Elektronika i Elektrotechnika

Prąd stały Natężenie, Napięcie, i ładunek Opór Prawo Ohma, Moc, Energia Obwody z oporami Prawa prądu stałego. Analiza obwodów Pojemność i prąd chwilowy

Literatura S. Bolkowski „Teoria obwodów” Wyd. Techn. Z.Cichowska,M.Pasko,E.Litwinowicz „Przykłady i zadania z elektrotechniki teoretycznej” S.Bolkowski „Teoria obwodów” zbiór zadań I wiele wiele innych

Układ jednostek SI Długość l Prąd I, i Temperatura T Masa m Czas t Wielkość SYMBOL Długość l Prąd I, i Temperatura T Masa m Czas t Jednostka Skrót. metr m amper A kelvin K kilogram kg sekunda s The solution of technical problems requires the use of units. The SI (Systeme International) system combines the MKS metric units and the electrical units into one unified system. Base units are defined units, i.e. based on physical law of nature e.g. 1 s = period of a cesium-based atomic clock 1 m = distance travelled by light in a vacuum in 1/299 792 458 of a second 1 kg = mass of a specific platinum-iridium cylinder preserved at the International Bureau of weights & Measures in France.

Jednostki pochodne SI Napięcie U, u, E, e Ładunek Q, q Opór R Moc P, p Pojemność C Indukcyjność L Częstotliwość f Strumień magnetyczny F Natężenie pola magnet. B volt V coulomb C ohm W watt W farad F henry H hertz Hz weber Wb tesla T Other derived units are: Force (F ) in newton (N) Energy (W ) in joule (J) SI derived units are obtained by combining the based units e.g. 1 C = 1A x 1s; 1 V = 1 J / 1 C Note: some symbols and abbreviations use capital letters while others use lower case letters.

Wielkośći wyrażane w jednostkach potęgi 10 Wielkości w elektrotechnice zmieniają się w dużym zakresie, wyrażamy je w jednostkach potęgi 10; np. 83.5 x 105 Hz. Można je zapisać w notacji naukowej tzn. wyrażając je stosując wielokrotności potęgi np: 8.35 x 106 Hz. Lub stosując notację używającą prefixu - notacja inżynierska; np. 8.35 MHz.

Prefixy w zapisie inżynierskim Potęgi 10 PREFIX SYMBOL 1012 tera T 109 giga G 106 mega M 103 kilo k 10-3 mili m 10-6 micro m 10-9 nano n 10-12 piko p Power of 10 Notation To handle a range of electrical values that vary tremendously, we need the power of 10 notation. e.g. 4,759,000 Hz = 4.759 x 106 Hz ; 35 x .0846 - 160 x 10-3= 2.801 Scientific Notation If power of 10 numbers are written with one significant digit to the left of the decimal place, they are said to be in scientific notation. e.g. 3.71 x 10-4 is, but 0.371 x 10-5 is not in scientific notation. Engineering Notation In engineering notation, prefixes are used to represent certain powers of 10. e.g. 440 x 103 V = 440 kV. Significant Digits & Numerical Accuracy The # of digits in a number that carry actual information are termed significant digits . e.g. .035 has 2 significant digits; 7.960 m has 4 significant digits. The # of significant digits as a result of multiplication, division, addition or subtraction is the same as the # of significant digits in the number with the least number of significant digits. e.g. 3.5 V + 159 mV = 3.7 V rather than 3.659 V.

Teoria Atomowa jądro K L M N elektrony Uproszczony schemat atomu Jądro składa się z protonów i neutronów (dodatni ładunek) The hydogen atom, the simplest of all atoms, has 1 proton and 1 electron. Cu atom has 29 electrons, 29 protons, and 35 neutrons. Si, which is important because of its use in transistors and other electronic devices, has 14 electrons, 14 protons, and 14 neutrons. Only certain no. of electrons can exist within any given shell. e.g. max.2 for K, 8 for L, 18 for M, 32 for N. Cu has all three inner shells completely filled but its outer shell (N) has only 1 electron. Valence Shells No element can have more than 8 valence electrons. When a valence shell has 8 electrons, it is filled. Free Electrons The force of attraction keeps the electrons in orbit. Coulomb’s Law states that F = kQ1Q2/r2, where F is in N, Q in coulombs, r in meters, and k = 9 x 109. The fewer the number of electrons in the valence shell, the smaller the amount of energy that is required for the electrons to escape. These escaped electrons wander from atom to atom and are called free electrons . Elektrony krążą po orbitach zwanych powłokami (K, L, M, N, etc.) Atom jest obojętny N. protonów = N.elektronów Elektrony znajdujące się na najbardziej zewnętrznych powłokach walencyjnych nazywają się walencyjnymi

Przewodniki, Izolatory, półprzewodniki Przewodniki to materiały które przewodzą (np. miedź, aluminium, złoto) -mają dużą ilość wolnych elektronów. Izolatory nie przewodzą (np. plastik, guma, porcelana) ponieważ mają prawie całkowicie zajęte poziomy walencyjne. Półprzewodniki mają zajęte do połowy pasma walencyjne (np. krzem, german).

Ładunek elektryczny Ciało jest naładowane gdy posiada nadmiar lub niedobór elektronów. Jednostką ładunku jest coulomb; 1 C = 6.24 x 1018 elektronów (1e=1.6x10-19 C) Prawo Coulomba: F = kQ1Q2 / r2 k = 9 x 109 [N*m2/C2], Q1 i Q2 są ładunkami w coulombach, a r (odległość) w m. If a body has an excess of electrons, it is negatively charged; if there is a deficiency, it is positively charged. Commonly encountered in daily live as static electricity. The charge on 1 electron is Qe = 1/(6.24 x 1018) = 1.6 x 10-19 C. e.g. 2.5 x 1012 electrons are added to a neutral body initially. Later, 7 x 1012 electrons are removed. What is the body’s final charge? 2.5 x 1012 electrons = .4 mC of -ve charge; 7 x 1012 electrons = 1.12 mC of positive charge Therefore, final charge, Qf = 1.12 - .4 = +0.72 mC. e.g. A positive charge Q1 = 3 mC is at a distance of 5 mm from a negative charge Q2 = 5 mC. Compute the force between them. Is the force repulsive or attractive? Attractive F = 9 x 109 x 3 x 5 x 10-12 / (5 x 10-3)2 = 5,400 N or 1,214 lb !

- - Napięcie Komórka Bateria Aby ładunek przemieszczał się pomiędzy dwoma ciałami musi istnieć pomiędzy nimi, różnica potencjałów lub napięcie . Napięcie pomiędzy dwoma punktami wynosi 1 V jeżeli potrzeba 1 J energii aby przenieść1 C ładunku charge z jednego punktu do drugiego U Q = E Symbole napięcia stałego (DC): Several thousand volts could be developed in static electricity and several hundred million volts could be created in a thunderstorm. A battery (dry cell) is a common example of dc (direct current) voltage sources. It is based on the chemical reaction inside the cell. The inner (+ve) electrode is a carbon rod and the outer (-ve) electrode is a Zn case. Ammonium-chloride/manganese-dioxide paste is the electrolyte. Work is required to move positive and negative charges apart. This gives them potential energy; hence voltage, V = W/Q e.g. The voltage between two points is 10 V. How much energy is required to move 3 x 1018 electrons from one point to the other? W = 10 x 3 x 1018 / (6.24 x 1018) = 4.81 J. A pracitcal way to measure voltage is to use a voltmeter. + - + - Komórka Bateria

Prąd - Lamp + E I Kierunek (umowny) przepływu prądu Elektrony przepływają od ujemnego potencjału do dodatniego ale umowny prąd przepływa w kierunku odwrotnym (tak jakby przepływał ładunek dodatni) 1 A jest ładunkiem 1 C przepływającym przez dany punkt obwodu w ciągu 1 s, tzn. I = Q / t lub Q = I x t. In a metal,such as Cu, there are large numbers of free electrons moving randomly throughout it but the net movement in any given direction is zero. If a battery is connected as shown above, electrons are attracted to the positive pole, passing through the wire and the lamp. The movement of charge is called an electric current. e.g. How many electrons pass through a circuit in 10 ms if the dc current is 2 A? 10 x 10-3 x 2 x 6.24 x 1018 = 1.248 x 1017 electrons or 20 mC of charges. Current is measured with an ammeter. Conventional current flow will be used throughout this course. AC or alternating current is current that changes direction cyclically, i.e. charges alternately flow in one direction then in the other.

Praktyczne źródła napięcia DC Pierwotne baterie są nie ładowalne Wtórne baterie są ładowalne. Baterie są w różnych kształtach, rozmiarach , typach ( np. alkaliczne, węglowo-cynkowa, litowa, NiCad, kwasowo-ołowiana) oraz pojemnościach i napięciach. Pojemność baterii (Ah) = pobór prądu x czas życia. The voltage of a battery, its service life, and other characteristics depend on the material from which it is made. Alkaline battery is one of the most widely used, general purpose primary cell available. It provides 50 % - 100 % more total energy for the same size unit than carbon-zinc cell. Cell voltage is 1.5 V. Rechargeable ones available. Carbon-Zinc battery (dry cell) was the most popular. Sizes and voltage are the same as alkaline. Lithium battery features small size, low cost, and long shelf life. Cell voltages are available from 2 - 3.5 V. NiCads are the most popular rechargeable batteries. They have long service lives, operate over wide temperature ranges, and have many styles and sizes. Lead acid commonly found in automotive batteries. Cell voltage is 2 V. They can deliver large current (> 100 A) for short periods, e.g. for starting an automobile. Battery Capacity is specified in Ah. It is affected by discharge rates, operating schedules, temperature, and other factors. Life is shortened at low as well as at high temperatures. Cells are often put in parallel or in series to increase their current or voltage respectively. Other dc voltage sources Electronic power supplies rectify ac to dc. Their voltage may be fixed (for equipment, TV, etc.) or variable (for lab). Solar/photovoltaic cells convert sunlight to electrical energy for remote areas or space applications. DC generators convert mechanical energy to electrical energy by rotating a coil of wire through a magnetic field.

Inne źródła napięcia DC Elektroniczne zasilacze napięcia prostują prąd zmienny (AC) do prądu stałego. Baterie słoneczne zamieniają energię słoneczną na energię elektryczną. Generatory DC zamieniają energię mechaniczną obracającej się ramki w zewnętrznym polu magnetycznym magnesu (stojana).

Pomiar napięciaV i prądu I + + + E _ R2 E _ R2 V _ _ A + a) Pomiar Napięcia b) Pomiar prądu Umieścić Woltomierz równolegle do urządzenia na którym mierzymy napięcie. Aby zmierzyć natężenie prądu Amperomierz musi być umieszczony szeregowo z urządzeniem przez które płynie prąd .

Przełączniki, bezpieczniki, & wyłączniki Switches are mechanical devices to close or open the current path e.g light switches at home. Types of switches: SPST, SPDT, DPST, DPDT, rotary, NO & NC pushbutton. Fuses & Circuit Breakers are protective devices that interrupt excessive current flow. Fuses use a metallic element that melts when current exceeds a preset value. They are available as fast-blow and slow-blow types. Slow-blow fuses do not blow on small, momentary overloads. An ohmmeter can be used to check whether the fuse has blown. Circuit breakers depend on the magnetic field produced by the excessive current to operate a mechanism that trips open a switch. After the fault or overload condition has been cleared, the breaker can be reset and used again. Their operation is slower than that of a fuse.

Oporność Oporność jest podaje odwrotność mobilności ładunku i zależy tylko od rodzaju materiału i wymiarów opornika (długość, przekrój poprzeczny): R = rl / A gdzie r jest opornością własną (W-m), l długością (m), a A jest przekrojem poprzecznym (m2) . Uwaga: r zależy od temperatury!. Dla przewodników ta zależność jest liniowa i podane przez współczynnik temperaturowy (a). Zmianę oporu w zależności od temperatury podaje: R2 = R1 [1 + a1 ( T2 - T1 )] Free electrons flow in a conductor when a voltage source is applied to it. The electrons collide with atoms and other electrons and these collisions give rise to resistance to the current flow. R depends on the type of material, the length of the conductor, the x-sectional area and temperature. e.g. Determine the resistance of 100 m of AWG 20 solid copper wire. R = 1.732x10-8x100/(px(0.813x10-3)2/4) = 3.34 W. AWG is still widely used in the U.S. and many English-speaking world. The higher the AWG #, the higher the R, and the lower the current handling capacity (see Table 3.2). Most applications do not use solid conductor sizes beyond AWG 10 because they are difficult to bend and easily damaged by mechanical flexing. Large-diameter cables are nearly always stranded rather than solid. AWG is based on a unit called the circular mil (CM) which is defined as the area of a circle of 1 mil (=.001 in.). Since not all conductors have circular x-sections, it is occasionally necessary to convert areas in square mil (sq. mil) to CM: 1 sq. mil = 4/p CM, or 1 CM = p/4 sq. mil. Temperature Effectsb As T increases, more electrons will escape their orbits, causing additional collisions within the conductor. For most materials, R increases almost linearly with T. They are said to have a +ve temperature coefficient. Semiconductor materials, however, have -ve a . At absolute zero (T=-273.15o C), R approaches 0. e.g. A copper wire has a resistance of 10 W at 20o C. Calculate its resistance at 100o C and -30o C. R100 = 10[1+.00393(100-20)] = 13.14 W; R-30 = 10[1+.00393(-30-20)] = 8.04 W.

Typy oporników (rezystorów) Stałe oporniki np.: sprasowane związki węgla, cienkie warstwy metalu, tlenku metalu, zwoje drutu cienkiego, & oporniki półprzewodnikowe Zmienne oporniki: potencjometr & reostaty Resistance is added to an electronic/electrical circuit to control the voltage or current. Fixed resistors have essentially constant resistance values. Carbon composition resistors consist of a carbon core mixed with an insulating filler. The ratio of carbon to filler determines the resistance. The entire resistor is encapsulated with an insulated coating. Values range from 1 W to 100 MW and have power ratings from 1/8 W to 2 W. Carbon core resistors are inexpensive & easy to produce but they have wide tolerances due to temperature changes. Where precision is important, film resistors are usually employed. If close tolerances are required, films of alloys such as NiCr, constantum, or manganin, are used. Wire-wound resistors made of a metal alloy wound around a hollow porcelain core and covered with a thin layer of porcelain are used for dissipating large quantity of heat. IC resistor packages containing many individual resistors are used to conserve space on pcb. Variable resistors consist of 3 terminals, two of which are fixed to the ends of the resistive material. A wiper enables the resistance between the central terminal and either terminal to change. Potentiometers are used to adjust the amount of potential provided to a circuit. Rheostats are used to adjust the amount of current within a circuit. Their connection is similar to that of potentiometers except one of the end terminals is not connected. Stały Opornik Potencjometr Reostat

Kod kolorów rezystorów Kolor: CZ, Br, Czer, Pom, Ż, Ziel, Bl, Fiol, Sz, Bi, Zł , Sr , Brak Pasm 1: 1 2 3 4 5 6 7 8 9 Pasm 2: 0 1 2 3 4 5 6 7 8 9 Pasm 3: 1 10 102 103 104 105 106 107 .1 .01 Pasm 4: 5% 10% 20% Large resistors such as the wire-wound resistors or the ceramic-encased power resistors have their resistor values and tolerances printed on their cases. Smaller resistors use coloured bands. e.g. Determine the resistance of a resistor having the colour codes: Y, V, O. What are the possible maximum and minimum values of this resistor? R = 47 kW +20% = 37.6 kW to 56.4 kW. Resistance is measured with an ohmmeter which is generally part of a multimeter. Remember to isolate the resistor and switch off all power supplies from the circuit when measuring R. Ensure the ohmmeter is on the correct range to provide the most accurate reading. If the resistor is open, an analog ohmmeter will read infinity while a digital one will read OL or 1---. Turn off the ohmmeter when finished to conserve battery. If a device has a nonlinear V-I characteristic, it is referred to as a nonohmic device. Diodes and varistors are such devices. When the voltage across a varistor exceeds its rated value, its resistance suddenly becomes very small. Thus varistors are used as overvoltage protection devices. The resistance of thermistors drops with temperature. They are suitable for use in circuits to control current and to measure or control temperature. Conductance Conductance, G, is defined as the measure of a material’s ability to allow the flow of charge and is assigned the unit of siemens (S). G = 1/R.

Prawo Ohma I U 6 + + E V R 3 + 2 4 I (mA) Prawo Ohma mówi że prąd (I) w obwodzie z rezystorem jest proporcjonalny do napięcia (E lub U) i odwrotnie proporcjonalny do wartości rezystancji (R). Równanie: Doubling the voltage, E, doubles the current, I. Doubling R halves I. e.g. What is the value of R in the circuit above? R = 3 / (2x10-3) = 1.5 kW. e.g. A 20 V source is connected across a 100 W resistor. Determine the current through the resistor. I = 20/100 = .2 A For loads (& other components), V = IR. These voltages are sometimes referred to as IR drops. Voltage & Current Conventions (pg 104 -107) For voltage across a resistor, place the plus sign at the tail of the current arrow. If the actual current is in the direction of its reference arrow, it will have a positive value, while if it is opposite to its reference arrow, it will have a negative value. Voltage exists between points, and when we place a + at one point and a - at another point, we define this to mean that we are looking at the voltage at the point marked + with respect to the point marked -. In complex circuits with unknown voltage polarities and current directions, proper application of the above principles will ensure correct results.

Moc Moc jest zdefiniowana jako ilość pracy lub Transferowanej energii w jedn. czasu (watt, W) gdzie W jest pracą (lub energią) w jednostach Jule’a (J) a t czasem w sekunach (s). The higher the watt rating of a device, the more energy we can get out of it per unit time. Since V = W/Q and I = Q/t, we can easily derive that P = W/t = QVxI/Q = VI. For a voltage source, P = EI. e.g. A 50V source is connected across a 40 W resistor in series with a 60 W resistor. Determine the power dissipated in each resistor. I = 50/100 = .5 A; P40 = .52x40 = 10 W; P60 = .52x60 = 15 W. Power Rating of Resistors Standard power rating of resistors are: 1/8, 1/4, 1/2, 1, and 2 W. To provide a safety margin, it is customary to select a resistor that is capable of dissipating 2 or more times the computed power. Max. I and V are computed by the equations: I = o(P/R), and V = o(PR). e.g. What is the max. voltage that can be applied across a 68 W,3 W resistor in seires with a 150 W, 2 W resistor? I68 = o(.25/68) = 60.6 mA, I150 = o(.5/150) = 57.7 mA, ˆVmax = .0577x218 = 12.57 V Dla prądu elektrycznego odpowiada to:

Energia & Wydajność Energia jest podana w: W = P x t [ J] Jednostka używana w elektrotechnice to kWh = 3.6 MJ or 1000 Wh. Wydajność urządzenia lub systemu jest zdefiniowana przez stosunek użytej mocy do całkowitej dostarczonej mocy., h = (Pout / Pin) x100 %. Całkowita wydajność to iloczyn poszczególnych wydajności . e.g. Determine the cost of leaving a 1.5 kW heater and five 100 W lamps on for 8 hours at $0.10 per kWh. cost = .1 x 2 x 8 = $1.60 Efficiency h = (Pout / Pin) x 100 % or (Wout / Win) x 100 % Since Pin = Pout + Plosses , therefore, h = Pout / (Pout + Plosses) x 100 %. Losses are generally in the form of heat. Large power transformers have efficiencies of 98% or better, while electronic amplifiers have efficiencies as low as 50% or less. e.g. A 120 V electric motor draws 11 A and develops an output power of 1.5 hp. What is its efficiency and how much power is wasted? h = 1.5 x 746 x 100 / (120 x 11) = 84.8 %; Plosses = 1320 - 1119 = 201 W For systems with subsystems or components in cascade: hT = h1 x h2 x --- x hn

Połączenia Szeregowe Dwa elementy są połączone szeregowo jeżeli są połączone w jednym punkcie i nie ma żadnych połączeń doprowadzających lub odprowadzających prąd w tym punkcie. Prąd (I) jest taki sam w każdym elemencie obwodu R1 R2 Punkt połączenia R1 E + I R2 R3

Poł. szeregowe & P.Kirchoffa Prawo Kirchoffa o napięciu dla zamkniętej pętli (oczka): Suma Vwzrostów = Suma Vspadków lub U=0 Całkowita rezystancja n rezystorów szeregowych: RT = R1 + R2 + . . . + Rn Całkowita moc : PT = P1 + P2 + . . . + Pn 2 elements are said to be in series if they are connected at a point & if there are no other current-carrying connections at this point. The current is the same through every element in the closed loop. KVL The summation of all voltages around a closed loop is equal to zero or the summation of voltage rises is equal to the summation of voltage drops around a closed loop. Resistors Applying KVL around a closed loop with n resistors in series gives IRT = I(R1 + R2 + . . . + Rn). Therefore RT = SR. Similarly, the power delivered by the voltage source is equal to the total power dissipated by all the resistors. e.g. E1 = 5 V, E2 = 10 V, R1 = 200 W, R2 = 100 W, R3 = 500 W. Determine RT , V2 , and P3. RT = 800 W; I = 15/800 = 18.75 mA; V2 = 18.75 x 10-3 x 100 = 1.875 V; P3 = .018752 x 500 = .176 W Voltage Sources in Series (pg 138) If the polarities of the voltage sources don’t result in voltage rises in the same direction, the magnitude of the resultant source will be the sum of the rises in one direction minus the sum of the rises in the opposite direction. The polarity of the equivalent voltage source will be the same as the polarity of whichever direction has the greater rise. The order of series components may be changed without affecting the operation of the circuit.

Zasada dzielnika napięcia Napięcie przyłożone do to połączenia szeregowego będzie spadać na rezystorach proporcjonalnie do wielkości poszczególnych rezystorów : Ux = (Rx / RT) E The total voltage dropped across the resistors must equal the applied voltage source(s) by KVL. e.g. E = 12 V, R1 = 3 kW , R2 = 5 kW , R3 = 2 kW . Determine the voltage drop across each resistor. V1 = 3x12/10 = 3.6 V; V2 = 5x12/10 = 6 V; V3 = 2x12/10 = 2.4 V. Note that SV = E. If there is an open circuit, I = 0, and voltage drop across each resistor = 0 V. On the other hand, if one or more of the resistors is short circuited, the voltage drop across the shorted resistor will also be 0 but I will be higher than the calculated value since RT is lower. Therefore, the voltage drop across the other normal resistors will be higher than the value calculated by the voltage divider formula. As a general rule, if a series resistor is more than 100 times larger than another series resistor, then the effect of the smaller resistor(s) may be effectively neglected.

Połączenia szeregowe Otwarty obwód będzie powodować brak spadów napięcia na opornikach ponieważ natężenie prądu płynącego w obwodzie I = 0. Rezystor zastąpiony zwarciem będzie powodował spadki napięcia na pozostałych rezystorach większe od oczekiwanych. Efekt obwodu na wartość napięcia na obciążeniu którego rezystacja jest > 100 większa niż wartość pozostałych rezystorów może być zaniedbany.

Ziemia obwodu Ziemia jest dowolnym punktem odniesienia lub wspólnym dla danego układu. Ziemia obwodu jest zwana ziemią obudowy wjeżeli jest ona połączona do metalowej obudowy układu, urządzenia. Dla bezpieczeństwa, ziemia obudowy jest połączona do uziemienia całego zasilania i ziemi sieci (właściwej ziemi) poprzez kabel

Symbole ziemi Ziemia obwodu Ziemia obudowy Symbole ziemi f Ground is an arbitrary electrical point of reference or common point in a circuit. Very often, the metal chassis of an appliance is connected to the circuit ground which is then referred to as the chassis ground. The chassis ground is usually further connected to the earth ground through a connection provided at the electrical outlet box for safety. Voltage Subscripts Double Subscripts: In the circuit above, Vab is positive which means the potential at point a is greater than the potential at point b; Vab = -Vba. Also, by KVL, Vac = Vab + Vbc. Single Subscripts: In the circuit with a reference point (or ground point), most voltages will be expressed with respect to the ground point. Thus, the voltage of point d with respect to ground is Vd. If the voltage at various points in a circuit is known with respect to ground, then the voltage between the points may be easily determined as follows: Vde = Vd - Ve When a voltage source is given with respect to ground, it may be simplified in the circuit as a point source. (pg 149) f Obwody równoważne

Opór wewnętrzny źródeł napięcia Napięcie obciążenia In an ideal voltage source, Rint = 0, and the terminal voltage will remain constant regardless of the load connected. An ideal voltage source will be able to provide as much current as the circuit demands. In a practical voltage source, the current in the circuit is limited by the combination of the internal resistance and the load resistance. Only under no-load condition (RL = open circuit), will the terminal voltage be equal to the voltage of the ideal source. e.g. Eideal = 12 V, Rint = 1 W, RL = 10 W. What is the terminal voltage? Vterminal = 10 x12/11 = 10.9 V. e.g. The voltage of a battery is 24 V when no load is connected across it, and 23 V when it is supplying 5 A to a load. Calculate the internal resistance of the battery. Rint = 1/5 = .2 W Loading effects: All instruments, regardless of types, will load the circuit to some degree. Dostępne napięcie

Obwód równoległy IT + RT R1 R2 R3 E I1 I2 I3 Ix = E / Rx; II P.Kirchoffa: IT = I1 + I2 + I3 = E / RT Elementy obwodu są równoległe kiedy mają tylko 2 węzły wspólne. Napięcie na wszystkich równoległych elementach obwodu będą takie same. Źródła napięcia o różnych potencjałach nie powinny nigdy być połączone równolegle. Elements or branches are said to be in a parallel connection when they have exactly two nodes in common. KCL: The summation of currents entering a node is equal to the summation of currents leaving the node e.g. In the diagram on the right Ib + Id + Ie = Ia + Ic and for the diagram on the left, IT = I1 + I2 + I3. Resistors in parallel Applying KCL to n resistors connected in parallel, IT = I1 + I2 + . . . + In or E/RT = E/R1 + E/R2 + .. + E/Rn i.e. 1/RT = 1/R1 + 1/R2 + . . . +1/Rn or RT = 1/ (1/R1 + 1/R2 + . . . + 1/Rn) W. e.g. Calculate the combined resistance of 4 resistors: 300 W, 500 W, 100 W, and 400 W, connected in parallel. RT = 1/ (1/300 + 1/500 + 1/100 + 1/400) = 56.1 W n Equal Resistors in Parallel RT = R/n or GT = nG 2 Resistors in Parallel RT = R1 R2 / (R1 + R2) Voltage sources of different potentials should never be connected in parallel. (pg 168) Current Divider Rule: Since IT = E/RT, and Ix = E/Rx. Therefore, Ix = (RT / Rx)IT . Hence, the current through the resistor with the lowest value is highest, and if all R’s are the same, IT will divide equally between them. Power dissipated by any parallel element is simply Px = E2 /Rx or IxE.

Obwód równoległy i P.Kirchoffa Prawo prądów Kirchoffa: I = 0, lub S Iin = S Iout Całkowita konduktancja: GT=G1 + G2 + . . + Gn =1/RT lub całkowita rezystancja, RT = 1/(1/R1 + 1/R2 + . . . + 1/Rn) Dla dwóch rezystancji równoległych: RT= R1R2 / (R1 + R2) Dla n identycznych rezystorów równoległych: RT = R/n gdzie R jest rezystancją każdego rezystora. Dzielnik natężenia: Ix = (RT/Rx)IT Całkowita moc wyemitowana:PT = P1 + P2 + . . . + Pn gdzie P1 = E2/R1 lub EI1; . . . . ; Pn = E2/Rn or EIn

Równoległo-szeregowe sieci RT1 R2 R3 (a) R2 R3 R4 R2 RT3 Many complex circuits may be separated into a combination of both series and/or parallel elements, while other circuits consist of even more elaborate combinations which are neither series nor parallel. For circuit (a), RT1 = R1 + R2 // R3; circuit (b), RT2 = R1//R2 + R3//R4; and circuit (c), RT3 = R1//R2 + (R3+ R5) // (R4 + R6) + R7. e.g. E = 20 V is applied to circuit (c) above with R1 = R2 = R7 = 1 kW, R3 = R4 = R5 = R6 = 500 W. Determine : IT , I2 and V6. RT3 = 1k//1k + (.5+.5)//(.5+.5) +1 = 2 kW; thus, IT = 20 / 2 k = 10 mA; I2 = 10 mA / 2 = 5 mA; V6 = (10 mA / 2) x 1 k = 5 V. Potentiometers (pg 234) An example of a variable resistor used as a pot. is the volume control on a receiver or amplifier. When the wiper is at the uppermost position, the voltage output is maximum. When it is at the lowermost position, the voltage is 0. The actual output voltage would depend on the load connected to the output terminals. R5 R6 R1 RT2 R3 R4 R7 (b) (c)

Źródło stałoprądowe i zamiana na napięciowe Idealne źródło stałoprądowe utrzymuje stały prąd niezależnie od wartości rezystancji . Idealne źródło stałoprądowe ma nieskończoną rezystancję bocznikującą Rs. Wiele źródeł prądowych połączonych równolegle może być zastąpionych przez jedno. Źródła prądowe nigdy nie powinny być łączone szeregowo. Idealne źródło prądowe I RS E I RS The direction of the current source arrow indicates the direction of current in the branch. The voltage across the current source depends on how the other components are connected. Source conversion e.g. E = 10 V, RS = 5 W, then I = 2 A. Although the sources are equivalent, currents and voltages within the sources may no longer be the same. They are only equivalent with respect to elements external to the terminals. Current Sources in parallel e.g. Current sources of 5 A, -2 A, -10 A, and 4 A are connected in parallel. The resultant current source is -3 A. (See also an example of a current source in parallel with a voltage source in series with an R). Current sources should never be placed in series since this would violate KCL. E = IRS I = E/RS Zamiana źródeł

Analiza oczek Ustalamy dowolny kierunek przepływu prądu (zgodny z kier. wskazówek zegara) dla każdego oczka Oznaczamy polaryzacje na rezystorach i źródłach. Napięcia przechodzone od – do + są brane z dodatnim znakiem od + do – z ujemnym znakiem. Napięcia na rezystorach przechodzących zgodnie z kierunkiem prądu są ujemne! Stosujemy I.Prawo Kirchoffa (napięciowe) do oczek W węzłach stosujemy II.Prawo Kirchoffa (natężeniowe) e.g. For the circuit above, E1 = 10V, E2 = 15 V, R1 = R4 = 1 kW , R2 = R3 = 500 W. Find the current through R2 , and the voltage across R3. 10 - 1k x I1 - .5k x I1 + .5k x I2 = 0 -15 + .5k x I1 - .5k x I2 - .5k x I2 + .5k x I3 = 0 .5k x I2 - .5k x I3 -1k x I3 = 0 Solving, I1 = .83 mA; I2 = -17.5 mA; I3 = -5.83 mA. Current through R2 is I1 - I2 = 18.33 (downward); and V3 = (I2 - I3)R3 = -5.84 V (+ve at bottom). If the circuit contains current sources, either convert each current source to a voltage source and then solving, or leave the current source as one of the loop currents. (pg 274 -276)

Zasada superpozycji + E E I I R1 R2 R1 R2 R1 R2 Całkowity prąd lub napięcie na rezystorze lub w gałęzi może być zastąpiony przez efekt spowodowany przez każde źródło z osobna. Zamieniamy wszystkie źródła napięciowe przez zwarcie a wszystkie źródła prądowe przez otwarty obwód, z wyjątkiem źródła które badamy. e.g. For the circuit above, let E = 50 V, I = 0.1 A, R1 = 2 kW , and R2 = 3 kW. Determine I2, P2. The current due to I is given by: 2x.1/(2+3) = .04 A The current due to E is given by: 50/(2+3) = -10 mA Therefore the resultant current, I2 = 40 -10 = 30 mA (downward), and P2 = (30 x10-3)23 k = 2.7 W Note that the superposition theorem does not apply to power. The advantage of this method over mesh analysis is that it is not necessary to use determinants or matrix algebra to analyse the circuit.

Twierdzenie Thevenin’a b b Jakikolwiek liniowy układ dwójników może być uproszczony do prostego układu składającego się obciążenia i z pojedynczego źródła napięcia, ETh i rezystancji wewnętrznej, RTh. ETh jest równowżne napięciu otwartego układu na zaciskach a i b, oraz RTh jest wypadkową rezystancją “widoczną” z punktu widzenia tych zacisków. Thevenin’s theorem is one of the most important theorems of electric circuits. It enables the most complicated circuit to be reduced to a single voltage source and a single resistance. It is particularly useful for analyzing the current that would flow in a load when the latter is variable. To convert any circuit into its Thevenin equivalent: 1. Remove the load from the circuit. 2. Set all sources in the circuit to zero. 3. RTh is obtained by calculating the resistance across the open-terminals ab. 4. To obtain ETh, replace the sources removed in step 2 and compute the open-circuit voltage across terminals ab. e.g. For the circuit above, E1 = 5 V, E2 = 10 V, R1 = R3 = 1 kW , R2 = 2 kW. Calculate the load current when RL = 100 W, and 5 kW. RTh = (1 + 2-1 + 1)-1 = 400 W; Open-circuit voltage caused by E1 = 5 x 1k//2k / (1k + 1k//2k) = 2 V; Open-circuit voltage caused by E2 = -10 x 1k / (1k + 1k//2k) = -6 V; therefore, ETh = -4 V When RL = 100 , IL = -4 / (100 + 400) = -8 mA; (-ve sign means current is upward) When RL = 5 k , IL = -4 / (5k + .4k) = -.74 mA

Twierdzenie Thevenin’a Procedura zamiany układu do układu ekwiwalentnego Thevenin’a : Usunąć obciążenie z obwodu. Ustalić wszystkie źródła na zero. RTh otrzymujemy wyliczając rezystancję zastępczą pomiędzy zaciskami ab. Umieścić z powrotem źródła z punktu #2 i obliczyć ETh jako napięcie otwartego układu na zaciskach ab.

Twierdzenie Nortona Jakikolwiek układ podwójnych połączeń można zastąpić poprzez układ składający się ze źródła prądowego, IN , i rezystora bocznikującego, RN. IN jest równoważnym prądem zwarcia pomiędzy punktami a i b, oraz RN jest równoważną rezystancją widzianą pomiędzy tymi punktami. Norton’s theorem is a circuit analysis technique similar to Thevenin’s theorem. To convert any circuit into its Norton equivalent: 1. Remove the load from the circuit. 2. Set all sources to zero. 3. Determine RN by calculating the resistance between terminals ab. 4. Replace the sources removed in step 2 and obtain IN by calculating the short-circuit current between terminals ab. e.g. With E1 = 5 V, E2 = 10 V, R1 = R3 = 1 kW, R2 = 2 kW for the circuit above, calculate the load current when RL = 100W, and 5 kW. RN = RTh = 400 W; Short-circuit current between ab for E1 is 5 / 1k = 5 mA; Short-circuit current between ab for E2 is -10 / (1k//2k) = -15 mA; therefore, IN = -10 mA. When RL = 100 , IL = -10 x 400 / (100 + 400) = -8 mA; When RL = 5 k , IL = -10 x 400 / (5k + 400) = -.74 mA. Note that the answers are the same as those computed by Thevenin’s theorem.

Zastosowanie Tw. Nortona Procedura zastępowania układu wg. Tw. Nortona: Opuszczamy rezystancję obciążenia . Wszystkie źródła ustalamy na zero. RN jest otrzymany przez obliczenie oporu zastępczego dla otwartego obwodu pomiędzy ab. Umieszczamy wszystkie źródła usunięte w punkcie 2 i obliczmy IN -jako prąd zwarcia pomiędzy punktami ab.

Maksymalna moc dostarczana b Obciążenie RL maksymalną otrzymuje moc ze źródła jeżeli rezystancja obciążenia jest dokładnie taka jak rezystancja Thevenin’a (lub Norton) obliczona patrząc w „tył” obwodu Ta maksymalna moc dostarczona do obciążenia wynosi: In amplifiers and in most communication circuits, it is often desirable that the load receives the maximum amount of power from a source. The maximum power transfer theorem states the condition for this to happen. When RL = RTh, VL = ETh / 2. Therefore, Pmax = VL2 / RL = ETh2 / (4RTh). Under the condition of maximum power transfer, the efficiency of the circuit is: h = Pout / Pin x 100 = IL2 RTh / (IL2 x 2RTh) x 100 = 50 %. For power transmission or supply applications, maximum power transfer is not a requirement. It is more important in this case to ensure that the efficiency is as close to 100% as possible. To meet this condition, RL must be kept much larger than the internal resistance of the voltage source (typically RL >> Rint) so that the voltage across the load will be very nearly equal to the maximum terminal voltage of the voltage source.

Uwagi do transmisji mocy We wzmacniaczach i urządzeniach telekomunikacyjnych, często chcemy aby moc która jest dostarczana obciążeniu była bliska maksymalnej możliwej dla danego źródła. Ale wtedy, dla takiego transferu mocy ( tzn. RL = Rźródła), wydajność wynosi tylko 50 %. Z drugiej strony dla transferu mocy lub zasilaczy mocy chcemy aby Rźródła << RL , czyli napięcie na obciążeniu jest bliskie napięciu źródła bo wtedy wydajność transferu wynosi blisko 100 %.

Kondensator Okładki Kon. płaski Symbol Pole elektryczne Kondensator składa się z 2 płyt przewodnika i izolatora pomiędzy nimi (dielectric) powietrze, olej, mika, plastik, ceramika, etc. Kiedy źródło dc jest przyłożone do kondensatora, jedna z płyt ładuje się dodatnio a druga ujemnie. Ilość ładunku zgromadzonego na kondensatorze: Q = CV (C) When a dc source is applied across the capacitor, electrons are pulled from the top plate and the same amount is deposited on the bottom plate. The capacitor is then said to be charged with the top plate positive and the bottom plate negative. Capacitors with little leakage can hold their charges for a considerable time. Always discharge capacitors after power has been removed if you want to handle them. The capacitance of a capacitor is one farad if it stores one coulomb of charge when the voltage across its terminals is one volt. Most practical capacitors range from pF to mF. e.g. How much charge is stored on a 47 mF capacitor when it is connected to a 50-V source? Q = 47 x 10-6 x 50 = 2.35 mC. Factors Affecting Capacitance C is directly proportional to plate area, inversely proportional to plate spacing, and directly proportional to the permittivity of the dielectric material. Air or vacuum has a value of eo = 8.85 x 10-12 F/m. For other materials, e is expressed as the dielectric constant or relative permittivity, er, times eo. Voltage Breakdown & Voltage Rating If the voltage across the capacitor is increased beyond a critical value, dielectric breakdown occurs and the electric field at breakdown is called the dielectric strength of the material. For air, breakdown occurs at about 3 kV/mm. Because of dielectric breakdown, capacitors are rated for maximum operating voltage and is indicated on the capacitor as WVDC or working voltage dc.

Kondensator (cd) Pojemność kondensatora płaskiego wynosi: C = e A / d (F), gdzie e jest przenikalnością dielektryka, A jest powierzchnia płytek a d jest odległością okładek. e = er eo gdzie er jest względną przenikalnością lub stałą dielektryczną dielektr. eo = 8.85 x 10-12 F/m jest przenikalnością powietrza. Energia zmagazynowana w kondensatorze wynosi W = 1/2 CU2 (J)

Typy kondensatorów Stałe: e.g. ceramiczne, plastikowe , mikowe, elektroliczne, montaż powierzchn. Elektrolityczne kondensatory są aluminiowe lub tantalowe i są spolaryzowane. Zmiennej pojemności: e.g. Zmienne pow. płyt . To suit different applications, capacitors are made in a wide variety of types and sizes. Fixed capacitors are often identified by their dielectric. Design variations include tubular & interleaved plates. The interleave design uses multiple plates to increase effective plate area. The tubular design uses sheets of metal foil separated by an insulator such as plastic film. Fixed capacitors are encapsulated in plastic, epoxy resin, or other insulaitng material. Ceramic capacitors: high permittivity type (er > 100) is compact but its capacitance varies widely with temperature & operating voltage, while low permittivity type has stable characteristics although physically larger. Plastic film capacitors: film/foil type uses metal foil separated by plastic film, while metallized-film type has foil material vacuum deposited directly onto plastic film. Film/foil type is generally larger, but has better capacitance stabiltiy & higher insulation resistance. Typical film materials: polyester, Mylar, polypropylene, & polycarbonate. Mica capacitors: low cost, low leakage, and good stability. Available values range from a few pF to about 0.1 mF. Electrolytic capacitors: provide large capacitance at a relatively low cost but leakage is relatively high & breakdown voltage relatively low. Basic aluminium capacitors are made with strips of Al foil separated by gauze saturated with an electrolyte. Chemical action creates a thin oxide layer that acts as the dielectric. Electrolytic capacitors have a shelf life of about 2 years. Tantalum capacitors are smaller, have less leakage, and are more stable. Surface mount capacitors: are extremely small and usually made of ceramic chip. Variable capacitors are commonly used in radio tuning circuits with a set of stationary plates & a set of movable plates ganged together & mounted on a shaft. Trimmer/padder capacitors are for fine adjustments.

Połączenie kondensatorów   + + + E C1 U1 U2 Un Un - - - C2   Dla kondensatorów połączonych równolegle powierzchnia okładek dodaje się Całkowita pojemność: CT = C1 + C2 + . . + Cn Napięcie wypadkowe: U1 = U2 = . . Un = E Całkowity ładunek: QT = Q1+ Q2 + . . + Qn For capacitors in parallel, the effective plate area is the sum of the individual plates. The charge on each capacitor is Q = CV. The total capacitance of capacitors in parallel is the sum of their individual capacitances. e.g. A 10 mF capacitor is in parallel with a 20 mF capacitor. Determine the total capacitance and total charge stored if a 12 V source is applied across them. CT = 30 mF; QT = 30 mF x 12 = 360 mC. For capacitors in series, the same charge appears on each. Thus, VT = Q/CT = Q/C1 + ... + Q/Cn i.e. CT = 1 / (1/C1 + ... + 1/Cn). e.g. A 47 mF capacitor is in series with the parallel capacitors above. Calculate CT and QT. CT = 47 x 30 / (47 + 30) = 18.31 mF; QT = 18.31 x 12 = 219.7 mC. Voltage Divider Rule for Series Capacitors Since QT = CTVT = Qx = CxVx, therefore, Vx = (CT / Cx)VT. Similarly, it can be shown that V1 = (C2 / C1)V2, V1 = (C3 / C1)V3, V2 = (C3 / C2)V3, etc. e.g. Three capacitors 5 nF, 10 nF, and 20 nF are in series. If the voltage across the 5 nF capacitor is 2 V, what is the voltage across the 10 nF capacitor? What is the voltage across all 3 capacitors? CT = (5-1 + 10-1 + 20-1)-1 = 2.86 nF, therefore V2 = 5 x 2 / 10 = 1 V; and VT = 5 x 2 / 2.86 = 3.5 V Energy Stored by a Capacitor : W = CV2 / 2 (J). This energy is stored in the capacitor’s electric field.

Połączenie szeregowe - - - UT U1 U2 Un + + + C1 C2 Cn E Całkowita pojemność, CT = 1/(1/C1 + 1/C2 + . . . + 1/Cn) Ładunki są takie same, np. Q1 = Q2 = . . . = Qn Całkowite napięcie, UT = U1 + U2 + . . . + Un = E Zasada dzielnika napięcia na kondensatorach:

Ładowanie kondensatora vR = Ee-t/t ładowanie Stałe napięcie Transient Region Assume the capacitor is initially uncharged and the switch is open. When the switch is moved to position a, the current jumps to E/R amps, then decays to zero, while the voltage rises from zero to E exponentially. Hence, an uncharged capacitor looks like a short circuit at the instant of switching. The capacitor voltage cannot change instantaneously. vc = E (1 - e-t/t); i = (E / R) e-t/t, where t = RC is known as the time constant of the circuit. If the capacitor has an initial voltage of Vo , then vc = E + (Vo - E)e-t/t ; i = (E-Vo)e-t/t / R. e.g. Given E = 25 V, Vo = 5 V, R = 5 kW, and C = 10 nF, determine vc, vR and i, 25 ms after the switch closes. t = 5kx10nF = 50 ms; vc = 25+ (5-25)e-25/50= 12.87 V ; vR =(25-5)e-0.5 = 12.13 V ; i = ((25-5)/5k)e-0.5 = 2.43 mA Note that after 1t, vc =.632E, and after 5t, vc =.993E (i.e. the transient period lasts for about 5t). Steady State When the capacitor voltage and current reach their final values and stop changing, the circuit is said to be in the steady state. In this state, no current flows through the capacitor. Thus a capacitor looks like an open circuit to steady state dc. For more complex circuits, first reduce it to its series equivalent, then apply the equations above. vc= E(1-e-t/t) i = (E/R)e-t/t t = RC

Ładowanie kondensatora (cd) Obszar ładowania: Kiedy przełącznik ustawimy w pozycji a, prąd skacze do E/R amperów (jak przy zwarciu), potem opada eksponencjalnie do zera, natomiast napięcie rośnie eksponencjalnie od zera do E woltów. Uwaga: po t =1, vC = 0.632E a po t =5 , vC = 0.993E. Obszar ustalony: Napięcie i natężenie nie ulegają zmianie. VC = E i IC = 0 ; a zatem kondensator po naładowaniu wygląda jak rozwarty obwód.

Rozładowywanie kondensatora vR = -Voe-t/t Assuming the capacitor voltage initially is Vo, and the switch is now placed in position b, the voltage across the capacitor would start to drop from Vo to zero exponentially while the current jumps to -Vo/R amps initially before decaying to 0. vc = Vo e-t / t ; and i = -(Vo / R) e-t / t , where t = RC is the time constant. Note that if the capacitor was fully charged to E, then substitute E for Vo in the equations. e.g. For the circuit above, assume the capacitor is charged to 50 V before the switch is closed, R = 2 kW , and C = 10 mF, determine the voltage and current at 2t . vc = 50 e-2 = 6.77 V; i = -(50 / 2k) e-2 = 3.39 mA vc = Voe-t/t i = -(Vo/R)e-t/t t = RC

Krzywe ładowania i rozładowywania The Universal Time Constant curves can be used to estimate the voltage and current of a charging or discharging capacitor. e.g. For a series circuit with R = 500 W, C = 5 mF, and E = 30 V, determine vc, and i after the switch is closed for 3.75 ms. t = 500 x 5 m = 2.5 ms; therefore t = 3.75/2.5 = 1.5 t. From the graph, vc = .77 x 30 = 22.1 V; and i = .23 x 30/500 = 13.8 mA. Stała czasowa