Welcome! In the last installment on capacitors, we discussed what capacitors are, what they do, and the different types of capacitors available. At the end of the lesson we took a very brief look at how capacitors charge and discharge.
In this installment, we’ll take a much deeper look at how capacitors behave in DC circuits to include both their transient and steady state response.
Transient vs. Steady State
Recall from our last lesson, that when a voltage is applied across a capacitor, current flows as the capacitor charges. Once the capacitor is fully charged current stops flowing.
While the capacitor is charging, the voltage across the capacitor is changing. The current through the capacitor is also changing. To be more specific, the voltage across the capacitor is increasing and the current through the capacitor is decreasing. During the time these changes are occurring the capacitor is said to be in a “transient” state. It is exhibiting a transient (changing) response.
Once the capacitor reaches it’s final voltage value, which is equal to the voltage that’s applied across the capacitor, and the capacitor is fully charged, the voltage and current stop changing. At this point, the voltage and current stay the same, they remain steady. This is called the “Steady State”.
Capacitor Charge & Discharge Rates
If we apply a voltage across a capacitor circuit, the capacitor starts to charge and continues to charge until it reaches its final voltage. But how quickly does it charge? Or, given the opposite condition, where we put a charged capacitor into a circuit, how quickly does the capacitor discharge? The rate of charge or discharge for a capacitor is a function of the capacitance (in Farads [f]) and the resistance/impedance of the charge/discharge path (in Ohms). The formula for the rate of charge or discharge is given by:
Ꞇ = C x R Where Ꞇ is the Greek letter Tau (in seconds)
So, for example, if we have a 100uf capacitor, in series with a 100kΩ resistor, then the rate of charge is:
Ꞇ = 0.0001f x 100,000Ω = 10 seconds.
Note that we didn’t consider the voltage when we calculated Ꞇ. That’s because the rate of charge or discharge is independent of the voltage. So how do we determine the voltage across the capacitor over time?
Actually, it’s fairly simple, by definition, the capacitor charges to 66.6% of the final voltage in Ꞇ (seconds). It charges another 66.6% of the remainder in 2x Ꞇ, and so on and so forth.
Capacitor Charge Example
In the following circuit, assuming the capacitor starts at 0V, the capacitor eventually charges to 10V according to the following:
| Time (n x Ꞇ) | Time (s) | Capacitor Voltage (start = 0V) | Formula |
| 1 x Ꞇ | 10 | 6.66V | [0V + (10V * 0.666)] |
| 2 x Ꞇ | 20 | 8.88V | [6.66V + ((10V – 6.66V) * 0.666)] |
| 3 x Ꞇ | 30 | 9.63V | [8.88V + ((10V – 8.88V) * 0.666)] |
| 4 x Ꞇ | 40 | 9.88V | [9.63V + ((10V – 9.63V) * 0.666)] |
| 5 x Ꞇ | 50 | 9.96V | [9.88V + ((10V – 9.88V) * 0.666)] |
But if the capacitor only charges by 66.6% of the remainder each Ꞇ (9.96V + 0.04 x 0.666), then it can never reach the full 10V. However, for engineering purposes, we define that the capacitor is fully charged after 5 x Ꞇ. So in the above case, we define the capacitor to be fully charged at 5 x 10s = 50 seconds.
Capacitor Discharge Example
It works exactly the opposite way for capacitor discharge. The capacitor discharges by 66.6% of its value each Ꞇ. Using the same capacitor and resistor values as above, and assuming the capacitor starts out being charged to 10V, then we see the following discharge rates:
| Time (n x Ꞇ) | Time (s) | Capacitor Voltage (start = 10V) | Formula |
| 1 x Ꞇ | 10 | 3.33V | [10V – (10V * 0.666)] |
| 2 x Ꞇ | 20 | 1.11V | [3.33V – (3.33V * 0.666)] |
| 3 x Ꞇ | 30 | 0.73V | [1.11V – (1.11V * 0.666)] |
| 4 x Ꞇ | 40 | 0.24V | [0.73V – (0.73V * 0.666)] |
| 5 x Ꞇ | 50 | 0.08V | [0.24V – (0.24V * 0.666)] |
Again, mathematically, the capacitor never actually reaches 0V. However, just like for charging, we assume the capacitor is fully discharged after 5 x Ꞇ. In the above example, that’s 50 seconds.
Here is how the charge and discharge for our above example would look on a graph of voltage change over time.
A More Complex Example
Let’s look at another example. Based on our previous lessons, you should recognize a voltage divider in the example circuit.
In the above circuit, the capacitor is across a voltage divider. When we close the switch, current flows and, the capacitor starts to charge.
In order to figure out the charge time and the voltage that the capacitor eventually charges to, we need to simplify the circuit. At the instant the switch is closed, the capacitor acts like a short circuit. When the capacitor is fully charged, no current flows through the capacitor. In essence, it behaves like an open circuit.
Looking at the equivalent circuit above, it becomes easy to figure out the final voltage by solving the voltage divider circuit.
V = 15V x 50k/(50k + 100k) =15V x 1/3 = 5V
The voltage across the capacitor once it’s fully charged is 5V. It is the voltage that the capacitor will eventually charge to when we close the switch. It charges through the 100K resistor.
The rate of charge: Ꞇ = R x C = 100,000 x 0.00001 – 1 second. The capacitor is fully charged in 5Ꞇ or 5 seconds.
If we allow the capacitor to fully charge and then open the switch…The capacitor discharges through the 50k resistor until it reaches 0V.
The rate of discharge is: Ꞇ = R x C = 50,000 x 0.00001 = 0.5 seconds. The capacitor, by definition, is completely discharged in 5Ꞇ or 2.5 seconds.
Conclusion
But what happens when the voltage across a capacitor is continuously changing? We’ll cover that, along with a little bit of AC theory as it relates to capacitors in the next installment.
Until then, if you found this useful, please like, share and subscribe. As always, thoughtful, on topic comments are always appreciated.
Cheers
Dominick
