Capacitors
Table of Contents
1. Capacitors
In its simplest form, a capacitor is an insulating material sandwiched between two metal electrodes, with the most common insulator used in electronics being silicon dioxide (in some form of glass).
A capacitor stores charge (\(Q\)), with the same amount of oppositely polarized charge on its two plates. Then, a capacitor, by virtue of storing charges, also has an electric field:
The capacitance of a capacitor is defined as:
\begin{align} \boxed{C = \frac{\text{d}Q}{\text{d}V}} \end{align}In other words, capacitance quantifies the rate of change in charge with respect to voltage.
Most commonly (for the insulators that we use), we have a linear capacitor, which is a capacitor where the charge changes linearly with applied voltage; linear capacitors have constant capacitance. Then, for linear capacitors, we have:
\begin{align} C = \frac{Q}{V} \notag \end{align}From \(E = \frac{V}{d}\), we get:
\begin{align} C = \frac{Q}{Ed} \notag \end{align}We can calculate the total charge on a capacitor by defining the charge density, \(Q_A\), and the area of a plate, \(A\). Then \(Q=Q_AA\), so we have:
\begin{align} C = \left(\frac{Q_A}{E}\right)\frac{A}{d} \notag \end{align}\(\frac{Q_A}{E}\) depends on the material used, and is known as the permittivity of the material, \(\epsilon\). Thus, we can write the capacitance of a linear capacitor as:
\begin{align} \boxed{C = \frac{\epsilon A}{d}} \end{align}Often, we measure permittivity relative to the permittivity of free space, \(\epsilon_0 = 8.854 \times 10^{-12} \text{ } \frac{\text{F}}{\text{m}}\), called the relative permittivity \(\epsilon_r\), and \(\epsilon = \epsilon_0\epsilon_r\).
1.1. Capacitor Charging
Consider the following circuit:
Initially, the capacitor has no charge, and at \(t=0\) we close the switch. Then current starts to flow as \(I = \frac{V_0-V_C}{R_S} = \frac{V_0}{R_S}\).
This will bring positive charges to the positive plate, and negative charges to the negative plate. Because of the insulator, charges cannot flow through the capacitor.
Say at \(t=t_1\), \(Q=Q_1\). Then \(V_C = \frac{Q_1}{C}\), as \(I = \frac{V_0-V_C}{R_S}\) is smaller than before. As charge builds up more and more, at some point \(V_C=V_0\), and \(I=0\). Then current stops and no more charges will pile up. Therefore, the total charge we can store in the capacitor is:
\begin{align} \boxed{Q=CV_0} \end{align}This process is called the charging of the capacitor.
From the definition of current, we have:
\begin{align} I &= \frac{\text{d}Q}{\text{d}t} \notag \\ &= \frac{\text{d}}{\text{d}t}(CV_C) \notag \end{align}Therefore, current can be related to capacitance and voltage like so:
\begin{align} \boxed{I = C \frac{\text{d}V_C}{\text{d}t}} \end{align}1.2. Capacitors in Series
We want to find the equivalent capacitance for capacitors in series:
We can write two equations for current for the left side:
\begin{align} I &= C_1\frac{\text{d}V_{C1}}{\text{d}t} \Rightarrow \frac{\text{d}V_{C1}}{\text{d}t} = \frac{I}{C_1} \notag \\ I &= C_2\frac{\text{d}V_{C2}}{\text{d}t} \Rightarrow \frac{\text{d}V_{C2}}{\text{d}t} = \frac{I}{C_2} \notag \end{align}Then, for the right side:
\begin{align} I = C_{eq}\frac{\text{d}V_1}{\text{d}t} \Rightarrow \frac{\text{d}V_1}{\text{d}t} = \frac{I}{C_{eq}} \notag \end{align}We see that \(V_1 = V_{C1} + V_{C2}\). Therefore, we have:
\begin{align} \frac{I}{C_{eq}} &= \frac{\text{d}V_{C1}}{\text{d}t} + \frac{\text{d}V_{C2}}{\text{d}t} \notag \\ \frac{I}{C_{eq}} &= \frac{I}{C_1} + \frac{I}{C_2} \notag \\ \frac{1}{C_{eq}} &= \frac{1}{C_1} + \frac{1}{C_2} \notag \end{align}Therefore, for series capacitors, we have:
\begin{align} \boxed{\frac{1}{C_{eq}} = \sum_{n=1}^N \frac{1}{C_N}} \end{align}Notice that the middle section between the two capacitors is isolated from the battery on both sides. Therefore, there is no way for it to get charged, so it must have neutral charge.
What happens is that the charges on one plate of a capacitor induces the opposite charge on the other plate. This allows the charges to spatially separate in the middle section while still keeping the net charge to be zero. It follows, then, that the charge of the two capacitors must be equal:
\begin{align} \boxed{Q_1 = Q_2} \end{align}Indeed, the charges on any set of series capacitors must be equal.
1.3. Capacitors in Parallel
We want to find the equivalent capacitance for capacitors in parallel:
For the left side, we have:
\begin{align} I &= I_1 + I_2 \notag \\ &= C_1\frac{\text{d}V_1}{\text{d}t} + C_2\frac{\text{d}V_1}{\text{d}t} \notag \\ &= (C_1+C_2)\frac{\text{d}V_1}{\text{d}t} \notag \end{align}Then, for the right side:
\begin{align} I = C_{eq}\frac{\text{d}V_1}{\text{d}t} \notag \end{align}Therefore, for parallel capacitors, it is clear that:
\begin{align} \boxed{C_{eq} = \sum_{n=1}^N C_N} \end{align}1.4. Capacitive Voltage Divider
The following is a capacitive voltage divider:
From (6), we know that the charges in both capacitors must be equal. Thus:
\begin{align} C_1(v_{12} - v) &= C_2v \notag \\ C_1v_{12} &= C_1v + C_2v \notag \\ C_1v_{12} &= (C_1+C_2)v \notag \end{align}Therefore, we have:
\begin{align} \boxed{v = \frac{C_1}{C_1+C_2}v_{12}} \end{align}2. RC Circuits
An RC circuit is composed of a resistor and a capacitor, like so:
2.1. Charging
2.1.1. Capacitor Voltage
When the capacitor is charged, we can say that:
\begin{align} IR + V_C = V_0 \notag \end{align}By (4), we replace \(I\) to be:
\begin{align} RC\frac{\text{d}V_C}{\text{d}t} + V_C &= V_0 \notag \\ \frac{\text{d}V_C}{\text{d}t} + \frac{V_C}{RC} &= \frac{V_0}{RC} \notag \end{align}Solving the differential equation, we get a solution of the form:
\begin{align} V_C(t) = Ae^{-\frac{t}{RC}} + V_0 \notag \end{align}The first part is known as the natural response of the circuit, and the second \(V_0\) term is known as the forced response. Since \(V_C\) at \(t=0\) is zero, we know that \(A=-V_0\). Thus, the general solution for a charging capacitor is:
\begin{align} V_C(t) = V_0\left( 1- e^{-\frac{t}{RC}}\right) \end{align}We see then that the voltage response of a charging capacitor looks like:
Typically we say that the charging cycle is complete when \(t=5\tau\).
2.1.2. Capacitor Current
We can also find the current response while the capacitor is charging:
\begin{align} I &= C\frac{\text{d}V_C}{\text{d}t} \notag \\ &= C\frac{\text{d}}{\text{d}t}\left[V_0-V_0e^{-\frac{t}{RC}}\right] \notag \\ &= C\frac{V_0}{RC}e^{-\frac{t}{RC}} \notag \end{align}Thus, we get:
\begin{align} I=\frac{V_0}{R}e^{-\frac{t}{RC}} \end{align}Which looks like:
2.1.3. Capacitor Energy
Power is given by:
\begin{align} P(t) &= IV_C(t) \notag \\ &= \frac{V_0}{R}e^{-\frac{t}{RC}}V_C(t) \notag \end{align}Then, energy is:
\begin{align} U = \int_0^{t'} \frac{V_0}{R}e^{-\frac{t}{RC}}V_C(t)\text{ d}t \notag \end{align}However, we know that \(\text{d}V_C = \frac{V_0}{RC}e^{-\frac{t}{RC}}\text{ d}t\), so we can do a change of variables:
\begin{align} U = \int_0^{V_0} CV_C(t)\text{ d}V_C \notag \end{align}Thus, the total energy stored by a capacitor is given by:
\begin{align} \boxed{U = \frac{1}{2}CV_0^2} \end{align}2.2. Discharging
Say that after charging the capacitor, we turn off the battery at \(t=0\). Then we have:
\begin{align} IR + V_C &= 0 \notag \\ RC\frac{\text{d}V_C}{\text{d}t} + V_C &= 0 \notag \\ \frac{\text{d}V_C}{\text{d}t} &= - \frac{V_C}{RC} \notag \end{align}Then, we get the general solution to be:
\begin{align} V_C(t) = V_0e^{-\frac{t}{RC}} \end{align}Which looks like: