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.