Final Project: Ant-Weight Combat Robot

Please click on the following link to redirect to the team's report:

http://engineering44atly.blogspot.com/


**Special thanks to the Alan Ly whom wrote the report for the team.

Impedence & AC Analysis I

In this lab, we will study the real porperties of an iductor, and the impedence in an AC circuit.

 Procedure:
We start with a real inductor connected in series with an external resistance.
2 Multimeters are taking measurements.

The resistor RL is the resistance in the inductor.

(note that a real inductor is consider an ideal inductor in series to a resistor)

 The measurement for circuit elements:
RL = 8.3 Ω
Rext = 67.2 Ω
Vrms = 5V at 1000 Hz.

The original circuit.

Vin,rms = 4.93 V
Iin,rms = 65.0 mA 
The voltage reading differs from the FG because internal resistane drop 0.7 V
Vin/Iin = ZL = 75.8 Ω

substituding ZL in the following equation:
ZL = √((Rext + RL)^2 + (ωL)^2)

we get:
ω = 6280 rad/s
L = 1.16 mH

Adding a capacitor to cancel inductance impedance
1/ωC = ωL
C = 1/(ω^2*L) = 2.19 *10^-5 F

The remodified circuit

The signal of the circuit displayed in a oscilloscope.

Vpp, Channe1 = 1.5V
Vpp, Channel2 = 20mV
ΔT = 0.4 ms
Phase difference = (ΔT) *6240* 360 ° = 184.32 °

Data & Data Analysis:

Frequency (kHz)	V_in (V)	I_in (A)	|Z_in| (Ω)
5	4.6	0.073	63.0137
10	4.2	0.0719	58.41446
20	3.51	0.0691	50.79595
30	3.51	0.0653	53.75191
50	4.5	0.0558	80.64516

Questions:
1) It is not the largest at 20 kHz as we used 1 kHz for calculations. The largest current should be at 1 kHz as imaginary part cancels out.

2)
VL = ZL/Z*Vin
VL = 2.8574 +  1.2737i
|VL| = 3.1284
Phasor = 24.0243 °

3) The circuit looks more inductive as imaginary part is always positive.

4) The circuit looks more inductive as imaginary part is always positive.

Conclusion:
The experiment was quite a success. Yet, some answers to the questions were off from the lab's original expectations as the original lab suggested a frequncy of 20kHz instead of 1 kHz (the actual frequency used) in the circuit, and in the calculations.

Solving Complex Number with FreeMat

In this activity, we will be practicing to solve complex number math problems by using the FreeMat solfware.

Some practice exercises:

let A = 3 + 4j
     B = 3 - 2j
     C = 2 <50˚

and compute D = (A+C)/B

D = 0.14+1.94j

 

 

let A = 2 + 2j

      B = -1 + 3j

      C = 2 + j

 

and compute D = (A*B)/C

        compute E = (A+B)*C

        find the magnitude and the phase angle for E.

        find the read component and the imaginary component for E.

 

D= -2.4 + 3.2j         E = -3 + 11j

phase angel = 105.26˚ , magnitude = 11.41

imaginary component = 11

real component = -3

 

Assignments:

 

let A1 = 3+ 2j

      A2 = -1+4j

      B = 2-2j

 

solve for C = (A1*B)/A2



 



hand calculation, C = -1.06 -2.24j

 

using FreeMat, C = -1.06 -2.24j

 

the magnitude and the phase angle for C

 

Solve for D = (A1+B)*A2

hand calculation

calculation with FreeMat, D = -5 +20j

 

Solving Matrix with complex numbers:

 

Matrix calculation with complex number has the same inputting method as real numbers



 

 

Conclusion:

     Why waste time on inefficient hand calculation when you can solve for complex numbers easily on freeMat? The result will be the same but FreeMat is so much faster.





Controlling Electric Motor Using MOSFET

 The main purpose in the experimient was to achieve a stable speed control of an electric motor using a MOSFET.

Procedure:

 Part 1:
 Building the MOSFET voltage control circuit shown below.

note: the "FET N" is the MOSFET, and there is a diode between the positive ad the ground wire connecting to the electric motor. The external resistor is used to limit the current in the circuit.

A completed MOSFET voltage control circuit on breadboard.

By adjusting the resistance in the potentiometer, we can adjust the voltage through the motor, and therefore, control its speed.

Part 2:

This time, the potentiometer is replaced with a Function Generator.
The Function Generator would produce Square Waves, and its duality was set to be on, as well as its duty cycle.

The completed view of the modifited circuit.

Eventhough the potentiometer was left on the breadboard, it was no longer connected to the circuit.

The voltage on the motor is correspondance with the Oscilloscope.

Displaying the speed control by increasing and decreasing duality.

 Answer to questions for discussion:
The motor rotates faster with a larger duty cycle.
The graph is the on/off times of the square function.
The time required to decelerate is 0.7s.
The voltage is 0.16 V at 30%

The voltage of motor at 30% of its maximum.

 The converter allows a smooth speed control.
T = 1/110 = 9.1ms

Conclusion:
 We had successfully controlled the voltage across the motor with both the potentiometer and FG, but the FG allowed for better speed control than the potentiometer.

2nd Order Circuit Tutorial

In this lab, an online tutoring program will be used to learn how to solve 2nd order circuits step by step. Each screen shot will be a step.

 

Conclusion:  2nd order circuits are hard to solve because they have so many steps ( so easy to make a

                     small error that will ruin the whole problem). However, they may not necessary be

                     difficult to solve.

 

Remarks: Hope i can keep myself calm whiling frustrating with all the math in these questions when

                 they are presented in exams and hw.

Oscilloscope 101

The objective of the lab is to learn, use, and anaylze an oscilloscope.

Procedure:
We connect the oscilloscope with a frequency generation.

Exercise 1: Sinusoid

f = 5 kHz, V = 5V

Once the trigger was set, we took measurements.
Period = 0.2 ms
Peak to Peak = 11.4 V
Zero to Peak = 5.8 V
Anticipated RMS = 3.53 V

Using DMM to read the Voltage values
VDC = 0.026 mV
VAC = 3.35 V
The VAC is close to our anticipated RMS value.

Exercise 2: Include DC Offset
We add an offset of 2.5 V, and another one at 5V

DC Coupling at 5V

AC Coupling at 5V

The difference is that we can see the offset in DC coupling while nothing changes in AC.
2.5V offset measurements:
VDC = 2.51 V
VAC = 3.37 V
The VDC shows the offset in the output like the graph while the offset does not affect VAC.

Exercise 3: Square Wave with offset

VDC = 10 mV
VAC = 5.34 V
The measured value was close to the theoretical VAC = 5 V.

Exercise 4: Mystery Signal

Mystery Signal

DC Voltage = 448 mV
f = 70.42 Hz
Pk-Pk = 940 mV

Conclusion:
 A digital oscilloscope was more easy to opertate than a traditional one since a digital one displays all reading of the properties of the wave on the screen. Meanwhile, the digital oscilloscpe also the screen to be printed and saved into a computer connected with a USB cable.

Charging and Discharging Capacitros

The purpose of the experiment is to learn how to control capacitor charge and discharge times using a nonideal case.

Prelab:
We first build the Thevenin expressions for the charge and discharge circuits below to use later.

Charging: Rth = Rleak*Rcharge/(Rleak + Rcharge), Vth = Rleak*Vs/(Rcharge + Rleak)
Discharge: Rth = Rleak*Rdischarge/(Rleak + Rdischarge), Vth = 0.

The setup is to use 9V to charge for 20s with 2.5 mJ, and discharge the 2.5 mJ in 2s.
Doing some math, we find:
C = 2*U/V^2 = 0.0617 mF
Charging: 5τc = 20s
Rc = 4/C = 64.8 kΩ
The power Pc = V^2/R_c =1.25 mW which is under the limit of 1W.

Discharging: 5tau_d = 2s
Rd = 2/(5*C) = 6.48 kΩ
The power Pd = 12.42 mW which is under the limit of 1W.

Procedure:

 Build the circuit according to the pre-calculated values in the prelab questions.

Capacitors combined in parallel to reach desired value

Entire circuit with resistor boxes

Analysis:
Logger pro was used to measure the change in voltage over time.

Capacitor charging

The peak voltage V_f = 8.25 V due logger pro's limit of 8Vs and the effect of the Rleak.
The time is estimated to be around 18s.
Rleak = Rcharge /(V_s/V_f - 1) = 712.8 kΩ

Capacitor discharge

Starting from Vf, the discharge time is around 2s.

Questions:
Using the Thevenin equations,
1. Rcth = 59.4 kΩ, Vcth = 8.25V
 2. Rdth = 6.42 kΩ, Vdth = 0V

3. 0.6321*V_f = 5.215 V
5.215 V is around 4.6s
 

τc = RC = 4.6s
R = 4.6/C = 74.55 kΩ
Error = 15%

Practical Question:
1)       U = (1/2)CV^2
     Ceq = 2*U/V^2

           = 2*160*10^6/(15*10^3)^2
           = 1.42 F

2)    1/2C + 1/2C + 1/2C + 1/2C = 2C = Ceq
        C = 1/2Ceq = 0.71 F

Conclusion:
 We had successfully reach a charging time of about 20s, and a discharging time of about 2s. The experiment was successful, and the pre-lab calculations were correct. The result in this lab proved that it was possible to control the times.