ECE 220 Network Analysis I

Lesson 7. Inductors and Capacitors and Their Uses


Capacitors

Capacitors are usually nothing more than 2 pieces of foil (the "plates") separated from each other by an insulating material (the dielectric).

In order to save space, the foil is often wrapped up into a tube as shown. One piece of the foil is clearly on the inside while the other is on the outside. Any "noise" or interference in the environment, be it electric, magnetic, or electromagnetic, would strike the outside foil. For this reason, the outside foil is frequently grounded (or connected as electrically close to ground as possible) so that such noise will be shunted to ground and will not be present in the output of the circuit.

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A rough sketch of a capacitor is shown with an indication of which lead is connected to the outside foil.

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The circuit symbol for a capacitor is shown below.  Note that the "passive sign convention" is used:  Current is shown flowing into the positive side of the capacitor.  This is the same convention as used for the resistor and inductor.

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The curved side of the capacitor in the circuit symbol used to have a special meaning.  It was the side to be connected to the more negative potential, as in the electrolytic capacitor described below.  This special significance has been lost, and capacitors now may be shown in any orientation.

Capacitors are used in circuits to smooth voltages, reduce hum, dampen spikes, and to link circuits.  For example, a capacitor might be used between two amplifier sections to block DC but allow time-varying signals to pass.  Another common use of capacitors is in frequency-selection circuits (filters).

The units of capacitance are farads (F).  As the farad is quite large, capacitors are frequently rated in microfarads (µF).

Capacitors are rated by capacitance and voltage rating.  For example, a capacitor may be rated at 25 F and 80 V.  The voltage rating indicates the maximum voltage that the capacitor can tolerate without breaking down.  There is a tradeoff between capacitance and voltage.  Manufacturers can make capacitance higher by making the dielectric thinner, but that lowers the voltage rating lower.

The basic equation for the capacitor is:

i = C dv
dt

This equation has an important implication:  You cannot change the voltage across a capacitor instantaneously.  An abrupt change in voltage (for example, from 3 V to 5 V) would result in an infinite value of dv/dt, and therefore an infinite value of the current i.  Since this is physically impossible, we must conclude that the voltage on a capacitor can change only gradually from one value to another.

Here's the equation for the capacitor in integral form:

v(t) =  1
C
i dt + K

 K is a constant of integration.  This form of the equation is usually not convenient, as it is not clear how to find the value of K.  The definite integral is a more useful form:

v(t) =     t
 1
C
   0 
i(s) ds + V(0)

The voltage across the capacitor at time t = 0 must be known.  The integral is from s = 0 to present time t.  Note that s is a dummy variable that stands in for time t in the integral; this s is a placeholder that disappears in the final result.

Another important equation for the capacitor relates voltage to charge:

Q = CV

Click this link: http://micro.magnet.fsu.edu/electromag/java/capacitor/ for a simulation on charging a capacitor.

The energy stored in a capacitor is given by the following equation:

W = CV2/2

The energy is stored in the electric field that develops between the plates.

The electrolytic capacitor is a special type of capacitor. It provides more capacitance in a given space and at a lower cost per microfarad than most others. Unlike other types, they may only be used with substantially direct voltage, and they must be correctly connected with regard to polarity. Electrolytics are usually rated at rather low voltages compared to others and have a greater leakage current than others (this means that they are less ideal than others - resistance may become a factor in analyzing them.) Electrolytics usually have "+++" on one end. This indicates the end that is to be connected to the higher DC potential. If connected backwards, it could explode!

A new breed of capacitor is the ultra capacitor.  These capacitors have extremely high capacitance.  Learn more at http://www.ultracapacitors.org/index.php?option=com_content&Itemid=57&id=37&task=view.

Capacitors (even electrolytics) are very nearly ideal.  This means that the capacitor equations can be trusted.  A charged capacitor will leak, and the voltage across it will decay with time.  Nevertheless, the leakage current is small, and the voltage will take a long time to decay.

Capacitors can be connected in series and parallel to increase voltage or capacitance.  When connected in this way, they behave opposite to resistors in series parallel.  For example, a 10 F capacitor in parallel with a 5 F capacitor gives a total capacitance of 15 F.

Inductors

Inductors are basically coils of wire.  The coils may be wrapped around an iron or ferrite core to better contain the magnetic flux (and increase the inductance).

The circuit symbol for an inductor is shown below.  The "passive sign convention" is used:  Current is shown flowing into the positive end of the inductor.  This is the same convention as used for the resistor and capacitor.

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Inductors are used for one or both of these properties: inductance and magnetism. 

Inductance is measured in henries (H).  Many inductors have smaller inductance, measured in millihenries (mH).

The basic equation for the inductor is:

v = L di
dt

This equation has an important implication:  You cannot change the current through an inductor instantaneously.  An abrupt change in current (for example, from 3 A to 0) would result in an infinite value of di/dt, and therefore an infinite value of the voltage v.  Since this is physically impossible, we must conclude that the current through an inductor can change only gradually from one value to another.  For example, if you have current flowing through an inductor in series with a toggle switch, and you open the toggle switch, a high voltage will be developed across the switch contacts causing an arc as the current rapidly decreases to zero.

Integral forms of this equation are similar to the capacitor equations:

i(t) =  1
 L
v(s) ds + K

Again, tke K is difficult to determine, so the definite integral form is preferred:

i(t) =     t
 1
 L
   0 
v(s) ds + I(0)

The integral is from s = 0 to present time t.  The current at time t = 0 must be known.

The energy stored in an inductor is given by the following equation:

W = I2L/2

The energy is stored in the magnetic field.

Inductors are not particularly ideal.  Since they are (usually) made of copper wire, they will have some resistance.  For the purposes of this class, we will assume that the inductor is an ideal device, that is, that it has zero resistance.

Except for their magnetic properties, inductors are not often introduced in circuits.  Capacitors are much more common.

Inductors in series and parallel are treated in exactly the same way as resistors in series and parallel.

Example Problem

In the circuit below, v(t) = 5 + 12sin(300t) V, L = 200 mH, and C = 25 F.  The current in the inductor is 0.15 A at t = 0.  Find i as a function of time.

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Note that the voltage of the source (5 + 12sin(300t) V) is also the voltage across the inductor and the voltage across the capacitor.  We can therefore find iL and iC from the equations for the inductor and capacitor.  Then we will use Kirchhoff's Current Law to find i.  First, let's find the currents in the capacitor and the inductor.

iC = C dv
dt
= 25×10-6 d
dt
[5 + 12sin(300t)]

iC = 2510-6 [3600cos(300t)] = .09cos(300t) A

iL(t) =     t
 1
 L
   0 
v(s) ds + I(0) =     t
 1
.2
    0
[5 + 12sin(300s)]ds + .15 = 5[5s -  12 
300
cos(300s)  t
] + .15
 0

iL(t) = 5[5t - .04cos(300t) + .04] + .15 = 25t - .2cos(300t) + .35 A

When solving definite integrals such as the above, it's important to consider the lower limit, the evaluation of the function at s = 0.

Now we'll solve Kirchhoff's Current Law at the upper node.

-i + iL + iC = 0

i = 25t - .2cos(300t) + .35 + .09cos(300t) = 25t - .11cos(300t) + .35 A


Before going on to the homework, you should complete Tutorial 7 on inductors and capacitors.


Homework Problems

1. Label the diagram with currents and voltages.  Use Kirchhoff's laws to write equations to solve for the voltage across and the current through each element.  Do not solve.
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2. v(t) = 5t volts, L = 50 H, C = 450 µF, R = 6 kΩ.  The downward current in the inductor is 3 mA at t = 0.  Find the magnitude and direction of the current flowing in all four elements as a function of time.  Simplify the results.

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3. i(t) = 4 + e-150t amps, R = 5 Ω, L = 60 mH, C = 700 µF.  The voltage across the capacitor is 12 V at t = 0, with the left side of the capacitor being more positive than the right.  Find v as a function of time.  Simplify your answer.

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4. A 500 F capacitor has 150 V across it. If all its energy is put into kinetic energy of a .05 gram BB (a BB is a small metal pellet), how fast would the BB go?

5. The voltage on a 40 µF capacitor is found to be 12(1 - e-8t) volts. Find the voltage at t = 0, and the voltage at t = 0.1 second.  How much energy is stored on the capacitor as t approaches infinity?

BONUS (No partial credit.) A candle was burning in an elevator at the top floor of a tall building.  The cable snapped, and while the elevator was falling, the candle went out.  What caused the candle to go out?  This is not a trick question.  There is a perfectly logical explanation based on physics.