Electronics Components: Calculate Time Constants
Knowing exactly how much time it takes to charge a capacitor is one of the keys to using capacitors correctly in your electronic circuits, and you can get that information by calculating the RC time constant.
When you put a voltage across a capacitor, it takes a bit of time for the capacitor to fully charge. During this time, current flows through the capacitor. Similarly, when you discharge a capacitor by placing a load across it, it takes a bit of time for the capacitor to fully discharge.
When a capacitor is charging, current flows from a voltage source through the capacitor. In most circuits, a resistor is working in series with the capacitor as well.
The rate at which the capacitor charges through a resistor is called the RC time constant (the RC stands for resistor-capacitor), which can be calculated simply by multiplying the resistance in ohms by the capacitance in farads. Here’s the formula:
T = R C
For example, suppose the resistance is 10 kΩ and the capacitance is 100 μF. Before you do the multiplication, you must first convert the μF to farads. Since one μF is one-millionth of a farad, you can convert μF to farads by dividing the μF by one million. Therefore, 100 μF is equivalent to 0.0001 F. Multiplying 10 kΩ by 0.0001 F gives you a time constant of 1 second.
Note that if you want to increase the RC time constant, you can increase either the resistance or the capacitance, or both. Also note that you can use an infinite number of combinations of resistance and capacitance values to reach a desired RC time constant. For example, all the following combinations of resistance and capacitance yield a time constant of one second:
|Resistance||Capacitance||RC Time Constant|
|1 kÙ||1,000 ìF||1 s|
|10 kÙ||100 ìF||1 s|
|100 kÙ||10 ìF||1 s|
|1 MÙ||1 ìF||1 s|
It turns out that in each interval of the RC time constant, the capacitor moves 63.2% closer to a full charge. For example, after the first interval, the capacitor voltage equals 63.2% of the battery voltage. So if the battery voltage is 9 V, the capacitor voltage is just under 6 V after the first interval, leaving it just over 3 V away from being fully charged.
In the second time interval, the capacitor picks up 63.2%, not of the full 9 V of battery voltage, but 63.2% of the of the difference between the starting charge (just under 6 V) and the battery voltage (9 V). Thus, the capacitor charge picks up just over two additional volts, bringing it up to about 8 V.
This process keeps repeating: In each time interval, the capacitor picks up 63.2% of the difference between its starting voltage and the total voltage. In theory, the capacitor will never be fully charged because with the passing of each RC time constant the capacitor picks up only a percentage of the remaining available charge. But within just a few time constants, the capacity becomes very close to fully charged.
The following gives you a helpful approximation of the percentage of charge that a capacitor reaches after the first five time constants. For all practical purposes, you can consider the capacitor fully charged after five time constants have elapsed.
|RC Time Constant Interval||Percentage of Total Charge|