Electric and Electronic Engineering

Transcription

1 Electric and Electrnic Engineering Intrductin Cmputer engineering is cncerned with the integratin f circuits and systems nt small pieces f silicn tday. A typical cmputer engineer has a wrking knwledge f silicn devices, CMOS circuits, lgic design, and system architecture and is usually a specialist in ne r mre f these areas. The field f electric and electrnic engineering has made spectacular advances in recent years. Overall, the bjective has been t prvide the technlgy needed t build large infrmatin systems n tiny chips, and t build a system f such chips in a bard. T slve mdern cmputatinal prblems new cmplex micrprcessrs are designed which due t their advanced architecture prvide t slve these prblems in real time and with minimized energy cnsumptin, and with the speed f several billin instructins per secnd. The mdern micrprcessr cnsists f bth micrprcessr cre and a set f peripheral devices including memry units, data transfer prts, DSP engines, ADC, DAC, etc. Such a micrprcessr r a system f them is named as System-On-the-Chip, r shrtly SOC. The fact is that the mdern SOC is characterized by high wire resistivity, high level f signal interferences, the delays in wires which are higher than delays in gates, lw feeding vltage and high current cnsumtin. The frequencies in such circuits are higher than thusands f megahertz. Therefre, t design mdern SOC the cmputer engineer has t take many electical laws int accunt. He r she has t knw that any bard intercnnectin is a transmissin line by definitin. Therefre, t design mdern cmputer bards ne has t cnsider that the reflectins, interference, and nise in bard wires cause measurable changes in the appearance r behavir f signals at higher frequencies. And as a result, negleblible attentin t these features causes unwrkable prjects. Lectures in electrnic engineering include electric engineering basics, transistr basics, transistr circuit design questins, and MOSFET transistr basics, which are needed in develpment f mdern circuits. The lectures are based n the methd f quadriple analysis. This methd helps t investigate and design bth transmissin lines and transistr circuits. Besides, the quadriple methd use shws the way t slve the cmplex prblem f the circuit analysis by the apprach "divide and cncuer". This teach the students t develp the algrithms fr slving similar prblems.

2 Used literature. Атабеков Г.И. Основы теории цепей. М.: Энергия с.. Uyemura J.P. Digital System Design. An Integrated Apprach. New Yrk: Brks/Cle Pub p. 3. Мигулин И.Н., Чаповский М.З. Усилительные устройства на транзисторах.- Киев: Технiка с. 4. Гальперин М.В. Практическая схемотехника в промышленной автоматике. М.:Энергоатомиздат с. 5. Гоноровский И.С. Радиотехнические цепи и сигналы. : Учебник для вузов. М.: Радио и связь с. 6. Щербаков В.И., Грездов Г.И. Электронные схемы на операционных усилителях: Справочник. Киев: Техника с. Electric engineering basics. Basic electric circuits and cmpnents.. Electric measurements The electric current is the stream, r cntinuus mvement f electric charges as illustrated by the fig.. The electric current is the time rate f change f charge acrss the referenced area, as given by: dq i dt, where Q is charge, which is measured in units f culmbs. A culmb is equal t charge f apprximately 6,4 0 8 electrns I + Fig.. Illustratin f the electric current flw The current unit is ampere. It is the base SI unit. SI means Systėme Internatinal d'unitės. This is the internatinally agreed upn system f cherent units that is nw in use fr all scientific and mst technlgical purpses in many cuntries. In shrt, ampere is A. Milliampere (ma),

3 3 micrampere (µa), nanampere (na) and ther subunits are distinguished. These units depend n each ther: A 000 ma, ma 000 µa, µa 000 na. In the space the electric ptential e is distinguished. It means the wrk, which is necessary t bring a unit f psitive charge t a given pint f the space. The difference between ptentials in tw pints, say e, e is equal t the vltage v, i.e. ve e. The vltage v is the electrmtive frce, which is equal t the wrk, which is necessary t bring a unit f psitive charge frm ne pint t anther. Frm this pint f view the ptential e is equal t the vltage between this pint and sme abstract pint f zered ptential e 0. Often the pint with such prperties is called as a grund. The vltage, r ptential difference, between tw pints in a circuit indicates the energy required t mve charge frm ne pint t the ther. As will be presently shwn, the directin, r plarity, f the vltage is clsely tied t whether energy is being dissipated r generated in the prcess. Bth ptential and vltage are measured in vlts, shrtly, V. Such subunits like kilvlt (kv), millivlt (mv), micrvlt (µv) are frequently used, and kv000v; V000mV; mv000µv. In the lw-frequency electrnic devices the direct-current (DC), and alternating-current (AC) circuits are distinguished. Belw the current and vltage are cnsidered which are exchanged in time, mstly alternating-current and alternating-vltage. If fr sme shrt time perid sme charge Q was flwn thrug a subcircuit, i.e. the current i flws, and at its edges the vltage v is present. Then the elementary energy which was emitted there as warm, radi waves, chemical cmpund, etc, is equal t dw VdQ vidt. The speed f the energy incme in the circuit is the instantaneus pwer and is equal t dw vdq P vi. dt dt The energy which is emitted frm time t t time t is equal t W Pdt. The energy and pwer are measured in jules (J) and watts (W). The pwer als is measured in kilwatts (kw), miliwatts (mw), kw 000 W; mw 0.00 W. t t 3

4 4..Resistance and Ohm law Between vltage and current the fllwing dependence is present named the Ohm law: v i, R where v is the vltage between tw given pints; i is the current which is mved ut ne pint and flwed in anther pint with less ptential; R is the resistance between these pints. In general, the resistance R in this equatin is substituted by the impedance Z, which is mre cmplex nature (see belw). The resistance is measured in hms (Ω), kilhms (kω), megahms (MΩ), and kω 000 Ω; MΩ 000 kω. is equal t: The electr engineering unit with the given resistance, named resistr is depicked as:. Fr hmgeneus cnductrs with the steady intersectin area S (see Fig. ) its resistance where σ is the cnductr cnductivity; l is its length. l R, σ S l σ S Fig Resistance f the metallic bar The pwer dissipatin P f a resistr is given by P vi i R. ( ) In cmputer engineering, this type f pwer dissipatin leads t heating and can cause thermal instabilities and circuit failures. The frmula ( ) analysis shws that fr the same vltage a small value f R gives large current flw, and is accmpanied by large pwer dissipatin..3. Vltage and Current surces In the electr engineering the vltage surces are signified as:, and the current surces are signified as:. An ideal vltage surce prvides a prescribed vltage acrss its terminals irrespective f the current flwing thrugh it. The amunt f current supplied by the surce is determined by the 4

5 5 circuit cnnected t it. The ideal vltage surce has zered inner resistance r s 0. And ideal current surce has zered inner cnductivity, i.e. unlimited inner resistance r s. Really, such a surce can be accumulatr, transducer, r generatr utput, which has cncrete inner resistance r s > 0. The symbl f such vltage surce as accumulatr, r alkaline battery is. Vs Rs + VL is RL Vs Rs ismax + VL - - Fig. 3 Practical vltage surce under lading (a) and under shrting (b) Figure 3 depicts a mdel fr a practical vltage surce, cmpsed f an ideal vltage surce, v S, in series with a resistance, R S. Nte that by cnventin the directin f psitive current flw ut f a vltage surce is ut f the psitive terminal. The resistance R S in effect pses a limit t the maximum current the vltage surce can prvide: i S max v S /R S Nte, hwever, that its presence affects the vltage acrss the lad resistance: this vltage is n equal t the surce vltage. Since the current prvided by the surce is the lad vltage can be determined t be v i S vs R + R S L is RL vs. RS + RL The circuit in Fig. 3 suggests that the ideal vltage surce is required t prvide an infinite amunt f current t the lad, in the limit as the lad resistance appraches zer. Naturally, this is impssible; fr example, cnsider a cnventinal car battery: V, 450 A-h (ampere-hurs). This implies that there is a limit t the amunt f current a practical surce can deliver t a lad. The limitatins f practical surces can be apprximated by expliting the ntin f the internal resistance f a surce. S v L 5

6 6 is Rs + VL RL - Fig. 4 Practical current surce under lading A similar mdificatin f the ideal current surce mdel is useful t describe the behavir f a practical current surce. The circuit illustrated in Figure 4 depicts a simple representatin f a practical current surce, cnsisting f an ideal surce in parallel with a resistr. Nte that as the lad resistance appraches infinity (i.e., an pen circuit), the utput vltage f the current surce appraches its limit, v S max i S r S. A gd current surce shuld be able t apprximate the behavir f an ideal current surce. Therefre, a desirable characteristic fr the internal resistance f a current surce is that it be as large as pssible. The surces described s far have the capability f generating a prescribed vltage r current independent f any ther element within the circuit. Thus, they are termed independent surces. There exists anther categry f surces, hwever, whse utput (current r vltage) is a functin f sme ther vltage r current in a circuit. These are called dependent (r cntrlled) surces. A different symbl, in the shape f a diamnd, is used t represent dependent surces and t distinguish them frm independent surces. Fr example, vltage cntrlled current surce F ( v (VCCS) : X )..4. Time dependent signal surces Figure 5 illustrates the cnventin that will be emplyed t dente time-dependent signal surces. V I V V(t) I(t) a) b) c) SINE(0Vdc Vac KHz) Fig. 5 Generalized time-dependent surces (a,b), and sinusidal surce (c) 6

7 7 One f the mst imprtant classes f time-dependent signals is that f peridic signals. These signals appear frequently in practical applicatins and are a useful apprximatin f many physical phenmena. A peridic signal x(t) is asignal that satisfies the fllwing equatin: x(t) x(t + nt ) n,, 3,... where T is the perid f x(t). Figure 6 illustrates a number f the peridic wavefrms that are typically encuntered in the study f electrical circuits. Wavefrms such as the sine, triangle, square, pulse, and sawtth waves are prvided in the frm f vltages (r, less frequently, currents) by cmmercially available signal (r wavefrm) generatrs. Such instruments allw fr selectin f the wavefrm peak amplitude, and f its perid. Sinusidal wavefrms cnstitute by far the mst imprtant class f time-dependent signals. Figure 6 depicts the relevant parameters f a sinusidal wavefrm. A generalized sinusid is defined as fllws: x(t) Acs(ωt + φ) where A is the amplitude, ω the radian frequency, and φ the phase. If f natural frequency /T(cycles/s, r Hz) Then ω radian frequency πf (radians/s) The phase shift, φ, permits the representatin f an arbitrary sinusidal signal. Thus, the chice f the reference csine functin t represent sinusidal signals arbitrary as it may appear at first des nt restrict the ability t represent all sinusids..5. Average and RMS Values Nw that a number f different signal wavefrms have been defined, it is apprpriate t define suitable measurements fr quantifying the strength f a time-varying electrical signal. The mst cmmn types f measurements are the average (r DC) value f a signal wavefrm which 7

8 8 crrespnds t just measuring the mean vltage r current ver a perid f time and the rtmean-square (r rms) value, which takes int accunt the fluctuatins f the signal abut its average value. Frmally, the peratin f cmputing the average value f a signal crrespnds t integrating the signalwavefrm ver sme (presumably, suitably chsen) perid f time. We define the time-averaged value f a signal x(t) as x( t) T T 0 x( t) dt, where T is the perid f integratin. Figure 7 illustrates hw this prcess des. In fact, it crrespnds t cmputing the average amplitude f x(t) ver a perid f T secnds. Fig. 7 Time-averaged value f the signal x(t) Prblem Cmpute the average value f the signal x(t) 0 cs(00t). Analysis: The signal is peridic with perid T π/ω π/00, thus we need t integrate ver nly ne perid t cmpute the average value: x( t) T T x( t' ) dt' 0 cs(00t) dt sin(π ) sin(0) 0 00 π π / π 0 Cmments: The average value f a sinusidal signal is zer, independent f its amplitude and frequency. Very cnveniently, a useful measure f the vltage f an AC wavefrm is the rt-meansquare, r rms, value f the signal, x(t), defined as fllws: x rms T T 0 x ( t' ) dt' 8

9 9 Nte that if x(t) is a vltage, the resulting x rms will als have units f vlts. If yu analyze equatin 4.4, yu can see that, in effect, the rms value cnsists f the square rt f the average (r mean) f the square f the signal. Thus, the ntatin rms indicates exactly the peratins perfrmed n x(t) in rder t btain its rms value. Prblem Cmpute the rms value f the sinusidal current i(t) I cs(ωt). Analysis: Applying the definitin f rms value we cmpute: i rms I T T 0 ω + π i ( t' ) dt' π / ω 0 I ω π π / ω 0 I cs(ω t' ) dt' cs ( ω t' ) dt' I ω π π / ω 0 I 0,707I I + cs(ω t' ) dt' where I is the peak value f the wavefrm i(t). Cmments: The rms value f a sinusidal signal is equal t times the peak value, independent f its amplitude and frequency. The factr f / remember, since it applies t any sinusidal signal. is a useful number t.6. Phasrs and impedance In this sectin, we intrduce an efficient ntatin t make it pssible t represent sinusidal signals as cmplex numbers, and t eliminate the need fr slving differential equatins. Named after the Swiss mathematician Lenhard Euler, Euler s identity frms the basis f phasr ntatin. Simply stated, the identity defines the cmplex expnential e jθ as a pint in the cmplex plane, which may be represented by real and imaginary cmpnents: e jθ cs θ + j sin θ. Figure 8 illustrates hw the cmplex expnential may be visualized as a pint (r vectr, if referenced t the rigin) in the cmplex plane. 9

10 0 Fig. 8 Cmplex data in the cmplex plane Nte that the magnitude f e jθ is equal t : e jθ since cs θ + j sin θ (cs θ + sin θ ) Nte als that the Euler s identity crrespnds t equating the plar frm f a cmplex number t its rectangular frm. Fr example, cnsider a vectr f length A making an angle θ with the real axis. The fllwing equatin illustrates the relatinship between the rectangular and plar frms: Aejθ Acs θ + jasin θ A θ T see hw cmplex numbers can be used t represent sinusidal signals, rewrite the expressin fr a generalized sinusid in light f Euler s equatin: Acs(ωtφ)Re[Ae j(ωt+φ) ] We see, that it is pssible t express a generalized sinusid as the real part f a cmplex vectr whse argument, r angle, is given by (ωt + φ) and whse length, r magnitude, is equal t the peak amplitude f the sinusid. The cmplex phasr crrespnding t the sinusidal signal Acs(ωt + φ) is therefre defined t be the cmplex number jϕ Ae cmplex phasr ntatin fr Acs(ωt+φ) A θ Prblem Cmpute the phasr vltage resulting frm the series cnnectin f tw sinusidal vltage surces v(t) 5 cs(34t + π/4) V v(t) 5 cs(34t + π/) V Find: Equivalent phasr vltage vs (t). Analysis: Write the tw vltages in phasr frm: V (jω) 5 π/4 V V (jω) 5e jπ/ 5 π/ V 0

11 Cnvert the phasr vltages frm plar t rectangular frm: V (jω) j0.6 V V (jω) j3.88 Then V S (jω) V (jω) + V (jω) j e jπ/ π/6 V Nw we can cnvert V S (jω) t its time-dmain frm: v S (t) 8.98 cs(34t + π/6) V. Phasr ntatin is a very efficient technique t slve AC circuit prblems..7. Impedance We nw analyze the i-v relatinship f the three ideal circuit elements in light f the phasr ntatin. The result will be a new frmulatin in which nt nly resistr but any linear tw ple circuits will be described in this ntatin. A direct cnsequence f this result will be that the Ohm law and a set f therems are extended t AC circuits. In the cntext f AC circuits, any ne f the ideal linear circuit elements will be described by a parameter called impedance, which may be viewed as a cmplex resistance. Figure 9 depicts the circuit represented in phasr-impedance frm; the latter representatin explicitly shws phasr vltages and currents and treats the circuit element as a generalized impedance. Vs(jw) I(jw) Z Fig. 9 Simple circuit represented in phasr-impedance frm It will be shwn that each f the ideal circuit elements may be represented by ne such impedance element. Let the surce vltage in the circuit f Figure be defined by v S (t) Acs ωt r V S (jω) Ae j0 A 0. Then the current i(t) is defined by the i-v relatinship fr each circuit element. Let us examine the prperties f the resistr, inductr, and capacitr, which are the general elements f any circuit.

12 .8. Capacitance The capacitance is a parameter that describes hw a particular device can stre electric charge. The capacitr is frmed by tw metal plates that are separated by an insulatr layer. The circuit symbl fr a capacitr is:. If a vltage v is applied t the plates, then charges Q f ppsite signs will be induced n these plates. The amunt f charge Q stred n the capacitr and the applied vltage v are in the fllwing relatin Q Cv, where C is capacitance. The unit f capacitance is the farad (F). In the wrld f electrnics, realistic capacitrs have very small values and are measured in micrfarads (µf), nanfarads (nf), and picfarads (pf), and µf0-6 F; nf0-9 F; pf0 - F. When mving charge frm r t the capacitr then a current I flws with the value dq dv I C dt dt that is prprtinal t the time rate f change the vltage. The electric field energy in the capacitr in arbitrary time is equal t, Cv Q W. (3) C The frmulae (3) shws that the capacitance can stre the energy. But this energy in mst cases is the alternating ne, and culd nt be stred mre than fr secnds r minutes. The capacitrs in the mdern dynamic randm access memries (DRAMs) are used fr the data string. The psitive r negative charge in them means the bit, which is equal t r 0. T stre the bits fr a lng time these strage capacitrs have t be upladed (refreshed) autmatically in the time perid f sme millisecnds. With i C i and v C v S, the capacitr current may be expressed as: ( Acs( ω t) ) C( Aω sin( ω t) ) ωcacs( ω + / ) dvc ( t) d ic ( t) C C t π dt dt s that, in phasr frm, V S (jω) A 0, and I(jω) ωca π/ The impedance f the ideal capacitr, Z C (jω), is therefre defined as fllws: Z C VS ( jω) j ( jω) π / I( jω) ωc ωc jωc

13 3 where we have used the fact that /j e jπ/ j. Thus, the impedance f a capacitr is als a frequency-dependent cmplex quantity, with the impedance f the capacitr varying as an inverse functin f frequency; and s a capacitr acts like a shrt circuit at high frequencies, whereas it behaves mre like an pen circuit at lw frequencies..9. Inductance The inductance L is the entity f the electric net which has the prperties f the inductive cil in which the magnetic energy can be laded. If the vltage at the ends f the inductivity L is equal t V d L di Ψ dt dt, ( 4) where Ψ is the magnetic linkage in the inductance. Fr the inductance frmed by the cil with w windings the magnetic linkage is ΨwΦ, where Φ is the magnetic flux. The magnetic linkage and flux are measured in webers (Wb). The circuit symbl fr an inductance is:. An inductance f ne henry, abbreviated H, represents a ptential difference f ne vlt acrss an inductr within which the current is increasing r decreasing at ne ampere per secnd. Usually, inductances are expressed in millihenries (mh), micrhenries (µh), r even in nanhenries (nh). Then mh 0.00 H; µh 0.00mH; and nh 0.00µH. The magnetic energy in the inductance is equal t Li Ψ W. ( 5) L The inductances are widely used in the mdern switching AC-DC, and DC-DC cnverters, which serve as the vltage surces f the cmputers. The inductances are used in the cmputer circuits fr the current filtering as well. The frmulas ( 4), ( 5) shw, that the inductance can lad the high energy, and the vltage in it can have high figures when the current is high, and it is exchanged sharply. Therefre, care have t be taken t keep the current exchange in the inductances f such circuits t save the integral circuits frm the dramatic vltage surges. A set f cils, which have the cmmn magnetic flux is named as a transfrmer. In the transfrmer the alternating current in the primary cil induces the alternating magnetic flux, which generates the alternating vltage, named electrmtive frce (EMF) in the secndary cil. The transfrmer steps up r steps dwn the input vltage depending n the rate f the secndary and primary windings. Let v L (t) v S (t) and i L (t) i(t) Then the fllwing expressin may be derived fr the inductr current: 3

14 4 A il( t) i( t) vs ( t' ) dt' il( t) Acs( ω t' ) dt' sin( ω t) L L ωl Nte hw a dependence n the radian frequency f the surce is clearly present in the expressin fr the inductr current. Further, the inductr current is shifted in phase (by 90 ) with respect t the vltage. This fact can be seen by writing the inductr vltage and current in timedmain frm: v ( t) v ( t) Acs( ω t) S L A π i( t) il ( t) cs ω t ω L It is evident that the current is nt just a scaled versin f the surce vltage, as it was fr the resistr. Its magnitude depends n the frequency, ω, and it is shifted (delayed) in phase by π/ radians, r 90. Using phasr ntatin, equatin becmes V S ( jω ) A 0 A I ( jω) π / ω L Thus, the impedance f the inductr is defined as fllws: Z L VS ( jω) ( jω ) ωl π / I( jω) jωl Nte that the inductr nw appears t behave like a cmplex frequency-dependent resistr, and that the magnitude f this cmplex resistr, ωl, is prprtinal t the signal frequency, ω. Thus, an inductr will impede current flw in prprtin t the sinusidal frequency f the surce signal. This means that at lw signal frequencies, an inductr acts smewhat like a shrt circuit, while at high frequencies it tends t behave mre as an pen circuit..0. Impedance meanings The impedance parameter is extremely useful in slving AC circuit analysis prblems, because it will make it pssible t take advantage f mst f the netwrk therems develped fr DC circuits by replacing resistances with cmplex-valued impedances. The examples, that fllw, illustrate hw branches cntaining series and parallel elements may be reduced t a single 4

15 5 equivalent impedance. It is imprtant t emphasize that althugh the impedance f simple circuit elements is either purely real (fr resistrs) r purely imaginary (fr capacitrs and inductrs), the general definitin f impedance fr an arbitrary circuit must allw fr the pssibility f having bth a real and an imaginary part, since practical circuits are made up f mre r less cmplex intercnnectins f different circuit elements. In its mst general frm, the impedance f a circuit element is defined as the sum f a real part and an imaginary part: Z(jω) R(jω) + jx(jω) where R is called the AC resistance and X is called the reactance. The frequency dependence f R and X has been indicated explicitly, since it is pssible fr a circuit t have a frequency-dependent resistance. Nte that the reactances have units f hms, and that inductive reactance is always psitive, while capacitive reactance is always negative. Figure 0 depicts Z C (jω) in the cmplex plane, alngside Z R (jω) and Z L (jω). Fig. 0 Impedance f resistr, capacitr, and inductr in the cmplex plane.. Measuring devices The hmmeter is a device that, when cnnected acrss a circuit element, can measure the resistance f the element. Figure depicts the circuit cnnectin f an hmmeter t a resistr. Ω Symbl fr hmmeter is. R 5

16 6 Fig. Cnnectin f the hmmeter One imprtant rule needs t be remembered: The resistance f an element can be measured nly when the element is discnnected frm any ther circuit. The ammeter is a device that, when cnnected in series with a circuit element, can measure the current flwing thrugh the element Symbl fr ammeter is cnnectin f an hmmeter int a circuit. A. Figure depicts the R A A3 V I(jw) R R V Fig. Cnnectin f an ammeter int a circuit The ammeter must be placed in series with the element whse current is t be measured (e.g., resistr R r R). The ammeter shuld nt restrict the flw f current (i.e., cause a vltage drp), r else it will nt be measuring the true current flwing in the circuit. An ideal ammeter has zer internal resistance. The vltmeter is a device that can measure the vltage acrss a circuit element (see Fig. 3). Since vltage is the difference in ptential between tw pints in a circuit, the vltmeter needs t be cnnected acrss the element whse vltage we wish t measure. The symbl fr vltmeter V is. A vltmeter must als fulfill the fllwing requirements. The vltmeter must be placed in parallel with the element whse vltage it is measuring. The vltmeter shuld draw n current away frm the element whse vltage it is measuring, r else it will nt be measuring the true vltage acrss that element. Thus, an ideal vltmeter has infinite internal resistance. R Vs i R Fig. 3 Cnnectin f a vltmeter int a circuit All f the cnsideratins that pertain t practical ammeters and vltmeters can be applied t the peratin f a wattmeter, a measuring instrument that prvides a measurement f the pwer 6

17 7 dissipated by a circuit element, since the wattmeter is in effect made up f a cmbinatin f a W vltmeter and an ammeter. Symbl fr the wattmeter is. Figure 4 depicts the typical cnnectin f a wattmeter in the same series circuit used abve. In effect, the wattmeter measures the current flwing thrugh the lad and, simultaneusly, the vltage acrss it and multiplies the tw t prvide a reading f the pwer dissipated by the lad. R R Vs R Vs R Fig. 4 Cnnectin f a wattmeter int a circuit.. Linear and unlinear cmpnents and circuits The relatinship between current and vltage at the terminals f a circuit element defines the behavir f that element within the circuit. In this sectin we shall intrduce a graphical means f representing the terminal characteristics f circuit elements. Suppse nw that a knwn vltage were impsed acrss a circuit element. The current that wuld flw as a cnsequence f this vltage, and the vltage itself, frm a unique pair f values. If the vltage applied t the element were variedand the resulting current measured, it wuld be pssible t cnstruct a functinal relatinship between vltage and current knwn as the i-v characteristic (r vltampere characteristic). Such a relatinship defines the circuit element, in the sense that if we impse any prescribed vltage (r current), the resulting current (r vltage) is directly btainable frm the i-v characteristic. A direct cnsequence is that the pwer dissipated (r generated) by the element may als be determined frm the i-v curve. Figure 5 depicts the i-v characteristic f a tungsten filament light bulb. A variable vltage surce is used t apply varius vltages, and the current flwing thrugh the element is measured fr each applied vltage. 7

18 8 Fig 5 i-v characteristic f a tungsten filament light bulb We culd certainly express the i-v characteristic f a circuit element in functinal frm: i f (v) v g(i). The examples f the unlinear cmpnents are varistr (vltage dependent resistr), dide, cil inductance. Mre cmplex cmpnents are transistrs, cil transfrmers, triacs. Due t the presence f unlinear cmpnents in the electrical circuit the linear and unlinear circuits are distinguished. In general, any real circuit can be cnsidered as unlinear ne because any cmpnent is nt ideal ne. The analysis and synthesis f the unlinear circuits are much cmplex f that f linear circuits. Therefre, t deal with the unlinear circuits its unlinear cmpnents are usually linearized, and the circuit is cnsidered at its state where the cmpnents are represented as linear nes.. DC electrical circuits and netwrks.. Electrical circuit elements In the previus sectins we have utlined mdels fr the basic circuit elements: surces, resistrs, capacitrs, inductances and measuring instruments. In rder fr current t flw there must exist a clsed circuit. We have assembled all the tls and parts we need in rder t define an electrical netwrk. It is apprpriate t frmally define the elements f the electrical circuit; the definitins that fllw are part f standard electrical engineering terminlgy. A branch is any prtin f a circuit with tw terminals cnnected t it. A branch may cnsist f ne r mre circuit elements. In practice, any circuit element with tw terminals cnnected t it is a branch. A nde is the junctin f tw r mre branches (ne ften refers t the junctin f nly tw branches as a trivial nde). Figure 6 illustrates the cncept. In effect, any cnnectin that can be 8

19 9 accmplished by sldering varius terminals tgether is a nde. It is very imprtant t identify ndes prperly in the analysis f electrical netwrks. R3 R4 R R5 Nde R Nde A R6 R7 R8 Nde b Figure 7. Fig. 6 Nde f the electrical circuit A lp is any clsed cnnectin f branches. Varius lp cnfiguratins are illustrated in R3 R4 Lp R Lp R Lp 3 Fig. 7 Lps in the circuit A mesh is a lp that des nt cntain ther lps. Meshes are an imprtant aid t certain analysis methds. In Figure 7, the circuit cnsists f tw meshes: lps and are meshes, but lp 3 is nt a mesh, because it encircles bth lps and. Whenever we reference the vltage at a nde in a circuit, we imply an assumptin that the vltage at that nde is the ptential difference between the nde itself and a reference nde called grund, which is lcated smewhere else in the circuit and which fr cnvenience has been assigned a ptential f zer vlts. The chice f the wrd grund is nt arbitrary. In every circuit a pint can be defined that is recgnized as grund and is assigned the electric ptential f zer vlts fr cnvenience. Symbl fr grund is, r, r depending n sme prperties f the grund wire. Fr example, the first symbl is used as the digital signal grund, and the latter ne as the analg signal grund is. 9

20 0.3. Kirchhff s current law Nte that in the circuit f Figure 3, a the current, i, flwing frm the vltage surce t the resistr is equal t the current flwing frm the resistr t the surce. In ther wrds, n current (and therefre n charge) is lst arund the clsed circuit. This principle was bserved by the German scientist G. R. Kirchhff and is nw knwn as Kirchhff s current law (KCL). Kirchhff s current law states that because charge cannt be created but must be cnserved, the sum f the currents at a nde must equal zer. Frmally: N i n n 0 The significance f Kirchhff s current law is illustrated in Figure 8, where the simple circuit f Figure 3,a has been augmented by the additin f tw resistrs. In applying KCL, ne usually defines currents entering a nde as being negative and currents exiting the nde as being psitive. Thus, the resulting expressin fr nde f the circuit f Figure 8 is: i + i + i + i 3 0 Vs i i i3 R R R3 4V Fig. 8 Kirhff's law representatin Kirchhff s current law is ne f the fundamental laws f circuit analysis, making it pssible t express currents in a circuit in terms f each ther; fr example, ne can express the current leaving a nde in terms f all the ther currents at the nde. The ability t write such equatins is a great aid in thesystematic slutin f large electric circuits. Prblem: If the battery in the diagram supplies a ttal f 0mW t the three elements shwn and i ma and i.5 ma, what is the current i 3? If i ma and i 3.5 ma, what is i?.4. Kirchhff s vltage law The vltage, r ptential difference, between tw pints in a circuit indicates the energy required t mve charge frm ne pint t the ther. The principle underlying Kirchhff s vltage law is that n energy is lst r created in an electric circuit; in circuit terms, the sum f all vltages assciated with surces must equal the sum f the lad vltages, s that the net vltage arund a 0

21 clsed circuit is zer. If this were nt the case, we wuld need t find a physical explanatin fr the excess (r missing) energy nt accunted fr in the vltages arund a circuit. Kirchhff s vltage law may be stated in a frm similar t that used fr Kirchhff s current law: N v n n 0 where the vn are the individual vltages arund the clsed circuit. Prblem Knwn Quantities: Vltages acrss each circuit element; current in circuit. Find: Pwer dissipated r generated by each element. Analysis: Fllwing the passive sign cnventin, we first select an arbitrary directin fr the current in the circuit; the example will be repeated fr bth pssible directins f current flw t demnstrate that the methdlgy is sund.. Assume clckwise directin f current flw, as shwn in Figure 9. + Vs. V R v8v - i0.a + R v4v - Figure 9 Simple circuit t analyze. Label plarity f vltage surce, as shwn in Figure 9; since the arbitrarily chsen directin f the current is cnsistent with the true plarity f the vltage surce, the surce vltage will be a psitive quantity. 3. Assign plarity t each passive element, as shwn in Figure Cmpute the pwer dissipated by each element: Since current flws frm t + thrugh the battery, the pwer dissipated by this element will be a negative quantity: P B v B * i ( V)* (0. A). W that is, the battery generates. W. The pwer dissipated by the tw lads will be a psitive quantity in bth cases, since current flws frm + t : P v * i (8 V)*. (0. A) 0.8 W P v * i (4 V)*. (0. A) 0.4 W

22 .5. Example: The Wheatstne Bridge The Wheatstne bridge is a resistive circuit that is frequently encuntered in a variety f measurement circuits. The general frm f the bridge circuit is shwn in Figure 0, where R, R, and R 3 are knwn while Rx is an unknwn resistance, t be determined. The bjective is t determine the unknwn resistance, Rx. c Vs a R R3 va vb b R4 d Rx Fig. 0 Wheatstne bridge circuit. Find the value f the vltage v ab v ad v bd in terms f the fur resistances and the surce vltage, v S. Nte that since the reference pint d is the same fr bth vltages, we can als write v ab v a v b.. If R R R 3 k", v S V, and v ab mv, what is the value f Rx? Slutin. First, we bserve that the circuit cnsists f the parallel cmbinatin f three subcircuits: the vltage surce, the series cmbinatin f R and R, and the series cmbinatin f R 3 and Rx. Since these subcircuits are in parallel, the same vltage will appear acrss each f them, namely, the surce vltage, v S. Thus, the surce vltage divides between each resistr pair, R R and R 3 Rx, accrding t the vltage divider rule: v a is the fractin f the surce vltage appearing acrss R, while vb is the vltage appearing acrss Rx : v v R a S and R + R v b v S Rx R + R 3 x Finally, the vltage difference between pints a and b is given by: v ab v a v b v S R R + R Rx R + R 3 x This result is very useful and quite general.

23 3. In rder t slve fr the unknwn resistance, we substitute the numerical values in the preceding equatin t btain 0,0,000 Rx,000,000 + R x.6. Netwrk analysis The analysis f an electrical netwrk cnsists f determining each f the unknwn branch currents and nde vltages. It is therefre imprtant t define all relevant variables as clearly as pssible, and in systematic fashin. Once the knwn and unknwn variables have been identified, a set f equatins relating these variables is cnstructed, and these are slved by means f suitable techniques. The analysis f electrical circuits cnsists f writing the smallest set f equatins sufficient t slve fr all f the unknwn variables. The analysis f electrical circuits is greatly simplified if sme standard cnventins are fllwed. The first bservatin t be made is that the relevant variables in netwrk analysis are the nde vltages and the branch currents. This fact is a cnsequence f Ohm s law. Cnsider the branch depicted in Figure, cnsisting f a single resistr. Va R i Vb Fig. A single branch f a circuit Here, nce a vltage v R is defined acrss the resistr R, a current i will flw thrugh the resistr, accrding t v R ir. But the vltage v R, which causes the current t flw, is really the difference in electric ptential between ndes a and b: v R v a v b Cnsider the circuit n the Fig.. Let us identify the branch and nde vltages and the lp and mesh currents in the circuit. + vr - + vr3 - a R b R3 c Vs V ia - vr + R ib R4 + vr4 - d Fig. Analyzed circuit The fllwing nde vltages may be identified: 3

24 4.7. Nde vltage methd The nde vltage methd is based n defining the vltage at each nde as an independent variable. One f the ndes is selected as a reference nde (usually grund), and each f the ther nde vltages is referenced t this nde. Once each nde vltage is defined, Ohm s law may be applied between any tw adjacent ndes in rder t determine the current flwing in each branch. In the nde vltage methd, each branch current is expressed in terms f ne r mre nde vltages; thus, currents d nt explicitly enter int the equatins. Once each branch current is defined in terms f the nde vltages, Kirchhff s current law is applied at each nde: Σi 0 The systematic applicatin f this methd t a circuit with n ndes wuld lead t writing n linear equatins. Hwever, ne f the nde vltages is the reference vltage and is therefre already knwn, since it is usually assumed t be zer. Thus, we can write n independent linear equatins in the n independent variables (the nde vltages). Ndal analysis prvides the minimum number f equatins required t slve the circuit, since any branch vltage r current may be determined frm knwledge f ndal vltages. The ndal analysis methd may als be defined as a sequence f steps, as utlined belw: Nde Vltage Analysis Methd. Select a reference nde (usually grund). All ther nde vltages will be referenced t this nde.. Define the remaining n nde vltages as the independent variables. 3. Apply Kirhhff Current Law at each f the n ndes, expressing each current in terms f the adjacent nde vltages. 4. Slve the linear system f n equatins in n unknwns. As an illustratin f the methd, cnsider the circuit shwn in Figure 3. 4

25 5 Nde a R Nde b V R R3 Nde c Fig. 3 Example f the circuit slved by the Nde Vltage Analysis Methd The directin f current flws selected arbitrarily (assuming that i S is a psitive current). Applicatin f KCL at nde a yields: i S i i 0 whereas, at nde b, i i 3 0 It is instructive t verify that it is nt necessary t apply KCL at the reference nde. The equatin btained at nde c, i i 3 i S 0. is nt independent f bth previus equatins. Nw, in applying the nde vltage methd, the currents i, i, and i 3 are expressed as functins f v a, v b, and v c, the independent variables. Ohm s law requires that i, fr example, be given by i ( v a v c )/R since it is the ptential difference, v a v c, acrss R that causes the current i t flw frm nde a t nde c. Similarly, i ( v a v b )/R i 3 ( v b v c )/R 3 The presence f vltage surces actually simplifies the calculatins. T illustrate this pint, cnsider the circuit f Figure 4. Nte that ne f the nde vltages is knwn already. va R vb R3 vc Vs R R4 Is Fig. 4 Circuit t analyze by the nde vltage methd 5

26 6.8 Mesh current methd The secnd methd f circuit analysis, which is in many respects analgus t the methd f nde vltages, emplys mesh currents as the independent variables. The idea is t write the apprpriate number f independent equatins, using mesh currents as the independent variables. Analysis by mesh currents cnsists f defining the currents arund the individual meshes as the independent variables. Subsequent applicatin f Kirchhff s vltage law arund each mesh prvides the desired system f equatins. Cnsider the circuit n the Fig. 5. Let us identify the branch and nde vltages and the lp and mesh currents in the circuit. + vr - + vr3 - a R b R3 c Vs V ia - vr + R ib R4 + vr4 - d Fig. 5 Analyzed circuit In the mesh current methd a current flwing thrugh a resistr in a specified directin defines the plarity f the vltage acrss the resistr, as illustrated in Figure, and that the sum f the vltages arund a clsed circuit must equal zer, by Kirhhff vltage law. Once a cnventin is established regarding the directin f current flw arund a mesh, simple applicatin f Kirhhff vltage law prvides the desired equatin. Figure 4 illustrates this pint. The number f equatins is equal t the number f meshes in the circuit. All branch currents and vltages may subsequently be btained frm the mesh currents, as will presently be shwn. Since meshes are easily identified in a circuit, this methd prvides a very efficient and systematic prcedure fr the analysis f electrical circuits. The fllwing bx utlines the descriptin f the prcedure used in applying the mesh current methd t a linear circuit. Mesh Current Analysis Methd. Define each mesh current cnsistently. We shall always define mesh currents clckwise, fr cnvenience.. Apply Kirhhff vltage law arund each mesh, expressing each vltage in terms f ne r mre mesh currents. 6

27 7 variables. 3. Slve the resulting linear system f equatins with mesh currents as the independent In mesh analysis, it is imprtant t be cnsistent in chsing the directin f current flw. T avid cnfusin in writing the circuit equatins, mesh currents will be defined exclusively clckwise when we are using this methd. T illustrate the mesh current methd, cnsider the simple tw-mesh circuit shwn in Figure 4. This circuit will be used t generate tw equatins in the tw unknwns, the mesh currents i and i. It is instructive t first cnsider each mesh by itself. Beginning with mesh, nte that the vltages arund the mesh have been assigned in Figure 4 accrding t the directin f the mesh current, i. Recall that as lng as signs are assigned cnsistently, an arbitrary directin may be assumed fr any current in a circuit; if the resulting numerical answer fr the current is negative, then the chsen reference directin is ppsite t the directin f actual current flw. Thus, ne need nt be cncerned abut the actual directin f current flw in mesh analysis, nce the directins f the mesh currents have been assigned. Accrding t the sign cnventin, then, the vltages v and v are defined as shwn in Figure 4. Nw, it is imprtant t bserve that while mesh current i is equal t the current flwing thrugh resistr R (and is therefre als the branch current thrugh R ), it is nt equal t the current thrugh R. The branch current thrugh R is the difference between the tw mesh currents, i i. Thus, since the plarity f the vltage v has already been assigned, accrding t the cnventin, it fllws that the vltage v is given by: Finally, the cmplete expressin fr mesh is v (i i )R v S i R (i i )R 0 The mesh current i is als the branch current thrugh resistrs R 3 and R 4 ; hwever, the current thrugh the resistr that is shared by the tw meshes, R, is nw equal t (i i ), and the vltage acrss this resistr is and the cmplete expressin fr mesh is v (i i )R (i i )R + i R 3 + i R 4 0 Cmbining the equatins fr the tw meshes, we btain the fllwing system f equatins: (R +R )i R i v S R i +(R + R 3 + R 4 ) i 0 7

28 8 These equatins may be slved simultaneusly t btain the desired slutin, namely, the mesh currents, i and i. One can verify that knwledge f the mesh currents permits determinatin f all the ther vltages and currents in the circuit..9. Matrix equatins in electric and electrnic engineering.0. Ndal and mesh analysis with cntrlled surces The methds just described als apply, with relatively minr mdificatins, in the presence f dependent (cntrlled) surces. Slutin methds that allw fr the presence f cntrlled surces will be particularly useful in the study f transistr amplifiers, transfrmer circuits, etc. Recall frm the Sectin.3 that a dependent surce is a surce that generates a vltage r current that depends n the value f anther vltage r current in the circuit. When a dependent surce is present in a circuit t be analyzed by nde r mesh analysis, ne can initially treat it as an ideal surce and write the nde r mesh equatins accrdingly. In additin t the equatin btained in this fashin, there will als be an equatin relating the dependent surce t ne f the circuit vltages r currents. This cnstraint equatin can then be substituted in the set f equatins btained by the techniques f ndal and mesh analysis, and the equatins can subsequently be slved fr the unknwns. It is imprtant t remark that nce the cnstraint equatin has been substituted in the initial system f equatins, the number f unknwns remains unchanged. Cnsider, fr example, the circuit f Figure 6, which is a simplified mdel f a biplar transistr amplifier. Vcc RC ib Rs rb T Vs V Is ib B*ib Rs rb Rc + V - Fig. 6 Example f the circuit with the current cntrlled current surce In the circuit f Figure 6, tw ndes are easily recgnized, and therefre ndal analysis is chsen as the preferred methd. Applying KCL at nde, we btain the fllwing equatin: 8

29 9 i S v + RS R b KCL applied at the secnd nde yields: v β ib + 0 (*) R C Next, it shuld be bserved that the current ib can be determined by means f a simple current divider: i b i S / Rb / R + / R b S i S RS R + R b S which, when inserted in equatin (*), yields a system f tw equatins: + β S is v is RS R b Rb + RS RC R v which can be used t slve fr v and v. The techniques presented in this sectin and the tw preceding sectins find use mre generally than just in the analysis f resistive circuits. These methds shuld be viewed as general techniques fr the analysis f any linear circuit; they prvide systematic and effective means f btaining the minimum number f equatins necessary t slve a netwrk prblem. Since these methds are based n the fundamental laws f circuit analysis, Kirhhff vltage law and KCL, they als apply t any electrical circuit, even circuits cntaining nnlinear circuit elements... The principle f superpsitin Rather than a precise analysis technique, like the mesh current and nde vltage methds, the principle f superpsitin is a cnceptual aid that can be very useful in visualizing the behavir f a circuit cntaining multiple surces. The principle f superpsitin applies t any linear system and fr a linear circuit may be stated as fllws: In a linear circuit cntaining N surces, each branch vltage and current is the sum f N vltages and currents each f which may be cmputed by setting all but ne surce equal t zer and slving the circuit cntaining that single surce. An elementary illustratin f the cncept may easily be btained by simply cnsidering a circuit with tw surces cnnected in series, as shwn in Figure 7. 9

30 30 VB VB VB R R i ib + ib VB R Fig. 7 The representatin f a circuit by tw circuit superpsitin The circuit f Figure 7 is mre frmally analyzed as fllws. The current, i, flwing in the circuit n the left-hand side f Figure 7 may be expressed as: v + v R v R v R B B B B i + ib + i B Figure 7 als depicts the circuit as being equivalent t the cmbined effects f tw circuits, each cntaining a single surce. In each f the tw subcircuits, a shrt circuit has been substituted fr the missing battery. This shuld appear as a sensible prcedure, since a shrt circuit by definitin will always see zer vltage acrss itself, and therefre this prcedure is equivalent t zering the utput f ne f the vltage surces. If, n the ther hand, ne wished t cancel the effects f a current surce, it wuld stand t reasn that an pen circuit culd be substituted fr the current surce, since an pen circuit is by definitin a circuit element thrugh which n current can flw (and which will therefre generate zer current). The principle f superpsitin can easily be applied t circuits cntaining multiple surces and is smetimes an effective slutin technique. 3. One-prt netwrks and equivalent circuits 3.. One-prt netwrk Yu may recall that, in the discussin f ideal surces in Chapter.3, the flw f energy frm a surce t a lad was described in a very general frm, by shwing the cnnectin f tw black bxes labeled surce and lad (see Figure 3,a). Each blck surce r lad may be viewed as a tw-terminal device, described by an i-v characteristic. The general circuit representatin is called a ne-prt netwrk and is particularly useful fr intrducing the ntin f equivalent circuits. Nte that the ne-prt netwrk is cmpletely described by its i-v characteristic; this pint is best illustrated by the next example. Prblem. Determine the surce (lad) current i in the circuit f Figure 8 using equivalent resistance ideas. 30

31 3 Vs i + V i i i3 R R R3 Req - Fig. 8 Finding the equivalent resistance ne-prt circuit Analysis: Insfar as the surce is cncerned, the three parallel resistrs appear identical t a single equivalent resistance f value R EQ + + R R R Thus, we can replace the three lad resistrs with the single equivalent resistr R EQ, as shwn in Figure 8, and calculate v i R S EQ Th evenin and Nrtn Therems In studying nde vltage and mesh current analysis, yu may have bserved that there is a certain crrespndence (called duality) between current surces and vltage surces, n the ne hand, and parallel and series circuits, n the ther. This duality appears again very clearly in the analysis f equivalent circuits: it will be shwn that equivalent circuits fall int ne f tw classes, invlving either vltage r current surces and (respectively) either series r parallel resistrs, reflecting this same principle f duality. The discussin f equivalent circuits begins with the statement f tw very imprtant therems, The Th evenin Therem As far as a lad is cncerned, any netwrk cmpsed f ideal vltage and current surces, and f linear resistrs, may be represented by an equivalent circuit cnsisting f an ideal vltage surce, v T, in series with an equivalent resistance, R T. 3

32 3 The Nrtn Therem As far as a lad is cncerned, any netwrk cmpsed f ideal vltage and current surces, and f linear resistrs, may be represented by an equivalent circuit cnsisting f an ideal current surce, i N, in parallel with an equivalent resistance, R N. Prblem Find the Th evenin equivalent resistance seen by the lad R6 in the circuit f Figure 9. Given Data: R 0; R 0; I 5 A; R 3 0; R 4 0; R 5 0. R3 R5 a R R I R4 R6 b Fig. 9 A netwrk fr which the Th evenin equivalent resistance is fund Analysis: Fllwing the Th evenin therem, we first set the current surce equal t zer, by replacing it with an pen circuit. Lking int terminal a-b we recgnize that, starting frm the left (away frm the lad) and mving t the right (tward the lad) the equivalent resistance is given by the expressin R T [((R R ) + R 3 ) R 4 ] + R 5 [((0 0) + 0) 0] Ohm. Nte that the reductin f the circuit started at the farthest pint away frm the lad. 3.3.Cmputing the Th evenin vltage The equivalent (Th evenin) surce vltage v T is equal t the pen-circuit vltage present at the lad terminals (with the lad remved). This states that in rder t cmpute v T, it is sufficient t remve the lad and t cmpute the pen-circuit vltage at the ne-prt terminals. Figure 30 illustrates that the pen-circuit vltage, v OC, and the Th evenin vltage, v T, must be the same if the Th evenin therem is t hld. One-prt netwrk + Vc - VT RT + Vc VT - Fig. 30 Equivalent pen circuit 3