ENT 172 Circuit Analysis and Design. 3 Credits
Analysis and design of reactive circuits, including use of phasor and J-operator techniques. Covers capacitors, inductors, transformers and filters, and use of electronic instrumentation. Prerequisites: ENT 171; MAT 104 or higher, completed or concurrent. 1 class/1 rec./5 lab hrs. (open-lab format).
ENT 172 assumes a comprehensive knowledge of the content of ENT 171. ENT 171 studied resistive circuits and the three fundamental electrical laws: Ohm's, and Kirchoff's Voltage and Current Laws. Please review parts 1 and 2 (below) and study thoroughly textbook chapters 2, 3, 9, 10, and especially 11. The better you prepare- the better you will do in this course.
Part 1 Review. Sinusoidal Waveforms- AC current, voltage, period, frequency and phase
AC Signals- Voltage and Current
Information is often expressed as voltage variation over time. Analog signals vary continuously, digital signals vary in discreetly (0, 1 states- low, higher voltage states). Sinusoidal waveforms are of a single frequency f (measured in cycles per second = Hz). The frequency indicates who many complete cycles (360 degrees) are completed in one second of time. The ac Function Generator (FG) in the lab generates ac voltage waveforms that can be adjusted (amplitude, and frequency). The Period of a waveform "T" is the time it takes to complete one cycle (360 degrees). T is the inverse of f. T = 1/ f
RMS voltage and current values of an ac waveform will be used most of the time throughout this course. RMS is the effective value which means that a 1v rms signal will "heat a resistor" the same as a 1v DC signal.
Two or more AC voltages or currents can be out of step with each other (in time). The two waveforms are not synchronized: their peaks and zero points do not match up at the same points in time. The following graph illustrates an example of this:
The two waves shown above (A versus B) are of the same amplitude and frequency, but they are out of step with each other. In technical terms, this is called a phase shift. Earlier we saw how we could plot a "sine wave" by calculating the trigonometric sine function for angles ranging from 0 to 360 degrees, a full circle. The starting point of a sine wave was zero amplitude at zero degrees, progressing to full positive amplitude at 90 degrees, zero at 180 degrees, full negative at 270 degrees, and back to the starting point of zero at 360 degrees. We can use this angle scale along the horizontal axis of our waveform plot to express just how far out of step one wave is with another:
The shift between these two waveforms is about 45 degrees, the "A" wave being ahead of the "B" wave. A sampling of different phase shifts is given in the following graphs to better illustrate this concept:
Because the waveforms in the above examples are at the same frequency, they will be out of step by the same angular amount at every point in time. For this reason, we can express phase shift for two or more waveforms of the same frequency as a constant quantity for the entire wave, and not just an expression of shift between any two particular points along the waves. That is, it is safe to say something like, "voltage 'A' is 45 degrees out of phase with voltage 'B'." Whichever waveform is ahead in its evolution is said to be leading and the one behind is said to be lagging.
Phase shift, like voltage, is always a measurement relative between two things. There's really no such thing as a waveform with an absolute phase measurement because there's no known universal reference for phase. Typically in the analysis of AC circuits, the voltage waveform of the power supply is used as a reference for phase, that voltage stated as "xxx volts at 0 degrees." Any other AC voltage or current in that circuit will have its phase shift expressed in terms relative to that source voltage.
REVIEW Part 2:
Ohm's & Kirchoff's Laws
The symbols V and are often used interchangeably. Technically, represents a voltage rise, and V a voltage drop (however, this distinction has been lost over the years)
Here is a simple simulated experiment to
re-enforce understanding of Ohm's Law. Note that the red box represents
the circuit voltage source (EMF), and the gray boxes are meters to
measure voltage (voltmeter) and current (ammeter). You are to
click on the buttons to increase/decrease the applied voltage and resistance,
and observe the resulting current. Does Ohm's Law accurately predict the result?
(YES it does!).Use your browser's "Back" button to return. Press here
to begin the experiment.
Example 1: The Simple Loop Circuit and application of Ohm's Law
An EMF source of 6.0V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-free. What is the resistance of the lamp?
Schematic diagram of the circuit in this problem.
|The gain of
potential energy occurs within the battery
(EMF adds energy)
No energy is lost in the wires, since they are assumed to be resistance-free (V = I x R = 2A X 0 = 0v). The potential that was gained (6.0V) must be consumed by the resistor. So, according to Ohm's Law:
V = I R
R=V / I = 6v/ 2A = 3ohm
R = 3.0
Voltage is the potential difference between two points. This potential difference is expressed by a POLARITY marker. Polarity ( + and - or red and black) indicates this difference in energy. Consider the positive point to be at A HIGHER electrical energy level relative to the negative point. It is obvious that you must have TWO points to express a difference in potential.
Voltage does not exist at a single point. (V = 0 because there is no difference between a point and itself)
If you track through any loop of an electrical circuit or system, you will discover that the sum of all voltage RISES exactly equals the sum of all the voltage DROPS in the loop. This is true for ALL loops in all circuits and systems. As you track through the loop, if you travel through a component from the positive point (terminal) through the component to the negative terminal- you have encountered a VOLTAGE DROP. Traveling from negative to positive constitutes a voltage RISE for that component. The polarity marker is not arbitrarily set- it is determined by current.
It is vital that you remember that the DIRECTION of current flow determines the polarity of every component in every loop of a circuit or system.
Example: Assume that you have four resistances and a 10v power supply connected in a single loop (series connection). If the first three resistance drop a collective 7v, what must the voltage drop be across the fourth resistance? (Ans. 3v).
A voltage diagram of a circuit loop (conducting current) shows how and where each voltage rise and drop occurs. Pay close attention to when a potential difference is positive (+) or negative (-) and the difference in a voltage from one arbitrary circuit point to another and in the reverse ( second point back to the first point). Also note that the direction of current flow (clockwise vs counterclockwise) determines the polarity of the voltage developed across all circuit components (except for the voltage source).
From the example above, determine Vad, Vce, Vda
Ans. Vad = Vae = - 12v
Vce = Vcd = - 5v
Vda = Vdc + Vcb + Vba = (+5v) + (+3v) + (+4v) = + 12v
Kirchoff's Laws: Current (KCL) & Voltage (KVL). Review chapters 9, and 10 in the textbook.
Kirchhoff's Current Law
|This fundamental law results
from the conservation of charge. It applies to a junction or node in a
circuit -- a point in the circuit where charge has several possible paths to
In Figure 1, we see that IA is the only current flowing into the node. However, there are three paths for current to leave the node, and these current are represented by IB, IC, and ID.
Once charge has entered into the node, it has no place to go except to leave (this is known as conservation of charge). The total charge flowing into a node must be the same as the the total charge flowing out of the node. So,
IB + IC + ID = IA
Bringing everything to the left side of the above equation, we get
(IB + IC + ID) - IA = 0
KCL applied. Ia = Ib + Ic + Id
Then, the sum of all the currents is zero. Note the convention we have chosen here: current flowing into the node are taken to be negative, and currents flowing out of the node are positive.
There are two laws necessary for solving circuit problems. For simple circuits, we have been applying these equations almost instinctively.
The voltages around a closed path in a circuit must sum to zero. (Kirchoff's Voltage Law), the voltage drops being negative (following a current through a resistor), while the gains are positive (going through a battery from the negative to the positive terminal).
The sum of the currents entering a node must equal the sum of the currents exiting a node. (Kirchoff's Current Law)
The first law is a simple statement of the meaning of potential. Since every point on a circuit has a unique value of the potential, travelling around the circuit, through any path must bring you back to the potential. Using the analogy to elevation: If one hikes from a starting point of a mountain, taking several paths, then finishes at the same point, the sum of the elevation changes of each path had better add to zero.
|The second law is the statment of current conservation mentioned before in the Ohm's law lecture. For the node on the right, i1=i2+i3. If all currents had been defined as enterning the node, then the sum of the currents would be zero.|
When to use Kirchoff's laws. Series-parallel circuits. (Refer to Chapter 11 in the textbook).
Although Kirchoff's laws were used to derive the simple relations for adding resistors in series and parallel, not all problems can be broken down into such simple pieces. Below are three diagrams which may need to be solved (e.g. given the resistances and voltages of the batteries, find all the currents).
The left-side figure can be treated by breaking up the circuit into pieces and applying the rules for adding resistances in parallel and series. The resistors R3, R4 and R5 can be treated as a single resistor with resistance R345, which can then be added in parallel with R2, to give an effective resistance R2345. The battery sees a net resistance R = R1 + R2345.
Now write down Kirchoff's laws then solve the equations.
instance consider the right-side figure. The 3 currents through the 3
i3 are unknown.
The three unknowns can be found by solving 3 equations:
The first equation describes current conservation into the the node. Note that i1 is defined as flowing into the node (to the right) while i2 and i3 are defined as flowing out of the node (down and to the right).
The second equation expresses the requirement that voltage losses cancel voltage gains for the loop on the left.
The third equation expresses the sum of voltages around the right loop. Note that if one makes a counter-clockwise loop, one goes against the current through R3 and the sign of that term is therefore opposite the normal one.
Remember that one gains the voltage of the battery when traveling from the small hash mark (minus terminal) and leaving from the large hashmark (positive terminal) of the battery. One could also imagine writing equations for a third loop, that travels all the way around the circuit, through both batteries. However, such an equation is merely a linear combination of Eq. 2 and Eq. 3 above.
Find the currents through all the resistors in the circuit below:
DATA: Va = 12 V, Vb = 12 V, R1 = 10 W, R2 = 15 W, R3 = 20 W
Summing the voltages around the left and right loops gives the following two equations:
One can then reduce the problem to "2 equations 2 unknowns" by substituting for i1 and obtaining 2 new equations.
One can add these two equations together to eliminate the the i3 term since all three resistors are the same, which will be now noted by R.
Once i2 is known, Eq. (4) can be used to get i3 and i1 can be found by summing i2 + i3.
i2 = 0.554 amps, i1= .369 amps, i3 = -.185 amps
ENT 172 The Capacitor:
Electric Field Energy Storage Device
A capacitor is simply the combination of two metal plates separated by an insulative "dielectric" material. The diagram shows a DC
voltage source connected in series with a capacitor (on the right)
A charge flows onto the plates & remains there when the battery is removed. The amount of charge stored depends on on the capacitance of the capacitor, which is measured on Farads (F). Capacitance is a measure of a capacitor's ability to store charge. A large capacitance means that more charge can be stored. The larger the charge differential across the plates- the larger the capacitor voltage will develop (up to the power supply voltage). The capacitor will charge up to the applied voltage level.
The separation of charge generates an "electric force field"- proportional to the established capacitor voltage level. This E Field is a fundamental "force field" that is capable of storing energy (in the dielectric material).
Capacitance is measured in farads, symbol F. However 1F is very large, so prefixes are used to show the smaller values. Below is a table of common capacitances:
|Units in shorthand||Value in Farads||Scientific notation||Written as|
The dielectric maintains a uniform separation of the plates and an efficient low-loss medium for the electric field and the storing of electric field energy. An electric field exists wherever there is a separation of electrical charge over distance- the smaller the distance and the greater the voltage difference of the plates- the higher the electric field intensity and more energy is stored. Once charged (via a charge circuit path) the capacitor will hold the charge until it is provided a means to discharge the electric field energy (discharge path). This charged state can be used to power a circuit for a while (until all the stored energy is depleted). Capacitors are often used in delay, timing, filtering, or switching circuits.
Interactive Activity: To understand better the relationship between dielectric constant, plate area, plate separation and the resulting value of capacitance, browse to: http://www.micro.magnet.fsu.edu/electromag/java/capacitance/ and experiment with the interactive utility. Then answer the following question: What is the capacitance of a mica dielectric capacitor with a plate area of 0.1 sq meters, separated a distance of 1 milli meter? (ans. 3825 pF).
"electrolytic" polarized, and regular "un-polarized".
TYPES OF CAPACITORS
There are many types of capacitor but they can be split into two groups,
Examples: Circuit symbol:
Electrolytic capacitors are polarized and they must be connected to match the marked polarity, at least one of their leads will be marked + or -
Examples: Circuit symbol:
Small value capacitors are un-polarized and may be connected either way round. They are not easily damaged by heat when soldering, except for one unusual type (polystyrene). They have high voltage ratings of at least 50V, usually 250V or so. It can be difficult to find the values of these small capacitors because there are many types of them and several different labeling systems!
Many small value capacitors have their value printed but without a
multiplier, so you need to use experience to work out what the multiplier
should be! For example 0.1 means 0.1µF = 100nF. Sometimes the
multiplier is used in place of the decimal point:
For example: 4n7 means 4.7nF.
A number code is often used on small capacitors where printing is difficult:
For example: 102 means 1000pF = 1nF (not 102pF!)
For example: 472J means 4700pF = 4.7nF (J means 5% tolerance).
A color code was used on polyester capacitors for many years. It is now obsolete, but of course there are many still around. The colors should be read like the resistor code, the top three color bands giving the value in pF. Ignore the 4th band (tolerance) and 5th band (voltage rating).
For example: brown, black, orange means 10000pF = 10nF = 0.01µF.
Note that there are no gaps between the color bands, so 2 identical bands actually appear as a wide band.
For example: wide red, yellow means 220nF = 0.22µF.
Sometimes, especially with capacitors of low capacity values, capacity may be represented with colors, similar to four-ring system used for resistors. The first two colors (A and B) represent the first two digits, third color (C) is the multiplier, fourth color (D) is the tolerance, and the fifth color (E) is the working voltage. With disk-ceramic capacitors and tubular capacitors the working voltage is not specified, because these are used in circuits with low or no DC voltage. If tubular capacitor does have five color rings on it, then the first color represents the temperature coefficient, while the other four specify its capacity value in the previously described way.
The capacity of miniature tantalum electrolytic capacitors is marked by colors. The first two colors represent the first two digits and have the same values as with resistors. The third color represents the multiplier, which the first two digits should be multiplied by, to get the capacity value expressed in µF. The fourth color represents the maximal working voltage value.
Tantalum electrolytic capacitors
One important note on the working voltage: capacitor voltage mustn't exceed the maximal working voltage as capacitor may get destroyed. In case when the voltage between nodes where the capacitor is about to be connected is unknown, the "worst" case should be considered. There is the possibility that, due to malfunction of some other component, voltage on capacitor equals the power supply voltage. If, for example, the power supply is 12V battery, then the maximal working voltage of used capacitors should exceed 12V, for security's sake.
Electrolytic capacitors represent the special type of capacitors with fixed capacity value. Thanks to the special construction, they can have exceptionally high capacity, ranging from one to several thousand µF. They are most frequently used in transformers for leveling the voltage, in various filters, etc.
There are two designs of electrolytic capacitors; axial where the leads are attached to each end and radial where both leads are at the same end. Radial capacitors tend to be a little smaller and they stand upright on the circuit board.
It is easy to find the value of electrolytic capacitors because they are clearly printed with their capacitance and voltage rating. The voltage rating should always be checked when selecting an electrolytic capacitor. It the project parts list does not specify a voltage, choose a capacitor with a rating which is greater than the project's power supply voltage. 25V is a sensible minimum for many circuits.
Electrolytic capacitors are polarized components, meaning that they have positive and negative connector, which is of outmost importance when connecting the capacitor into a circuit. Positive connector has to be connected to the node with a high voltage than the node for connecting the negative connector. If done otherwise, electrolytic capacitor could be permanently damaged due to electrolysis and eventually destroyed.
Explosion may also occur if capacitor is connected to voltage that exceeds its working voltage. In order to prevent such instances, one of the capacitor's connectors is very clearly marked with a + or -, while working voltage is printed on capacitor body.
Several models of electrolytic capacitors, as well as their symbols, are shown on the picture below.
Variable capacitors are capacitors with variable capacity. Their minimal capacity ranges from 10 to 50pF, and their maximum capacity goes as high as few hundred pF (500pF tops). Variable capacitors are manufactured in various shapes and sizes, but common feature for all of them is a set of immobile, interconnected aluminum plates called stator, and another set of plates, connected to a common axis, called rotor. In axis rotating, rotor plates get in between stator plates, thus increasing capacity of the device. Naturally, these capacitors are constructed in such a way that rotor and stator plates are placed consecutively. Insulator (dielectric) between the plates is a thin layer of air, hence the name variable capacitor with air dielectric. When setting these capacitors, special attention should be paid not to band metal plates, in order to prevent short-circuiting of rotor and stator and ruining the capacitor.
Bellow is the photo of the variable capacitor with air dielectric
a, b, c. Variable capacitors d. Trimmer capacitors
Beside the capacitors with air dielectric (a), there are also variable capacitors with solid insulator. With these, thin insulator foil occupies the space between stator and rotor, while capacitor itself is contained in a plastic casing. These capacitors are much more resistant to mechanical damage and quakes, which makes them very convenient for portable electronic devices. One such one-fold capacitor is shown on the figure 2.5b.
This type is rarely used now. Their value (in pF) is normally printed without units. Polystyrene capacitors can be damaged by heat when soldering (it melts the polystyrene!) so you should use a heat sink (such as a crocodile clip). Clip the heat sink to the lead between the capacitor and the joint.
Capacitors are not available with every possible value, for example 22µF and 47µF are readily available, but 25µF and 50µF are not!
To produce a sensible range of capacitor values you need to increase the size of the 'step' as the value increases. The standard capacitor values are based on this idea and they form a series which follows the same pattern for every multiple of ten.
The E3 series (3 values for each multiple of
10, 22, 47, ... then it continues 100, 220, 470, 1000, 2200, 4700, 10000 etc.
Notice how the step size increases as the value increases (values roughly double each time).
The E6 series (6 values for each multiple of
10, 15, 22, 33, 47, 68, ... then it continues 100, 150, 220, 330, 470, 680, 1000 etc.
Notice how this is the E3 series with an extra value in the gaps.
The E3 series is the one most frequently used for capacitors because many types cannot be made with very accurate values.
|Variable Capacitor Symbol|
Variable capacitors are mostly used in radio tuning circuits and they are sometimes called 'tuning capacitors'. They have very small capacitance values, typically between 100pF and 500pF (100pF = 0.0001µF). The type illustrated usually has trimmers built in (for making small adjustments - see below) as well as the main variable capacitor.
Many variable capacitors have very short spindles which are not suitable for the standard knobs used for variable resistors and rotary switches. It would be wise to check that a suitable knob is available before ordering a variable capacitor.
Variable capacitors are not normally used in timing circuits because their capacitance is too small to be practical and the range of values available is very limited. Instead timing circuits use a fixed capacitor and a variable resistor if it is necessary to vary the time period.
|Trimmer Capacitor Symbol|
Trimmer capacitors (trimmers) are miniature variable capacitors. They are designed to be mounted directly onto the circuit board and adjusted only when the circuit is built.
A small screwdriver or similar tool is required to adjust trimmers. The process of adjusting them requires patience because the presence of your hand and the tool will slightly change the capacitance of the circuit in the region of the trimmer!
Trimmer capacitors are only available with very small capacitances, normally less than 100pF. It is impossible to reduce their capacitance to zero, so they are usually specified by their minimum and maximum values, for example 2-10pF.
Charging a capacitor
Regulate charge/discharge time by putting a resistance in series with the capacitor. The larger the resistance- the longer it will take to charge the capacitor to the level of the voltage source. Once established, this charge time is predictable and can be used to provide a "timing" control for integrated circuits and other similar circuitry.
A good example is the intermittent feature of the automobile windshield wiper control. As you increase the wait time (between wipes) you are increasing the resistance of an DC RC series circuit and thereby increasing the charge time required. Once charged, the capacitor triggers the wiper relay switch which activates the wiper motor.
When a capacitor is connected across a voltage source, such as a battery, the voltage forces electrons onto one plate resulting in a negatively charged plate. The electrons of the other plate are pulled off by the battery resulting in a positively charged plate. Because the dielectric between the plates is an insulator, current cannot flow through it. A capacitor has a finite amount of capacity to store charges. When a capacitor reaches its capacity it is fully charged.
The following diagrams illustrate the charging of a capacitor. Figure 2 shows a circuit containing a conductor connecting a battery, an open switch, and a capacitor. The capacitor below is not charged. There is no potential difference between the plates.
When the switch is closed (below) there is a momentary surge of current through the conductor to and from the plates of the capacitor. When the current reaches the negative plate of the capacitor, it is stopped by the dielectric.
The surge of electric current to the capacitor deposits charge on the capacitor plates (extra electrons on the NEGATIVELY charged plate, and removed electrons on the POSITIVELY charged plate). This continues until the potential difference across the capacitor equals the applied voltage of the source. Then there is no further flow of current. When the capacitor is fully charged (five time constants = 5RC), the switch may be opened and the capacitor will retain its charge (Figure 4). Because of the difference of charges on the plates an electric force field is created that stores potential energy in the dielectric of the capacitor. The energy stored is the energy that was required to charge the capacitor (delivered from the voltage supply).
Discharging of a Capacitor
The charged capacitor (below) is now a source of potential energy. This potential energy is now available for its intended electronic application. If the switch is closed to the right side current will immediately begin to flow through from the negative plate to the positive plate. The capacitor is discharging.
The charged capacitor is the source of voltage for the current flow. The current will cease flowing when the charges of the two plates are again equal, meaning that the capacitor is completely discharged.
The charge/discharge time is five time constants = 5t = 5 x R x C
The time constant is what you get if you multiply the resistance with the capacitance in the following circuit. The actual time, however, will not be this value, due to the production spread. For this exercise, we are seeing how long it takes for the capacitor to charge up to 5V. The switch is in the circuit to drain the capacitor. Note that below we are dealing in microfarads. Note also that s is the symbol for seconds.
|Capacitance (mF)||Resistance (M)||Time (s)||Time Constant = RxC|
The capacitor charges rapidly at first, then slows down. The time to taken to charge is proportional to capacitance * resistance, also known as C*R.
For example, at t = 1RC, the capacitor will charge up to 63.2% of the maximum. At five time constants- the capacitor charges to over 99% (but never reaches 100%).
Capacitor discharge: In one time constant (1RC), the capacitor voltage falls to 36.8% of its fully charged voltage level (100% - 63.2% = 36.8%). At two time constants, the capacitor voltage is down to 14.5% (100% - 86.5% = 14.5%).
To practice the calculation of capacitor charge duration, browse to: http://www.micro.magnet.fsu.edu/electromag/java/timeconstant/ and select R = 10 ohms, and C = 3 uF.What is the charge time constant, and how long will it take to fully charge?
(ans. 30us, 150us)
To interactively view charging and discharging of a capacitor, browse to:
Circuits in which current is proportional to voltage are called linear circuits. The ratio of voltage to current in a resistor is its resistance. Resistance does not depend on frequency, and in resistors the two are in phase.
In most situations and for most devices, the ratio of voltage to current does depend on frequency and in general there is a phase difference. So impedance is the general name we give to the ratio of voltage to current (opposition from chapter 1). It has the symbol Z. Resistance is a special case of impedance. Another special case is that in which the voltage and current are out of phase by 90°: this is an important case because when this happens, no power is lost in the circuit. In this case where the voltage and current are out of phase by 90°, the ratio of voltage to current is called the reactance, and it has the symbol X.
We will learn how to add the impedances of various devices when connected in series, and use the "reciprocal of the sum of reciprocals" for parallel-connected devices. The impedances can only be directly added (scalar addition) when the phase angle is the same- otherwise "vector addition" is required.
The voltage on a capacitor depends on the amount of charge you store on its plates. The current flowing onto the positive capacitor plate (equal to that flowing off the negative plate) is by definition the rate at which charge is being stored. So the charge Q on the capacitor equals the integral of the current with respect to time. From the definition of the capacitance,
vC = q/C
Capacitor current is proportional to the capacitance "C" of the capacitor times the "rate-of-change" of capacitor voltage over time. In other words, if you want the capacitor voltage to increase (or decrease) quickly- you must provide more current than if the capacitor voltage changes slowly over time.
If the voltage waveform is a sine function, the the current is a cosine function- 90 degrees ahead. Therefore: capacitor current leads capacitor voltage by 90 degrees (capacitor voltage lags capacitor current by 90 degrees).
Capacitive reactance XC (as per Ohm's Law) is the ratio of the magnitude of the capacitor voltage to magnitude of the capacitor current
Xc can be calculated using the following formula:
f representing the frequency in Hz and C representing the capacity in Farads.
For example, 5nF-capacitor's reactance at f=125kHz equals:
while, at f=1.25MHz, it equals:
Capacitor has infinitely high reactance for direct current, because f=0. Therefore a capacitor "blocks" DC current (asks as an OPEN for DC). Each frequency is impeded differently by the capacitor. XC decreases over increasing frequency.
Capacitors are used in circuits for filtering signals of specified frequency. They are common components of electrical filters, oscillator circuits, etc.
The voltage is proportional to the current, and the voltage and current are related by
V = (I)XC
Recall that reactance is the name for the ratio of voltage to current when they differ in phase (If they are in phase, the ratio is called resistance.) Another difference between reactance and resistance is that the reactance Xc is frequency dependent.
Capacitors in Parallel Circuits
Capacitance can be increased in a circuit by connecting capacitors in parallel as shown in the following diagram:
We know that capacitance of a capacitor can be increased by increasing the size of its plates. Connecting two or more capacitors in parallel in effect increases plate size. Increasing plate area makes it possible to store more charge and therefore creating greater capacitance. To determine total capacitance of several parallel capacitors, simply add up their individual values. The following is the formula for calculating total capacitance in a circuit containing capacitors in parallel:CT = C1 + C2 + C3 . . .
Capacitors in Series Circuits
Capacitance can be decreased in a circuit by capacitors in series as shown in the following diagram:
We know that capacitance of a capacitor can be decreased by placing the plates further apart. Connecting two or more capacitors in series in effect increases the distance between the plates and thickness of the dielectric, thereby decreasing the amount of capacitance.
The following is the formula for calculating total capacitance in a circuit containing three capacitors in series:CT = 1/(1C1/ + 1/C2 + 1/C3). (NOTE: This is similar to the calculation for resistance in parallel).
For two capacitors in parallel the formula reduces to: CT = (C1 x C2) / (C1 + C2)
Browse to the following links for applications using capacitors:
Now research and share some of your own links.
The kind of information that is expressed as a single dimensional quantity, such as the resistance of a resistor, is called a scalar quantity. Scalar numbers give only the MAGNITUDE value of a quantity- not simultaneous PHASE information. What is needed is a two-dimensional system (complex numbers) to quantify fully a parameter such as Impedance, Voltage, and/or Current.
Complex numbers are graphical in nature and can be expressed as a VECTOR or as a J Operator expression. This section discusses the use of Vectors to express both magnitude and direction (phase). Here are some examples of common vectors.
Like distances and directions on a map, there must be some common frame of reference for angles to make sense relative to each other. We will use the mathematical angle convention. That means that directly right is considered to be 0o, and angles are counted in a positive direction going counter-clockwise, and negative running from 0 clockwise.
The idea of representing a number in graphical form is use of the "number line"
Addition and subtraction (in a single dimension) is accomplished by extending/reducing one value by a second. Only two quantities can be handled at a time.
These fields of numbers (integer, fractional, rational, irrational, real, etc.) are all one-dimensional. The number line illustrates this graphically. You can move up or down the number line (positively or negatively), but all "motion" along that line is restricted to a single axis. One-dimensional, scalar numbers are perfectly adequate for resistances, and same phase voltages and currents- but can't handle differing phase information AT THE SAME TIME. To represent both magnitude and phase simultaneously, we need multidimensional representations- phase vectors or PHASORS. Reactance, impedance, voltage and current are all vector quantities- having both a simultaneous length (magnitude) and a direction (phase). Phase will be expressed in degrees relative to the 0 degree axis in the two dimensional complex number plane. Horizontal axis is magnitude (minus infinity to plus infinity) and the vertical axis represents phase (0 to 360 degrees).
The length of a vector represents the magnitude (or amplitude) of the waveform
The greater the amplitude of the waveform, the greater the length of its corresponding vector. The angle of the vector, however, represents the phase shift in degrees between the waveform in question and another waveform. Remember that phase is always a relative measurement between two waveforms rather than an absolute property.
The greater the phase shift in degrees between two waveforms, the greater the angle difference between the corresponding vectors.
Remember that phasors are mathematical objects: they can be added, subtracted, multiplied, and divided. Addition is perhaps the easiest vector operation to visualize, so we'll begin with that. If vectors with common angles are added, their magnitudes (lengths) add up directly. An example of this is the addition of two phasors- both of phase = 0 degrees.
Similarly, if AC voltage sources with the same phase angle are connected together in series, their voltages add just as you might expect with DC voltages.
Note the (+) and (-) polarity marks next to the leads of the two AC sources. Even though we know AC doesn't have "polarity" in the same sense that DC does, these marks are essential to knowing how to reference the given phase angles of the voltages. If vectors directly opposing each other (180o out of phase) are added together, their magnitudes (lengths) subtract just like positive and negative scalar quantities subtract when added:
Similarly, if opposing AC voltage sources are connected in series, their voltages subtract as you might expect with DC sources connected in an opposing fashion:
Determining whether or not these voltage sources are opposing each other requires an examination of their polarity markings (phase angles). See the example below. Opposing voltages subtract when connected in series.
The resultant voltage can be expressed in two different ways: 2 volts at 180o with the (-) symbol on the left and the (+) symbol on the right, or 2 volts at 0o with the (+) symbol on the left and the (-) symbol on the right. A reversal of wires from an AC voltage source is the same as phase-shifting that source by 180o.
If vectors with uncommon phase angles are added, their magnitudes (lengths) add up quite differently than that of scalar magnitudes:
If two AC voltages -- 90o out of phase -- are added together by being connected in series, their voltage magnitudes do not directly add or subtract as with scalar voltages. Instead, these voltage quantities are complex quantities, and just like the above orthogonally-oriented phasors (which add up in a trigonometric fashion) a 6 volt ac source at 0o added to an 8 volt ac source at 90o (same frequency) results in 10 volts at a phase angle of 53.13o:
Compared to DC circuit analysis, this is very strange indeed. Note that it's possible to obtain voltmeter indications of 6 and 8 volts, respectively, across the two AC voltage sources, yet only read 10 volts for a total voltage- NOT 14v ! Magic... How does this work?
There are two basic forms of complex number notation: polar and rectangular.
Polar form is where a complex number is denoted by the length (magnitude) and the angle of its vector. Here are two examples of vectors and their polar notations:
Standard orientation for vector angles in AC circuit calculations defines 0o as being to the right (horizontal), making 90o straight up, 180o to the left, and 270o straight down. Please note that vectors angled "down" can have angles represented in polar form as positive numbers in excess of 180, or negative numbers less than 180. For example, a vector angled ∠ 270o (straight down) can also be said to have an angle of -90o. The above vector on the right (5.4 ∠ 326o) can also be denoted as 5.4 ∠ -34o.
Rectangular form, on the other hand, is where a complex number is denoted by its respective horizontal and vertical components. In essence, the phasor is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. Rather than describing a phasor's length and direction by denoting magnitude and angle, it is described in terms of "how large left/right" and "how large up/down."
These two dimensional figures (horizontal and vertical) are symbolized by two numerical figures. In order to distinguish the horizontal and vertical dimensions from each other, the vertical is prefixed with a lower-case "j". These lower-case letters do not represent a physical variable (such as instantaneous current, also symbolized by a lower-case letter "i"), but rather are mathematical operators used to distinguish the vector's vertical component from its horizontal component. As a complete complex number, the horizontal and vertical quantities are written as a sum:
COMPLEX NUMBER PLANE
The horizontal component is referred to as the real component, since that dimension is compatible with normal, scalar ("real") numbers. The vertical component is referred to as the imaginary component.
Either method of notation is valid for complex numbers. The primary reason for having two methods of notation is for ease of longhand calculation, rectangular form lending itself to addition and subtraction, and polar form lending itself to multiplication and division.
Conversion between the two notational forms involves simple trigonometry.
To convert from polar to rectangular, find the real component by multiplying the polar magnitude by the cosine of the angle, and the imaginary component by multiplying the polar magnitude by the sine of the angle. This forms a right triangle, the hypotenuse of the triangle representing the vector itself (its length and angle with respect to the horizontal constituting the polar form), the horizontal and vertical sides representing the "real" and "imaginary" rectangular components, respectively.
To convert from rectangular to polar, find the polar magnitude through the use of the Pythagorean Theorem (the polar magnitude is the hypotenuse of a right triangle, and the real and imaginary components are the adjacent and opposite sides, respectively), and the angle by taking the arctangent of the imaginary component divided by the real component:
Complex numbers can be added, subtracted, multiplied, divided, squared, inverted, integrated, etc.
Addition and subtraction with complex numbers in rectangular form is easy. For addition, simply add up the real components of the complex numbers to determine the real component of the sum, and add up the imaginary components of the complex numbers to determine the imaginary component of the sum
When subtracting complex numbers in rectangular form, simply subtract the real component of the second complex number from the real component of the first to arrive at the real component of the difference, and subtract the imaginary component of the second complex number from the imaginary component of the first to arrive the imaginary component of the difference
When multiplying complex numbers in polar form, multiply the polar magnitudes of the complex numbers to determine the polar magnitude of the product, and add the angles of the complex numbers to determine the angle of the product
Division of polar-form complex numbers is also easy: simply divide the polar magnitude of the first complex number by the polar magnitude of the second complex number to arrive at the polar magnitude of the quotient, and subtract the angle of the second complex number from the angle of the first complex number to arrive at the angle of the quotient
To obtain the reciprocal, or "invert" (1/x), a complex number, simply divide the number (in polar form) into a scalar value of 1, which is nothing more than a complex number with no imaginary component (angle = 0)
These are the basic operations you will need to know in order to manipulate complex numbers in the analysis of AC circuits. Operations with complex numbers are by no means limited just to addition, subtraction, multiplication, and division, Virtually any arithmetic operation that can be done with scalar numbers can be done with complex numbers.
Here is an example of phasor addition of two out of phase voltages (but same frequency):
The polarity markings show these two voltage sources aiding each other, so to determine the total voltage across the resistor we must add the voltage figures of 10 V ∠ 0o and 6 V ∠ 45o together to obtain 14.861 V ∠ 16.59o.
Let's connect three AC voltage sources in series and use complex numbers to determine the resultant total applied voltage. All the rules and laws learned in ENT 171 apply: (Ohm's Law, Kirchhoff's Laws, network analysis methods), The only qualification is that all variables must be expressed in complex form, taking into account phase as well as magnitude, and all voltages and currents must be of the same frequency (in order that their phase relationships remain constant).
The polarity marks for all three voltage sources are oriented in such a way that their stated voltages should add to make the total voltage across the load resistor. Notice that although magnitude and phase angle is given for each AC voltage source, no frequency value is specified. If this is the case, it is assumed that all frequencies are equal. KVL yields:
Graphically, the vectors add up in this manner:
The sum of these vectors will be a resultant vector originating at the starting point for the 22 volt vector (dot at upper-left of diagram) and terminating at the ending point for the 15 volt vector (arrow tip at the middle-right of the diagram):
In order to determine what the resultant vector's magnitude and angle are without resorting to graphic images, we can convert each one of these polar-form complex numbers into rectangular form and add. Remember, we're adding these figures together because the polarity marks (phase) of the three voltage sources are oriented in an aiding manner:
In polar form, this equates to 36.8052 volts ∠ -20.5018o. What this means in real terms is that the voltage measured across these three voltage sources will be 36.8052 volts, lagging the 15 volt (0o phase reference) by 20.5018o. A voltmeter connected across these points in a real circuit would only indicate the polar magnitude of the voltage (36.8052 volts), not the angle. An oscilloscope could be used to display two voltage waveforms and thus provide a phase shift measurement, but not a voltmeter. The same principle holds true for AC ammeters: they indicate the polar magnitude of the current, not the phase angle.
COMPUTER Assignment: L210 Lab Breadboard Analysis software:
We can use BREADBOARD software (in the L210 lab) to verify our results. In the test circuit above: a 10 kΩ resistor value at a power frequency of 60 Hz has been selected. A ground reference is required as well. The 1, 2, 3, numbers specify branch values (required by the software).
v1 from 1-0 ac 15 0 v2 from 2 1 ac 12 35 v3 from 3 2 ac 22 -64 r1 from 3 0 10K freq v(3-0) angle 6.000E+01 3.681E+01 -2.050E+01
This result confirms our calculations (above).
The resultant (sum) vector should begin at the upper-left point (origin of the 22 volt vector) and terminate at the right arrow tip of the 15 volt vector:
The connection reversal on the 12 volt supply can be represented in two different ways in polar form: by an addition of 180o to its vector angle (making it 12 volts ∠ 215o), or a reversal of sign on the magnitude (making it -12 volts ∠ 35o). Either way, conversion to rectangular form yields the same result:
The resulting addition of voltages in rectangular form
In polar form, 30.4964 V ∠ -60.9368o.
RC series Phasor Diagram
When we connect components together, Kirchoff's laws apply at any instant of time. So the voltage v(t) across a resistor and capacitor in series is just
vseries(t) = vR(t) + vC(t)
Note that Vr has zero phase (resistor voltage and current are in phase), and Vc has a -90 degree phase angle (see above- Vc leads Ic by 90 degrees). Since It = Ir = Ic (series circuit), Vr leads Vc by 90 degrees.
and w =
From Pythagoras' theorem:
Vt = V2RC = V2R + V2C
V is proportional to I. Their ratio is the series impedance, Zseries and so for this series circuit,
Note the frequency dependence of the series impedance ZRC: at low frequencies, the impedance is very large, because the capacitive reactance 1/wC is large (the capacitor is open circuit for DC). At high frequencies, the capacitive reactance goes to zero (the capacitor doesn't have time to charge up) so the series impedance goes to R.
Remember how, for two resistors in series, you could just add the resistances: Rseries = R1 + R2 to get the resistance of the series combination. That simple result comes about because the two voltages are both in phase with the current, so their phasors are parallel. Because the phasors for reactances are 90° out of phase with the current, the series impedance of a resistor R and a reactance X are given by Pythagoras' law:
Zseries2 = R2 + X2 .
We can rearrange the equations above to obtain the current flowing in this circuit. Alternatively we can simply use the Ohm's Law analogy and say that I = Vsource/ZRC.
From simple trigonometry, the phase angle by which the current leads the voltage is
tan-1 (VC/VR) = tan-1 (IXC/IR) = tan-1 (XC/R)
= tan-1 (1/wRC) = tan-1 (1/2pfRC).
REMEMBER: As frequency increases, the total circuit phase angle becomes increasingly positive (decreasingly negative).
At low frequencies, the impedance of the series RC circuit is dominated by the capacitor, so the voltage is 90° behind the current. At high frequencies, the impedance approaches R and the phase difference approaches zero. The frequency dependence of Z and f are important in the applications of RC circuits. The voltage is mainly across the capacitor at low frequencies, and mainly across the resistor at high frequencies. Of course the two voltages must add up to give the voltage of the source, but they add up as vectors.
V2RC = V2R + V2C.
At the frequency w = wo = 1/RC, the phase f = 45°
Thus for a capacitor, the ratio of voltage to current decreases with frequency (Xc decreases, and therefore It increases for a constant Vt). This can be used for filtering different frequencies in electronic systems.
Examples of RC Series Circuit Analysis
The resistor will offer 5 Ω of resistance regardless of frequency, while the capacitor will offer 26.5258 Ω of reactance at 60 Hz. Because the resistor's resistance is a real number (5 Ω ∠ 0o, or 5 + j0 Ω), and the capacitor's reactance is an imaginary number (26.5258 Ω ∠ -90o, or 0 - j26.5258 Ω), the combined effect of the two components will be an opposition to current equal to the complex sum of the two numbers. The term for this complex opposition to current is impedance, its symbol is Z, and it is also expressed in the unit of ohms. In the exampleabove, the total circuit impedance is:
As with the purely capacitive circuit, the current wave is leading the voltage wave (of the source), although in this example the phase angle is 79.325o (instead of a full 90o).
Let's place out known figures for this series circuit into a table. Ohm's Law (E=IR) can be applied to determine voltage across the resistor and capacitor:
Notice how the voltage across the resistor has the exact same phase angle as the current through it, telling us that E and I are in phase (for the resistor only). The voltage across the capacitor has a phase angle of -10.675o, exactly 90o less than the phase angle of the circuit current. This tells us that the capacitor's voltage and current are still 90o out of phase with each other.
If we were to build this series RC circuit and measure the voltage across the resistor, our voltmeter would indicate 1.8523 volts, not 343.11 mv (real rectangular) or 1.8203 volts (imaginary rectangular). Instruments connected to real circuits provide indications corresponding to the vector length (magnitude) of the calculated figures. The rectangular form of complex number notation is useful for performing addition and subtraction and determining an "equivalent circuit".
Since this is a parallel circuit, the applied voltage is shared equally by all components. The table for this circuit is:
Now we can apply Ohm's Law (I=E/Z) calculating current through the resistor and the capacitor:
Branch currents in a parallel circuit add up to form the total current (KCL):
Finally, total impedance can be calculated by using Ohm's Law (Z=E/I). Parallel impedance can also be calculated by using a reciprocal formula identical to that used in calculating parallel resistances. The parallel impedance rule holds true regardless of the kind of impedances placed in parallel.
Capacitors tend to be smaller and lighter weight than other components. Since the electric field energy is almost totally contained between the plates (unlike inductors, whose magnetic fields extends beyond the dimensions of the core), capacitors are less prone to transmitting or receiving electromagnetic "noise" to/from other components. For these reasons, designers tend to favor capacitors over inductors wherever a design permits either alternative.
The ideal capacitor is a purely reactive device, containing absolutely zero resistive (power dissipative) effects. In the real world nothing is so perfect. However, capacitors are "reasonably pure reactive" components (over resistors or inductors). It is a lot easier to design and construct a capacitor with low internal series resistance than it is to do the same with an inductor. The practical result of this is that real capacitors typically have impedance phase angles more closely approaching the ideal (-90o) than other components exhibit. Capacitors to have a higher quality factor (Q) than inductors or resistors. Capacitors with significant resistive effects are said to be lossy, in reference to their tendency to dissipate ("lose") power- like a resistor. The source of capacitor loss is usually the dielectric material rather than any wire resistance, as wire length in a capacitor is minimal.
Dielectric materials tend to react to changing electric fields by producing heat. This heating effect represents a loss in power, and is equivalent to resistance in the circuit. The effect is more pronounced at higher frequencies. Dielectric resistivity manifests itself both as a series and a parallel resistance with the pure capacitance.
The stray equivalent resistances are usually insignificant (low series resistance and high parallel resistance), much less significant than the stray resistances present in an inductor.
Electrolytic capacitors, known for their relatively high capacitance and low working voltage, are also known for their notorious lossiness, due to both the characteristics of the microscopically thin dielectric film and the electrolyte paste. Unless specially made for AC service, electrolytic capacitors should never be used with AC unless it is offset with a sufficiently large DC voltage preventing the capacitor from ever being reversed biased.
A coil of wire is an inductor. The turns should be insulated so as not to short out electrically. When an inductor (coil) is connected to an external circuit and carries a current, a magnetic field develops (through the center of the inductor) that is proportional to the magnitude of the current. The direction of current and its relationship to the circulating magnetic field is illustrated by the "right hand rule" where the thumb points in the direction of current and the wrapped fingers show the magnetic field direction.
Browse to the following link for an interactive illustration of magnetic lines of force: http://www.micro.magnet.fsu.edu/electromag/java/magneticlines/index.html
Inductors oppose changes in current through them, by dropping a voltage directly proportional to the rate of change of current. In accordance with Lenz's Law, this induced voltage is always of such a polarity as to try to maintain current at its present value. That is, if current is increasing in magnitude, the induced voltage will act as a voltage drop to "push against" the electron flow; if current is decreasing, the polarity will reverse and "push with" the electron flow to oppose the decrease. It is for this reason that inductors are sometimes called "chokes" (reduce current changes). The inductor's opposition to current change is called inductive reactance (XL), NOT resistance. The impedance of an inductor is XL. The inductance (L) of an inductor, is measured in Henries (H).
Inductor voltage is proportional to the rate of change of current.
To view an interactive illustration of Lenz's Law browse to:
To view an interactive illustration of Faraday's Law, browse to:
To experiment with charging/discharging an inductor, browse to:
A plot of the inductor current (i) and voltage (e)
The voltage dropped across an inductor is a reaction against the change in current through it (counter emf). Therefore, the instantaneous voltage is zero whenever the instantaneous current is at a peak, and the instantaneous voltage is at a peak wherever the instantaneous current is at maximum change (where it crosses the zero line). This results in a voltage waveform that is 90o leading the current waveform: the voltage "leads" the current, and the current "lags" behind the voltage. It should be noted that inductor current CANNOT CHANGE INSTANTANEOUSLY (otherwise Vc would become infinite).
An inductor's opposition to change in current translates to an opposition to alternating current in general, which is by definition always changing in instantaneous magnitude and direction. This opposition to alternating current always results in a phase shift between current and voltage, and it dissipates zero power. Inductive reactance is:
If we expose a 10 mH inductor to frequencies of 60, 120, and 2500 Hz, it will manifest the following reactances:
Frequency (Hertz) Reactance (Ohms) ---------------------------------------- | 60 | 3.7699 | |--------------------------------------| | 120 | 7.5398 | |--------------------------------------| | 2500 | 157.0796 | ----------------------------------------
In the reactance equation, the term "2πf" has a special meaning unto itself (omega). It is the number of radians per second that the alternating current is "rotating" . A radian is a unit of angular measurement: there are 2π radians in one full circle, just as there are 360o in a full circle. Thus, the reactance formula XL = 2πfL could also be written as XL = ωL.
However, we need to keep in mind that voltage and current are not in phase here. As was shown earlier, the voltage has a phase shift of +90o with respect to the current. If we represent these phase angles of voltage and current in the form of complex numbers, we find that an inductor's opposition to current has a phase angle, too:
The phase angle of an inductor's opposition to current is +90o
The resistor has a real impedance R, the inductor's reactance is a positive imaginary impedance
XL = jwL
in Series Circuits
A series circuit is a circuit in which the current has only one path. In a series circuit, all of the current passes through each of the components in the circuit. The circuit below has three inductors in series.
The total inductance of the circuit is cumulative. The total inductance of such a circuit is the sum of all the inductors in the circuit. The total inductance of a series circuit:
LT = L1 + L2 + L3 . . .
where LT is the total inductance in the circuit, and L1 through L3 . . . are the inductance ratings of the individual inductors in the circuit.
Using this formula, the total inductance of the series circuit can be calculated as follows:
LT = 50 + 40 + 20
LT = 110 mh
Inductors in Parallel Circuits
A parallel circuit is a circuit in which components are arranged so that the path for the current is divided. The circuit below has three inductors in parallel.
Placing inductors in parallel always decreases the total inductance of the circuit. The total inductance of the circuit can be calculated using the following formula:
LT = 1 ÷ (1/L1 + 1/L2 + 1/L3 . . .)
where LT is the total Inductance in the circuit, and L1 through L3 . . . are the inductance ratings of the individual inductors in the circuit.
The total or inductance of the parallel circuit above can be calculated as follows:
LT = 1 ÷ (1/5 + 1/15 + 1/30)
LT = 1 ÷ (0.2 + 0.066 + 0.033)
LT = 1 ÷ 0.299
LT = 3.344 mh
Some Radio Frequency chokes have their values indicated by a color code similar to that of resistors:
RLC Phase Chart
The resistor will offer 5 Ω of resistance to AC current regardless of frequency, while the inductor will offer 3.7699 Ω of reactance to AC current at 60 Hz. Because the resistor's resistance is a real number (5 Ω ∠ 0o, or 5 + j0 Ω), and the inductor's reactance is an imaginary number (3.7699 Ω ∠ 90o, or 0 + j3.7699 Ω), the combined effect of the two components will be an opposition to current equal to the complex sum of the two numbers. The total circuit impedance is:
As with the purely inductive circuit, the current wave lags behind the voltage wave (of the source), although this time the lag is not as great: only 37.016o as opposed to a full 90o as was the case in the purely inductive circuit.
In an RL series circuit, the voltage across the inductor leads the current by 90°, and the inductive reactance is XL = wL.
It is straightforward to use Pythagoras' law to obtain the series impedance and trigonometry to obtain the phase.
For the resistor and the inductor, the phase relationships between voltage and current haven't changed. Across voltage across the resistor is in phase (0o shift) with the current through it; and the voltage across the inductor is +90o out of phase with the current going through it.
The voltage across the resistor has the exact same phase angle as the current through it, telling us that E and I are in phase (for the resistor only).
The voltage across the inductor has a phase angle of 52.984o, while the current through the inductor has a phase angle of -37.016o, a difference of exactly 90o between the two. This tells us that E and I are still 90o out of phase (for the inductor only). We can also prove that these complex values add together to constitute the total voltage, just as Kirchhoff's Voltage Law would predict:
In tabular form:
Zt = Zr + ZL
Ohm's Law (I=E/Z) applied yields the total current:
The voltage drop across the resistor and inductor is done through the use of Ohm's Law (E=IZ)
If you were to connect a voltmeter across the resistor in this circuit, it would indicate 7.9847 volts, not 6.3756 (real rectangular) or 4.8071 (imaginary rectangular) volts. To describe this in graphical terms, measurement instruments indicate the phasor magnitude (voltage or current).
Create the table:
Now we can apply Ohm's Law (I=E/Z)
Branch currents in a parallel AC circuit add to form the total current (Kirchhoff's Current Law)
Total impedance can be calculated by using Ohm's Law (Z=E/I). Parallel impedance can also be calculated by using a reciprocal formula identical to that used in calculating parallel resistances.
An ideal inductor is as a purely reactive device. Current is based on inductive reaction to changes in current, and not internal resistance. However, inductors are not quite so pure in reality. Inductors are made of wire, and all wire possesses some measurable amount of resistance. This built-in resistance acts as though it were connected in series with the perfect inductance of the coil. The design impedance of any real inductor will always be a complex combination of resistance and inductive reactance.
Compounding this problem is something called the skin effect, which is the tendency to flow through the outer areas of a conductor's cross-section rather than through the middle. When electrons flow in a single direction, they use the entire cross-sectional area of the conductor to move. Electrons switching directions of flow, on the other hand, tend to avoid travel through the very middle of a conductor, limiting the effective cross-sectional area available. The skin effect becomes more pronounced as frequency increases. The skin effect and radiation losses apply just as well to straight lengths of wire in an AC circuit as they do a coiled wire. Usually their combined effect is too small to notice, but at radio frequencies they can be quite large. A radio transmitter antenna, for example, is designed with the express purpose of dissipating the greatest amount of energy in the form of electromagnetic radiation.
Also, the alternating magnetic field of an inductor energized with AC may radiate off into space as part of an electromagnetic wave, especially if the it is of high frequency. This radiated energy does not return to the inductor, and so it manifests itself as resistance (power dissipation) in the circuit.
Added to the resistive losses of wire and radiation, there are other effects at work in iron-core inductors which manifest themselves as additional resistance between the leads. When an inductor is energized, the alternating magnetic fields produced tend to induce circulating currents within the iron core known as eddy currents. These electric currents in the iron core have to overcome the electrical resistance offered by the iron, which is not as good a conductor as copper. Eddy current losses are primarily counteracted by dividing the iron core up into many thin sheets (laminations), each one separated from the other by a thin layer of electrically insulating varnish. With the cross-section of the core divided up into many electrically isolated sections, current cannot circulate within that cross-sectional area and there will be no (or very little) resistive losses from that effect. Eddy current losses in metallic inductor cores manifest themselves in the form of heat. The effect is more pronounced at higher frequencies. In high-frequency service, eddy currents can even develop within the cross-section of the wire itself, contributing to additional resistive effects. To counteract this tendency, special wire made of very fine, individually insulated strands called Litz wire (short for Litzendraht) can be used. The insulation separating strands from each other prevent eddy currents from circulating through the whole wire's cross-sectional area.
Additionally, any magnetic hysteresis that needs to be overcome with every reversal of the inductor's magnetic field constitutes an expenditure of energy that manifests itself as resistance in the circuit. Some core materials (such as ferrite) are particularly notorious for their hysteretic effect. Counteracting this effect is best done by means of proper core material selection and limits on the peak magnetic field intensity generated with each cycle.
Altogether, the stray resistive properties of a real inductor (wire resistance, radiation losses, eddy currents, and hysteresis losses) are expressed under the single term of "effective resistance:"
Effective resistance in an inductor can be a serious consideration for the designer. To help quantify the relative amount of effective resistance in an inductor, another value exists called the Q factor, or "quality factor" which is calculated as follows:
The higher the value for "Q," the "purer" the inductor is. Because it's so easy to add additional resistance if needed, a high-Q inductor is better than a low-Q inductor for design purposes. An ideal inductor would have a Q of infinity, with zero effective resistance.
Because inductive reactance (X) varies with frequency, so will Q. However, since the resistive effects of inductors (wire skin effect, radiation losses, eddy current, and hysteresis) also vary with frequency, Q does not vary proportionally with reactance. In order for a Q value to have precise meaning, it must be specified at a particular test frequency.
The multiple turns of wire comprising inductors are separated from each other by an insulating gap (air, varnish, or some other kind of electrical insulation) which creates capacitance to develop between turns. This capacitance "shunts" the equivalent inductor in the model shown above.
Transformer Inductive Circuits
A transformer is a clever utilization of two (or more) independent sets of inductors- sharing a common circulating magnetic energy field. Energy delivered to the transformer input (primary) can be directed to output(s). Electrons DO NOT FLOW from the transformer input to output, so a transformer provides electrical isolation. Only energy is exchanged via the circulating magnetic force field.
A changing primary current in the transformer INDUCES a voltage across the secondary proportional to the rate of change of the primary current (180 degrees out of phase). This is because the changing primary current creates a changing magnetic field that circulates through the secondary windings and induces a voltage across the secondary turns (generator principle).
RMS "true" power is the product of voltage and current. Primary power = Secondary power (100% efficiency). So (Vp)(Ip) = (Vs)(Is) and Vp/Vs = Is/Ip. Also the turns ratio is proportional to the voltage ratio (higher the secondary turns count- the higher the secondary voltage) Np/Ns = Vp/Vs
Because transformers can step voltage and current to different levels, and because power is transferred equivalently between primary and secondary windings, they can be used to "convert" the impedance of a load to a different level. A transformer designed to increase voltage from primary to secondary is called a step-up transformer. A transformer designed to reduce voltage from primary to secondary is called a step-down transformer.
EXAMPLE 1:Design a transformer circuit to interface an arbitrary voltage source to a specified load.
The purpose of a load (usually) is to do something productive with the power it dissipates. Consider these two 1000 watt resistive elements:
Both circuits dissipate exactly 1000 watts of true power, but they do so at different voltage and current levels (either 250 volts and 4 amps, or 125 volts and 8 amps). Using Ohm's Law to determine the necessary resistance of these heating elements (R=E/I), we arrive at figures of 62.5 Ω and 15.625 Ω, respectively.
If we desired to operate the 250 volt element directly on a 125 volt power system, we would end up being disappointed. With 62.5 Ω of impedance (resistance), the current would only be 2 amps (I=E/R; 125/62.5), and the power dissipation would only be 250 watts (P=IE; 125 x 2), or one-quarter of its rated power. The impedance of the heater and the voltage of our source would be mismatched, and we couldn't obtain the full rated power dissipation from the heater.
Using a step-up 1:2 transformer as an interface, we could operate the 250 volt heater element on the 125 volt power system
The ratio of the transformer's windings provides the voltage step-up and current step-down we need for the otherwise mismatched load to operate properly on this system. As far as the power supply "knows," it's powering a 15.625 Ω (R=E/I) load at 125 volts, not a 62.5 Ω load! The voltage and current figures for the primary winding are indicative of 15.625 Ω load impedance, not the actual 62.5 Ω of the load itself. In other words, not only has our step-up transformer transformed voltage and current, but it has transformed impedance as well.
The transformation ratio of impedance is the square of the voltage/current transformation ratio, the same as the winding inductance ratio:
Illustrations of transformers
Now try the interactive transformer practice session by browsing to the following link
EXAMPLE 2 Design transformer to provide impedance matching
With an internal impedance of 500 Ω, the amplifier can only deliver full power to a load (speaker) also having 500 Ω of impedance. Such a load would drop higher voltage and draw less current than an 8 Ω speaker dissipating the same amount of power. If an 8 Ω speaker were connected directly to the 500 Ω amplifier as shown, the impedance mismatch would result in very poor (low peak power) performance. Additionally, the amplifier would tend to dissipate more than its fair share of power in the form of heat trying to drive the low impedance speaker.
To make this system work better, we can use a transformer to match these mismatched impedances. Since we're going from a high impedance (high voltage, low current) supply to a low impedance (low voltage, high current) load, we'll need to use a step-down transformer:
To obtain an impedance transformation ratio of 500:8, we would need a winding ratio equal to the square root of 500:8 (the square root of 62.5:1, or 7.906:1). With such a transformer in place, the speaker will load the amplifier to just the right degree, drawing power at the correct voltage and current levels to satisfy the Maximum Power Transfer Theorem and make for the most efficient power delivery to the load. The use of a transformer in this capacity is called impedance matching.
Now let's put a resistor, capacitor and inductor in series. At any given time, the voltage across the three components in series, vseries(t), is the sum of these:
vseries(t) = vR(t) + vL(t) + vC(t),
From phasor analysis, the total applied voltage is the phasor sum of the series component voltages V2series = V2R + (VL - VC)2
Now VR = IR, VL = IXL = wL and VC = IXC= 1/wC. Substituting and dividing out the common factor I gives:
Zseries2 = R2 + Xtotal2 = R2 + (XL - XC)2. and Zseries = sq root [ R2 + (XL - XC)2] The series impedances add (vector addition).
The inductive and capacitive phasors are 180° out of phase, so their reactances tend to cancel.
The angle by which the voltage leads the current is
f = tan-1 ((VL - VC)/VR) = tan-1 ((XL - XC)/ R)
Because vL(t) and vC are 180° out of phase, this means that the two reactive voltages oppose (subtraction of magnitudes)..
EXAMPLE RLC series circuit
The first step is to determine the reactances for the inductor and the capacitor
Inductive reactance exhibits a positive angle for impedance (impedance at +90o). Capacitive reactance exhibits a negative phase angle (impedance at -90o). Resistance has a purely "real" impedance (polar angle of 0o)
An analysis table for this circuit (total voltage, and the impedances of the resistor, inductor, and capacitor) yields
Since this is a series RLC circuit, the total circuit impedance is equal to the sum of the individual impedances
Inserting this figure for total impedance into the table
Ohm's Law (I=E/R) is used to find total current for this series circuit which is equal to the R, L, and C currents
Ohm's Law (E=IZ) can be used to determine the R, L, and C voltage drops:
Although the supply voltage is 120 volts, the voltage across the capacitor is 137.46 volts! How can this be? The answer lies in the interaction between the inductive and capacitive reactances: the inductor opposes current in a manner precisely opposite that of the capacitor. Expressed in rectangular form, the inductor's impedance has a positive imaginary term and the capacitor has a negative imaginary term. When these two contrary impedances are added (in series), they tend to cancel each other out. Although they're still added together to produce a sum, that sum is actually less than either of the individual (capacitive or inductive) impedances alone. If the total impedance in a series circuit with both inductive and capacitive elements is less than the impedance of either element separately, then the total current in that circuit must be greater than what it would be with only the inductive or only the capacitive elements there. With this abnormally high current through each of the components, voltages greater than the source voltage may be obtained across some of the individual components
KVL tells us that the algebraic sum of the voltage drops across the resistor, inductor, and capacitor should equal the applied voltage from the source.
We can also use L210 BREADBOARD analysis software to verify our calculations
Breadboard output (simulated)
v1 1 0 ac 120 sin r1 1 2 250 ohm l1 2 3 650mH c1 3 0 1.5uF ac 60Hz freq v(1,2) v(2,3) v(3) i(v1) 6.000E+01 1.943E+01 1.905E+01 1.375E+02 7.773E-02
Notes and observations:
Resonance occurs when XL = Xc. The frequency where this occurs is the resonant frequency:
At resonance Xc = XL so when added to form Zt, they cancel therefore Zt = R.
The current equals (Vt / Zt) = Vt/ R (Ohm's Law). Ic = (Vc/ Xc) = IL = Ir = (Vt / R). Therefore Vc = (Vt x Xc) / R = QVt where Q is the quality factor = (X / R).
As R -- 0 ohms, then I -- infinity and so does Q, Vc, and VL (infinity here refers to being unboundedly large).
When a state of resonance is reached (capacitive and inductive reactances equal), the two impedances cancel each other out and the total impedance drops to the series resistance alone.
Note that the VL waveform (green) is of equal magnitude (but opposite phase) as Vc waveform (blue).
Let's assume (hypothetically) that a circuit has a total R = 0 ohms.
With the total series impedance equal to 0 Ω at the resonant frequency of 159.155 Hz, the result is a short circuit across the AC power source at resonance. This is not advisable as it would "load down" the voltage source (Vt = 0v). There is always some resistance in a circuit, so let's consider a more practical situation to see the effect.
Now solve the following circuit at resonance. Use Breadboard software in the L210 lab to plot out the current from 100 to 200 Hz.
series RLC at resonance v1 1 0 1 sin r1 1 2 1 c1 2 3 10uF l1 3 0 100mH plot ac i(v1) freq i(v1) 3.162E-02 1.000E-01 3.162E-01 1.0 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1.000E+02 1.038E-02 * . . . . 1.053E+02 1.176E-02 . * . . . . 1.105E+02 1.341E-02 . * . . . . 1.158E+02 1.545E-02 . * . . . . 1.211E+02 1.804E-02 . * . . . . 1.263E+02 2.144E-02 . * . . . . 1.316E+02 2.611E-02 . * . . . . 1.368E+02 3.296E-02 . .* . . . 1.421E+02 4.399E-02 . . * . . . 1.474E+02 6.478E-02 . . * . . . 1.526E+02 1.186E-01 . . . * . . 1.579E+02 5.324E-01 . . . . * . 1.632E+02 1.973E-01 . . . * . . 1.684E+02 8.797E-02 . . * . . . 1.737E+02 5.707E-02 . . * . . . 1.789E+02 4.252E-02 . . * . . . 1.842E+02 3.406E-02 . .* . . . 1.895E+02 2.852E-02 . *. . . . 1.947E+02 2.461E-02 . * . . . . 2.000E+02 2.169E-02 . * . . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The above plotted circuit current amplitude increases from left to right, while frequency increases from top to bottom. The peak is seen to occur at 157.9 Hz, The spread of the plot reflects the bandwidth of the response, where BW = fr / Q. As Q increases, the BW decreases.
Bandwidth and Q factor: High Q and Low BW versus Low Q and high BW
At resonance, the impedances of the capacitor and the pure inductance cancel out, so the series impedance takes its minimum value: Zo = R. If we keep the voltage constant, the current is a maximum at resonance. The current goes to zero at low frequency, because XC becomes infinite (the capacitor is open circuit for DC). The current also goes to zero at high frequency because XL increases with w (the inductor opposes rapid changes in the current).
The graph shows I(w) for circuit with a large resistor low Q circuit (lower curve) and for one with a small resistor high Q circuit (upper curve). A circuit with low R, for a given L and C, has a sharp resonance. Increasing the resistance makes the resonance less sharp. The high Q circuit is more selective: it produces high currents only for a narrow bandwidth, (a small range of w or f). The low Q circuit responds to a wider range of frequencies and so has a larger bandwidth. The bandwidth Dw (indicated by the horiztontal bars on the curves) is defined as the difference between the two frequencies w+ and w- at which the circuit converts power at half the maximum rate. BW occurs at the half power frequencies where Vc = 0.707 of Vc at resonance.
It is possible to produce dangerously high voltage drops across the capacitor and the inductor at the resonant frequency,
Computer Simulation- Breadboard
Without actually connecting the circuit, simulate resonance on the circuit above (using Breadboard) and plot out the component voltages from 100 Hz to 200Hz.
*: i(v1) +: v(2,3) =: v(3) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (*)----------- 1.000E-02 3.162E-02 1.000E-01 0.3162 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (+)----------- 1.000E+00 3.162E+00 1.000E+01 31.62 100 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (=)----------- 1.000E-01 1.000E+00 1.000E+01 100 1000 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - freq i(v1) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1.000E+02 1.038E-02 * + = . . . . 1.053E+02 1.176E-02 . * + =. . . . 1.105E+02 1.341E-02 . * + = . . . 1.158E+02 1.545E-02 . * + .= . . . 1.211E+02 1.804E-02 . * + . = . . . 1.263E+02 2.144E-02 . * +. = . . . 1.316E+02 2.611E-02 . *+ = . . . 1.368E+02 3.296E-02 . .*+ = . . . 1.421E+02 4.399E-02 . . *+ = . . . 1.474E+02 6.478E-02 . . *+= . . 1.526E+02 1.186E-01 . . .=*+ . . 1.579E+02 5.324E-01 . . . = . x . 1.632E+02 1.973E-01 . . . = x . . 1.684E+02 8.797E-02 . . x = . . 1.737E+02 5.707E-02 . . +* = . . . 1.789E+02 4.252E-02 . . + * = . . . 1.842E+02 3.406E-02 . +.* = . . . 1.895E+02 2.852E-02 . + *. = . . . 1.947E+02 2.461E-02 . + * . = . . . 2.000E+02 2.169E-02 . + * . = . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The voltage across the capacitor and inductor (plotted with "+" and "=" symbols, respectively) reach a peak somewhere between 100 and 1000 volts (marked by the "x" where the graphs overlap).
This is quite impressive for a power supply that only generates 1 volt. Needless to say, extreme caution is in order when experimenting with circuits such as this and this is the reason that we simulate this circuit rather than actually build it in the lab.
The shunt impedances at 60Hz are:
Since the components are connected in parallel, each component has identical voltage and different currents. The analysis table is as follows:
Ohm's Law (I=E/Z) applied to determine current through each component:
There are two strategies for calculating total current and total impedance. First, we could calculate total impedance from all the individual impedances in parallel (ZTotal = 1/(1/ZR + 1/ZL + 1/ZC), and then calculate total current by dividing source voltage by total impedance (I=E/Z). The second way to calculate total current and total impedance is to add up all the branch currents to arrive at total current (total current in a parallel circuit is equal to the sum of the branch currents), then use Ohm's Law to determine total impedance from total voltage and total current (Z=E/I).
Extensive use of J Operator technique to analyze general-case situations.
The values of impedance (Z) for all components (based on the frequency of the AC power source):
Component impedance table:
Being a series-parallel combination circuit, we must reduce it to a total impedance in more than one step. The first step is to combine L and C2 as a series combination of impedances, by adding their impedances together. Then, that impedance will be combined in parallel with the impedance of the resistor, to arrive at another combination of impedances. Finally, that quantity will be added to the impedance of C1 to arrive at the total impedance. Calculating these new (combination) impedances will require complex addition for series combinations, and the "reciprocal" formula for complex impedances in parallel.
The total impedance (818.34 Ω ∠ -58.371o) is divided into the total voltage (120 volts ∠ 0o) via Ohm's Law (I=E/Z), to arrive at total current:
C1 and the parallel combination R // (L--C2) share the same (total) current. The total impedance is composed of the two sets of impedances in series. Thus, we can transfer the figure for total current into both columns:
Ohm's Law applied yields:
The resistor (R) and the combination of the inductor and the second capacitor (L--C2) share the same voltage, because those sets of impedances are in parallel with each other.
Applying Ohm's Law (I=E/Z)
Since the L and C2 are connected in series, and since we know the current through their series combination impedance, we can distribute that current figure to the L and C2 columns
Ohm's Law (E=IZ) can be used to calculate voltage drops
In a pure parallel (tank) LC circuit (R = 0), the total circuit impedance will be infinite at resonance. This property is the reason that parallel resonant circuits are used- to filter out (or filter in) a certain bandwidth of signals.
Zt ~ QXc ( quite large). Here Q acts as an impedance multiplier (series resonance Q was a voltage multiplier factor).
The circuit current is therefore at a minimum at resonance. The resonant frequency still occurs when XL = Xc:
BW = (fr / Q) as it was for series resonance.
Let's now consider an example.
The resonant frequency (below) operates at 159.155 Hz. Calculate the expected frequency response, and confirm your results using Breadboard software (L210 lab).
Parallel resonant circuit v1 1 0 1 sin c1 1 0 10u r1 1 2 100 l1 2 0 100m
freq i(v1) 7.079E-03 7.943E-03 8.913E-03 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1.000E+02 7.387E-03 . . * . . 1.053E+02 7.242E-03 . . * . . 1.105E+02 7.115E-03 . .* . . 1.158E+02 7.007E-03 . *. . . 1.211E+02 6.921E-03 . * . . . 1.263E+02 6.859E-03 . * . . . 1.316E+02 6.823E-03 . * . . . 1.368E+02 6.813E-03 . * . . . 1.421E+02 6.830E-03 . * . . . 1.474E+02 6.874E-03 . * . . . 1.526E+02 6.946E-03 . * . . . 1.579E+02 7.044E-03 . *. . . 1.632E+02 7.167E-03 . .* . . 1.684E+02 7.315E-03 . . * . . 1.737E+02 7.485E-03 . . * . . 1.789E+02 7.676E-03 . . * . . 1.842E+02 7.886E-03 . . *. . 1.895E+02 8.114E-03 . . . * . 1.947E+02 8.358E-03 . . . * . 2.000E+02 8.616E-03 . . . * . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Filters select one range of frequencies out of a spectrum of simultaneously occurring frequencies in a circuit. A good example in the use of filter circuits is in audio systems- where certain ranges of audio frequencies need to be amplified or suppressed (enhanced) for best sound quality and power efficiency.
Another practical application of filter circuits is in the power supply circuitry. Some devices are sensitive to the presence of sinusoidal harmonics generated by the system's power supply (esp. from switching power supplies) and require power conditioning (filtering) for proper operation.
Low-pass filters pass low-frequency signals (below fc) and attenuatehigh-frequency signals. There are two basic kinds of circuits capable of accomplishing this objective, and many variations of each one:
The inductor's impedance increases with increasing frequency. This high impedance in series tends to block high-frequency signals from getting to the load. This can be demonstrated with a
Breadboard simulation design analysis:
v1 1 0 ac 1 sin l1 1 2 3 rload 2 0 1k freq 1 200
freq v(2) 0.2512 0.3981 0.631 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1.000E+00 9.998E-01 . . . * 1.147E+01 9.774E-01 . . . *. 2.195E+01 9.240E-01 . . . * . 3.242E+01 8.533E-01 . . . * . 4.289E+01 7.776E-01 . . . * . 5.337E+01 7.050E-01 . . . * . 6.384E+01 6.391E-01 . . * . 7.432E+01 5.810E-01 . . * . . 8.479E+01 5.304E-01 . . * . . 9.526E+01 4.865E-01 . . * . . 1.057E+02 4.485E-01 . . * . . 1.162E+02 4.153E-01 . .* . . 1.267E+02 3.863E-01 . *. . . 1.372E+02 3.607E-01 . * . . . 1.476E+02 3.382E-01 . * . . . 1.581E+02 3.181E-01 . * . . . 1.686E+02 3.002E-01 . * . . . 1.791E+02 2.841E-01 . * . . . 1.895E+02 2.696E-01 . * . . . 2.000E+02 2.564E-01 .* . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Load voltage decreases with increasing frequency
The capacitor's impedance decreases with increasing frequency. This low impedance in parallel with the load resistance tends to short out high-frequency signals, dropping most of the voltage gets across series resistor R1.
Breadboard Simulation Design
v1 1 0 ac 1 sin r1 1 2 500 c1 2 0 7u rload 2 0 1k freq 30 150
freq v(2) 0.3162 0.3981 0.5012 0.631 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 3.000E+01 6.102E-01 . . . . *. 3.632E+01 5.885E-01 . . . . * . 4.263E+01 5.653E-01 . . . . * . 4.895E+01 5.416E-01 . . . . * . 5.526E+01 5.180E-01 . . . .* . 6.158E+01 4.948E-01 . . . *. . 6.789E+01 4.725E-01 . . . * . . 7.421E+01 4.511E-01 . . . * . . 8.053E+01 4.309E-01 . . . * . . 8.684E+01 4.118E-01 . . .* . . 9.316E+01 3.938E-01 . . *. . . 9.947E+01 3.770E-01 . . * . . . 1.058E+02 3.613E-01 . . * . . . 1.121E+02 3.465E-01 . . * . . . 1.184E+02 3.327E-01 . .* . . . 1.247E+02 3.199E-01 . * . . . 1.311E+02 3.078E-01 . * . . . . 1.374E+02 2.965E-01 . * . . . . 1.437E+02 2.859E-01 . * . . . . 1.500E+02 2.760E-01 .* . . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Load voltage decreases with increasing frequency
All low-pass filters are rated at a certain cutoff frequency (fc) that is the frequency above which the output voltage falls below 70.7% of the input voltage (3dB point). Fc is the frequency at which capacitive reactance equals resistance given by:
Inserting the values of R and C from the last simulation into this formula, we arrive at a cutoff frequency of 45.473 Hz.
Capacitive filter designs are generally preferred over inductive because:
However, the inductive low-pass filter is often preferred in power supplies to filter out the AC "ripple" waveform created when AC is converted (rectified) into DC. The primary reason for this is the requirement of low filter resistance for the output. A capacitive low-pass filter requires an extra resistance in series with the source, whereas the inductive low-pass filter does not. In the design of a high-current circuit like a DC power supply where additional series resistance is undesirable, the inductive low-pass filter is the better design choice.
LC LP examples:
Now practice your skill at the following weblink:
A high-pass filter passes high frequencies (above fc) and attenuates low-frequencies. The inductive and capacitive versions of the high-pass filter are just the opposite (duals) of the low-pass filter designs:
The capacitor's impedance increases with decreasing frequency. This high impedance in series tends to block low-frequency signals from getting to load.
Breadboard Design simulation
v1 1 0 ac 1 sin c1 1 2 0.5u rload 2 0 1k freq 1 200
freq v(2) 1.000E-03 1.000E-02 1.000E-01 1.000E+00 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1.000E+00 3.142E-03 . * . . . 1.147E+01 3.602E-02 . . * . . 2.195E+01 6.879E-02 . . * . . 3.242E+01 1.013E-01 . . * . 4.289E+01 1.336E-01 . . . * . 5.337E+01 1.654E-01 . . . * . 6.384E+01 1.966E-01 . . . * . 7.432E+01 2.274E-01 . . . * . 8.479E+01 2.574E-01 . . . * . 9.526E+01 2.867E-01 . . . * . 1.057E+02 3.152E-01 . . . * . 1.162E+02 3.429E-01 . . . * . 1.267E+02 3.698E-01 . . . * . 1.372E+02 3.957E-01 . . . * . 1.476E+02 4.207E-01 . . . * . 1.581E+02 4.448E-01 . . . * . 1.686E+02 4.680E-01 . . . * . 1.791E+02 4.903E-01 . . . * . 1.895E+02 5.116E-01 . . . * . 2.000E+02 5.320E-01 . . . * . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Load voltage increases with increasing frequency
The inductor's impedance decreases with decreasing frequency. This low inductive impedance (in parallel) shunts low-frequencies and prevents them from being conducted to the load resistor. High frequencies are conducted through the load.
Breadboard Simulation Design
v1 1 0 ac 1 sin r1 1 2 200 l1 2 0 100m rload 2 0 1k freq 1 200
freq v(2) 1.000E-03 1.000E-02 1.000E-01 1.000E+00 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1.000E+00 3.142E-03 . * . . . 1.147E+01 3.601E-02 . . * . . 2.195E+01 6.871E-02 . . * . . 3.242E+01 1.011E-01 . . * . 4.289E+01 1.330E-01 . . . * . 5.337E+01 1.644E-01 . . . * . 6.384E+01 1.950E-01 . . . * . 7.432E+01 2.248E-01 . . . * . 8.479E+01 2.537E-01 . . . * . 9.526E+01 2.817E-01 . . . * . 1.057E+02 3.086E-01 . . . * . 1.162E+02 3.344E-01 . . . * . 1.267E+02 3.591E-01 . . . * . 1.372E+02 3.828E-01 . . . * . 1.476E+02 4.053E-01 . . . * . 1.581E+02 4.267E-01 . . . * . 1.686E+02 4.470E-01 . . . * . 1.791E+02 4.662E-01 . . . * . 1.895E+02 4.845E-01 . . . * . 2.000E+02 5.017E-01 . . . * . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Load voltage increases with increasing frequency
Capacitors are favored over inductors in HP filter design .High frequencies cause "skin effect" and electromagnetic core losses in inductive devices
High-pass filters also have a cutoff frequency:
Using a simple audio cross-over system as an example of filtering, a capacitor connected in series with the tweeter speaker will serve as a high-pass filter, imposing a high impedance to low-frequency bass signals. The inductor connected in series with the woofer speaker will serve as a low-pass filter for the low frequencies. In this simple example circuit, the midrange speaker is subjected to the full spectrum of frequencies from the stereo's output.
Now practice your skill at the following weblink:
Band-pass filters allow a designable band of frequency to pass to an output, and attenuate the frequencies above/below this band. BP filters integrate both a LP (to set the top frequency of the band) and a HP (to set the bottom of the pass band).
Breadboard Design Simulation
v1 1 0 ac 1 sin r1 1 2 200 c1 2 0 2.5u c2 2 3 1u rload 3 0 1k freq 100 500
freq v(3) 4.467E-01 5.012E-01 5.623E-01 6.310E-01 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1.000E+02 4.703E-01 . * . . . 1.211E+02 5.155E-01 . . * . . 1.421E+02 5.469E-01 . . * . . 1.632E+02 5.676E-01 . . .* . 1.842E+02 5.801E-01 . . . * . 2.053E+02 5.865E-01 . . . * . 2.263E+02 5.882E-01 . . . * . 2.474E+02 5.864E-01 . . . * . 2.684E+02 5.820E-01 . . . * . 2.895E+02 5.755E-01 . . . * . 3.105E+02 5.676E-01 . . .* . 3.316E+02 5.585E-01 . . *. . 3.526E+02 5.487E-01 . . * . . 3.737E+02 5.384E-01 . . * . . 3.947E+02 5.277E-01 . . * . . 4.158E+02 5.169E-01 . . * . . 4.368E+02 5.060E-01 . .* . . 4.579E+02 4.951E-01 . *. . . 4.789E+02 4.843E-01 . * . . . 5.000E+02 4.736E-01 . * . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Load voltage peaks within narrow frequency range
Examples of BP resonant filters:
Also called band-elimination, band-reject, or notch filters, this kind of filter passes all frequencies above and below a particular range and rejects those unwanted frequencies in the stop-band. A LP passes below the bottom of the reject band, and a HP passes above the top of the reject band.
Using two capacitive filter sections:
The low-pass filter section is comprised of R1, R2, and C1 in a "T" configuration. The high-pass filter section is comprised of C2, C3, and R3 in a "T' configuration as well. Together, this arrangement is commonly known as a "Twin-T" filter, giving sharp response.
Given these component ratios, the frequency of maximum rejection (the "notch frequency") can be calculated as follows:
Band reject ability of this filter is illustrated by the following
v1 1 0 ac 1 sin r1 1 2 200 c1 2 0 2u r2 2 3 200 c2 1 4 1u r3 4 0 100 c3 4 3 1u rload 3 0 1k freq 200 1.5k
freq v(3) 1.000E-02 3.162E-02 1.000E-01 3.162E-01 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2.000E+02 5.400E-01 . . . . *. 2.684E+02 4.512E-01 . . . . * . 3.368E+02 3.686E-01 . . . . * . 4.053E+02 2.946E-01 . . . *. . 4.737E+02 2.290E-01 . . . * . . 5.421E+02 1.707E-01 . . . * . . 6.105E+02 1.185E-01 . . . * . . 6.789E+02 7.134E-02 . . * . . . 7.474E+02 2.832E-02 . *. . . . 8.158E+02 1.126E-02 .* . . . . 8.842E+02 4.796E-02 . . * . . . 9.526E+02 8.222E-02 . . * . . . 1.021E+03 1.144E-01 . . . * . . 1.089E+03 1.447E-01 . . . * . . 1.158E+03 1.734E-01 . . . * . . 1.226E+03 2.007E-01 . . . * . . 1.295E+03 2.267E-01 . . . * . . 1.363E+03 2.515E-01 . . . * . . 1.432E+03 2.752E-01 . . . * . . 1.500E+03 2.980E-01 . . . *. . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Other examples of resonant BR filters:
Other examples of filters. Identify each type of filter illustrated below. (Hint: Consider the response curves).
Advanced Network Analysis
Network theorems applied to RLC systems- Use J Operator to perform the calculations.
|Identify and label the nodes in the circuit and then choose one of these nodes as a reference point (ground). This is shown in below.|
|Once the nodes are identified, we can write the nodal equations for each node.|
|Node 1:||V1 = (50v @ 0°)|
|Node 2:||(V2-V1)/(4-j3) + V2/5 + (V2-V3)/(8+j8) = 0|
|Node 3:||(V3-V2)/(8+j8) + V3/20 - (5A @ 0°) = 0|
|After a considerable amount of J Operator math, the following results indicate the node voltages:|
|V1 = 50v @ 0°|
|V2 = 30v @ 8°|
|V3 = 57v @ 16°|
|The first step is to identify and label the loop currents in the circuit and then write the equations for these loops. The loops are shown above.|
|Loop 1:||-50v @ 0° + (4-j3)I1 + 5(I1+I2) = 0|
|Loop 2:||20(I2-I3) + (8+j8)I2 + 5(I2+I1) = 0|
|Loop 3:||I3 = 5A @ 0°|
|By solving the above equations, the following results are found:|
|I1 = 3.85A @ 25°|
|I2 = 2.4Av @ -20°|
|I3 = 5A @ 0°|
|With the current source replaced
by its ideal internal impedance (an open circuit), the voltage source "sees"
all the components in series.
The total impedance would then be:
|Total Impedance:||ZT = 6 + j8 + 8 - j8 = 14ohm @ 0°|
|Current:||IT = V/Z = 28/14 = 2A @ 0°|
|With the voltage source replaced by its ideal internal impedance (a short circuit), the current source "sees" the two branches in parallel. The current through each branch would then be:|
|i1:||i1 = 2 * (8-j8)/[(6+j8)+(8-j8)] = 1.62A @ 45°|
|i2:||i2 = 2 * (6+j8)/[(6+j8)+(8-j8)] = 1.43A @ 53.13°|
|The actual value of current through each branch would then be the algebraic sum of the two individual currents. (iL will be the current through the inductor and 6ohm resistor and iC will be the current through the capacitor and 8ohm resistor).|
Putting it all together:
|iL:||iL = 2AÐ0° - 1.62AÐ45°|
|iL:||iL = (2+j0) - (1.15+j1.15)|
|iL:||iL = (.85-j1.15)|
|iL:||iL = 1.43A @ -53.5°|
|iC:||iC = 2AÐ0° + 1.43AÐ53.13°|
|iC:||iC = (2+j0) + (.86+j1.14)|
|iC:||iC = (2.86+j1.14)|
|iC:||iC = 3.08A @ 21.7°|
|Thevenin Equivalent Circuit @ RL.|
|Thevenin Impedance (ZTH) . Replace the source with its ideal internal impedance (0 ohms):|
|Thevenin Impedance calculation|
|The complex "ohmmeter" (@ RL) would see the parallel combination of 8ohm and -j6 ohm in series with the parallel combination of 3 ohm and +j4ohm:|
|ZTH:||ZTH = [(8 * -j6)/(8-j6)] + [(3*j4)/(3+j4)] = 5.37ohm @ -26.6° = 4.8 -j2.4 ohm|
|The Thevenin Voltage (VTH) would be the difference between the voltage across the capacitor and the voltage across the inductor:|
|Thevenin Voltage calculation|
|VTH:||VTH = [(20 * -j6)/(8-j6)] - [(20*j4)/(3+j4)] = 20V @ 73.7°|
|The Thevenin Circuit appears below. If
the source frequency is known, the capacitor value could be found instead of
|Thevenin Equivalent Circuit|
|Norton Equivalent Circuit for the branch between terminals A and B in the circuit below.|
|Norton Impedance (ZN) . Replace the sources with their ideal internal impedances:|
|Norton Impedance calculation|
|The complex "ohmmeter" would measure the parallel combination of 30 and 60ohm resistances.|
|ZN:||ZN = 30*60/(30+60) = 1800/90 = 20 ohms|
|The Norton Current (IN) would be the current through the shorted terminals.|
|Norton Current calculation|
|1A from the voltage source plus the 1A
from the current source (the 60 ohm resistor is
shorted, so that no current will flow through it). Therefore, IN
is 2A @ 0°
The Norton equivalent circuit is shown below:
|Norton Equivalent Circuit|