Electrical – How to find the charging time of a capacitor given a frequency and an amplitude

If you want a "simple" equation, and it seems that you do, you could start with definition of current.

First, let's start with the farad. It is usually expanded as \$F=\frac {As}{V}\$.

Now let's write that with symbols for capacitance, current, voltage and time:

\$C=\frac {It}{U}\$

Since we have constant current and voltage and we need time, we'll divide the equation with current and multiply with voltage so that we can get time.

That gives us \$\frac{UC}{I}=t\$.

If this is just a school problem, then we have a solution.

In real life things will work differently. As the capacitor charges, the voltage on the capacitor will drop resulting in drop of current and the time will therefore be longer.

Here's an example: Let's assume that at the beginning, the capacitor is discharged. First we have the voltage on the resistor which is \$U_r=Ri\$. Then we have voltage on the capacitor which is \$U_c=\frac{1}{C} \int {i \mbox{ }dt} \$.

So we know that \$E=Ri+\frac{1}{C} \int {i \mbox{ }dt}\$. To solve this, we need to turn it into differential equation.

\$(E=Ri+\frac{1}{C} \int {i \mbox{ }dt}) / \frac{d}{dt}\$

Since \$E\$ is constant, it will turn into zero. The integration and differentiation will cancel each-other out and we'll get:

\$R\frac{di}{dt}+\frac{i}{C}=0\$ Next we divide everything with \$R\$ and get
\$\frac{di}{dt}+\frac{i}{RC}=0\$

After that we move the \$\frac{1}{RC}i\$ to the other side and multiply everything with \$dt\$ and divide everything with \$i\$ and we get:

\$\frac{di}{i}=-\frac{1}{RC}dt\$

Now we integrate everything and get \$\int {\frac{di}{i}} = -\int {\frac{1}{RC}dt}\$ As a result, we get:

\$\ln{i}=-\frac{t}{RC}+C_1\$

Now to get rid of the logarithm, we raise everything to \$e\$
\$i=C_1 e^{-\frac{t}{RC}}\$

Now we have the general solution and we need to determine the constants. So first we look at what's happening when the time is equal to zero:

\$i=C_1 e^{-\frac{0}{RC}} = C_1\$.

We also know that the initial current is \$i_{(0)}=\frac{E}{R}\$. From that we can determine that \$C_1=\frac{E}{R}\$.

The complete equation for the current is:
\$i_{(t)}=\frac{E}{R} e^{-\frac{t}{RC}}\$

This is a classical capacitor charging equation and it is available on many sources on the Internet.

The \$RC\$ is also called the time constant, so \$\tau=RC\$. It is usually considered that five time constants are enough to charge a capacitor.

With a 800 Volts DC supply, and a 200 kΩ resistor in series with the capacitor, the maximum current the charging will draw is 4 mA.

How to calculate this:

• Assume an ideal capacitor i.e. Zero Equivalent Series Resistance. Real capacitors and their leads can only increase this ESR.
• At start of charging the current draw is maximum, as potential difference is maximum
• At this time, I = V / R, V = 800 Volts and R = 200 kΩ
• Thus I_max = 4 mA
• As the capacitor charges up, this current will reduce further as per the normal RC curve.

As this is lower than the 20 mA supply limit, the supply-limited maximum current will not come into play at all.

EDIT: Note that the time constant of the resistor and capacitor is 47 seconds, which means that it will take that long to reach 63% of its final voltage. It will take 3× that long (2:21) to reach 95% and 5× that long (3:55) to reach 99%. You can reduce those numbers by a factor of 5 by reducing the resistor to 40KΩ, which would just touch the source limit of 20mA at the beginning of the charge cycle. If you simply charge the capacitor with a constant current of 20mA, it will take 9.4 seconds.