Capacitor Charging/Discharging

Purpose
The purpose of this lab is to understand the concept of RC charging and discharging circuit and then see how a real capacitor works in the circuit.

Background
In the RC charging circuit, the voltage and current across capacitor can be express as V(t)=Vs(1-e^(-t/tau)) and I(t)=Vs/R*e^(-t/tau), where tau is time constant which is Rcharge*C. In addition, in the discharging circuit the voltage and current across the capacitor are V(t)=Vic*e^(-t/taud) and I(t)=-Vic/Rdiscahrge*e^(-t/taud), where taud is time constant in discharging circuit. However, in real capacitor model the capacitor connected to resistor Rleak in parallel, so we must apply Thevenin's theorem to find out the real value of resistance.

Problem
We use a 9V power supply to employ a charging interval of about 20 s with a result stored energy of 2.5mJ, and then discharges that 2.5mJ in 2s.

Approach
So, the value of capacitance can be determined by applying w=0.5*cV^2, which is 61.7 uF. For R(charging), we can use 5 R(charging)*C=20, so R(charging)=64.8k ohms. From the values above, we can determine peak current and power in the charging circuit which are 0.139 mA and 1.25mW. For discharging circuit, 5 tau = 2, so R(discharging)=6.48k ohms. The peak discharge current and power are 1.39 mA and 12.5mW respectively.

Assemble
R(charging) = 65k ohms
R(discharging) = 6.47k ohms
C=63.6 uF
All values above are measured.

Figure1. Build-up Circuit on Breadboard

Figure 2. Build-up Circuit Connected Logger Pro and MAC

Graphs

• Charing

Figure 3. Vc v.s Time When Charging

• Discharging

Figure 3. Vc v.s Time When Discharging

Data Analysis

Vs = 9.01 V

Vfinal = 8.562 V = Vs*Rleak/(Rleak+65k), so R(leak)=1.24 Mohms

The Thevenin equivalent voltage and resistance seen by the capacitor during charging are

Vth=9.01*1.24M/(1.24M+65k)=8.55V and Rth=1.24M and 65k in parallel which is 61.8k ohms, and the Thevenin equivalent voltage and resistance seen by the capacitor during discharging are

Vth=9.01*1.24M/(1.24M+65k)=8.55V and Rth=1.24M and 6.47k in parallel which is 6.44k ohms 

Practical Question

Suppose we wish to scale our result to the rail gun problem we worked in a previous example. the rail gun requires a stored electrical energy of 160 MJ. The capacitor charging voltage is selected to be 15 kV DC.

1. Determine the required equivalent capacitance

160MJ=0.5*C*V^2

==>C=1.42F

2. If we assume that the capacitance will be achieved in the manner shown below, via a series and parallel combination, calculate the required value of individual C

0.5C*4=1.42F

==>C= 0.71F