright. R1 = 220 ohms, R2 = 10,000 ohms, R3 = 680 ohms, Q1 = 2N3904 Transistor, D1 = LED.







2) R2 was removed, and we performed finger tip switching. Although not as bright, the circuit could be closed just by using our fingertip. Video is available in the following link: http://www.youtube.com/watch?v=Ok8e_V7VmqQ









3) According the following scheme, a circuit was built. R1 = R3 = 220 ohms, R2 = R4 = 1000 ohms, P1 = 10k potentiometer, Q1 = 2N3904 Transistor. A 6V power supply was used in this experiment. After collecting 10 data points for Ib and Ie, the data was plotted, and the slope (β value) was determined to be 7.8060.

Conclusion: The maximum amount of Ib that could be achieved in the lab was 0.308 mA (instead of the expected 4 mA) while the minimum was 0.119 mA. Therefore, 10 points were picked between 0.308 and 0.119 mA. The β value of 7.8060 was far off from the expected 50. This might be because the transistor has not reached its active phase therefore the slope is not that steep. In future experiment, the values of all circuit elements must be carefully calculated to reach higher values of Ib.

LAB 3 - NODAL ANALYSIS

Goal: To analyze a circuit using nodal analysis and compare the theoretical value to the measured value


1) We calculated theoretical values for the following circuit.

2) We built the circuit and measured the values.

3) We calculated theoretical values for Vbat1 and Vbat 2 if we want to set V2 = V3 = 9 V in two different ways.


First, we treated the circuit as two different circuits (since no current will be flowing between V2 and V3). We determined V2 = 9.9 V and V3 = 10.98 V.

Second, we used the nodal analysis equations from the experiment and plug in 9 V for V2 and V3 solving for Vbat1 and Vbat 2. The results agreed as if we treated the circuit as separate circuits.

We set Vbat1 = 9.9 V and Vbat2 = 11 V and measured values. Here is a table comparing the theoretical and measured values.

Conclusion: The nodal analysis can be used to analyze a circuit quite accurately. We have compared the theoretical and measured values, and the results agreed to each other in an acceptable range. However, when we designed the circuit to have V2 = V3 = 9 V, I have thought and thought about reasons why the measured values were very off, but I still cannot figure it out. I guess the world of electrical engineering is pretty magical.

MATLAB - INTRODUCTION TO NUMERICAL COMPUTATION

Assignment 1

I through R3 was found to be -0.1857 A

Assignment 2

The circuit with tau = 100 ms (blue plot) has the lowest output sooner.

After the circuit was redesigned to give output = 2*(1-e^(-t/tau))

Adding 2 sinusoids 3sin(2t+10 degree) and 5cos(2t-30 degree)

Script for adding 2 sinusoids

Adding the same 2 sinusoids with f = 10 Hz



Assignment 3

Assignment 4

LAB 2 - INTRODUCTION TO BIASING

Goal: To design a parallel circuit to power two different Light-Emitting Diodes (LEDs) with different ratings (LED1 is rated for 5V and 22.75 mA while LED2 is rated for 2V and 20 mA). Different resistors (R1 and R2) were used to adjust the right amount of current and voltage to power the LEDs.

Experiment:

Scheme of the circuit that was built in this experiment:



Calculations for the circuit:

Instead of using 176 ohms and 350 ohms resistors for R1 and R2, 220 ohms and 470 ohms resistors were used due to availability (measured resistances were 217 ohms and 457 ohms). We decided to use higher resistances to keep voltage supplied to LEDs below their maximum capacity.





Circuit was built on breadboard and experimental values were recorded:

Questions:

Conclusion:
Battery with higher voltage would give lower efficiency as higher resistances must be added to dissipate extra power from the battery (which becomes energy lost). In this experiment, 9V battery was used, and the circuit had a 42.8% efficiency. With lower battery voltage, lower resistances would be required, which would lower energy lost and raise circuit efficiency. Based on the available resistors, the most efficient circuit would have a 4V battery and use 100 ohms resistors for both R1 and R2. The efficiency of this circuit could reach up to 89.9% theoretically.

LAB 1 - INTRODUCTION TO DC CIRCUITS

Goal: To determine the maximum cable resistance so that the battery source would still be able to deliver enough voltage to operate an electronic device (load) normally from a distance. After the maximum resistance is determined, the maximum distance would also be calculated assuming the AWG #30 wire is used.

Experiment:

From the given information, resistance of the load (Rload) can be calculated:

Using the calculated Rload, a circuit was assembled. A 1000 ohms resistor was used to represent Rload. A resistor box was used to represent resistance of the cable (Rcable). Setting Rcable to 0 ohms, the measured Vbattery was 11.91 V and Ibattery was 12.23 mA (0.01223 A).

After Rcable was adjusted, the maximum Rcable was determined to be 80 ohms. The measured Vbattery was 11.02 V and Ibattery was 11.32 mA (0.01132 A).

After data was collected, time to discharge, power to the load (Pout), power to the cable (Plost), efficiency, and the maximum distance between the battery and the load were calculated assuming the AWG #30 wire with resistance of 0.3451 ohm/m was used.

Data Calculations:

Conclusion: The maximum cable resistance to still be able to operate a distant electronic equipment normally was found to be 80 ohms. Depending on the wire used, the maximum distance may vary. In this case, AWG #30 wire with resistance of 0.3451 ohm/m was used, and the maximum distance was determined to be 116 m.