continued from above post...
Energy In Capacitors
Storing energy is very important. You count on the energy stored in your gas tank if you drove a car to school or work today. That's an obvious case of energy storage. There are lots of other places where energy is stored. Many of them are not as obvious as the gas tank in a car. Here are a few.
You're reading this on a computer, and the computer keeps track of the date and time. It does that by keeping a small part of the computer running when you think that the computer is turned off. There's a small battery that stores the energy to keep the clock running when everything else is turned off.
If you own a stereo or television that you have to plug into the wall plug, then you should realize that the wall plug voltage becomes zero 120 times a second. When that happens, the system keeps running because there are capacitors inside the system that store energy to carry you through those periods when the line voltage isn't large enough to keep things going!
Capacitors can't really be used to store a lot of energy, but there are many situations in which a capacitor's ability to store energy becomes important. In this lesson we will discuss how much energy a capacitor can store.
Capacitors are often used to store energy.
When relatively small amounts of energy are needed.
Where batteries are not desired because they might deteriorate.
For larger power/short duration applications - as in power supply filters, or to keep power up long enough for a computer to shut down gracefully when the line power fails.
To calculate how much energy is stored in a capacitor, we start by looking at the basic relationship between voltage and current in a capacitor.
i(t) = C dv(t)/dt
Once we have this relationship, we can calculate the power - the rate of flow of energy into the capacitor - by multiplying the current flowing through the capacitor by the voltage across the capacitor.
P(t) = i(t)v(t)
Given the expression for the power:
P(t) = i(t)v(t)
And given the expression for the current:
i(t) = C dv(t)/dt
We can use the expression for current in the power expression:
P(t) = (C dv(t)/dt) v(t)
We can recognize that power is simply rate of energy input.
P(t) = dE/dt = (C dv(t)/dt) v(t)
Now, the derivative of energy can be integrated to find the total energy input.
P(t) = dE/dt = (C dv(t)/dt) v(t)
gives
Now, assuming that the initial voltage is zero (there is no energy stored in the capacitor initially, we find that the energy stored in a capacitor is proportional to the capacitance and to the square of the voltage across the capacitor.
Ec = (1/2)CV2
The expression for the energy stored in a capacitor resembles other energy storage formulae.
For kinetic energy, with a mass, M, and a velocity, v.
EM = (1/2)MV2
For potential energy, with a spring constant, K, and an elongation, x.
ESpring = (1/2)Kx2
Since the square of the voltage appears in the energy formula, the energy stored is always positive. You can't have a negative amount of energy in the capacitor. That means you can put energy into the capacitor, and you can take it out, but you can't take out more than you put in.
Power in to the capacitor can be negative. Voltage can be positive while current is negative. Imagine a capacitor that is charged. You could charge a capacitor by putting a battery across the capacitor, for example. Then, if you placed a resistor across the capacitor, charge would leave the capacitor - current would flow out of the capacitor - and the energy in the capacitor would leave the capacitor only to become heat energy in the resistor. When energy leaves the capacitor, power is negative.
When you use capacitors in a circuit and you analyze the circuit you need to be careful about sign conventions. Here are the conventions we used, and these conventions were assumed in any results we got in this lesson.
Frequency Dependent Behavior For A Capacitor
We start with a capacitor with a sinusoidal voltage across it.
where:
vC(t) = Voltage across the capacitor
iC(t) = Current through the capacitor
C = Capacitance (in farads)
We will assume that the voltage across the capacitor is sinusoidal:
vC(t) = Vmax sin(wt)
Knowing the voltage across the capacitor allows us to calculate the current:
iC(t) = C dvC(t)/dt = wC Vmax cos(wt) = Imax cos(wt)
where Imax = wC Vmax
Comparing the expressions for the voltage and current we note the following.
The voltage and the current are both sinusoidal signals (a sine function or a cosine function) at the same frequency.
The current leads the voltage. In other words, the peak of the current occurs earlier in time than the peak of the voltage signal.
The current leads the voltage by exactly 90o. It will always be exactly 90o in a capacitor.
The magnitude of the current and the magnitude of the voltage are related:
Vmax/Imax = 1/ wC
Now, with these observations in hand, it is possible to see that there may be an algebraic way to express all of these facts and relationships. The method reduces to the following.
If we have a circuit with sinusoidally varying voltages and currents (as we would have in a circuit with resistors, capacitors and inductors and sinusoidal voltage and current sources) we associate a complex variable with every voltage and current in the circuit.
The complex variable for a voltage or current encodes the amplitude and phase for that voltage or current.
The voltage and current variables can be used (using complex algebra) to predict circuit behavior just as though the circuit were a resistive circuit.
We need to do two things here. First, we can illustrate what we mean with an example. Secondly, we need to justify the claim above. We will look at an example first, and we will do two examples. The first example is jsut the capacitor - all by itself. The second example will be one that you have considered earlier, a simple RC low-pass filter.
Example 1 - The Capacitor
In a capacitor with sinusoidal voltage and currents, we have:
where:
vC(t) = Voltage across the inductor
vC(t) = Vmax sin(wt)
iC(t) = Current through the inductor
iC(t) = wC Vmax cos(wt) = Imax cos(wt)
C = Capacitance (in farads)
We represent the voltage with a complex variable, V. Considering this as a complex variable, it has a magnitude of Vmaxand and angle of 0o. We would write:
V = Vmax/0o
Similarly, we can get a representation for the current. However, first note:
iC(t) = wC Vmax cos(wt) = Imax cos(wt) = Imax sin(wt + 90o)
(Here you must excuse the mixing of radians and degrees in the argument of the sine. The only excuse is that everyone does it!) Anyhow, we have:
I = Imax/90o = j Imax = jwC Vmax
Where j is the square root of -1.
Then we would write:
V/I = Vmax/jwC Vmax = 1/jwC
and the quantity 1/jwC is called the impedance of the capacitor. In the next section we will apply that concept to a small circuit - one you should have seen before.
Before moving to the next section, a little reflection is in order. Here are some points to think about.
A phasor summarizes information about a sinusoidal signal. Magnitude and phase information are encoded into the phasor. Frequency information is not encoded, and there is a tacit assumption that all signals are of the same frequency, which would be the case in a linear circuit with sinusoidal voltage and current sources.
We looked at a case where we encoded a signal Vmax sin(wt) into a phasor of Vmax/0o. That was completely arbitrary, and many others would have encoded Vmax cos(wt) into a phasor of Vmax/0o.
Phasors are intended only to show relative phase information, and it doesn't matter which way you go.
Using Impedance
In the last section we began to talk about the concept of impedance. Let us do that a little more formally. We begin by defining terms.
A sinusoidally varying signal (vC(t) = Vmax sin(wt) for example) will be represented by a phasor, V, that incorporates the magnitude and phase angle of the signal as a magnitude and angle in a complex number. Examples include these taken from the last section. (Note that these phasors have nothing to do with any TV program about outer space.)
vC(t) = Vmax sin(wt)
is represented by a phasor V = Vmax/0o
iC(t) = Imax sin(wt + 90o)
is represented by a phasor I = Imax/90o
va(t) = VA sin(wt + f)
is represented by a phasor Va = VA/f
Next, we can use the relationships for voltage and current phasors to analyze a circuit. Here is the circuit.
Now, this circuit is really a frequency dependent voltage divider, and it is analyzed differently in another lesson. However, here we will use phasors. At the end of this analysis, you should compare how difficult it is using phasors to the method in the other lesson.
We start by noting that the current in the circuit - and there is only one current - has a phasor representation:
I = Imax/0o
We will use the current phase as a reference, and measure all other phases from the current's phase. That's an arbitrary decision, but that's the way we will start.
Next we note that we can compute the voltage across the capacitor.
VC = I/jwC
This expression relates the current phasor to the phasor that represents the voltage across the capacitor. The quantity 1/jwC is the impedance of the capacitor. In the last section we justified this relationship.
We can also compute the phasor for the voltage across the resistor.
VR = IR
This looks amazingly like Ohm's law, and it is, in fact, Ohm's law, but it is in phasor form. For that matter, the relationship between voltage and current phasors in a capacitor - just above - may be considered a generalized form of Ohm's law!
Now, we can also apply Kirchhoff's Voltage Law (KVL) to compute the phasor for the input voltage.
VIN = VR + VC = IR + I/jwC = I(R + 1/jwC)
You should note the similarities in what happens here and what happens when you have two resistors in series.
If you have a resistor, R, and a capacitor, C, in series, the current phasor can be computed by dividing the input voltage phasor by the sum of R and 1/jwC.
If you have two resistors in series (call them R1and R2), the current can be computed by dividing the input voltage by the sum of R1and R2.
