Introduction: Analysis Of Electronic Circuits
1/30/2008Introduction1/2Introduction: Analysis ofElectronic CircuitsReading Assignment: KVL and KCL text from EECS 211Just like EECS 211, the majority of problems (hw and exam) inEECS 312 will be circuit analysis problems. Thus, a key todoing well in 312 is to thoroughly know the material from211!!So, before we get started with 312, let’s review 211 and seehow it applies to electronic circuits.Q: I aced EECS 211 lastsemester; can I just skipthis “review”?A: Even if you did extremely well in 211, you will want to payattention to this review. You will see that the concepts of 211are applied a little differently when we analyze electroniccircuits.Both the conventions and the approach used for analyzingelectronic circuits will perhaps be unfamiliar to you at first—I thus imagine that everyone (I hope) will find this review tobe helpful.Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008Introduction2/2Electronic Circuit NotationKVL and Electronic Circuit NotationAnalysis of Electronic CircuitsEven the quantities of current and resistance are a littledifferent for electronic circuits!Q: You mean we don’t useAmperes and Ohms?A: Not exactly!Volts, Milli-Amps, Kilo-OhmsNow let’s try an example!Example: Circuit Analysis using Electronic Circuit NotationJim StilesThe Univ. of KansasDept. of EECS
1/30/2008Electronic Circuit Notation1/9Electronic Circuit NotationThe standard electronic circuit notation may be a littledifferent that what you became used to seeing in in EECS 211.The electronic circuit notation has a few “shorthand”standards that can simplify circuit schematics!Consider the circuit below:i1R1 1K v1 -Vs 5V -i2R2 4K v2-i3R3 4K v3-Note the voltage values in this circuit (i.e., Vs ,v1 ,v2 ,v3 ) providevalues of potential difference between two points in thecircuit.For example, from the voltage source we can conclude:Vs 5VJim Stiles -The electric potential at thispoint in the circuit is 5 voltsgreater than:the electric potential at thispoint in the circuit.The Univ. of KansasDept. of EECS
1/30/2008Electronic Circuit Notation2/9Or the resistor voltage v3 means:R3 4K v3-The electric potential at thispoint in the circuit is v3 voltsgreater than:the electric potential at thispoint in the circuit.But remember, v3 could be a negative value!Thus, the values of voltages are comparative—they tell us thedifference in electric potential between two points with in thecircuit.JohnSallyAs an analogy, Say John,Sally, and Joe work in a verytall building. Our circuitvoltages are little likesaying:“John is 5 floors above Joe”“Sally is 2 floors above Joe”JoeJim StilesFrom this comparativeinformation we can deducethat John is 3 floors aboveSally.The Univ. of KansasDept. of EECS
1/30/2008Electronic Circuit Notation3/9What we cannot determine is on what floor John, Sally, orJoe are actually located. They could be located at the highestfloors of the building, or at the lowest (or anywhere inbetween).Similarly, we cannot deduce from the values Vs ,v1 ,v2 ,v3 theelectric potential at each point in the circuit, only the relativevalues—relative to other points in the circuit. E.G.:“ Point R has an electric potential 5V higher than point B”“ Point G has an electric potential v3 higher than point B”Q: So how do we determine thevalue of electric potential at aspecific point in a circuit?A: Recall that electric potential at somepoint is equal to the potential energypossessed by 1 Coulomb of charge iflocated at that point.Thus to determine the “absolute” (as opposed torelative) value of the electric potential, we first mustdetermine where that electric potential is zero.The problem is similar to that of the potential energypossessed by 1.0 kg of mass in a gravitational field.We ask ourselves: Where does this potential energyequal zero?Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008Electronic Circuit Notation4/9The answer of course is when the mass is located on theground!But this answer is a bit subjective; is the “ground”:A. where the carpet is located?B.where the sidewalk is located?C. The basement floor?D. Sea level?E. The center of the Earth?The answer is—it can be any of these things!Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008Electronic Circuit Notation5/9We can rather arbitrarily set some point as the location ofground. The potential energy is therefore described inreference to this ground point.For tall buildings, the groundfloor is usually defined as thefloor containing the front door(i.e. the sidewalk)—but it doesn’thave to be (just look at EatonHall!).Now, having defined a ground reference, if we add to ourearlier statements:“Joe is 32 floors above ground”We can deduce:JohnSally“ John is 5 floors aboveJoe—therefore John is onthe 37th floor”“ Sally is 2 floors aboveJoe—therefore Sally is onthe 34th floor”JoeJim StilesThe Univ. of KansasDept. of EECS
1/30/2008Electronic Circuit Notation6/9Q: So, can we define a groundpotential for our circuit?A: Absolutely! We just pick a point on thecircuit and call it the ground potential. Wecan then reference the electric potential atevery point in the circuit with respect tothis ground potential!Consider now the circuit:i1R1 1K v1 -Vs 5V -i2R2 4K v2-i3R3 4K v3-Look at this!Note we have added an “upside-down triangle” to thecircuit—this denotes the location we define as our groundpotential!Now, if we add the statement:“Point B is at an electric potential of zero volts (withrespect to ground).”Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008Electronic Circuit Notation7/9We can conclude:“Point R is at an electric potential of 5 Volts (withrespect to ground).”“Point G is at an electric potential of v3 Volts (withrespect to ground).”Note that all the points within the circuit that reside atground potential form a rather large node:i1R1 1K v1 -Vs 5V -i2R2 4K v2-i3R3 4K v3-Look at this!Standard electronic notation simplifies the schematic byplacing the ground symbol at each device terminal:i1R1 1K v1 -Vs 5VJim Stiles -i2R2 4K v2-The Univ. of Kansasi3R3 4K v3-Dept. of EECS
1/30/2008Electronic Circuit Notation8/9Note that all terminals connected to ground are likewiseconnected to each other!Now, in the case where one terminal of a device is connectedto ground potential, the electric potential (with respect toground) of the other terminal is easily determined:Vs 5VThe electric potential at thispoint in the circuit is 5 voltsgreater than ground (i.e., 5volts). -This point is at groundpotential (i.e., zero volts).For this example, it is apparent that the voltage source simplyenforces the condition that the terminal is at 5.0 Voltswith respect to ground. Thus, we often simplify ourelectronic circuit schematics as:i1Vs 5VR1 1K v1 -i2R2 4KJim Stiles v2-The Univ. of Kansasi3R3 4K v3-Dept. of EECS
1/30/2008Electronic Circuit Notation9/9Finally, we find that:The electric potential at thispoint in the circuit is v2 voltsgreater than ground potential(i.e., v3 ). v3-R3 4KThis point is at groundpotential (i.e. zero volts).Thus, we can simplify our circuit further as:i1Vs 5VR1 1K v1 -i2i3 v2-R2 4Kv3R3 4KThis circuit schematic is precisely the same as our originalschematic:i1R1 1K v1 -Vs 5VJim Stiles -i2R2 4K v2-The Univ. of Kansasi3R3 4K v3-Dept. of EECS
1/30/2008KVL and Electronic Notation1/12KVL and ElectronicCircuit NotationConsider this circuit:iVs 4V -R1 3K vR1-R2 2K v R2-R3 1K vR3-- v4 - v5 Vs 2V -We can apply Kirchoff’s Voltage Law (KVL) to relate thevoltages in this circuit in any number of ways.Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008KVL and Electronic Notationi Vs 4V-R1 3K vR1-R2 2K v R2-R3 1K vR3-- v4 - v5 Vs 2V-2/12For example, the KVLaround this loop is: 4 vR 1 vR 2 vR 3 2 0We could multiply bothsides of the equation by 1 and likewise get a validequation:4 vR 1 vR 2 vR 3 2 0Q: But which equation iscorrect? Which one do we use?Which one is the KVL result?A: Each result is equally valid; both willprovide the same correct answers.Essentially, the first KVL equation is constructed using theconvention that we add the circuit element voltage if we firstencounter a plus ( ) sign as we move along the loop, andsubtract the circuit element voltage if we first encounter aminus (-) sign as we move along the loop.Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008KVL and Electronic Notation3/12 v1 -For example: 5 v1 " -5VBut, we could also use the convention that we subtract thecircuit element voltage if we first encounter a plus ( ) sign aswe move along the loop, and add the circuit element voltage ifwe first encounter a minus (-) sign as we move along the loop!For example: 5 v1 " v1 - -5VThis convention would provide us with the secondof the two KVL equations for our original circuit:4 vR 1 vR 2 vR 2 2 0Q: Huh?! What kind of sense does this conventionmake? We subtract when encountering a ? Weadd when encountering a - ?Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008KVL and Electronic Notation4/12A: Actually, this second convention is more logical than thefirst if we consider the physical meaning of voltage!Remember, “the voltage” is simply a measure of potentialenergy—the potential energy of 1 Coulomb of charge. v1 - - 5V1CIf 1 C of charge were to be transported aroundthe circuit, following the path defined by ourKVL loop, then the potential energy of thischarge would change as is moved through eachcircuit element.In other words, its potential energy would go up, or it wouldgo down.The second convention describes this increase/decrease! -Jim Stiles v1 -5VFor example as our 1C charge moves throughthe voltage source, its potential energy isincreases by 5 Joules (the potential is 5 Vhigher at the terminal than it was at theminus terminal)!The Univ. of KansasDept. of EECS
1/30/2008KVL and Electronic Notation v1 - -5V5/12 But when it moves through the resistor, itspotential energy drops by v1 Joules (thepotential at the minus terminal is v1 Volts lessthan that at the plus terminal).Thus, the second convention is a more accurate “accounting”of the change in potential! v1 - -5V 5 v1 "This convention is the one typically used for electroniccircuits. You of course will get the correct answer eitherway, but the second convention allows us to easily determinethe absolute potential (i.e., with respect to ground) at eachindividual point in a circuit.To see this, let’s return to our original circuit:Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008KVL and Electronic NotationVs 4V -iR1 3K vR1-R2 2K v R2-R3 1K vR3-- v4 - v5 Vs 2V -6/12The KVL from these loops are thus: 4 vR 1 v 4 0 v 4 vR 2 v 5 0 2 v 5 vR 3 0 4 vR 1 vR 2 v 5 0 v 4 vR 2 vR 3 2 0Q: I don’t see how this new convention helps us determinethe “absolute” potenial at each point in the circuit?Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008KVL and Electronic Notation7/12A: That’s because we have not defined a ground potential!Let’s do that now:iVs 4V -R1 3K vR1-R2 2K v R2-- v4 - v5 Vs 2V -R3 1KVs 4V vR3-We can thus rewrite thiscircuit schematic as:Remember that all groundterminals are connected toeach other, so we can performKVL by starting and ending at aground node:Jim Stiles v5-The Univ. of Kansas iR1 3K vR1-R2 2K v R2-R3 1K vR3-Vs 2V v4- Dept. of EECS
1/30/2008KVL and Electronic Notation8/12 4 vR 1 vR 2 vR 3 2 0 4 vR 1 v 4 0Vs 4V i v 4 vR 2 v 5 0R1 3K vR1-R2 2K v R2-R3 1K vR3-The same results as before!Now, we can furthersimplify the schematic: v5-4ViR1 3KR2 2KR3 1K vR1 v R2 vR3-v4Vs 2V v4- v5-2VJim StilesThe Univ. of KansasDept. of EECS
1/30/2008KVL and Electronic Notation9/12Note that we were able to replace the voltage sources with adirect, simple statement about the electric potential at twopoints within the circuit.4ViThe electricpotential heremust be 4 V!Vs 4V R1 3KR2 2KThe electricpotential heremust be -2 V!Vs 2V R3 1K vR1-v4 v R2 vR3-v5-2VNote the KCL equation we determined earlier: 4 vR 1 vR 2 vR 3 2 0Let’s subtract 2.0 from both sides: 4 vR 1 vR 2 vR 3 2This is the same equation as before—a valid result from KVL.Yet, this result has a very interesting interpretation!Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008KVL and Electronic Notation10/12The value 4.0 V is the initial electric potential—the potentialat beginning node of the “loop” .The values vR 1 ,vR 2 , and vR 3 describethe voltage drop as we move througheach resistor. The potential is thusdecreased by these values, and thusthey are subtracted from the initialpotential of 4.0.When we reach the bottom of the circuit, the potential atthat point wrtg (with respect to ground) must be equal to: 4 vR 1 vR 2 vR 3But we also know that the potential at the “bottom” of thecircuit is equal to -2.0 V! Thus we conclude: 4 vR 1 vR 2 vR 3 2Our KVL equation!In general, we can move through a circuit written with orelectronic circuit notation with this “law”:The electric potential at the initial node (wrtg),minus(plus) the voltage drop(increase) of each circuitelement encountered, will be equal to the electricpotential at the final node (wrtg).Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008KVL and Electronic Notation11/12For example, let’s analyze our circuit in the oppositedirection!4VHere, the electric potential atthe first node is -2.0 volts (wrtg)and the potential at the last is4.0.iR1 3KR2 2KR3 1K vR1 v R2 vR3-2Vv4v5Note as we move through theresistors, we find that thepotential increases by vR : 2 vR 3 vR 2 vR 1 4Note this is the effectively thesame equation as before: 4 vR 1 vR 2 vR 3 2Both equations accurately stateKVL, and either will the samecorrect answer!Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008KVL and Electronic Notation12/12Now, we can use our new found knowledgeto come to these correct conclusions,see if these results make sense to you!4V4 vR 1 v 4iv 4 vR 2 v 5R1 3K 2 vR 3 v 54 vR 1 vR 2 v 5R2 2K 2 vR 3 vR 2 v 4R3 1K vR1 v R2 vR3-v4v5-2VJim StilesThe Univ. of KansasDept. of EECS
1/30/2008Analysis of Electronic Circuits1/7Analysis ofElectronic CircuitsIn EECS 211 you acquired the toolsnecessary for circuit analysis.Fortunately, all those tools are stillapplicable and useful when analyzingelectronic circuits!Ohm’s Law, KVL and KCL are all still valid, but (isn’t therealways a but?) the complicating factor in electronic circuitanalysis is the new devices we will introduce in EECS 312.In EECS 211 you learned about devices such as voltagesources, current sources, and resistors. These devices allhad very simple device equations: i vRv VsivIs v i RJim Stilesiv VsThe Univ. of Kansasi IsDept. of EECS
1/30/2008Analysis of Electronic Circuits2/7But (that word again!), in EECS 312 we will learn aboutelectronic devices such as diodes and transistors. The deviceequations for these new circuit elements will be quite a bitmore complicated!iD DG vDvDS vGS SiD iD K 2 (vGS Vt )vDS vDS 2 vD nVT iD IS e 1 As a result, we often find that both node and mesh analysistools are a bit clumsy when analyzing electronic circuits. Thisis because electronic devices are non-linear, and so theresulting circuit equations cannot be described by as set oflinear equations. 2 3 i1 2 i2 1 i3 2 3 2 1 i1 1 2 11 2 i1 1 i21 i2 0 4 i1 2 i2 2 i3 0 4 2 2 i3 Jim StilesThe Univ. of KansasNot from anelectronic circuit!Dept. of EECS
1/30/2008Analysis of Electronic Circuits3/7Instead, we find that electronic circuits are more effectivelyanalyzed by a more precise and subtle application of:1. Kirchoff Voltage LawCircuitEquations2. Kirchoff Current Law3. Ohm’s Law4. Electronic device equationsDeviceEquationsNote the first two of these are circuit laws—they eitherrelate every voltage of the circuit to every other voltage ofthe circuit (KVL), or relate every current in the circuit toevery other current in the circuit.I1 I 2 I 3 0V1 V2 V3 0The last two items of our list are device equations—theyrelate the voltage(s) of a specific device to the current(s) ofthat same device. Ohm’s Law of course describes thecurrent-voltage behavior of a resistor (but only the behaviorof a resistor!).V2 I2 R2So, if you:1. mathematically state the relationship between all thecurrents in the circuit (using KCL), and:Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008Analysis of Electronic Circuits4/72. mathematically state the relationship between all thevoltages of the circuit (using KVL), and:3. mathematically state the current-voltagerelationship of each device in the circuit, then:you have mathematically described your circuit—completely!At this point you will find that thenumber of unknown currents andvoltages will equal the number ofequations, and your circuit analysissimply becomes an algebra problem!But be careful! In order to get thecorrect answer from your analysis,you must unambiguously define eachand every voltage and current variablein your circuit!!!!!!!!!We do this by defining the direction of a positive current(with and arrow), and the polarity of a positive voltage (with a and - ).Placing this unambiguousnotation on your circuit isan absolute requirement!Q: An absolute requirement in order to achieve what?Jim StilesThe Univ. of KansasDept. of EECS
1/30/2008Analysis of Electronic Circuits5/7A: An absolute requirement in order to:1. determine the correct answers.2. receive full credit on exams/homework.Q: But why must I unambiguously define each current andvoltage variable in order to determine the correct answers?A: The mathematical expressions (descriptions) of the circuitprovided by KVL, KCL and all device equations are directlydependent on the polarity and direction of each voltage andcurrent definition!For example, consider a three current node, with currents I1 ,I2 , I3 . We can of course use KCL to relate these values, butthe resulting mathematical expression depends on how wedefine the direction of these currents:I1I1I1Jim StilesI3I3I3I2I1 I 2 I 3 0I2I1 I 2 I 3 0 I1 I 2 I 3I2I1 I 2 I 3 0 I1 I 2 I 3The Univ. of KansasDept. of EECS
1/30/2008Analysis of Electronic Circuits6/7Q: But that’s the problem! How do I know which directionthe current is flowing in before I analyze the circuit? Whatif I put the arrow in the wrong direction?A: Remember, there is no way to incorrectly orient thecurrent arrows of voltage polarity for KCL and KVL. If thecurrent or voltage is opposite that of your convention, thenthe numeric result will simply be negative.For example, say that in a 3-wire node there is:3 mA flowing toward the node in wire 12 mA flowing toward the node in wire 25 mA flowing away from the node in wire 3Depending on how you define the currents, the numericalanswers for I1 , I2 and I3 will all be different, but therephysical interpretation will all be the same!I1I1I1Jim StilesI3I3I3I2I2I2I1 3mA, I2 2mA, I3 5mAI1 3mA, I2 2mA, I3 5mAI1 3mA, I2 2mA, I3 5mAThe Univ. of KansasDept. of EECS
1/30/2008Analysis of Electronic Circuits7/7Remember, a negative value of current(or voltage) means that the current isflowing in the opposite direction (orpolarity) of that denoted in the circuit.So, without current arrows and voltagepolarities, there is no way to physicallyinterpret positive or negative values!Now we know that with respect to KCL or KVL, thecurrent/voltage conventions are arbitrary (it up to you todecide!).However, we will find that the voltage/current conventions ofelectronic devices are not generally arbitrary, but insteadhave required orientations.Q: Why is that?iD DG vDS vGS S 2 vD K 2 (vGS Vt )vDS vDSJim StilesA: The conventions are coupled toelectronic device equations—theseequations are only accurate when using thespecific voltage/current conventions!Thus, you must know both the deviceequation and the current/voltageconvention for each electronic device.Furthermore, you must correctly labeland uses these current/voltageconventions in all circuits that containthese devices!The Univ. of KansasDept. of EECS
1/25/2008Volts milliamps kiloohms1/4Volts, Milli-Amps,and Kilo-OhmsLet’s determine the voltage across a 7 KΩ resistor if acurrent of 2 mA is flowing through it:v ( 0.002 )( 7000 ) 1.4 VOr the resistance of a resistor if a current of 2 mA resultsin a voltage drop of 20 V:R 20 1000 Ω0.002Or the current through a 2 KΩ resistor if the voltage dropacross it is 4.0 V:i 4 0.2 mA2000There’s just one big problem with this analysis, and thatproblem is:Jim StilesThe Univ. of KansasDept. of EECS
1/25/2008Volts milliamps kiloohms2/4The correct answers are 14 Volts, 10 KΩ, and 2.0 mA.The problem of course is all those decimal places! It is easyto get incorrect answers when resistances are in the kiloohms (or higher) and the currents are in the milli-amps (orsmaller).Unfortunately, that’s exactly the situation that we have todeal with in electronic circuits!Frequently, we find that in electronic circuits:1. Voltages are in the range of 0.1 to 50 Volts.2. Currents are in the range of 0.1 to 100 mA.3. Resistances are in the range of 0.1 KΩ to 50.0 KΩ.Fortunately, there is an easy solution to this problem.In electronic circuits, the standard unit of voltage isvolts, the standard unit of current is milli-amps, andthe standard unit of resistance is kilo-ohms.This works well for Ohm’s Law, because the product ofcurrent in milli-amps and resistance in KΩ is voltage in volts:Jim StilesThe Univ. of KansasDept. of EECS
1/25/2008Volts milliamps kiloohms3/4v [V ] i [mA ] R [K Ω ]And so:i [mA ] v [V ]R [K Ω ]R [K Ω ] v [V ]i [mA ]The trick then is not to numerically express currents in Amps,or resistances in Ohms, but instead to leave the values in mAand KΩ !!!For example, let’s recompute our earlier examples in this way:The voltage across a 7 KΩ resistor if a current of 2 mA isflowing through it:v 2 ( 7 ) 14 VOr the resistance of a resistor if a current of 2 mA results ina voltage drop of 20 V:R 20 10 K Ω2Or the current through a 2 KΩ resistor if the voltage dropacross it is 4.0 V:Jim StilesThe Univ. of KansasDept. of EECS
1/25/2008Volts milliamps kiloohmsi 4/44 2.0 mA2Not that these are all obviously the correct answers!!!!Jim StilesThe Univ. of KansasDept. of EECS
1/25/2008Example Circuit Analysis using Electronic Circuit Notation1/5Example: Circuit Analysisusing ElectronicCircuit NotationConsider the circuit below:vA10.0 VR1 1 K5.0 VR2 1 KR3 1 K2.0 mADetermine the voltage vA , and the current through each ofthe three resistors.SolutionOur first task is to unambiguously label the currents andvoltages of this circuit:Jim StilesThe Univ. of KansasDept. of EECS
1/25/2008Example Circuit Analysis using Electronic Circuit NotationvA10.0 V5.0 V v1 i1 v2 1K2/5i21Ki4i3 v3 1K1.0 mANow lets relate all the currents using KCL:vA5.0 Vi410.0 Vi11Ki21Ki31K1.0 mAi1 i2 1i2 i4 i3And relate all the voltages using KVL:vA10.0 V5.0 V v1 v2 v3 1K1K1K1.0 mAJim StilesThe Univ. of KansasDept. of EECS
1/25/2008Example Circuit Analysis using Electronic Circuit Notation10 v1 vA v1 10 vAvA v2 5 v2 vA 55 v3 03/5 v3 5.0 VAnd finally, a device equation for each resistor:i1 v1R1i2 v2R2i3 v3R3The equations above provide a complete mathematicaldescription of the circuit.Note there are eight unknown variables (i1 ,i2 ,i3 ,i4 ,v1 ,v2 ,v3 ,vA ),and we have constructed a total of eight equations!Thus, we simply need to solve these 8 equations for the 8unknown values. First, we insert the KVL results into ourdevice equations:Jim StilesThe Univ. of KansasDept. of EECS
1/25/2008Example Circuit Analysis using Electronic Circuit Notationi1 v1 10 vA 10 vAR11i2 v2 vA 5 vA 5R21i3 v3 5 5.0 mAR3 14/5And now insert these results into our KCL equations:i1 i2 110 vA (vA 5 ) 1and:i2 i4 i3(vA 5 ) i4 5Note the first KCL equation has a single unknown. Solving thisequation for vA :10 vA (vA 5 ) 1 vA 10 5 1 14 7.0 V22And now solving the second KCL equation for i4 :Jim StilesThe Univ. of KansasDept. of EECS
1/25/2008Example Circuit Analysis using Electronic Circuit Notation(vA 5 ) i45/5 5 i4 5 vA 5 10 7 3.0 mAFrom these results we can directly determine the remainingvoltages and currents:v1 10 vA 10 7 3.0 Vv2 vA 5 7 5 2.0 Vi1 10 vA 10 7 3.0 mAi2 vA 5 7 5 2.0 mAvA 7.0 3.0 10.0 V3.01K 2.0 2.0 1 K5.0 V3.0 5.0 5.0 1 K1.0 mAJim StilesThe Univ. of KansasDept. of EECS
1/30/2008 Electronic Circuit Notation 1/9 Jim Stiles The Univ. of Kansas Dept. of EECS Electronic Circuit Notation The standard electronic circuit notation may be a little different that what you became used to seeing in in EECS 211. The electronic circuit notation has a few “shorthand” standards that can simplify circuit schematics! Consider the circuit below:
Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008 Class Notes Ch. 9 Page 1 Strangeway, Petersen, Gassert, and Lokken CHAPTER 9 Series–Parallel Analysis of AC Circuits Chapter Outline 9.1 AC Series Circuits 9.2 AC Parallel Circuits 9.3 AC Series–Parallel Circuits 9.4 Analysis of Multiple-Source AC Circuits Using Superposition 9.1 AC SERIES CIRCUITS
Bowden's Hobby Circuits - Collection of circuits, for everyone. Circuit Exchange International - Andy's website. Good selection of excellent circuits Electronic Tutorials - Collection of electronics tutorials. Dolbowent.Com - Electronic Surplus and Engineering Support. Jordan's Electronics Page - Lots of good circuits here also. LED Webpage.
Lab Experiment 7 Series-Parallel Circuits and In-circuit resistance measurement Series-Parallel Circuits Most practical circuits in electronics are made up combinations of both series and parallel circuits. These circuits are made up of all sorts of components such as resistors, capacitors, inductors, diodes, transistors and integrated circuits.
1 Introduction to RL and RC Circuits Objective In this exercise, the DC steady state response of simple RL and RC circuits is examined. The transient behavior of RC circuits is also tested. Theory Overview The DC steady state response of RL and RC circuits are essential opposite of each other: that is, once steady state is reached, capacitors behave as open circuits while inductors behave as .
/ Voltage Divider Circuits Voltage Divider Circuits AC Electric Circuits Question 1 Don’t just sit there! Build something!! Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of samp
circuits, all the current flows through one path. In parallel circuits, current can flow through two or more paths. Investigations for Chapter 9 In this Investigation, you will compare how two kinds of circuits work by building and observing series and parallel circuits. You will explore an application of these circuits by wiring two switches .
CIRCUITS LABORATORY EXPERIMENT 1 DC Circuits – Measurement and Analysis 1.1 Introduction In today's high technology world, the electrical engineer is faced with the design and analysis of an increasingly wide variety of circuits and systems. However, underlying all of these systems at a fundamental level is the operation of DC circuits. Indeed,
Introduction to Analog Circuits Prerequisites: PHYS 102 or PHYS 108. Introduction to analog electronics. Theory, design, and application of circuits using passive and active components. or how to understand and design circuits like these Small print: If you haven’t had some circuits in an introductory physics course
