Electrical Circuits HE 13.11.018 1. Electrical Components hese are tabulated below Component Name Properties esistor Simplest passive element, no dependence on time or frequency Capacitor eactive element, impedance changes with frequency nductor eactive element, impedance changes with frequency DC Source Output does not change with time AC Source Output changes with time Diode Simplest pn junction ransistor Basic building block for electronic circuits ntegrated Circuit Collection of several circuit elements on one chip Symbol Offers Units esistance, ohms C Capacitance F, Farads L nductance H, Henry E, oltage, Current, olts, A, Amps E, oltage, Current, olts, A, Amps D Signal shaping r Amplification, switching C All asks able 1.1 List of electrical components. he circuit symbols of these components are drawn in Fig. 1.1. EEM 13 HE Kasım 018 Sayfa 1
esistor Capacitor C F nductor L H DC Source (oltage source) ransistor E,, A b DC Source (Current source) c ntegrated circuit C AC Source E ( t ), ( t ), A D e Diode Fig. 1.1 Schematic drawings of circuit symbols of the electrical components listed in able 1.1.. mportant Measuring nstruments and Signal Generators a) oltmeter : Measures the voltage across a resistor, inserted in parallel as shown in Fig..1. Looking at Fig..1, it is easy to conclude that in order to get a reliable voltage measurement, the internal resistance of the voltmeter must be much higher than 1 E oltmeter Fig..1 he way of connecting a voltmeter in a circuit. b) Ammeter : Measures the current in a circuit loop. A typical configuration that uses the ammeter is given in Fig... As seen from Fig.., the use of ammeter requires that previously connected part of the circuit is broken. his disturbance may not be convenient in all cases. herefore, mostly we obtain the value of current by first measuring the voltage, then applying ohms law, i.e., / to arrive at current. EEM 13 HE Kasım 018 Sayfa
1 E Ammeter Fig.. he use of ammeter in a circuit. c) Oscilloscope : Displays the variation of a signal against time, that is st against its variable t. d) Spectrum Analyser : Displays the frequency spectrum of a signal, i.e., F s t S f. e) Signal Generator : Generates all types of time signals, st discontinuous. 3. esistors, against f, periodic, aperiodic, continuous, Now we concentrate on the most basic element, resistor. esistance of a wire is related to its resistivity and its physical dimensions as follows l A (3.1) where stands for resistivity (the resistivity figure for each element is different), l is the length of the wire, A is the cross section. hese parameters have the following units l A Ω, ohm, Ωm, ohm meter, m, meter, m, meter (3.) When these units are used in (3.1), it is easy to see that both sides of the equation in (3.1) are balanced. From (3.1), we see that resistance is directly proportional to resitivity and length of wire (or the element in question) but inversely proportional to cross sectional area. Physically these relations seem to be meaningful. he resistivity is important in order to obtain the desired resistance value with reasonable dimensions and further to classify elements as conductors, semiconductors and insulators. esistivity figures of some materials are listed below EEM 13 HE Kasım 018 Sayfa 3
Material name Units Classification Copper 8 1.7 10 Ωm Conductor Gold 8.4 10 Ωm Conductor ron 8 1.3 10 Ωm (Poor) Conductor Aluminium 8.65 10 Ωm Conductor Carbon 8 3500 10 Ωm esistor Germanium 50 Ωcm Semiconductor Silicon 3 00 10 Ωcm Semiconductor Gallium Arsenide 6 70 10 Ωcm Semiconductor China, Porcelain 15 10 Ωcm nsulator able 3.1 esistivity of some materials. As understood from able 3.1, copper is a good conductor, is therefore used in metallic cable industry. On the other hand carbon has relatively high resistivity and is mostly used in resistor manufacturing. Germanium, silicon and gallium arsenide are perfect examples of semiconductors and widely used in the fabrication of electronic components. Lastly china and porcelain are known to be good insulators. Exercise 3.1 : Write the resistivity figures of Germanium, silicon, gallium arsenide, china and porcelain in units of Ωm. here are colour code related to numeric values of resistors and tolerances as shown in Fig. 3.1. Fig. 3.1 esistor colour codes. EEM 13 HE Kasım 018 Sayfa 4
Finally we introduce the inverse of resistance, called conductance, G and define it as stated below 1 A G l (3.3) he units of conductance is Siemens. 4. Ohm s Law and Conventions We take a simple circuit shown in Fig. 4.1 and define the current flowing through the resistor as Potential difference acrros total E otal resistance 1 1 (4.1) Node A A + DC voltage source E + 1 Node B B + 1 1 Ground Fig. 4.1 Simple circuit for definitions of current and voltage directions. Our convention is that he current is assumed to flow from the positive (+) terminal of the DC voltage source to the negative () terminal. his way, as marked in Fig. 4.1, it is possible to define the currents and voltages in the positive and negative directions. hey are simply related as,,, (4.) 1 1 For a circuit, where it is ambiguous which direction is positive, we arbitrarily choose any direction and place our arrow along this chosen direction. f the calculated quantity comes out to be positive, then we say the chosen direction is correct, but if the result is negative, then we reverse the chosen direction. Sometimes the negative terminal of the DC voltage source is grounded as drawn in Fig. 4.1. his means that unless indicated otherwise by the voltage arrow, the voltages (potentials) are measured with respect to (wrt) the ground. n this manner, for the voltages marked in Fig. 4.1, we can write EEM 13 HE Kasım 018 Sayfa 5
A : oltage of node A with respect to ground : oltage drop across resistor 1 1 : oltage drop across resistor B : oltage of node B with respect to ground A 1 E, 1 A, B = (4.3) By rewriting (3.1) (4.4) he implication in (4.4) is that, the relationship between voltage and current is a linear one, thus it can be plotted in the form of lines in two ways as illustrated in Fig. 4.. Low resistance High conductance ) High resistance Low conductance ) d ) High resistance Low conductance d ) Low resistance High conductance ), cot ( ) = 0 0 d ), tan ( ) = d Fig. 4. llustration for (4.4). he low and high resistance effects shown in Fig. 4. are created by turning of Fig. 4.1 into a variable resistor. As seen from Fig. 4., an alternative definition of resistance is d d (4.5) n the case of a simple resistor, where the relationship between voltage and current is linear (as seen in Fig. 4.), then d d (4.6) here are also cases, where we have to apply (4.6) individually to different regions of curves. Such characteristics is exhibited by the diode. hus by using the circuit in Fig. 4.3, we get the response shown in Fig. 4.4. EEM 13 HE Kasım 018 Sayfa 6
D oltmeter D E L L oltmeter D Fig. 4.3 Circiut to measure the diode characteristics. Z D egion of low conductance High resistance Forward bias Low resistance 0 D everse bias Low resistance = 0.7 olts for Si = 0.3 olts for Ge Fig. 4.4 Diode voltage current characteristics. As seen from Fig. 4.4, in terms of resistance (or conductance), diode has three regions. hese are everse bias (Low resistance) : Low conductance region : < D Z D Forward bias (Low resistance) : (4.7) his way, a diode will be operated in forward bias mode if we want to cut out for instance the negative cycles of an AC signal. On the other hand, it can also be operated in reverse bias mode for voltage regulation. 5. Power and Energy As stated before power is found from the relation (assuming no time dependence of voltage and current) D Z EEM 13 HE Kasım 018 Sayfa 7
P (5.1) f the second and third expressions are considered, we can say that power is directly proportional to current or voltage squared provided that the resistance is set to unity, i.e., 1 Ω. his way we speak about normalized power given by P U n (5.) n this case, it becomes unimportant which quantity (i.e. current or voltage) we take to measure power. Energy on the other hand, is power used over a time interval, thus P (5.3) As stated in (3.5) of the notes, ECE 107_Assessment of Quantities_June 013_HE, if there is time variation in power, then (5.3) turns into 0 Pt dt, P av (5.4) where P indicates the average power consumed over a time interval av. 6. Series and Parallel Circuits he simplest series circuit consisting of two resistors connected in series is displayed in Fig. 6.1. A 1 DC voltage source E 1 B Fig. 6.1 Simplest series circuit and related circuit parameters. Looking at Fig. 6.1, it is possible to define the followings EEM 13 HE Kasım 018 Sayfa 8
otal resistance seen looking into the circuit across AB : 1 1 E E, E,, 1 1 1 E E E E, 1 1 1 1 1 (6.1) he last line in (6.1) is known as the voltage dividing rule, which means that the total voltage, E is divided into and 1 in direct proportion to value of the resistor. Since there no to time dependence of the quantities in (6.1), the respective powers will be P P 1 1 1 1, 1 otal power delivered by the DC voltage source : P P P E (6.) s 1 Extending the results of (6.1) and (6.) to the case of N connected in series, we get he rules of finding the total voltage, 6. pictorially. N N N, E, P P (6.3) n n s n n1 n1 n1 E of DC voltage sources connected in series are shown in Fig. + + + + E 1 E E 3 E = E 1 + E + E 3 + + + + E E E E = E 1 + E + E 3 1 3 f E + E 3 > E 1 Fig. 6. ules for the total voltage, E of DC voltage sources connected in series. Now we turn to the simplest parallel circuit consisting of two resistors connected in parallel. Such a configuration is depicted in Fig. 6.. EEM 13 HE Kasım 018 Sayfa 9
A DC voltage source E 1 1 B Fig. 6.3 Simplest parallel circuit and related circuit parameters. From Fig. 6.3, we define the followings 1 1 otal resistance seen looking into the circuit across AB : 1// 1 1 1 E E E 1 E, 1, 1, 1 1 1 1 1 1, 1 1 1 (6.4) he last line in (6.1) is known as the current dividing rule, which means that the total current delivered by the DC voltage source, is divided into 1 and in inverse proportion to value of the resistor. Exercise 6.1 : Write the equivalent of (6.) for the parallel circuit of Fig. 6.3. n a manner similar to (6.3), we can express the total resistance, total current and the total power consumed in the circuit ( = total power delivered by the DC voltage source) as N N 1 1// // N, n, Ps Pn (6.5) 1 1 1 n1 n1 1 N 7. Kirchhoff s oltage Law (KL) and Kirchhoff s Current Law (KCL) KL states that the algebraic sum of voltage rises (due to sources) and voltage drops (due to resistors or other circuit elements) around a closed circuit loop is zero. n this notation voltage rises and drops are taken positive if they are in the direction of the loop arrow. f on the other hand, voltage rises and drops are taken negative if they are in the opposite direction to the loop arrow. Here the positive and negative directions are as defined in Fig. 4.1. A simple case of KL is illustrated in Fig. 7.1, EEM 13 HE Kasım 018 Sayfa 10
+ 1 1 + + E KL L Fig. 7.1 llustration of KL for a simple circuit of one loop and two resistors. We see in Fig. 7.1 that, the E is in same the direction as that of KL arrow, while 1 and are in the opposite. hus the KL equation for the circuit in Fig. 7.1 becomes E E E 0, E (7.1) 1 L 1 L 1 1 For this simple circuit of Fig. 7.1, the defined loop current L is equal to. he expression on the right hand side of (7.1) simply means that with the signs or the directions of voltage rise and drops as defined in Fig. 7.1, the voltage rise is equal to voltage drops. KCL states that the algebraic sum of currents around a node is zero, where the entering currents are taken positive, the leaving currents are taken negative. Alternatively, KCL states that the currents entering and leaving the nodes are equal. A simple illustration for KCL is given in Fig. 7.. Node E (entering the node) 1 1 (leaving the node) (leaving the node) Fig. 7. Simple illustration of KCL. EEM 13 HE Kasım 018 Sayfa 11
Application of KCL to Fig. 7. will lead to 0, (7.) 1 1 8. Examples We solve various examples in this section highlighting the use of KL and KCL. Example 8.1 : For the circuit given in Fig. 8.1, find all the indicated currents and voltages. A 1 1 = 9 B C s 3 6 E = 16.8 = 6 3 = 4 4 5 6 = 3 4 = 6 5 = 3 Fig. 8.1 Circuit for Example 8.1. Solution : Although it is possible to solve the circuit as drawn in Fig. 8.1, finding the parallel equivalent of 1 1 // and 4// 5 and adding 3 to 4// 5, that is 345 3 4 // 5 and redrawing with these rearrangements will convert the circuit into a simpler form as shown in Fig. 8.. 1 A s B C E = 16.8 1 = 3.6 KL1 L1 345 345 = 6 3 6 KL L 6 = 3 6 Fig. 8. he circuit of Example 8.1, after rearrangement. Now we write for KL1 and KL as follows EEM 13 HE Kasım 018 Sayfa 1
After rearrangement, (8.1) will become KL1 : E E 0 s 6 3 1 345 L1 1 L1 L 345 KL : 0 (8.1) 345 6 L1 L 345 L 6 KL1 : E L1 1 345 L 345 L1 345 L 345 6 KL : 0 (8.) Although (8.) can be solved by hand, below we illustrate the solution in Matlab >> A = [9.6 6;6 9]; %%% [KL1;KL] >> B = [16.8;0]; %%%% [E;0] >> C = A\B C = 3 %%%% _L1 %%%% _L his way, 3 A, A, 1 A L1 s L 6 3 L1 L E 16.8, 6, 10.8 (8.3) A B C 3 345 1 s 1 Additionally from current dividing rule of (6.4), for the configuration of the circuit in Fig. 8.1, we have 1. A, 1.8 A, 1 1 s s s 1 1 1 1 A, A, (8.4) 5 4 4 3 5 3 3 4 5 4 5 3 4 5 3 t is possible to formulate an alternative solution as follows aking the circuit in Fig. 8. one stage further, we get the simplified model in Fig. 8.3, where 1 345 6 is // 5.6 Ω (8.5) From (8.5), we find initially s and the other currents from current dividing rule and the voltages at the nodes as follows E, 1 A, A, 6 345 s 3 s 6 s s 3 6 345 6 345 6 E 16.8, 6, 10.8 (8.6) A B C 3 345 1 s 1 EEM 13 HE Kasım 018 Sayfa 13
A s E = 16.8 Fig. 8.3 he most simplified model of the circuit in Fig. 8.1. Finally computation of the remaining currents can be carried out as shown in (8.4). he circuit to be solved may not always be in the layout or form as introduced in the previous example. n such cases, it is helpful to redraw the given circuit as illustrated in Example 8.. Example 8. : For the circuit given in Fig. 8.4, find all the indicated currents and voltages. E = + 18 A E 1 = 6 B = 6 1 = 6 1 s 3 = 5 3 4 = 7 Fig. 8.4 he circuit for Example 8.. o facilitate an easier solution, the circuit of Fig. 8.4 can be redrawn as shown in Fig. 8.5. EEM 13 HE Kasım 018 Sayfa 14
B E = 18 1 = 6 s A KL1 L1 E 1 = 6 1 = 6 3 KL3 KL L L3 3 = 5 4 = 7 Fig. 8.5 he circuit of Fig. 8.4 after rearrangement. he three KLs defined in Fig. 8.5 can be written as s KL1 : E E 0 E E 6 4 L1 1 1 L1 1 1 L1 KL : E 0 E 6 6 6 1 L L3 L L3 1 L L3 3 KL3 : 0 0 6 18 0 (8.7) L3 L 3 4 L3 L 3 4 L3 L L3 (8.7) can be solved by the following Matlab operations >> A = [6 0 0;0 6 6;0 6 18]; >> B = [4;6;0]; >> C = A\B C = 4.0000 %%%% _L1 1.5000 %%%% _L 0.5000 %%%% _L3 From the above results, we get the currents in the branches and the voltages in the nodes of the circuit in Fig. 8.5 as 4 A, 5.5 A, 1 A, 0.5 A s L1 1 L1 L L L3 3 L3 E 18, E E 6 (8.7) A B s 1 1 EEM 13 HE Kasım 018 Sayfa 15
hese notes are based on 1) MALAB m files. ) My own Lecture Notes. EEM 13 HE Kasım 018 Sayfa 16