Capacitor Charging/Discharging

 Purpose

         Capacitor are components that enable the storage of electrical energy via electric field. The purpose of this experiment was to design a capacitor circuit to store energy of 2.5mJ, charge in 20s and discharge in 2s.

Pre-lab

       As the output voltage is 9V, to store 2.5mJ energy, the capacitor should 62 microF.

       To charge in 20s, the time constant should one fifth of the total charge time. So R_charge should be  66k Ohms.

       To discharge in 2s, the time constant should one fifth of the total charge time. So R_discharge should be 6.48k Ohms.

Experiment





Data and Analysis

 

     The curve measured by "Logger Pro" during charging: 

 

    

    The capacitor was charged around 18 seconds, which is closed to 20 seconds. Final voltage was measured to be around 8v.

 

    According to the data,

 

    R_leak = R_charge /(V_s/V_final-1) = 264k Ohms

 

     The curve measured by "Logger Pro" during discharging:

 

 

  capacitor was discharged around 2 seconds.

Questions: 1.  Calculate the Thevenin equivalence voltage and resistance values seen by the capacitor during charging:
R_th = 61.4 kΩ

V_th  = 5.707 V

2 . Calculate the Thevenin equivalence voltage and resistance values seen by the capacitor during discharging:

R_th = 6.42 kΩ

V_th­ = 5.707 V

3.  When t equal one time constant, e^-t/T_­ = e^-1 =0.3679. The charging voltage equal  V_f(1-0.3679) = 0.6321*V_final = 0.6321*5.707V = 3.607V
Look at the graph, time for charging voltage to reach 3.607V is about 4s:
T= t = 4s =RC -> R= 4/C = 4/62µF = 64.5 kΩ

Practice Questions: We want to scale our result to the rail gun problem we worked in previous exercise. The rail gun requires a stored energy of 16MJ, and the capacitor charging is 15kV.

1. Find the required equivalence capacitance:

E=V²C/2 = 160MJ

C= 2(160 MJ)/(15 kV)² =1.4F

2.  If the capacitance will be achieved in the manner shown below, the required value of individual capacitance C= 0.7 F

Op Amp

Operational Amplifiers I

The purpose of the lab is to learn how to use an Op Amp. And use it to design the circuits we want.

Pre Lab
 
V_cc = 12 V
V_ee = -12 V
V_in = 0 V to +1 V
V_out = 0 V to -10 V
The sensor outputs at most 1 mA
The op-amp power supplies supply at most 30 mW each
 

Solve for R_i and R_f
R_i = 1/1 V/mA = 1 kΩ
R_f = 10*R_x = 10 kΩ

However, the lab is limited to two power supplies so V_cc is also going to provide V_in using the appropriate voltage divider like the circuit below.

V = 12 V because it is the worst case where R_y is zero while using V_cc.
Using half of the power rating of a 1/4 W resistor, R_x = (V^2)*8 = (12^2)*8 = 1152 Ω.
With the voltage divider rule, R_y = R_x/11 = 104.7 Ω.

Thevenin Equivalents

R_th = R_x*R_y/(R_x + R_y) = 96 Ω.
V_th = 1 V.
However, R_th is NOT 20 times under R_i.
Thus, the new R_i = 20*96 = 1.92 kΩ.
The new R_f = 19.2 kΩ.

Procedure:
Measure the parts and build the circuit according to the two Lab Manual.

Component	Nominal Value	Measured Value	Power or Current Rating
R_i	2000 Ω	1957 Ω	1/8 W
R_f	20000 Ω	19760 Ω	1/8 W
R_x	1152 Ω	1786 Ω	1/4 W
R_y	104.7 Ω	7020 Ω	1/4 W
V_1 = V_cc	12 V	12.12 Ω	2 A
V_2 = V_ee	12 V	12.08 Ω	2 A

Set up the circuit

 

Connect the power supply

V_in	V_out (Measured)	GAIN (Calculated)	V_Ri (Measured)	I_Ri (Calculated)	V_Rf (Measured)
0.0 V	0.00 V	0.00	0.00 V	0.00 A	0.00 V
0.25 V	-2.55 V	-10.20	0.25 V	0.125 mA	2.57 V
0.50 V	-5.07 V	-10.14	0.50 V	0.250 mA	5.11 V
0.75 V	-7.62 V	-10.16	0.75 V	0.375 mA	7.65 V
1.00 V	-10.16 V	-10.06	1.01 V	0.505 mA	10.18 V

I_v1 = 2.31 mA

I_v2 = -1.645 mA

Data Analysis

P_v1 = V*I = 28.00 mW

P_v2 = -19.87 mW
 

I_v1 + I_v2 = 0.665 mA

I_f = 0.509 mA

Error = 30.6%

Conclusion

   We had about 30% error which was not very good. We found that the current_in and current_out are not zero because the op amp is not ideal (R_in is not infinite). Therefore, we should not calculate the op amps by assuming they are ideal in this case.

Thevenin Equivalent

Objective
Learn how to use the Thevenin Theorem and calculate the maximum power.

Procedure 

Calculate the R_th

Calculate the V_th

Calculate the minimum of R_L2

Measure the true values

Measure the true values 

V_L2 when R_L2 is at min and max

Set up the circuit

Data for different R_L2

Conclusion

The Thevenin Equivalent is a great method to simplify complicated circuits. From the results we can tell that the error of this method is significantly small. Also, from the last part of this experiment, we noticed that we have maximum power when R_l2=R_th, which satisfied the maximum power theorem.

PSpice

Objective
Learn how to use PSpice.

Part 1
Build and simulate the circuit:

Part2 
Find the Thevenin equivalent of the circuit:

Graph

plot V v.s I and the slope is the Thevenin resistance.

From the graph we can see m=(10-7)/(-0.9)=3.3Ohm

Then, replace the current supply by a changing resistor. And Plot Power v.s Resistance. From the graph, find max power supply.















Since max power occurs when R_load =R_thevenin, the theoretical R should be 3.3Ohm. And the result from the graph also shows the max power occurs when R is at about 3.3Ohm. We verify the max power theorem by graphing on PSpice.

Conclusion
PSpice is an awesome tool for simulating circuits!!!

FreeMat

Objective
The purpose of this lab is to learn how to use FreeMat.

Assignments

1.

1-1

Plot 2e^(-t/100) and 2e^(-t/200)

1-2

Compare 2(1-e^-t/tau) and 2e^(-t/tau)

 2-1

Graph 3sin(2t+10), 5cos(2t-30), and the sum of them.

2-2

Change the angular frequency

New result

Maximun Power Transfer

Objective
Use the potentiometer and logger pro to verify the the Maximum Power Theorem
(Max power occurs when R_load = R_thevenin)

Procedure 

        Part A

Measured the voltages across the potentiometer at different resistances. (15 trials)

Find the max power transfer from there.

           Data

       Part B

Connect the circuit as showed with the sensor

 Use the logger pro to graph the voltages and currents on the sensor

From the graph, we can't really see the at which point is the max power transfer.

The reason why this experiment failed was because the surrounding was too noisy, and it effected the sensor by a lot.

Conclusion
The method of sensor-logger pro fails because the environment was too noisy. We concluded that the method of Part A was the better way to find the maximum power. Although the max power did not occur at R_load = R_thevenin, we can still see the max power happened to be around that range. We conclude that the reason is because the powers supply has internal resistance and it is continuously changing. Therefore, R_load + R_r =R_ thevenin.

Transistor Switching

Objective
Get a brief idea of transistor and learn how to use it.

Fingertip Switching

Procedure

Connect the circuit as the picture shows

Connect it to the power supply but open circuit

A closer look 

Use our fingers to switch on/off

 With Potentiometer

Set up the circuit and potentiometer

Adjust the potentiometer and get some different values

Data

Conclusion

The gain is the slope of the graph, which is about 80.

At about 0.75 mA, the transistor saturates.