In the midst of trying to explain inductors, I realized that to be able to have a really good understanding of inductors, you need to have a really good understanding of capacitors. And, well, capacitors are quite tricky to properly understand, but they are somewhat easier to understand due to their similarity to batteries. But, the point is, amidst my attempt to explain inductors, I needed to have a clear explanation of capacitors. So here I am.
First of all, a very brief explanation of batteries as they relate to capacitors. Often times, a battery is modeled in a circuit simply as a fixed voltage source. The current in the circuit can then be computed using Ohm’s Law:
I = V / R
20200319/https://en.wikipedia.org/wiki/Ohm%27s_law
However, true batteries don’t last forever. Their lifetime is based off of how much electricity is consumed from them. Which is what? Intuitively, this is energy. For electricity, these are the pertinent equations:
E = P * t
P = V * I
A battery, of course, supplies a “fixed voltage,” though when it runs low, the voltage sags. But first, let’s assume the voltage is fixed. That means that the only unknown left to computing the energy consumption is the current draw.
For purely resistive loads such as an incandescent light bulb, the current draw is proportional to voltage. By extension, an LED, by virtue of using a resistor in series with the LED, can also be treated as a purely resistive load. So, now it should be obvious that it is straightforward to compute the current draw from a battery with a purely resistive load attached: just use Ohm’s Law.
Okay, so let’s do an example. How much power does an LED in series with a 220 Ohm resistor consume when attached to a 5 V battery?
I = V / R
I = 5 V / 220 = 0.022727 A = 23 mA
If the resistance of the LED on its own is assumed to be near zero, that means we consume 23 mA of current. Otherwise, we might actually consume something closer to 10 mA or so. Let’s work with a 20 mA figure for simplicity in the subsequent calculations.
P = V * I
P = 5 V * 0.020 A = 0.1 W
If our LED can run for 200 hours (8.3 days) on the battery, how much energy does the battery store?
E = P * t
E = 0.1 W * 200 * 3600 s
E = 72000 J = 72 kJ
So our battery stores 72 kJ of energy.
Now, as it turns out, since we assumed that the battery supplies a constant voltage, we really have no need for carrying the voltage number through our energy calculations. So we can eliminate that variable and still get a number indicative of the battery’s energy storage and run time. Of course, we still use voltage in the calculation of current draw.
Q = I * t
Q = E / V
Q = 72000 J / 5 V = 14400 C = 14.4 kC
The unit of electric charge is Coulombs, whose base units are amperes times seconds. It’s a measure of how much current is transported over time. The primary unit for measuring how much energy is stored inside a battery is milliamp-hours (mAh). You can see this is also a measure of electric charge, since it is current times time.
20200319/http://en.wikipedia.org/wiki/Coulomb
Due to the fact that real batteries have a voltage sag as they run low, they do in fact behave somewhat similar to capacitors. The equation of capacitance is as follows:
C = Q / V
This means that a capacitor’s output voltage will sag as the available charge runs low, due to the following:
V = Q / C
Here’s another key to remember. A battery’s energy storage can only be specified in milliamp-hours at a particular voltage… or an effective capacitance which is determined by its voltage sag profile. And, a battery can only have power drawn from it at a voltage different than its nominal supply voltage via a switched-mode power supply. So to determine the runtime at a different voltage draw, you must first convert milliamp-hours to watt-hours, then you can compute the power draw of the connected device and compute the subsequent runtime. For example:
7083 mAh @ 12 V = 7083 mAh * 1 A / 1000 mA * 12 V = 85 Wh
Device power = 5 V @ 1 A = 5 W
85 Wh / 5 W = 17 h
Device power = 12 V @ 1.25 A = 15 W
85 Wh / 15 W = 5.67 h
Laptop battery calculations… you gotta love them.
How do we know how long it takes to charge a capacitor? The following differential equation is given, based off of the information in the two Wikipedia articles.
I = -C * dV/dt
20200325/https://en.wikipedia.org/wiki/Inductor 20200325/https://en.wikipedia.org/wiki/RC_circuit
What the heck does all this mean? First of all, we want to know how long it takes to charge up a capacitor to a voltage. That, ultimately, depends on the capacitance. Larger capacitance, longer time, assuming a resistor of known size is used when charging.
But capacitors, like I said, are kind of like batteries, so this is the key to remember. When charging a capacitor, it develops a difference in potential, just like a battery. Current only flows when there is a difference in potential. If you have a path wired between two +3 V sources, no current will flow since the difference in potential is zero. Once you know the difference in potential, you can compute the amount of current that will flow using Ohm’s law with that computed voltage. This means that as a capacitor charges, the difference in potential between the capacitor and the battery decreases, so the current that flows into the capacitor also decreases. And, therefore, the rate at which it charges slows down as it approaches the voltage of the battery. Once it is equal to the voltage of the battery, it cannot accept any more charge and develop any higher of a voltage.
This means that the equation that defines the rate of charging of a capacitor is an exponential growth equation, because the rate at which a capacitor accepts current and increases its charge/voltage is proportional to its existing charge/voltage level.
As it turns out, charge and voltage are directly proportional to each other for capacitors, but since it is often times more useful and easier to calculate using voltage, the exponential growth and decay equations for capacitors are all specified using voltage. I tried my own hand at calculating the exponential growth and decay equations using charge and I got stymied at solving the limit for the rate of proportionality as the delta-time slice approaches zero. Alas, Wikipedia has solved the differential equations in terms of voltage, so we will present them here and use them in our discussions.
So, let’s solve this differential equation. An exponential growth differential equation has this form:
dy/dt = ky
The only solutions to these exponential growth differential equations are of the following form:
y(t) = y(0) * e^(k * t)
We can solve using the capacitor equations as follows:
I = V / R
I = -C * dV/dt
V / R = -C * dV/dt
V / (-R * C) = dV/dt
let k = 1 / (-R * C)
kV = dV/dt
V(t) = V(0) * e^(k * t)
V(t) = V(0) * e^(-t / (R * C))
This equation can then be substituted and solved to get an equation to determine how long it will take to charge or discharge a capacitor to a specific voltage with the given resistance.
V_e = V_0 * e^(-t / (R * C))
V_e / V_0 = e^(t / (-R * C))
ln (V_e / V_0) = t / (-R * C)
t = -R * C * ln (V_e / V_0)
This equation actually tells us the time it takes to discharge a capacitor. To get an equation that tells us the time to charge a capacitor, simply remove the negative sign.
Does this make sense? Indeed it does. A simple test circuit for a capacitor connects a capacitor in parallel to a purely resistive load such as an incandescent light bulb or LED. Once the capacitor is charged, the battery voltage source is removed and the capacitor discharge powers the light. As the capacitor discharges, the light intensity decreases, but there is an non-linear curve such that the light lasts longer at the lower charge than it does at the higher charge. This is consistent with the underlying equations that state there should be an exponential growth/decay curve with both capacitor charge and discharge: as the capacitor nears the end of charging, its charge rate slows down. Likewise, as the capacitor nears the end of its discharging, its discharge rate slows down.
Yeah, that’s nice and correct and all, but where did that exponential growth differential equation come from out of the blue? Can you prove by solving for it using the base equations of capacitors? Indeed we can… the question is am I able to do it? We’ll see.
The primary equation of a capacitor:
C = Q / V
Now, we want to start by defining a step-wise approximation equation explaining how the voltage of a capacitor decreases based off of how much current has flowed out of it. We are discharging this capacitor using a constant-value resistor. Previously, I defined the equations in terms of charge and got stymied.
V = Q / C
Q = I * t
I = V / R
Q = V * t / R
V_0 = Q_0 / C
V_n = Q_n / C
delta_t = t_n - t_0
delta_Q = Q_n - Q_0
I_0 = V_0 / R
delta_Q ~= -I_0 * delta_t
delta_Q ~= -V_0 * delta_t / R
Q_n = Q_0 + delta_Q
Q_n = Q_0 - V_0 * delta_t / R
Q_n = V_0 * C - V_0 * delta_t / R
Q_n = V_0 * (C - delta_t / R)
V_n * C = V_0 * (C - delta_t / R)
V_n = V_0 / C * (C - delta_t / R)
V_n = V_0 * C / C - V_0 * delta_t / (R * C)
V_n = V_0 - V_0 * delta_t / (R * C)
Okay, now I’m playing my trick. We can rephrase this as a differential equation by describing how change in voltage and change in time are related. The factor in the relation is the constant of proportionality, and we can then solve the differential equation using our knowledge on the required form of the known solution.
V_n = V_0 - V_0 * delta_t / (R * C)
V_n - V_0 = -V_0 * delta_t / (R * C)
delta_V = -V_0 * delta_t / (R * C)
delta_V / delta_t = -V_0 / (R * C)
Alright, so now that we have the deltas of both quantities, we can
rephrase into the final differential equation. By taking the limit as
the deltas approach zero, V_0 = V_n, thus we can substitute V for
V_0. And with the limit as the deltas approach zero, delta_V /
delta_t is the instantaneous rate of change, thus we can replace it
with dV/dt.
dV/dt = -V / (R * C)
let k = -1 / (R * C)
kV = dV/dt
And then we’re right back to the earlier presented equations that were left unsolved.
And, alas, I must also note. Using charge instead of voltage wasn’t
the real reason I got stymied, I got stymied because I did not look at
the problem at hand and realize it would be better to rephrase it as a
differential equation. The key to the solution was to break away from
trying to solve V_n or Q_n based off of V_0 or Q_0 and instead
also change those variables into a delta.
So, the final question. How do we get the differential equation with current instead of resistance? Simply use Ohm’s Law to replace resistance with current.
I = V / R
R = V / I
dV/dt = -V / (R * C)
dV/dt = -V * I / (V * C)
dV/dt = -I / C
I = -C * dV/dt