Some devices require rapid application of power in short bursts. Providing this energy from a source like a motor or a battery is impractical because it must have high capacity but be used infrequently. A capacitor can store electrical energy, and release it rapidly, which makes it ideal for this application.
In the figure above, the switch initially connects C1 to V1 and R1, in this state the capacitor will begin to charge. The voltage on the capacitor grows as a function of (1-e^(-kt)) similar to the following free mat plot.
When the switch is moved to the second position, the voltage across C1 will begin to bleed through R2. The voltage during discharge will resemble a exponential decay plot, like the following graph.
The rate of increase and decrease are limited by resistors R1 and R2 respectively. If we would like to slowly charge the capacitors with a low power source we can make R1 large and if we would like to discharge the capacitor quickly we can make R2 very small.
Lab Problem:
Design build and test a charge/discharge system that uses a 9V power supply, has a charging interval of about 20s with a stored energy of 2.5mJ and can discharge is 2s.
1. Determine Capacitance
The capacitance can be found by solving for the energy stored on a capacitor, which is one half the capacitance multiplied by the square of the voltage. This corresponds to a capacitance of 61.7uF (ideal) or 64.5uF (real). (the difference between the real and ideal is the concept of leakage current through the dielectric material used to insulate the capacitive elements from each other.)
2. Determine R1
The charging time is approximately 5tau, where tau is the time constant for the RC circuit. In RC circuits, tau is equal to the product of R and C. In order to solve for the resistance, one would take the capacitance found in step one and multiply it with the unknown R1 and then multiply by 5. This expression is then equated to 20s and solved for the unknown.
3. Determine R2
As the discharge time is one tenth the charge time, the capacitor is shared in each instance, and the resistance is proportional to the time constant, one can just assume R2 = R1/10, yielding a value of 6.48kOhms.
Analysis of Circuit Measurements:
1. What could account for the difference between the supply voltage and the maximum capacitor voltage recorded?
Every capacitor has some leak resistance, which means that the maximum voltage seen across its terminals is controlled by the voltage divider formed by the input resistance and the leak resistor.
In out experiment the max voltage was found to differ from the supply voltage by approximately 6 percent. This corresponds to a leak resistance of 872 kOhms.
A cursory examination of this result would allow the engineer to record that data point and then move on, however there is a variable which needs to be accounted for. The typical input resistance of a scope probe is 1 MOhms, and as we were recording our voltage measurements on a scope the calculated resistance may be greatly in error.
Once compensated, the leak resistance is found to be approximately 6.8 Mohms.
2. Does the capacitor fully discharge within 2 seconds?
Yes, the discharge time is approximately 2 seconds.
3. Calculate the Thevinin Voltage 'seen' by the capacitor during charging.
Trivial. In the steady state condition a capacitor has the Thevinin voltage across its input terminals, so the Vth is the measured value. If the scope resistance was not interfering, the terminal voltage would be less than 1 percent different than the source voltage.
The measured voltage was found to be 8.45 volts.
4. Calculate Vth and Rth 'seen' during discharge.
Using the cursory solution, Rth is found to be 6432 Ohms, a .7 percent variance from the ideal circuit. Using the compensated solution the resistance is found to be 6473 ohms, a .1 percent variance.
Vth is zero in the discharging state as there is no source in the circuit.
5. Determine the time constant for your circuit by measuring how long it takes to reach 99 percent of Vth.
Charge time:
Actual 22.1 seconds
Theoretical 20 seconds.
Time constant:
Actual 4.42 seconds
Theoretical 4 seconds
Practical questions:
If this system was used to power a rail gun, 160MJ per discharge, and is charged at 15kVDC find:
1. Capacitance
(1.42F)
2.If the capacitance was provided by 8 capacitors arranged such that 2 are in series in four parallel branches, what should the individual capacitance ratings be?
(.711F)
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