Capacitor Charging/Discharging Lab

In this lab we studied the function of a capacitor, and how voltage and current behave when they are connected to this passive element. A capacitor is a device that stores energy supplied by a power source. The power supply initially charges the capacitor, which can later discharge and release the stored energy.

Task1: Taking a look at the following two circuits, we can develop expressions for Vth and Rth:

a). Rth = Rcharge*Rleak/(Rcharge + Rleak)
     and
     Vth =Vcap = Rleak(Vs)/(Rleak + Rcharge)

b). Rth = Rdischarge*Rleak/(Rdischarge + Rleak)

Problem: Design, build and test a charge/discharge system with Vs = 10V power supply, charging interval 20s with resulting energy of 2.5mj, and discharges in 2s.

Step 1. Required Capacitance

w = 1/2*CV^2;        C = 2W/v^2 = 2*0.002/(9)^2 = 61.728 microF

Step 2a. Estimate value of charging resistance (ideal capacitance).
It takes about 5 time constants (tau = RC) to charge up to source voltage.

5Tau =5 RchargeC = 20s;          Rcharge = 20/5C = 20/(5 * 61.728microF) = 64.7kohms

Step 2b. Calculate peak current charge and peak power.

Icap,charge = (Vs/Rcharge)e^(-t/RchargeC) = 0.1389 A

Power = 1.2501 W

Step 3a. Estimate value of discharging capacitance.

5Tau = 5RdischargeC = 2;        Rdischarge = 2/(5C) = 2/(5*6.173*10^-5) = 6.5kohms

Step 3b. Calculate peak discharge current and power in resistor.

Icap,discharge = -(Vic/R)e^-t/RdischargeC = -1.389mA

Power = 0.0125 W

Step 4: Build the circuit.

Step 5. For this experiment we used the LoggerPro interface. We used a current measuring device and a voltage prove, and hooked them up to take readings.

Charging readings:

From the graph, we can see that the final voltage achieved in the capacitor while charging was;
Vfinal = 8.523 +/- .001 V at around 18 seconds.

It's worth noting that the final voltage may not be definitive because Logger Pro saturates in this experiment, meaning that the time only goes up to 18 seconds. In reality the voltage keeps rising as time increases even if the increase is small.

Having Vfinal, we can now calculate Rleak;
Vf = Vs(Rleak/Rcharge + Rleak) = 1.15 Mohms


Discharging readings:

Based on the obtained graph, the capacitor discharged and achieved a value of 0.9 -0.7 V at around 3.1-3.3 seconds.

Step 7.
1). Calculate Thevenin equivalent voltage and resistance values seen by capacitor during charging

Vf = Vth = Vs(Rleak/(Rcharge + Rleak)) = 9*1.15*10^6/(64.7*10^3 + 1.15*10^6) = 8.521 V

Rth = Rcharge * Rleak / (Rcharge + Rleak) = 1.15*10^6 * 64.7*10^3 (1.15*10^6 + 64.7*10^3) = 61.3 kohms.


2). Calculate Thevenin equivalent voltage and resistance values seen by capacitor during discharging.
Vth = Vcap = 8.521 V


Rth = Rcharge * Rleak / (Rcharge + Rleak) = 1.15*10^6 * 64.7*10^3 (1.15*10^6 + 64.7*10^3) = 61.3 kohms.

3). At t = tau; e^-1 = 0.3679. When t = tau during charging, the voltage should be equal to 0.6321(Vf). Estimate value of tau when charging. 

0.6321*8.521 = 9(1-e^-20/tau)

0.5985 = 1-e^-20/tau

-20/tau = -0.9124

tau = 21.92

R = 21.92/(61.728*10^-6) = 355 kohms

Practical Question:

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

1. Determine required eq. capacitance

E = 1/2 CeqV^2

160*10^6 = 1/2 (15*10^3)^2 Ceq

Ceq = 1.422F

2. If the capacitance is achieved as shown, calculate the required value of individual capacitance C.

Ceq = C/2 *4 = 2C
1.422 = 2C,
C = 0.711 F