Capacitor Charging/Discharging

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: R_th = R_leak*R_charg/(R_leak + R_charg), V_th = R_leak*V_s/(R_charg + R_leak)
Discharge: R_th = R_leak*R_dischar/(R_leak + R_dischar), V_th = 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: 5tau_c = 20s
R_c = 4/C = 64.8 kΩ
The power P_c = V^2/R_c =1.25 mW which is under the limit of 1W.

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

Procedure:

We build the circuit according to our values.

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 R_leak.
The time is estimated to be around 18s.
R_leak = R_charge /(V_s/V_f - 1) = 712.8 kΩ

Capacitor discharge

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

Questions:
Using the Thevenin equations,
1. R_cth = 59.4 kΩ, V_cth = 8.25V
 2. R_dth = 6.42 kΩ, V_dth = 0V

3. 0.6321*V_f = 5.215 V
5.215 V is around 4.6s
tau_c = RC = 4.6s
R = 4.6/C = 74.55 kΩ
Error = 15%

Practical Question:
1. U = (1/2)CV^2
C_eq = 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 = C_eq
C = 1/2C_eq = 0.71 F

Conclusion:
The experiment was a success. The charge time was close to 20s, and discharge time was about 2s proving that it was possible to control the times.