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Part 1: DC Analysis

1. 166 101 Getting started What is Symbulator Symbulator Symbolic Circuit Simulator is a small Tl Basic program widely regarded as the best simulator of linear circuit ever made for a calculator It can perform direct current alternating current transient time domain and complex frequency domain analyses of numerical and symbolic linear circuits It can find the Th venin Norton and two port equivalent of a circuit It accepts elements such as resistors inductors mutual inductances capacitors independent and dependent current and voltage sources ideal operational amplifiers ideal transformers and six types of two ports It can also do Bode plots Where does it run Symbulator 6 runs in the TI 89 calculator and compatible devices Symbulator does not run in the TI Nspire machines since they are incompatible with the TI 89 platform Why would I use it Symbulator is extremely useful in solving a wide variety of circuits theory problems such as those taken by engineering students in Circuits and Circuits II classes Its main advantage is that Symbulator allows the student to focus in the conceptual understanding of circuit analysis rather than the mathematical technics used for their solution Symbulator is a fantastic tool when you need a symbolic approach to a small or medium sized circuit composed of ideal linear elements You should use Symbulator whenever you have to manage symbo
2. sidc ei a 0 24 ro a b 10 r12 b 0 12 r4 b c 4 r24 a c 24 ed c 0 4iro approx iro The answer 1 5 A is correct Bo2 s Drill Exercise 1 12 Determine i v and ig We are given an unnecessary piece of information the 4V drop in the 20 resistor My solution is shown below the schematic My simulation took 18 seconds 135 The answer 4 6 1 is correct i 4 v 6 and ig 1 AS2 s Example 3 2 Determine the voltages at the nodes My solution below s dc ji 0 1 3 jd 3 0 2ir2 r2 1 2 2 r4a 1 3 4 r8 2 3 8 r4b 2 0 4 approx v1 v2 v3 The answer 4 8 2 4 2 4 is correct AS2 s Example 3 4 Find the node voltages in the circuit My solution below 3Q s dc r2 1 0 2 e 1 2 20 0 2 10 r6 2 3 6 rx 1 4 3 r4 3 0 4 ed 3 4 3vrx r1 4 0 1 approx v1 v2 v3 v4 The answer 26 67 6 67 173 33 46 67 is correct Bo2 s Drill Exercise 2 6 Determine the voltage in each node Notice that in the schematic the resistors values are given in siemens Symbulator has to be fed the resistors with values in ohms This means that when we simulate them we have to convert them from siemens to ohms by dividing 1 over the siemens value 136 My solution below 5T D Wo ele ee s dc j 0 1 6 e1 3 1 6 ed 2 3 3v1 1 r4 3 0 1 v1 v2 v3 3 2 0 17 37 1 2 5 The answer 1 2 5 is correct Bo2 s Example 2 7 Determine the voltages in all nodes My solution below the schematic
3. moan of Step 1 Describe the circuit The first step is to describe the circuit Circuit description starts with naming the nodes You can call the nodes anything you want be it a number or a letter as long as the name is unique You must always have one node called 0 zero this is your ground node and has a voltage of OV labeled the nodes starting in the ground and moving clockwise 0 1 2 and 3 It helps me to pencil the names in the schematic itself After naming the nodes am ready to describe the circuit in Symbulator notation When I only have one voltage source like here enjoy naming it with a single letter e describe the voltage source as follows e 1 0 36 since its name is e its positive node is called 1 its negative node is called 0 and its value is 36 volts between these nodes Now describe the resistors The first resistor described as follows r1 1 2 1k because name it r1 its first node is called 1 its second node is called 2 and its value is 1k ohms The second resistor we describe similarly r2 2 3 3k and likewise for the third resistor r3 3 0 2k Since we want to use k set the SI flag to true by typing true gt s si 107 A numerical analysis is possible when we have the values of all the elements in the circuit First step is to describe the circuit Start by naming the nodes Your ground should always be node 0 The circuit is passed to Symbulator as a string
4. s dc e 1 0 vs r1 1 2 r1 r2 2 0 r2 v2 vs r2 r1 r2 Which is correct for this part Then simulate the right half s dc e 2 0 1 0 2 3 0 r3 0 3 r3 r4 3 0 r4 vo v2 r3 r4 r4 Which is correct for this part The product of these two partial answers produces the same expression shown above after the big simulation and is the right answer AS2 s Figure 5 17 Voltage Follower Find Vo 172 s dc e 1 0 vi 0 1 0 0 vo The answer vi is correct TR5 s Figure 4 32 Voltage Follower The circuits in Figure 4 32 have v 1 5 V R 2 KQ and R 1 KQ Compute the maximum power available from the source Compute the power absorbed by the load resistor in the direct connection in Figure 4 32 b and in the voltage follower circuit in Figure 4 32 a Discuss any differences Answers 281 pW 250 pW 2250 uW 4 NS 2 e FIGURE 4 32 a S load interface with a voltage fo Ry VS Ry b Interface without the volta lower O a b Rs Ip qi PCT Alright so first we simulate the b circuit to find its maximum power and load power true gt s si s th e 1 0 1 5 rs 1 2 2k 2 0 omax prL L 1000 The answers we get 2 8125E 4 2 5 4 are correct Now we simulate the a circuit s dc e 1 0 1 5 rs 1 2 2k 0 2 0 0 rl 0 0 1k prL The answer 00225 is correct The explanation to the apparent paradox that the load in a is drawing more power than the source in b seems able to provide evapor
5. Derive an expression for vs in terms of the inputs v4 and v2 z lOkO y 40 kQ S 20 kf 7 40 KQ k 7 4 E 179 true gt s si s dc e1 3 0 v1 r1 3 4 10k 01 0 4 5 r2 4 5 40k r3 5 7 20k e2 6 0 v2 r4 6 7 10k r5 7 0 40k 02 0 7 0 expand vo The answer 8 v1 4 v2 is correct AS2 s Example 5 10 Cascade If v4 1V and v 2V find vo in the circuit below true gt s si s dc e1 1 0 1 e2 2 0 2 r2 1 3 2k r4 2 4 4k r6 3 a 6kK r8 4 6 8k r5 a c 5k r15 b c 1 5k r10 c 0 10k 0a 0 3 a 00 0 4 b 0C 0 c 0 approx vo The answer 8 667 is correct TR5 s Example 4 17 Cascade Derive an expression for vo in terms of the inputs vi and v2 s dc r1 5 0 r1 r2 5 a r2 r3 a 6 3 r4 6 0 r4 e1 3 0 v1 e2 4 0 v2 01 3 5 a 02 4 6 0 vo We get the right answer Let s compare it with the book s answer which is below EIE d oe Vv R R Our answer expanded via expand vo is shown below R R U2 3 r2r4v1 r4vi r4 v2 rlr3 r3 r3 v2 Playing with it by hand we get a form that in my opinion is prettier than the book s 2 1 SJ E 1 AT wes let 180 TR5 s Example 4 18 Multiple Derive an expression for vo in terms of the inputs vi and v2 s dc e2 a 0 v2 e1 f 0 v1 02 a c b 01 f d e r1 b c r1 r2 c d r2 r3 d e r3 To find vo we ask for vb ve We get an expression that can easily be rearranged to look like this r2 r3 Joi v2 This is exact
6. Find Norton equivalent of the shaded part of the circuit 10 Q s th e1 1 0 7 r1 1 a 4 j a 0 8 r2 a 0 6 a 0 Choose DC Via ino we find Ino 6 25 A Via req we find Rea 2 4 Q 155 Some circuit theory books and some professors find it entertaining or instructive to surprise unsuspecting students with tricky problems like the two we solve below Bo2 s Example 3 11 Tricky Find the Norton equivalent of the circuit below 1V Since you do not know beforehand that this is a tricky problem you go for the usual s th r2 0 b6 2 r8 0 2 8 r3 2 a 3 r1 1 a 1 1 0 1 j 2 6 3ir8 a b You choose DC press Enter and wait Symbulator reports the calculator was unable to solve the equations This often means there is a division by zero somewhere Clean the variables from the MAIN folder and try again this time following Symbulator s advise using a symbolic value We chose to use x instead of 3 in the dependent source s th r2 0 b 2 r8 0 2 8 r3 2 a 3 r1 1 a 1 e 1 0 1 j 2 b x ir8 a b Now the calculator can solve just fine Evaluate the answers the expression for ino is fine but the expression for req 9 x 35 4 x 3 will result in a division by zero Via Define x 3 ino req we find that Ino 1 A and Req is undefined This means the equivalent resistance is for practical purposes infinite Your idea of fun right Bo2 s Drill Exercise 3 13 Tricky Below is another tricky problem F
7. 00154 00111 is correct Ir2 1 54 mA and las 1 11 mA B11 s Example 6 3 Hidden source Determine Vs and l4 The problem presents a seemingly hanging node with 20V To simulate this imagine a hidden 20V voltage source connected to the node 61id6 10 1 5 1 2 212 1 2 1 5 2 0 20 Iv T art We get the answer 24 2 which is correct Vs is 24V and l is 2A HK5 s Figure 1 24c Expert Determine ix and v in the following circuit s ex e 1 0 60 r8 1 2 8 r10 2 0 10 r4 2 3 4 r2 3 0 2 0 3 1x ix v3 Select DC Add unknown ix Add equation ir8 5 Run the simulation and you will get 1 8 This is correct lx is 1A and that Vx is 8V 130 B11 s Example 6 22 Hidden source Determine l4 I 212 mA 12 mA 22 kQ Although no source is shown the circuit has a current use a hidden current source true s si sNdc jt 0 1 12m r1 1 0 1K r2 1 0 10k r3 1 0 22k approx ir1 We get I1 10 48 mA which is correct B11 s Example 6 21 Hidden source Expert Determine ls l and la This hidden source problem is perfect for Expert My solution true gt s si s ex js 0 1 is r1 1 0 6 r2 1 0 3 r3 1 0 1 approx is ir1 ir3 Select DC and press Enter Add unknown is Add equation ir2 2m Run the simulation The answer 009 001 006 is correct since the currents are as follows ls is 9mA lh is LMA and ls is 6mA Resistors in parallel S
8. Choose DC Wait for Done Evaluate req to find the equivalent resistance is 2 O 147 148 AS2 s Example 4 10 Find the equivalent resistance of the circuit below s er r4 a 0 4 rx 0 a 2 j a 0 2irx a 0 Choose DC Wait for Done Evaluate req The equivalent resistance is 4 O It may be surprising to have a negative resistance This is the result of the dependent sources AS2 s Practice Problem 4 10 Find the equivalent resistance of the circuit below a WA s er r15 a 0 15 e 1 a 4vrx r10 1 x 10 rx x 0 5 a 0 Choose DC Wait for Done Evaluate req we find the equivalent resistance is 7 5 Q HK5 s Figure 2 29 Find the equivalent resistance of the circuit below s 32 2 1 5 i 22 O s er e 3 0 1 5is r3 3 2 3 r2 2 0 2 8 2 1 1 0 Choose DC Wait for Done Evaluate req The equivalent resistance is 0 6 Q HK5 Drill Problem 2 9d Find the equivalent resistance of the circuit below 10 Q SQ s er r10 1 2 10 r5 2 3 5 r1 2 0 30 e 1 0 20ir1 3 0 Choose DC Wait for Done Evaluate req The equivalent resistance is 20 Q Bo2 s Example 3 12 Find the equivalent resistance of the circuit below 6i V s er r1 1 0 6 r4 a 0 4 e a 1 6ir1 a 0 Choose DC Wait for Done Evaluate req The equivalent resistance is 3 Q Th venin Norton Just like a passive circuit can be reduced to an equivalent resistance an active circuit e g one with independent sources can be red
9. Part 1 DC Analysis OT CIC TES PNE E s RR 103 RUNNING A SIMULATION NER 104 ELE NIENTI S E ANDR rinra TE A EE E E A A TE TEE E E OO 105 VOLTAGE SOURCE AND RESIST ORS setnuctcicacnacisaduieacamaasundtaieleccntadedamadnnssleawanaesicmedaaeausdmaeste i aai 105 VMTN TS TR TN EXAMPLE ET EEEN ETENEE EEEE EAO ETE Ek 107 PRACTICE PROBLEM CNRC neen EREE A E E Ei 112 ELENIENT Jorre E E E E E A A E E E E E 123 CORRENT OUN Eeen E E A E E AAE A EAN 123 E TRANE A APEE eE EEE AE 124 dT ecd c8 NS SNO E EE EEE EEE 125 PRES TOF IN ER ERNEUT 131 Bis EZ MESE ec NO 132 az x XC mm 143 SHORT er CUIT NR T 143 EQUITY ACCENTS 144 OUI ALENT RESISTING RR RET TL E 144 s 2 28 ER CT v M ee 144 TE VE NINE NORTON NR 149 SRT T uu 149 Hz sppiee tte c m 165 O meri Q 165 SO OT ROBEN NET
10. m re L 1 I t Ry VT My answer starts by defining v as va vb This is done thus Define vx va vb Now I run the th script with the circuit description shown below false gt s si s th eli a 0 vs ed 1 0 u vx ro b 1 ro b 0 vth req 157 th saves the current voltage drop and power inaloadasa function of load value Lin variables irL vrL and prL respectively 158 The answers we get vs u u 1 ro u 1 are correct as can be seen by comparing them to those in the book H Us p 1 Ro kd Ur and R am not sure there is any other circuit simulator for calculators that can do this Ri Problems One type of problems that books and professors like to present to students when teaching the Th venin Norton equivalents is what like to call R problems A typical RL problem statement goes like this First reduce the circuit as seen by resistor RL to its Th venin or Norton equivalent Then find the value of the voltage drop current and or power consumed in the load resistor R if its value is whatever ohms Since this is such a typical problem Symbulator helps you solve them Right after a Th venin or Norton equivalent is found Symbulator will store a series of variables related to a RL connected among the equivalent s terminals its current is stored in irL its voltage drop in vrL and its consumed power in prL all in function of the value of the load L These are g
11. Calculate the source current c Determine the current through each branch 100 R 2 2200 R Since the ground is the bottom node named it 0 named the top node 1 s dc e 1 0 24 r1 1 0 10 r2 1 0 220 r3 1 0 1 2k a Evaluating re gets the total resistance 9 49 Q b Evaluating ie gets us the source current 2 53 A c Evaluating ir1 gets l 2 4 A ir2 gets l2 0 11 A and ir3 gets l3 0 02 A B11 s Example 7 2 Determine l4 Is and V2 My solution below named the top node 1 and the other 2 truesisi s1dc e6 1 0 12 r1 1 2 6 8k 72 2 0 18k 103 2 0 2k 104 1 0 8 2 Answers v2 is 2 51 V ie e g Is is 2 86 mA and ir4 is 1 46 mA B11 s Example 7 7 114 a Find the voltages Vj V4 and V for the network b Calculate the source current For your benefit have labeled the node names used My solution s dc e1 0 1 6 e2 0 2 18 r1 1 a 5 r2 a 2 3 r3 1 6 6 r4 b 2 2 Answers vr1 is 7 5 V vr3 is 9 V For Vba vb va is 1 5 V For ls ie2 is 3 A B11 s Figure 7 32 Determine ls and Ve My solution below s idc e 1 0 240 r1 1 2 5 r2 2 0 6 r3 2 3 4 r4 3 0 6 r5 3 4 1 r6 4 0 2 Answers ir6 is 10 A and vr6 is 20 V B11 s Example 7 10 Calculate the indicated currents and voltages 2 1 12 kQ Ox o 6 kQ 12 KQ My solution below true gt s si s dc r2 2 3 8k r1 3 4 4k r3 1 2 12k r4 1 4 24k r5 1 0 12k e 4 0 72 r6 4 5 12k r7 5 0 9k r8 5 6 3k r9 0 6 6k Answers
12. dm 1 2 0 1502 o 2s amp ido LO T 1 r3 0 1 9 04 2 4 0 2vr4 v1 v2 v3 v4 r5 3 0 5 e1 3 4 1 5 ed The answer 1 5 5 2 5 4 is correct Bo2 s Example 2 6 Determine the voltages in all nodes 137 138 My solution below s dc e1 0 1 1 e2 3 4 5 ed 3 2 3vr4 j 0 4 2 r4 1 2 1 4 r1 2 0 1 r8 3 0 1 8 r2 2 4 1 2 v1 v2 v3 v4 The answer 1 2 1 5 is correct HK5 s Drill Problem 1 12 Find lA lg and lc My solution below s dc jl x 0 5 6 ra 0 x 18 jb 0 x 1vx r9 0 x 9 jr 0 x 2 approx ira ijb ir9 The answer 3 5 4 6 is correct Numerical from symbolic examples Bo2 s Drill Exercise 1 10 Determine i v i and Vs With one unknown value and one known solution this problem is a job for Expert Determine i v i and Vs s ex es 2 0 vs jd 0 3 2ir1 r7 0 1 7 r1 3 1 1 r3 3 2 3 r4 1 2 4 ir1 vjd ies vs Add vs to the unknowns and add and vr4 4 to the equations Run the simulation The answer 2 9 3 3 is correct i22 v 9 i 2 3 and v 3 Bo2 s Drill Exercise 1 11 Determine i v and vg Since all element values are known the tip we are given by the book namely that the voltage drop in the 60 resistor is 1 5V is totally superfluous s dc e 1 0 12 r1 1 2 1 r4 2 0 4 r10 2 3 10 r6 3 0 6 r2 3 4 2 4 0 vr10 15 approx vr10 ir2 vj The answer 7 5 5 5 is correct i 5 v 7 5 and vg 5 Symbolic examples TR5 s Ex
13. ir5 is 3 mA ie e g Is is 7 36 mA and vr7 is 19 6 V 115 116 B11 s Example 7 4 Determine the currents h l2 la lg and Ic and the voltage drop areas A B and C My solution below s dc e 1 0 16 8 r1 1 2 9 r2 1 2 6 r3 2 3 4 r4 3 0 6 r5 3 0 3 r6 2 0 3 Current l is found via ir1 1 2 A lo via ir2 2 1 8 A l4 via ie 3A lg via ir3 1 A and Ic via ir6 2 A The voltage drop in area A is vr1 10 8 V in both B and C it is v2 6 V B11 s Example 6 15 a Determine the total resistance Rz b Find the source current and the current through each resistor c Calculate the power delivered by the source My solution below s dc e 1 0 28 r1 1 0 1 6k r2 1 0 20k r3 1 0 56k a Evaluating re gets the total resistance 1 44 KQ b Evaluating ir1 gets 17 5 mA ir2 gets 1 4 mA and ir3 gets 0 5 mA c Evaluating pe1 gets the power 543 mW These are the correct answers B11 s Figure 7 40 Determine Vp and V E120 VR My solution below s dc e a 0 120 r1 a 6 10 r2 6 c 20 r3 c 0 30 riga 0 20112 0 0 20 113 0 0 20 Answers vb is 66 21 V and vc is 24 83 V B11 s Example 8 10 Determine the current through each resistor My solution below s dc e1 1 0 15 r1 1 a 4 e3 3 0 20 r3 3 a 10 e2 0 2 40 r2 a 2 5 Through an array and using the approx command we ask for all the three answers approx ir1 ir2 ir3 The calculator returns 4 77 7 18 2 41 meaning Ip 4 77A l
14. is correct the maximum transfer of power occurs when the load is 9Q At this point the power transferred is 13 44W Now let s solve another one AS2 s Practice Problem 4 13 Find the R value for maximum power transfer and the maximum power transferred s th ei 1 0 9 rx 1 2 2 r1 2 4 1 ed 4 0 3vrx r4 2 3 4 3 0 approx req pmax The logic of this problem is identical to the previous one The answer is 4 22 2 901 B11 s Example 9 17 Find the R value for maximum power transfer and the maximum power transferred E s th j 2 0 6 r2 2 0 10 r1 2 3 3 r3 0 1 2 e 3 4 68 4 1 approx req pmax Same thing Answer is 15 273 07 Let s now see one that is a little different B11 s Example 9 15 a Find the load resistance that will result in maximum power transfer to the load and find the maximum power delivered b Ifthe load were changed to 68 kQ would you expect a fairly high level of power transfer to the load based on the results of part a What would the new power level be Is your initial assumption verified c If the load were changed to 8 2 kQ would you expect a fairly high level of power transfer to the load based on the results of part a What would the new power level be Is your initial assumption verified 162 My one line solution to all three questions is presented below true gt s si s th j 0 1 10m r 1 0 40k 1 0 freq pmax prl I 68000 prl l 8200 The answer we obtain 4000
15. press F5 in the main menu 3 If you are not familiar with the solve command refer you to the calculator s manual 111 Variables r1 through r99 are reserved variable in the calculator This means you cannot store things into them If you want to get numerical answers you must have as many equations as you have unknowns 112 Set the SI flag and run the expert mode by typing this in the calculator true gt s si s ex cir When prompted select DC and press Enter Now you will see a prompt asking you to add equations variables and conditions You may recall from your algebra class that you need an equal number of equations and unknowns in order to solve a set of equations into numerical values The statement of the problem gives us the information we need to write the two additional equations In Add equations type re 12k and ir326m In Add unknowns type e rx Now we have six variables and six equations Press Enter and wait just a few seconds A few of other dialogs will appear In this and all the other Expert examples in this volume just press OK in these prompts without changing anything in them When Symbulator says Done go ahead and retrieve the answers rx You get 2000 the right answer You get 72 the right answer The speed advantage of the expert mode is not necessarily evident in this simple problem It does give you an idea of what the expert mode is all about you get to halt the s
16. 0 2 5m r2 2 0 2k 2 0 fino req irL L 0 irL L 2000 irL L 5000 The answer 0075 1500 0075 00321 00173 is correct Bo2 s Example 3 5 Find the Norton equivalent of the circuit external to the 10 resistor Then determine the voltage drop across this 10 resistor 159 My solution below s th e 1 0 24 r12 1 2 12 r20 2 0 4 r23 2 3 4 134 3 4 2 j 4 0 3 r40 4 0 5 3 0 ino req vrL L 1 The answer 9 7 7 2 1 is correct RM3 s Example 9 13 Use Millman s Theorem to simplify the circuit left of a b so that there is only one voltage and one resistor Then find the current in the load resistor R R 192 0 don t know Millman s Theorem but in my book this is called the Th venin equivalent s th r1 0 1 240 e1 2 1 96 r2 0 3 200 e2 3 2 40 r3 0 4 800 e3 2 4 80 2 0 vth req irL L 192 The answer 28 8 96 1 is correct Bo2 s Example 3 7 Find the Th venin equivalent for the circuit left of a b Then find the voltage across the 3 O resistor My answer is presented below s th e 1 0 20 r6 1 2 6 r1 1 3 1 r2 3 2 2 j32 3 2 15 j30 3 0 15 2 0 vth reg vrL L 3 vrL L 6 The answer we find 30 2 18 22 5 is correct Bo2 s Drill Exercise 3 7 Find the Th venin equivalent for the circuit left of a b Then find the voltage v My answer below s th e 1 0 8 r3 1 2 3 r12 1 3 12 r6 2 0 6 r2 2 3 2 3 0 vth req vrL L 5 The answer 6 3 3 75 is correct RM3 s Pra
17. 0 a 4 b a Choose DC Via vth we find Vr 6 V Via reg we find Reza 7 5 OQ 153 AS2 s Example 4 11 Find the Norton equivalent of the circuit below 8 Q 1 sXh j 0 2 2 06 4 0 12 71 2 4 4 02 2 3 8 73 0 1 8 74 3 1 5 3 1 Choose DC Via ino we find luo 1 A Via req we find Ra 7 4 O HK5 s Drill Problem 2 8a Find the Th venin equivalent of the circuit below 200 a wa 50V s th e1 1 0 100 r2 1 2 20 j 0 2 4 r1 2 3 10 e5 3 4 50 4 0 Choose DC Via vth we find Vr 130 V Via reg we find Req 30 O B11 s Example 8 7 Find Norton equivalent of the circuit below DO sge C s th j7 0 1 7 j3 1 0 3 r1 1 0 4 j4 0 1 4 1 0 Choose DC Via ino we find Ino 8 A Via req we find Rea 4 O AS2 s Practice Problem 4 9 Find the Th venin equivalent of the circuit below 5Q Ik 30 154 s th e 1 0 6 r5 1 2 5 rx 2 3 3 0 2 1 5irx r4 3 0 4 3 0 Choose DC Via vth we find Vy 5 33 V Via req we find Rea 0 44 Q AS2 s Example 4 9 Find the Th venin equivalent of the circuit below s th j b 0 5 rx 0 6 4 r1 0 1 2 r2 1 b 6 r3 1 a 2 e 1 0 2vrx a b Choose DC Via vth we find Vr 20 V Via req we find Req 6 O Bo2 s Drill Exercise 3 8 Find the Th venin equivalent of the circuit below Stn T0 T S r 1 0 1 1 66 1 2 0 710 1 3 10 08 2 0 0 12 2 9 2 3 0 Choose DC Via vth we find Vr 2 V Via req we find Ra 2 4 O B11 s Example 9 13
18. 27 3 27 You should know how to read these by now but here it is just in case V1 37 82V V2 32 73V h 3 27A lz 1 27A la 3 27A RM3 s Example 9 12 solve If R3 is to be replaced with R4 and l4 determine the value and direction of the source 20 V First make sure you understand what this problem is asking you to do The idea is that despite the change we keep the same voltage drop and current flow between nodes a and b We must first know what they are So we simulate the original circuit s dc e 0 0 20 r1 0 a 16 r2 a b0 40 r3 a b 60 va vb ir3 We find that the voltage drop is 12V and the current is 0 2A These are the currents and voltages that we have to keep once we do the replacement Simulate the circuit now replacing Rs with a resistor R4 of 2400 and a source j with value l4 Run this s dc e 0 6 20 r1 0 a 16 r2 a b 40 r4 a b 240 j a b i4 Notice that both the voltage drop given by va vb and the current given by ir4 ij are algebraic functions in terms of i4 Now you can find i4 solving by voltage drop solve va vb 12 i4 orbycurrent flow solve ir4 ij 0 2 14 The result is the same i4 15 A The required current source is 15A from a to b 129 RM3 s Example 8 13 Solve for the currents through R2 and Rs in the circuit shown My solution is below true gt s si s dc r1 a 0 10k r2 1 0 5k r3 b a 6k r4 0 2 16k j a o 2m e1 1 60 10 e2 b 2 8 approx ir2 ir3 The answer
19. 3 r1 1 2 66 r2 2 0 24 2 0 Pick DC Via vth find Vr 20 88 V Via ino find Ino 0 05 A Via req find Rea 17 6 Q B11 s Example 9 7 Find the Th venin equivalent of the circuit below as seen from the Rs resistor Sih o4 23 3 04 19 1 2 2530 Choose DC Via vth we find Vr 48 V Via req we find Req 6 O B11 s Example 9 11 Find the Norton equivalent of the circuit below as seen from the R resistor s th e 1 0 9 r1 1 2 3 r2 2 0 6 2 0 Choose DC Via ino we find Ino 3 A Via req we find Rea 2 0 AS2 s Practice Problem 4 12 Find the Norton equivalent of the circuit below s th r6 1 0 6 j 0 1 10 r2 x 0 2 e 1 x 2vx x 0 Choose DC Via ino we find Ino 10 A Via req we find Req 1 O B11 s Example 9 12 90 Find the Norton equivalent of the circuit as seen from the R resistor s th r2 1 0 4 r1 1 2 5 j 1 2 10 2 0 Choose DC Via ino we find Ino 5 56 A Via req we find Rea 9 O HK5 s Drill Problem 2 8b Find the Th venin equivalent of the circuit below s th j 0 2 0 01v1 r 0 2 20 e 1 2 100 1 0 Choose DC Via vth we find Vr 125 V Via reg we find Req 25 Q HK5 s Figure 2 27 Find the Th venin equivalent of the circuit below i 2kQ 2 3k2 true gt s si s th e 1 0 4 r2 1 2 2k r3 2 x 3k j 0 2 vx 4000 x 0 Via vth we find Vm 8 V Via req we find Rea 10 kO B11 s Example 9 8 Find the Th venin equivalent of the circuit below as seen
20. Differential Design an op amp circuit with inputs v and v such that Vo 5v1 3v2 My solution follows The problem statement is a fancy way to say for the same circuit of the previous problem find what values of resistors you need to use if you want to get an output v 5v1 3v2 Although this is not strictly a Symbulator problem present it here because it illustrates how Symbulator fits in such design problems First we take that part of vs that is a factor of v and make it equal to 5 Thus Define v121 Define v2 0 expand vo 5 We get r2 r1 5 Now make that part of v that is a factor of v equal to 3 Thus Define v1 0 Define v2 1 expand vo 3 We get r2 r4 r1 r3 r4 r4 r3 r4 3 Now since you have two equations you can solve for two unknowns Out of the four resistors whose values you can decide upon two can be whatever you want The book recommends you use R 10K and R3 20K Now let s find the values for the other two resistors Ro and Ra solve ans 1 and ans 2 r2 r4 r1 10000 and r3 20000 The expression above assumes that ans 1 and ans 2 are pointing to the two equations we found before We get r2 50000 and r4 20000 This is correct Now if this problem was part of a test I d like to verify that the answer is correct To confirm this simulate the circuit using the four values given above for the resistors true gt s si s dc e1 d 0 v1 e2 c 0 v2 r1 d a 10k r3 c 6 20k r2 a 0 50k r4 b 0 20k 0 b a
21. Q This is correct Below are many practice examples of this type Practice problems for resistive circuits B11 s Example 8 30 Find the equivalent resistance of the circuit below s er rac a c 6 rad a d 9 rab a b 6 rcd c d 9 rbc b c 6 rod b d 9 a c Choose DC When Done use approx req to find the equivalent resistance is 3 27 Q AS2 s Example 2 9 Find the equivalent resistance of the circuit below Please notice my use of p to reduce resistors in parallel and for resistors in series s er r12 1 2 4 r03 0 3 8 r23 2 3 0 1 5 2 p 6 3 1 0 145 Choose DC When Done use approx req to find the equivalent resistance is 14 4 O AS2 s Practice Problem 2 9 Find the equivalent resistance of the circuit below 2 30 In my circuit description below please notice the use of p to reduce resistors in parallel and to reduce resistors in series s er r12 1 2 2 r04 0 4 1 r24 2 4 6 r23 2 3 3 134 3 4 p 4 44 5 3 1 0 Choose DC When Done evaluate req to find the equivalent resistance is 6 Q AS2 s Example 2 10 Find the equivalent resistance of the circuit below I0Q IQ q IQ In my circuit description below please notice the use of p to reduce resistors in parallel and addition to reduce resistors in series s er rac a c 10 roc b c p 3 6 rcd c d 1 rod b d p p 12 4 1 5 a b Choose DC Wait for Done Evaluate approx req The equivalent resistance is 11 2 O AS2
22. Rr rl 4rrc rp rs Rs Rp Re Ri 1 Bo2 s Example 1 11 Symbolic Determine v2 My solution below In my solution named the value of the source vi to keep it similar to the book This required avoiding naming any node as 1 if there was a node 1 Symbulator would store in v1 the voltage of the node creating trouble There is no problem with using r1 as a value since nothing will be stored in that r1 value s dc e a 0 v1 r1 a 3 r1 rg 3 0 rg j 2 0 gm vrg rd 2 0 rd rl 2 0 rl v2 The simulation took 25 seconds The answer got is shown left the textbook s right am rd rar vil ee nR ri rg rd rl R R Ry Ry TR5 s Exercise 4 3 Symbolic Find vo in terms of the value in the circuit For resistors use their conductance value This is my solution Since we want to use p as a constant in the dependent source set the s si flag to false so that it is not confused with the SI unit for micro false gt s si s dc ei 1 0 vs ed 2 0 u vrx r1 1 2 1 g1 r2 2 0 1 g2 rx 1 0 1 gx rl 0 0 1 gl vo This is the answer we get It is correct Compare it with the textbook s answer _ta2 u gx vs 200 GtG g2 ut 1 91 ax B Ge GE DG The last four problems show Symbulator at its DC best don t know of any calculator based program that was able to provide this kind of purely symbolic answer to a circuit simulator back in 1999 when made Symbulator As a matter of fact ev
23. elements separated by colons There is more than one way to skin a cat 108 We pass along this description to Symbulator as a string e g we open with a quotation mark then enter the descriptions of each element separated with colons and then close with a quotation mark We can store this string in a variable like to call this variable cir short for circuit but you can call it something else if you prefer type this in my Symbulator ready calculator e 1 0 36 r1 1 2 1k r2 2 3 3k r3 3 0 2k gt cir Once press enter the circuit will be stored in that variable Step 2 Run the simulation We now ask Symbulator to simulate this circuit in direct current DC by typing this s dc cir After you press enter Symbulator will quickly simulate the circuit stored in cir Symbulator will let you know when it is Done It took 12 seconds in my calculator to get 16 answers and store them in variables in the current folder the voltage in each of the three nodes v1 v2 and v3 for each of the four elements in the circuit its voltage drop ve vr1 vr2 and vr3 current ie ir1 ir2 and ir3 and power consumed pe pr1 pr2 and pr3 and the equivalent resistance as seen by the source re Some alternatives We could also have given Symbulator the command to simulate and the circuit description in a single line using two possible methods The first is connecting the two steps above using a colon as shown below e 1 0 36 r
24. first is treated as an exact value sisi is true e Another example a value of 180mV can be entered as 180m 180 1000 0 180 18E 2 and 180E 3 The fraction will be treated as an exact value Simulation answers After a DC simulation for each voltage source in a circuit Symbulator will store a series of answers in the current folder of the calculator e The voltage of each of its nodes with reference to the ground is stored in a variable called v and the name of the node For example for a node called 1 its voltage with reference to ground is calculated and stored in a variable called v1 e The voltage drop in the source that is to say the voltage in the first node minus the voltage in the second node is stored in a variable called v and the name of the source For example for a source called e5 the voltage drop is stored in ve5 e The current through the source flowing from the first node towards the second node is stored in a variable called i and the name of the source For example the current flowing through a source called ex from its first node towards its second node is stored in iex The way the current direction is defined might seem counterintuitive for voltage sources it is such for consistency the same direction is applied to all other two node elements throughout Symbulator e The power consumed by the source is stored in a variable called p and the name of the source For example for a source called e12 the power
25. from the Rz resistor ignore the textbook s decision to call the nodes a and b since b is ground anyway 151 152 s th r1 2 0 6 r2 2 1 4 r3 1 0 2 e 0 1 8 2 0 Choose DC Via vth we get V 4 8 V Via req we get Reg 2 4 Q B11 s Example 8 6 Find the Norton equivalent of the circuit below s th j1 0 1 6 r1 1 0 3 j2 1 0 10 r2 1 0 6 1 0 Choose DC Via ino we find Ino 4 A Via req we find Reg 2 0 Bo2 s Drill Exercise 3 12 Find the Norton equivalent of the circuit below s th e 1 0 12 r6 1 2 6 j 0 2 3ir6 r3 2 0 3 2 0 Choose DC Via ino we find Ino 8 A Via req we find Reg 1 O Bo2 s Drill Exercise 3 9 Find the Th venin equivalent of the circuit below s th j bo 0 10 r1 0 b 1 e a 0 3ir1 r6 a b 6 a b Choose DC Via vth we find Vr 24 V Via reg we find Rea 2 4 Q AS2 s Example 4 12 Find Norton equivalent of the circuit below sXMh rx 0 b 4 e6 0 b 10 r5 0 3 5 j 0 a 2irx a b Choose DC Via ino we find Ino 7 A Via req we find Reqg 5 OQ B11 s Example 9 10 Hidden source Find the Th venin equivalent of the circuit below Ej 9 10V true gt s si s th e1 3 0 6 e2 4 0 10 r1 2 3 0 8k r2 2 4 4k r3 2 0 6k r4 2 1 1 4k 1 0 Choose DC Via vth we find Vr 3 V Via reg we find Reg 2 KQ B11 s Example 9 9 Find the Th venin equivalent of the circuit below as seen from the R resistor s th e 1 0 72 r1 1 6 6 r2 1 a 12 r3 0 6 3 r4
26. 0 176 Evaluating vo gives us the desired output 3 v2 5 v1 The resistor values are correct AS2 s Practice Problem 5 7 Difference or Differential Design a difference amplifier with gain 4 My solution follows The problem statement again is just a fancy way to say for the same circuit of the previous problem find what values of resistors you need to use if you want to get an output vo 4 v2 v1 or in other terms 4v1 4v2 Same as before 7 f you have no idea what I m talking about you should learn about ans in the user s manual Define v121 Define v2 0 vo 4 Define v1 0 Define v2 1 expand vo 4 This time the book asks that you use Ri 10K and Rs 10K So we do that solve ans 1 and ans 2 r2 r4 r1 10000 and r3 10000 We get r2 40000 and r4 40000 the correct values for the remaining resistors AS2 s Practice Problem 5 8 Instrumentation Find io 40 KQ true2s si sidc e1 1 0 8 62 2 0 8 01 01 1 3 3 02 2 4 4 01 3 5 20k r2 4 6 20k r3 5 0 40k r4 6 0 40k 03 6 5 0 r5 0 0 10k ir5 In the schematic the current io corresponds to ir5 The answer 2 E 6 is correct Bo2 s Example 3 3 Cascade Find vo in terms of the conductances and the applied voltage vs G4 s dc e 1 0 vs r12 1 2 1 g1 r14 1 4 1 g2 r40 4 0 1 g3 r20 2 0 1 94 r23 2 3 1 g 134 3 4 1 g 01 0 2 3 02 0 4 0 Evaluating vo we get gl B4 vs g3 g4 which is correct as can be seen by comparing it to the
27. 0 0 10k 22k provides the same answer we got before e g 11 10 48 mA Moving forward we will use the p tool whenever we feel it is appropriate A similar reduction of resistors in series is possible through simple addition whenever we do not need to know specific answers for each resistor such as the voltage drop or power consumed in each or the voltage in the node between them The current through series resistors is the same so even through an equivalent you can get the current Dependent sources One of my favorite scenes in cinema comes from The Dark Knight the Joker played by Heath Ledger is rolling on the floor of a Gotham City prison taking a bare knuckle beating from an ever more frustrated Batman Master of the situation and laughing hysterically the Joker says You have nothing Nothing to threaten me with Even though the movie had not been made yet remember feeling something along the same lines although maybe less hysteric back in 1999 when I realized that one of the consequences of having used a 10096 symbolic implementation for Symbulator meant that could make any element s value dependent on any answer of the circuit could simulate voltage or current sources that were dependent on any voltage current or combination thereof with the same ease that could simulate a 12V source Here is what you need to know for simulating dependent sources on Symbulator nothing There is nothing special to it
28. 0 1 93 57 is correct Let s deconstruct it Part a is answered by the first two values a 40K O resistor as load would receive 1W power Since this is the maximum this is the most that any load could receive ever Parts b is answered by the third value Making use of the variable prL which contains the power delivered by the circuit equivalent to the load as a function of the load value L we find that a load of 68kQ receives 93W which is less than the maximum Parts c is answered in similar manner by the fourth value A load of 8 2kO receives 57W which is less than the maximum Any resistance other than 40K gets less power Automatic circuit equivalent At risk of promoting vagrancy among EE students let me show you one more goodie th saves a of the th script once it has found the Norton equivalent of a circuit it will description of automatically write for you the description of the equivalent circuit and store it in a the equivalent variable eqcir to save you time just in case you need to run any simulation using it circuit in a The following example illustrates the use of this laziness promoting feature of th variable called eqcir RM3 s Example 9 8 Find the Norton equivalent of the circuit left of a b then find the current through R Let s first find the circuit equivalent true gt s si s th e 1 0 24 r1 1 2 120 r2 2 0 280 2 0 560m 2 0 ino req The answer 36 84 is correct Now to the second
29. 1 2000 ohms These are the circuit right answers Not many other circuit simulators allow this flexibility description Examples to practice this type of problems are found starting in page 116 solving a symbolic circuit with Expert Symbulator s true strength is seen in numerical from symbolic problems like the one we solved above but using its Expert mode of simulation which cracks these problems open even faster and can give you fully numerical answers to problems like this with an equal number of unknown values and of answers provided by the problem Learning to use the expert mode pays off handsomely in terms of additional power and speed Let s solve the same circuit from B11 s Example 5 6 this time using the expert mode B11 s Example 5 6 Redux We will use the same circuit description as before with a single change we will use rx for the value of resistor Ri instead of the r1 value we used before Like this e 1 0 e r1 1 2 rx r2 2 3 4k r3 3 0 6k cir The reason for this is that r1 is a reserved variable in the calculator The expert mode tries to store all the values of the solved unknowns into variables of the same name Trying to store a value into r1 would result in an error f you want to know how long it took Symbulator to run a simulation you have to make sure your calculator s clock is on by typing ClockOn As long as the clock is on you can see how long a simulation took by going to the Program IO screen
30. 1 1 2 1k r2 2 3 3k r3 3 0 2k gt cir s dc cir The second is giving the circuit description as an argument of the s dc gate s dc e 1 0 36 r1 1 2 1k r2 2 3 3k r3 3 0 2k Any of the three approaches yields the same result and I use them interchangeably Step 3 Get the answers Answer to question a The equivalent resistance seen by the source e is stored in re Evaluate this expression The calculator returns the right answer 6000 e g 6K Answer to question b Current ls is the current flowing through the source from node O to node 1 thus we can obtain it by evaluating ie Since this is a series circuit we could also find it as the current through any of the resistors ir1 ir2 or ir3 In either case the calculator provides the right answer 006 e g 6mA Answer to question c The voltage drop in resistor R1 is found evaluating vr1 the calculator returns 6 e g 6 V For R2 evaluate vr2 to obtain 18 V And for Rs vr3 gives us 12 V These are all the right answers Answer to question d We evaluate pe and get the right answer 216 e g 216mW Answer to question e The power consumed by the resistors are found evaluating ori pr2 and prs The calculator returns the right answers 036 e g 36mW 108 e g 108mW and 072 e g 72mMW respectively Answer to question f Let s ask the calculator if the sum of the consumed power in the resistors equals the power supplied by the source Evaluate this equ
31. 2r4 2 RR r1 r24 r3 r2r3 R R RR RR which is correct as can be seen by comparing it to the book s answer shown right AS2 s Practice Problem 5 4a Transresistance This is a current to voltage converter also called a transresistance amplifier Find Vo is 169 s dc j 0 1 is r 1 0 r 0 0 1 0 vo is The answer r is correct AS2 s Practice Problem 5 4b Transresistance This is another transresistance amplifier Again find vo is s dc j 0 1 is r1 1 2 r1 r2 2 0 r2 r3 2 0 r3 0 0 1 0 Asking expand vo is get r1 r3 r2 r1 r3 which is equivalent to the book s answer Uo R3 R3 3 as MT ide qe i Ri x Bo2 s Example 3 1 Non Inverting Amplifier Find vo vi s dc e p 0 v2 r1 1 0 r1 r2 1 0 r2 0 p 1 0 vo v1 We get 1 r2 r1 which is correct AS2 s Figure 5 16 Non Inverting Amplifier Find Vo s dc e 2 0 vi r1 0 1 r1 rf 1 0 rf 0 2 1 0 vo 170 We get the right answer below an expression equivalent to the book s answer rl crf yi rl TR5 s Example 4 13 Non Inverting Amplifier We can solve this problem in one simulation as shown below s dc e 1 0 vs r1 1 2 r1 r2 2 0 r2 0 2 3 0 r3 0 3 r3 r4 3 0 r4 vo vs r2 r3 r4 r1 r2 r4 which is correct as can be seen by comparing it to the book s answer shown below R g R R R R We can also solve it in stages as shown below First simulate the left half
32. 30 3 0 1 5 v1 v2 v3 The answer 1 2 3 is correct vi 1V v2 2V v3 3V B11 s Example 8 5 Determine the current l2 My solution below s dc j1 1 0 4 r1 1 0 3 e2 1 2 5 r2 0 2 2 approx ir2 This gives us a value for lz of 3 4 A This is correct Bo2 s Example 2 5 Conductances 2S vi Loe A U3 3A D 58 Determine the voltages in the nodes and the current through the voltage source 6001 0 T 3 6 9 2 9312 2 1 726r120 2 0 1 3 630 3 0 1 55 r13 1 3 1 2 approx v1 v2 v3 ie The answer 5 1 5 1 5 11 5 is correct vi 5 V2 1 5 v3 1 5 i 11 5 127 B11 s Example 8 22 Determine V4 and V2 My solution below s dc j1 0 1 6 r1 1 0 4 e 1 2 12 r3 1 2 10 r2 2 0 2 j2 2 0 4 approx v1 v2 The answer 10 67 1 33 tells us that Vi is 10 67V and V is 1 33V B11 s Example 8 19 Determine Vi h and lo My solution below s dc e 2 0 24 r1 1 2 6 r2 1 0 12 j 0 1 1 approx v1 ir1 ir2 The answer 20 667 1 67 tells us that V4 is 20V li is 667A and lz is 1 67V B11 s Example 8 14 Determine lI and ls s dc e1 1 0 20 r1 1 2 6 r2 2 a 4 j a 0 4 r3 a 3 2 e2 0 3 12 approx ir2 ir3 Answer 3 33 666 This is correct 128 B11 s Example 8 20 Determine Vi V5 lj l and l3 My solution is found below the circuit schematic s dc e 3 0 64 r1 3 1 8 r2 1 2 4 j 1 2 2 r3 2 0 10 approx v1 v2 ir1 ir2 ir3 Answer 37 82 32 73 3 27 1
33. 8 0 0 8 vo iro The answer 8 4 is correct vo 8 and io 4 HK5 s Example 1 3 Determine the power delivered by each source and consumed by both resistors 133 s dc ei 1 0 120 r1 1 2 30 ed 2 3 2vra ra 0 3 15 pei ped pri pra The answer 960 1920 2880 is right the independent source delivers 960W the dependent source delivers 1920W and the resistors consume 2880W together AS2 s Practice Problem P2 6 Find v and vo in the circuit My solution below 35v QD s dc ei x 1 35 rx x 0 10 ed 0 2 2vx ro 1 2 5 vx vro The answer 10 5 is correct vx 10 and vo 5 Bo2 s Example 1 10 Determine Vi v and i My solution below The simulation took 14 seconds s dc ei 1 0 2 r1 1 2 1 3 ed 3 2 4 ir1 r2 3 0 1 5 ir1 vr1 vr2 The answers 15 26 5 26 3 26 is correct vi 5 26 v2 3 26 and i 15 26 AS2 s Example 2 6 Determine vo and i in the circuit My solution is shown below sidc e12 1 0 12 ri 1 2 4 6d 2 3 2vo0 e4 0 3 4 r0 0 0 6 vo iri 134 The answer 48 8 is correct vo 248 and i 8 HK5 s Drill Problem 1 11 Find the power absorbed by each element in the circuit My solution below 12 V 30 Q ln s dc r1 x 0 30 e1 1 x 12 r2 1 2 8 r3 2 3 7 ed 3 0 4vx approx pr1 pei or2 pr3 ped The answer 768 1 92 2048 1792 3 072 is correct AS2 s Example 3 6 Determine the value of in the circuit My solution below 24v GR
34. 8 6k r8 8 0 8k r4 0 0 4k 0 0 8 0 vo io The answer 3 8 001425 is correct Again the current is going into the op amp Bo2 s Drill Exercise 3 3 Difference or Differential 174 Find v for the difference amplifier circuit shown in Fig DE3 3 when R R and R4 R R Answer v va R s dc ea 4 0 va eb 3 0 vb r1 4 1 r1 r2 1 0 r2 r3 3 2 r1 r4 2 0 r2 0 2 1 0 Evaluating vo we get the correct answer equivalent to the book s answer above vb va r2 r1 TR5 s Exercise 4 13 Difference or Differential Find Vo 10 kQ 40 KQ e s dc e1 3 0 v1 e2 4 0 v2 r1 3 5 10k r2 4 6 10k r3 5 0 40k r4 6 0 15k 0 6 5 0 Evaluating vo we get 3 v2 4 v1 which is the correct answer AS2 s Figure 5 24 Difference or Differential Find vo And keep it in the memory for you will use it in the next three problems sidc el d 0 v1 e2 c 0 v2 r1 d a rl r3 c b r3 r2 a 0 r2 r4 b 0 r4 0 b a 0 vo r1r4v2 r2 r3 v1 r4 v1 v2 r1 r3 r4 This expression is equivalent to the book s answer better formatted shown below R 1 R R2 R5 DU M R 1 R3 R4 Ri 175 AS2 s Figure 5 24 Subtractor For the same circuit of the previous problem find vo when Ri R2 and R3 Ra Since we already have the expression for vo stored in the memory we only do this expand vo r2 r1 and r3 r4 The answer we get v2 v1 is correct AS2 s Example 5 7 Difference or
35. Who made this thing did started writing it in April 1999 as a junior in Electrical Engineering at Universidad Tecnol gica de Panam UTP Version 5 was done by January 2001 Symbulator won 1 place in the 2k IEEE Student Paper Contest Latin America It got me a Distinguished Alum award from UTP in 2008 and was at least part of the reason for the Outstanding Young Person award that the Panamanian Chapter of the Junior Chamber International gave me in 2010 In June 2013 I released a beta of version 6 Get your calculator ready Symbulator relies on a software for Laplace transforms called DiffEq made by Lars Frederiksen You can download zip files for both Symbulator and DiffEq from this URL download symbulator com Inside the zips you will find two files you need to transfer to your TI 89 calculator For the transfer following the manufacturer s instructions Both Symbulator and DiffEq run faster when they are archived To archive them execute these two programs s install and dif install That is all You are now all set We d love to hear from you Please send us your message to email symbulator com Have fun Running a simulation You will learn how to simulate through examples For now let me just say that we give Symbulator the circuit description as a string of elements separated by semi colons The description can be stored in a variable or passed directly as an argument To runa DC simulation in Symbulator we us
36. an ask the calculator for the answers you need e Evaluating ir1 or ir2 gets the current in the resistors 2 A e Evaluating vr1 gets the voltage drop in the 30 resistor 60 V e Evaluating vr2 gets the voltage drop in the 156 resistor 30 V e Evaluating pr1 gets the power consumed in the 30 resistor 120 W e Evaluating pr2 gets the power consumed in the 15 resistor 60 W e Evaluating pe1 gets the power delivered by the 120V source 240 W e Evaluating pe2 gets the power delivered by the 30V source 60 W This means this source is actually consuming 60W e Evaluating pri pr2 pe1 pe2 gets the sum of powers 0 W As expected Wasn t that easy We could also have asked for all the answers with one array ir1 vri vr2 por1 pr2 pel1 pe2 pri pr2 pel pe2 B11 s Example 5 20 The practice problems will get progressively more complicated as we move on This will allow you to build up your symbulating skills with confidence Determine and the voltage across the 7 Q resistor V 70 named the nodes clockwise starting from the ground 0 1 2 3 and 4 Below is my circuit description given as an argument of the command to do the DC simulation 113 s dc e1 1 0 50 r1 1 2 4 e2 2 3 12 5 r2 3 4 7 r3 4 0 4 When it s done ask for the answers we need Evaluating ir1 gets the current I 2 5 A Evaluating vr2 gets the voltage drop in the 7Q resistor 17 5 W B11 s Example 6 13 a Find the total resistance b
37. ates once we remember that the ideal op amp shown in the schematic is only part of the truth the real op amp has its own source of power which provides the difference AS2 s Example 5 5 Inverting Find Vo true gt s si s dc e 1 0 6 r4 1 a 4k r10 a 0 10k e4 b6 0 4 0 b a 0 vo The answer 1 is correct AS2 s Practice Problem 5 5 Non Inverting Calculate Vo true gt s si s dc e 1 0 3 r4 1 2 4k r8 2 0 8k r2 3 0 2k r5 3 0 5k 0 2 3 0 The answer for vo 7 is correct Bo2 s Example 3 2 Adder or Summing Find Vo s dc ea 3 0 va eb 2 0 vb r31 3 1 r1 r21 2 1 r1 r10 1 0 r2 0 0 1 0 vo LET va c v zt which is correct as can be seen by comparing it to the book s answer shown below R Uo aR Va T Up AS2 s Figure 5 21 Adder or Summing Find Vo s dc e1 b 0 v1 e2 c 0 v2 e3 d 0 v3 r1 b a r1 r2 C a r2 r3 d a r3 rf a o rf 0 0 a o expand vo 173 which is correct as can be seen by comparing it to the book s answer shown below AS2 s Example 5 6 Adder or Summing Find Vo and io 2 5 kQ 10 kQ true gt s si s dc e1 1 0 1 e2 2 0 2 r1 2 a 5k r2 1 a 2 5k r3 a 0 10k r4 0 0 2k 0 0 a 0 vo io The answer 8 0048 is correct Notice a current of 4 8mA is going into the op amp AS2 s Practice Problem 5 6 Adder or Summing Find vo and io n Lb 2 20 KQ 8 kQ true gt s si s dc e2 2 0 1 5 e1 1 0 2 e6 6 0 1 2 r2 2 8 20k r1 1 8 10k r6 6
38. ation pri pr2 pr3 pe The calculator returns true This is the right answer and concludes the solution An alternative You can ask for all the answers above in a single line by using an array such as this TI E T re irt vr1 vr2 vr3 pe pr1 pr2 pr3 pr1 pr2 pr3 pe multiple answers using We evaluate this array and get the same answers as before also as an array Actually an array we could solve the whole problem using a single line of text as follows true gt s si s dc e 1 0 36 r1 1 2 1k r2 2 3 3k r3 3 0 2k approx re ir1 vri vr2 vr3 pe pri pr2 pr3 pri pr2 pr3 pe An easy way to read this is to identify the colons as delimiters between things that you would have asked in separate lines First set the SI flag to true another colon second the command to simulate in DC the circuit provided between quotation marks another colon and finally an array of all the answers we would like to evaluate By showing you this just want to make you aware of the possibilities But please do not rush into trying to solve a whole problem using a single line There are times when you may not be sure about your circuit description or about which variables to evaluate as answers In these cases it is easier to solve the problem step by step first describe the circuit then run the simulation and finally find the answers Practice problems of this type are found starting in page 112 A word on symbolic problems call a circui
39. book s answer shown below G U PCM Zy 7 X eds j 177 Bo2 s Drill Exercise 3 4 Cascade For the op amp circuit shown in Fig DE3 4 suppose that G 1 S G 2 S G 3 S G 4S and G 5S Find v in terms of Answer 0 75r Gy amp s dc e 1 0 vs r1 2 0 1 1 r2 1 2 1 2 r3 2 3 1 3 r4 4 0 1 4 r5 4 0 1 5 01 0 2 3 02 3 4 0 Evaluating vo we get 75 vs which is correct AS2 s Example 5 9 Cascade Find Vo and io 20 mV true gt s si s dc e 1 0 20m 01 1 2 a 02 a b 0 r0 0 0 10k r4 b 0 4k r2 a 2 12k r3 2 0 3k Evaluating approx vo iro we get the answer 35 2 5E 5 This is correct AS2 s Practice Problem 5 9 Cascade Determine va and io true gt s si s dc e 1 0 4 01 1 2 2 02 2 3 0 r0 3 0 4k r6 3 0 6k vo iro 178 The answer 10 1 1000 is correct To say a 1 1000 A current is the same as 1mA AS2 s Practice Problem 5 10 Cascade If v4 2V and v 1 5V find vo in the circuit below true gt s si s dc e1 1 0 2 e2 2 0 1 5 01 1 3 3 r1 2 5 10k r2 3 6 20k r5 5 4 50k 02 0 5 4 3 6 4 30k r6 6 0 60k 03 0 6 0 vo The answer 9 is correct TR5 s Example 4 16 Cascade Derive an expression for v in terms of the two inputs true gt s si s dc e1 2 0 v1 e5 3 0 5 r1 2 4 5k r2 3 4 10k 01 0 4 a r3 4 a 1 Ok r4 4 0 20k 02 a 5 0 r5 5 0 20k r6 5 0 10k expand approx vo The answer 2 4 v1 6 is correct TR5 s Exercise 4 14 Cascade
40. cal from symbolic problems solving a symbolic circuit with solve Let s do a numerical from symbolic problem by means of the solve command We will obtain numerical values from the simulation of a circuit with some unknown element values Later we will solve this same problem using the expert mode B11 s Example 5 6 Given Rzy and J calculate R and E 4kQ This is a very nice symbolic problem we should be able to solve it into numerical results because even though the problem hides two values from us e g the value of the source E and of the resistor R1 it gives us in exchange two answers e g the equivalent resistance Rr and the current I3 that we can use to solve for the unknowns Since this circuit is structurally identical to B11 s Example 5 7 we will use the same names for the nodes The circuit description is identical except for the element s values As values for the elements in the circuit Symbulator will accept numbers variables or even algebraic expressions For this example chose to use e for the value of source e and r1 for the value of the r1 resistor Make sure that the variables are empty either by emptying the current folder or by deleting them from the calculator s memory thus DelVar e r1 Set the SI flag true s si Below is how described this circuit pass along this description to Symbulator as a string and store it in a variable e 1 0 0 r1 1 2 r1 r2 2 3 4k r3 3 0 6k cir Giving a symb
41. consumed is stored in pe12 the power delivered is the negative of that and can be found via pe12 where is the negative sign not the subtraction operator For sources the choice to store the consumed power instead of the delivered powered may seem odd it is such for consistency for all elements the power given is the consumed power e The equivalent resistance of the rest of the circuit as seen by a source is stored in a variable called r and the name of the source For example the equivalent resistance of a circuit as seen from a source called e2 is stored in variable re2 106 Illustrative examples Solving a numerical circuit Exemplum docet Let s dive right in and solve a numerical linear circuit in direct current using the s dc gate A circuit is numerical if we know the numerical value of every element in the circuit have chosen Example 5 7 from Boylestad s Introductory Circuit Analysis 11e Moving forward will refer to that textbook as B11 For your benefit the problem statement and the circuit schematic were scanned and are reproduced below exactly as they appear in the textbook B11 s Example 5 7 Determine the total resistance R7 Calculate the current Z Determine the voltage across each resistor Find the power supplied by the battery Determine the power dissipated by each resistor Comment on whether the total power supplied equals the total power dissipated Pr I R R p Q 3kQ
42. ctice Problem 9 5 Find the Norton equivalent external to R in the circuit Solve for the current when R 0 10 kQ 50 kQ and 100 KQ R 2 R3 A Answers Ry 42 KQ Iy 1 00 mA For R 0 J 1 00 mA For R 10 kQ J 0 808 mA For R 50 kQ Ij 0 457 mA For Rp 100 kQ J 0 296 mA My solution below true gt s si s th e1 1 0 35 r1 1 2 15k r2 2 3 60k e2 3 0 70 r3 2 4 30k 4 0 ino reg irL L 0 irL L 1E4 irL L 5E4 irL L 1E5 The answer 001 42000 001 000808 000457 00296 is correct Power transfer problems Another type of problem that is often associated to the Th venin Norton equivalents th saves the is that regarding the power transfer to a load particularly the maximum power maximum transfer possible Maximum power is transfered when the load Ri is equal to the Reg of power in pMax the circuit s equivalent Symbulator stores in variable pMax the maximum power that and the power can be delivered by a Th venin Norton equivalent circuit and also stores in prL the as a function of power transferred to the load as a function of its value L Let s see some examples load L in prL AS2 s Example 4 13 Find the R value for maximum power transfer and the maximum power transferred 161 R for maximum transfer is rEq The maximum power is in pMax My solution below 1n e 1 0 12 r6 1 2 5 12 2 0 12 53 2 3 3 j 0 3 2 r2 3 4 2 4 0 approx req pmax The answer 9 13 44
43. e an access program or gate called s dc which takes one argument the circuit description in string form Below you will find plenty of examples of DC simulations recommend that you make MAIN your current folder and empty it before each simulation Symbulator runs faster and more accurately if you use integers and fractions in your element values instead of decimals It is preferable to simulate with exact numbers and then evaluate the answers using e g ENTER In this book when we say that we evaluate an expression what we mean is that we use either or as required Ok let s get right into it 104 Elements e andr Voltage source and resistors Describing a resistor In Symbulator an ideal resistor is described as follows first the name of the resistor This is how which must start with the letter r coma second the name of the first node of the resistors are resistor another coma third the name of the second node of the resistor another described in coma and fourth the value of the resistor in Q There is no coma at the end Symbulator For example an ideal resistor called r1 connected between nodes a and b with a When the ssi resistance of 3000 would be described thus r1 a 6 300 Since the value of resistors variable is true is often given in kilo ohms Symbulator will interpret any k in the value of a resistor as Symbulator an indication that the values are in kQ as long as the s si variabl
44. e is set to true For interprets any example a value of 8kO can be entered in many ways 8000 8k and 8E3 are k in the value equivalent The only difference is that the first will be treated as exact values while of the resistor the others will be treated as approximate to mean kilo e g x1000 Simulation answers After the simulation in DC is complete Symbulator stores a series of answers in the calculator s memory labeled with easy to remember names for your convenience e The voltage of each of its nodes with reference to the ground is stored in a variable called v and the name of the node For example for a node called 1 its voltage with reference to ground is calculated and stored in a variable called v1 e The voltage drop in the resistor that is to say the voltage in the first node minus the voltage in the second node is stored in a variable called v and the name of the resistor For example for a resistor called r5 the voltage drop is stored in vr5 e The current through the resistor flowing from the first node towards the second is stored in a variable called i and the name of the resistor l e the current flowing through resistor rx from the first node towards the second is stored in irx e The power consumed by the resistor is stored in a variable called p and the name of the resistor For example the power consumed by resistor r12 is stored in pr12 What about conductances As an ideal circuit element a conducta
45. e move on Practice problems B11 s Example 8 15 Determine the current through each resistor My solution below direction in blue s dc j6 0 1 6 r2 1 0 2 r6 1 2 6 r8 0 2 8 j8 2 0 8 approx ir2 ir6 ir8 We get these answers 1 25 4 75 3 25 So In is 1 25A Irs is 4 75A and lrg is 3 25A B11 s Example 8 21 125 Determine the voltage in each node and the current through each resistor My solution below s dc j1 0 1 4 r1 1 0 2 r3 1 2 12 r2 0 2 6 I12 0 0 2 0V 1 V2 Ir 1 2 r9 We get the following answers 6 6 3 1 1 So V126V V2 6V IrRi 3A and lg7lIn371A Bo2 s Drill Exercise 2 2 Conductances Given the circuit below determine the voltages in the nodes 2S Below my solution s ide r10 1 0 T 39 112 1 2 1 2 013 1 9 172 293 2 9 1 8 r20 2 0 1 8 j12 1 2 17 j03 0 3 2 approx v1 v2 v3 The answer 2 1 5 indicates v1 2V v2 1V v3 0 5V This is correct HK5 s Figure 1 24b Expert Determine ix and v in the following circuit My solution below s ex j6 0 1 6 r5 1 2 5 r2 2 0 2 r1 1 3 1 r3 0 4 3 10 2 3 10 r 3 4 rx ir1 vr Select DC Add equation ir2 4 Add unknown rx Run the simulation The answer 8 80 means that lx is 8A and that Vy is 80V 126 HK5 s Example 2 2 Conductances Determine the voltages in the nodes My solution below s dc j01 0 1 8 j30 3 0 25 j21 2 1 3 r12 1 2 1 3 r23 2 3 1 2 r13 1 3 1 4 r20 2 0 1 r
46. en today fifteen years later know of no other calculator based simulator that can do this TR5 Example 4 7 Symbolic Find Rin e g the resistance as seen by the current source My solution below lIN AJ 141 Ss dc ji 0 a is re a 0 re jd b a B is rl o O rl rji The answer we get re B 1 is correct as can be seen by comparing it to the textbook s answer shown below Rw B 1 Rg TR5 s Example 4 5 Symbolic Find ip My solution is shown below Be patient This simulation takes more than one minute s dc e1 1 0 vcc rb 1 b ro e2 e b vy re e 0 re rrc 1 c rrc j c e B irb irb Compare my answer left to the book s answer right ULC ug L Y Ys rh re 8 1 P Rg B DR 142 Element s Short circuits Describing an ideal short circuit Shorts are used mostly when we need to find out a current in a part of the circuit where there is no element already Otherwise we would just define it as a single node In Symbulator an ideal short circuit is described as follows first the name of the short which must start with the letter s and a coma second the name of the first node another coma and third the name of the second node There is no coma at the end For example a short circuit called s1 connected between nodes 3 and 5 e g current flowing through it from node 3 towards node 5 would be described thus 1 3 5 Simulation answers No power is consumed
47. er input node another coma and Symbulator fourth the name of the output node The polarity of the input nodes does not matter An ideal op amp called 01 whose input nodes are 2 and 3 and whose output node is node 5 would be described thus 01 2 3 5 And an ideal op amp called o whose input nodes are p and n and whose output node is also called o would be o p n o Simulation answers For each ideal op amp in a circuit Symbulator will store the following answers e The voltage of each of its nodes with reference to the ground is stored in a variable called v and the name of the node For example for node o the voltage is in vo e Thecurrent through the output node flowing from the output node outwards For example the current leaving the output node of an op amp called o is stored in io e The power consumed by the visible part of the op amp is stored in a variable called p and the name of the op amp For example for an op amp called 02 the power consumed is po2 the power delivered is the negative of that po2 A node by any other name Since was born to be bad like to play with how name nodes in order to get the answers as close to the book as possible So do not be surprised if call one op amp o and then also name its output node o That way can will ask for vo and io and get the output voltage and current But realize this in asking for vo we are asking for the voltage in node o And in asking for io t
48. ercise 4 2 Find Vo and io in terms of is Vy 500 Q This is my solution The simulation took 14 seconds true gt s si s dc ji 0 x is r1 x 0 1k r2 x 0 2k jd 0 0 vx 500 ro 0 0 500 vo iro The answer 1000 is 2 is is correct Vo 1000 is and io 2 is TR5 s Example 4 4 Find vo and the equivalent resistance Rin in terms of vs when Ri is 50 Rz is 1k R3 is 100 R4 is 5k and g is 100mA e g 100m Ri R gt lO This is my solution The simulation took 18 seconds true gt s si s dc e 1 0 vs r1 1 2 50 r2 2 0 1k r3 0 0 100 r4 0 0 5k j 0 0 100m vr2 approx vo re The answer 904 vs 10951 is correct Vo 904 vs and Rjy210 95KQ 139 140 TR5 s Figure 4 4 Find the voltage drop current and power consumed by the 5000 resistor and the ratio of that power to that delivered by the independent source all in terms of is have not labeled the nodes in the figure so you can practice doing it My solution s dc js 0 1 is r50 1 0 50 rx 1 0 25 jd 0 0 48irx r3 0 0 300 r0 0 0 500 iro vo pro pro pjs The answers we get are correct io 12is vo 6000is po 72000is and po ps 4320 TR5 s Example 4 1 Symbolic Find vo did not label the nodes so you can practice Remember not using rc it is reserved s dc ei 1 0 vs rs 1 2 rs rx 2 0 rp ed 0 3 r irx rrc 3 o0 rrc rl o 0 rl vo The answer shown left is correct The textbook s answer is shown right rl vs r
49. finite current when short circuited So the Th venin equivalent is given by Vra and no resistance or Rea OQ Evaluating vth results in r1 r2 vs r1 This is correct as can be seen by comparing it to the book s answer shown below R3 rA 5 eye The Thevenin resistance as explained above is OQ AS2 s Example 5 2 Find Vo and io lo I 20 KQ true gt s si s de e 2 0 1 r5 1 0 5k r4 1 0 40k r2 0 0 20k 0 2 1 0 vo io The answer 9 00065 is correct vo is 9V and i is 0 65mA Bo2 s Figure 3 3 Inverting Find the gain of the overall circuit vo vs sidc e 2 0 vs r1 2 1 r1 r2 1 0 r2 0 1 0 0 vo vs We get r2 r1 which is correct as can be seen in the book s answer above AS2 s Figure 5 10 Inverting Find Vo s dc e 2 0 vi r1 2 1 r1 rf 1 0 rf 0 0 1 0 vo This problem is almost identical to the one above The answer we get is correct 167 AS2 s Example 5 3 Inverting If vi is O 5V calculate the output voltage vo and the current in the 10kQ resistor 25 kQ true gt s si s dc e 2 0 5 r1 2 1 10K rf 1 0 25k 0 0 1 0 vo ir1 The answer 1 25 5 E 5 is correct TR5 s Exercise 4 11 Inverting Find Vo when vs is 2V 4V and 6V Notice the output of the op amp is limited to 15V 4 10kQ 2 33 kD e This may be the only non linear problem you will see in this book because solved it before I realized it included the 15V cons
50. g 7 18A and las 2 41A These are the correct answers We could get this answer in a single line command s dc e1 1 0 15 r1 1 a 4 e 3 3 0 20 r3 3 a 10 e2 0 2 40 r2 a 2 5 approx irl ir2 ir3 Moving forward we will often use this single line approach for getting our answers B11 s Example 7 6 117 My solution below s dc e 1 0 24 r1 1 2 6 r2 1 2 6 r3 1 2 2 r4 2 0 8 r5 2 0 12 Answer ie is ls 4 A ir2 is l22 8 A ir4 is 1422 4 A vr 1 is Vi 4 8 V vr5 is V5 19 2 V B11 s Example 7 11 V a Determine voltages V V and V a a b Find voltages V and V g c Find current X 100 d Find the source current ZNNSV R i V Ro Ve E 20 V o me o h 40 m c R 3 NU l N 52 8V Oo A My solution below s dc e1 a 0 20 e2 a 6 5 e3 c 0 8 r1 a c 10 r2 6 c 4 r3 6 0 5 Answers va 20 vo 15 vc28 va vc 12 vb vc 7 ir2 1 75 ls via ie3 2 95 B11 s Example 8 24 Find the voltage drop in the 3Q resistor My solution below s dc e8 1 0 8 r2 1 2 2 r4 2 0 4 r6 2 3 6 r3 3 0 3 r10 3 4 10 e1 0 4 1 Now evaluating vr3 via approx vr3 finds that Vaso is 1 1 V This is correct B11 s Example 8 18 Find the current through the 10 resistor in the network shown below 4 100 lion 43 2 My solution below s dc e15 1 0 15 r10 1 2 10 r8 1 3 8 r5 3 2 5 r3 3 0 3 r2 2 0 2 approx ir10 We find that ir10 1 22 A B11 s Example 8 26 Find the voltage drop in the 2Q resist
51. he o stands for the element o When deciding the name of nodes however you should be aware of the fact that for each node Symbulator will create a variable called v to store the voltage of that node If you are not careful you may inadvertently create a problem for Symbulator Particularly you should avoid describing a source thus e 0 v where is anything For example if you describe a source as e1 1 0 v1 Symbulator will define the voltage of node 1 namely v1 as having the value you provided v1 The resulting equation v1 v1 is discarded by the calculator which leaves Symbulator one equation short Read this wearing amber sunglasses and playing in your mind the beginning of the CSI Miami theme 165 Solved problems Bo2 s Drill Exercise 3 2 For the op amp circuit shown find vo and the power absorbed by the op amp 10 KQ true gt s si s dc e 1 0 1 r12 1 2 1k r20 2 0 10kK r30 3 0 1k r30 3 0 20k 0 3 2 0 vo po 166 The answer 2 1 00063 is correct The voltage in node o is 2 1V and the op amp is absorbing 63mW which is to say it is delivering 63mW of power to the circuit Bo2 s Drill Exercise 3 11 Th venin Find the Th venin equivalent s th e 1 0 vs r1 2 0 r1 r2 2 3 r2 0 1 2 3 3 0 The th script tells us it found the Th venin voltage but could not find the Norton current This is not a surprise since an ideal op amp has zero output resistance and a fixed voltage an in
52. he values of two variables irl and vr1 The calculator returns 01 200 meaning a 10mA current and a 200V voltage drop An example with conductances The following example is taken from the textbook Elementary Linear Circuit Analysis 2ed by Leonard S Bobrow From this point forward will refer to this book as Bo2 Bo2 s Example 2 2 Given the circuit below determine the voltages in the nodes 2S U PEN 2A 4S 3S As explained before in Symbulator all conductances are simulated as resistors So the 4 siemens conductance becomes a 1 4 resistor and so on Below my description s dc j10 1 0 2 r12 1 2 1 r20 2 0 1 4 r30 3 0 1 3 r13 1 3 1 2 32 3 2 3 approxt V1 V2 V3 The answer 1 3 34 1 12 indicates vi1 1 3V v2 34V v3 1 12V This is correct 124 An example with E J and R B11 s Example 8 2 Determine the values of Vs l and l2 In the line below we concatenate three commands using colons The first stores the circuit s definition in a variable The second asks Symbulator to run a DC simulation of the circuit described in that variable The third asks the calculator to provide us the values of three variables that given the circuit description answer the questions s dc j 0 1 7 e 1 0 12 r 1 0 4 v1 ie ir The calculator returns 12 4 3 meaning Vs is 12V l is 4A and lz is 3A These are the correct answers We will continue to use the single line instruction as w
53. hree equations gt In the expert mode we avoid using as values to be solved any variable between r1 and r99 which are reserved variables We also avoid using an element s name as its symbolic value 121 122 ir128m and ir2210m and ir322m Add these unknowns e rr2 rr3 Press Enter and Enter When Symbulator is Done find Is and E by asking approx ie1 e We get that Is is 0 2 A and E is 16 V B11 s Example 7 12 Expert Determine R R2 and Rs for the voltage divider supply Can 2W resistors be used mA This problem is also perfect for the expert mode because a the target is to obtain numerical values and b we have N unknown element values and in turn we are given N numerical answers Here is how solved it true2s si svex e a c 72 r1 a b rr1 r2 b 0 rr2 r3 0 c rr3 rl1 a 0 rrl1 r12 b 0 rr12 Choose DC Add these five equations irl1220m and vrl1260 and irl2 10m and vrl2 20 and ie 50m Add these five variables to the list of first level variables red tie frost ITI Enter When Done evaluate these variables to find the answers rr1 1 33 KO rr2 1 KO rr3 240 Q These are correct Since all powers consumed in the resistors prr1 prr2 and prr3 are smaller than 2W it is possible to use 2W resistors in the design Element J Current source Describing an ideal current source In Symbulator an ideal current source is described as follows first the name of the source which must
54. imulation in mid air and give Symbulator extra information Had this circuit been larger the benefit of the expert mode in computation time would be clear Examples to practice this type of problems are found starting in page 121 Practice problems Numerical problems The following are practice problems taken from several textbooks They were chosen because they apply only the concepts that you have learned so far This will allow you to practice these concepts and reinforce them before moving on to new ones The problem below comes from Figure 1 26 a in Hyat and Kemmerly s Engineering Circuit Analysis bed Moving forward we will refer to that textbook as HK5 We know the value for Rr and the value for Is in terms of the variables listed in the prompt because Rr is given by re which is a function of e and ie and Is is given by ir3 which is a function of ie HK5 s Figure 1 26 We are asked to determine the current voltage drop and power consumed in each resistor as well as the power delivered by each voltage source We are also asked to check that the powers in the circuit add up to zero Here is my solution 30 Q 30V p named the nodes thus the bottom node is named 0 the top nodes from left to right are named 1 2 and 3 My description of the circuit is given below followed after a colon by the DC simulation command s dc e1 1 0 120 r1 1 2 30 e2 2 3 30 r2 3 0 15 When the simulation is done you c
55. ind the Norton equivalent of the circuit below After trial and error we use x2 and x3 instead of 2 and 3 in the dependent sources s th ei 1 b 6 ed b 0 x2 ir2 r1 1 0 6 r2 0 a 2 jd a 0 x3 ir1 a b 156 Exploring the answers we see that the denominator of the expression for req x2 x3 6 is zero which results in a division by zero Via2 gt x2 3x3 ino req we find that Ino 3A and Req is undefined or for practical purposes infinite TR5 s Exercise 4 6 Symbolic Find the input resistance and output Th venin equivalent circuit of the circuit Ry 1 p Re T wri R4 Ro c 4 Il Rin vy RT First let s find the input resistance Rin e g the resistance as seen by the vs source We want to use p for the constant in the dependent source so we set the s si flag to false to avoid confusion with the SI unit for micro false gt s si s dc el 3 0 vs rf 3 2 rf ro 2 t ro ed 2 0 u vrt rel We get rf u 1 which is correct The textbook s answers are shown right of the circuit schematic Now we find the output Th venin equivalent circuit as seen by R s th ei 3 0 vs rf 3 2 rf ro 2 t ro ed 2 0 u vrf t 0 vth req We get vs u u 1 ro which is correct as can be seen in the textbook s answers for vr and Rr shown right of the circuit schematic above TR5 s Example 4 8 Symbolic Find the Th venin equivalent as seen by the load Load I ee we we Ow wr wr Or re ee
56. ir13 we get a value of 2 5A for current i The answers were easily found because there was a source connected between the two desired nodes But how can we find the equivalent resistance of a passive circuit that is to say a circuit that has no independent source in it One way to find this in Symbulator is to connect a 1A current source between the two nodes where we want to find the equivalent resistance and then read the voltage drop across the source An easier way is to run the er script which does the same thing for you automatically Script er The er script finds the equivalent resistance of a passive circuit and stores it in the case of a DC analysis in a variable called req The er script takes three arguments the first is the circuit description in the form of a string the second and third are the two nodes between which we need to find the equivalent resistance Let s learn by doing Example for a resistive circuit B11 s Example 8 29 Calculate the equivalent resistance of the circuit shown below Let me solve this problem step by step since it is the first time you see the er script After label the nodes describe the circuit In this case store it in a variable r4 0 a 4 r2 0 b 2 r6 a b 6 rb a c 3 ra b c 3 cir Run the er script giving it as arguments the circuit and the nodes s er cir 0 c When prompted choose DC as analysis type Once it is done evaluate approx req The value is 2 89
57. lic values or need symbolic results When compared to SPICE PSpice or Electronic Workbench this small software offers a simpler way to define dependent sources and includes special elements such as ideal transformers and six different two ports usually not included in other simulators When is it most useful Symbulator is most useful with linear circuits with symbolic values such as you find in undergraduate courses in circuit analysis Students of Circuits and Il may benefit a lot from this pocket sized circuit wizard for the numerical or symbolic solution of linear circuits It can help you with learning circuits theory doing your homework or taking that final exam It is not however a replacement for your brain or an excuse to not study your circuit theory classes you must understand circuits theory to use it Symbulator is a linear circuit simulator and does not simulate non linear elements such as diodes and transistors 103 Get the current versions of Symbulator and DiffEq online Install the programs so they run faster s dc is the gate to run a DC simulation How much does it cost Nothing It s free Has always been will always be You are welcome Furthermore since June 2013 Symbulator is offered under a Creative Commons BY NC SA license If you want to port it to other platforms you are free to do so under the CC license above look forward to ports to the TI Nspire platform and to Mathematica
58. ly the answer from the book R R E Vo Up VUE R 2 e vi If you are ever in doubt about whether two expressions are the same enter them both separately into the calculator and then ask the calculator to compare them using the equality sign If the answer is true then they are the same That is it for direct current DC analysis of linear circuits using Symbulator and marks the end of Part 1 181
59. nce is not different from a resistor they are the same element with the only difference that their values are given in different terms Resistance values are given in ohms and conductance values are given in siemens A conductance value can be converted into an ohms value by dividing 1 over the conductance in siemens Thus for example an 8S conductance between nodes a and b could be described thus as a 1 80 resistor r1 a b 1 8 105 Describing an ideal voltage source This is how In Symbulator an ideal voltage source is described as follows first the name of the voltage source which must start with the letter e a coma second the name of the positive sources are node of the source another coma third the name of the negative node of the source described another coma and fourth the value of the source in volts No coma at the end Symbulator For example an ideal voltage source called e1 connected between nodes 3 and 0 interprets any with a voltage of 12V given as voltage of node 3 with regards to node 0 would be korm in the described thus e1 3 0 12 Since the value of voltage sources is sometimes given in kV value of a or mV Symbulator will interpret any k or m in the value of a voltage source as an voltage source indication that the values are in kV or mV respectively as long as s si is set to true to mean kilo e For example a value of 4kV can be entered as 4000 4k and 4E3 are and milli respectively if equivalent The
60. nothing at all Just write the value as a function of the circuit s answers using the variables that by now you should know well and run the simulation like it s nobody s business With Symbulator instead of fearing them you will laugh in the face of dependent sources thinking You have nothing Booyah Without any more introduction let me show you examples so you can learn by doing Numerical examples AS2 s Example 2 7 Find is and vo in the circuit below My solution below named the top note o the letter and used the bottom node as ground so that when ask for vo get the voltage in node o matching the book s variable vo Likewise name the resistor ro so that can ask for iro to get current io None of this special labeling is necessary but it s kind of cool when the variable Symbulator gives you is called the same as the one the book is asking for Chalk it up to my nerdiness Notice that 5iro is interpreted by the calculator as 5 times iro s dc ji 0 0 3 r0 0 0 4 jd 0 0 5iro iro vo The answer 6 24 is correct io 6 and vo 24 Bo2 s Example 1 9 Determine h l2 and v This is my solution The simulation took 10 seconds s dc ji 0 1 2 r1 1 0 3 jd 0 1 4v1 r2 1 0 5 ir1 v1 ir2 The answer 5 26 15 26 3 26 is correct l1 5 26 l2 3 26 and v 15 26 AS2 s Practice Problem 2 7 Find vo and io in the circuit My solution below s dc ji 0 0 6 ro 0 0 2 jd 0 0 iro 4 r
61. olic element a value equal to its name is a useful trick to remember whose value it is However due to the potential for conflicts reserved names should be avoided in the expert mode which stores all solved unknowns into variables as the third example shows Ask Symbulator to simulate this circuit in direct current DC by typing this s dc cir A moment later 16 seconds in my calculator Symbulator is Done and we are ready to answer the questions Using the symbolic answers provided by Symbulator and the known answers given by the problem we can write two equations We can then solve these two equations for the two unknowns that interest us using the calculator s solve command Let s explore what we have here The problem says that l3 is 6m A In Symbulator l5 is called ir3 e g the current through resistor r3 If you evaluate ir3 you will see it produces an algebraic expression in terms of the two unknowns e and r1 This is what we call a symbolic answer The problem also says that Rr is 12KQ As we saw in the previous problem the equivalent resistance as seen by the source e is given by re which when evaluated gives another algebraic expression in terms of r1 We can then write two new equations re 12000 Notice the kilo and ir326 1000 and solve them for the two unknowns we want e and r1 nomenclature does not apply solve re 12000 and ir3 006 e r1 puni fe arb An instant later we get the answers e 72 volts and r
62. ood as long as the load is the only thing connected to the equivalent B11 s Example 9 6 Find the Th venin equivalent circuit for the network in the shaded area Then find the current through R for R values of 20 100 and 1000 My answer below First we find the Th venin equivalent s th e1 1 0 9 r1 1 2 3 r2 2 0 6 2 0 vth req The answer 6 2 is correct Now we find the values of irL for the different values es 2 le ell 1 0 Where is the given operator The answers 1 5 5 059 are correct AS2 s Example 4 8 Find the Th venin equivalent of the circuit shown to the left of terminals a b Then find the current through RL 6 16 and 360 My solution below Sin 6 1 0 32 04 1 2 27 512 2 0 12 0 2 2 71 2 9 1 9 0 7 vth req irL L 6 irL L 16 irL L 36 The answer 30 4 3 1 5 75 is correct Bo2 s Example 3 10 Find the Norton equivalent of the circuit left of the a b terminals and then find the voltage drop and the current through the O resistor My one line solution below 1 4 s th e 3 0 3 r31 3 1 1 2 r10 1 0 1 2 r12 1 2 1 4 j 2 0 v1 2 2 0 fino req vrL L 1 4 irL L 1 4 The book gives the answers as fractions We get it right 21 8 4 9 21 50 42 25 RM3 s Example 9 7 Find the Norton equivalent of the circuit external to R Then determine the load current IL when R 0 O 2k O and 5k O My one line solution below true gt s si s th e 1 0 15 r1 1 2 6k j
63. or My solution is below s dc e 1 0 240 r1 1 2 3 r2 2 3 4 r3 3 4 1 r4 4 5 2 r5 3 5 6 r6 2 5 6 r7 5 0 9 Evaluating approx vr4 we find the voltage drop in R4 the 2Q resistor it is 10 67 V Circuits with hidden source Sometimes the schematics of circuits are presented in such a way that sources of voltage are not shown explicitly yet their voltage is provided These are what call hidden source problems Below offer two examples of these types of problems Both are taken from the textbook Circuit Analysis Theory and Practice 3ed by Allan H Robbins and Wilhelm C Miller to which from this point on we will refer as RM3 119 RM3 s Example 7 5 Hidden source Find the indicated currents and voltages li 300 In my solution below notice introduced two sources one for 12V and one for 6V s dc e1 1 0 12 r1 1 b 10 r2 b a 10 r3 a 2 50 r4 6 2 30 e2 2 0 6 approx ir1 ir2 ir4 va vb The answer 6 2 4 2 indicates l1 6 l22 2 132 4 and Vap 2 This is correct RM3 s Figure 7 16 Hidden source Find the total circuit resistance and the indicated currents and voltages 10V My solution below This is how named the nodes the top 1 the bottom 2 and a and b as in the figure In my solution below notice introduced two sources one for 10V and one for 6V true gt s si s dc r1 1 6 4k r2 1 a 3k r3 b a 2k r4 b a 3k r5 1 6 1k rt a 2 6k e1 1 0 2 e2 2 0 10 app
64. or voltage is dropped in a short For each short in a circuit Symbulator stores only the current through it flowing from the first node towards the second in a variable called i and the name of the short For example the current flowing through a short called sx from the first to second node is stored in isx An example using short circuits HK5 s Drill Problem 1 13 Find l1 l2 l3 and la My solution below define the shorts in the same direction as the arrows s dc r1 1 0 25 jd 0 2 2v1 r2 2 3 10 ji 4 3 2 5 r3 4 5 100 s1 1 2 8s2 2 4 8s3 0 3 8s4 3 5 approx is1 is2 is3 is4 The answer 2 3 8 5 is correct This answer can only be found in Symbulator using short circuits 143 This is how short circuits are described in Symbulator The er script finds the equivalent resistance of a circuit and stores it in req 144 Equivalents Equivalent resistance As we saw before Symbulator has the ability to provide the equivalent resistance of a circuit as seen from any source This allows us to solve problems like the following AS2 s Practice Problem 2 15 Find the equivalent resistance as seen by the 100V source and the value of current i 130 10 Q 100 V 50 Q s dc e a 0 100 r13 a 1 13 r24 1 2 24 r10 1 3 10 r20 2 3 20 r30 2 0 30 r50 3 0 50 By evaluating re we get the value of the equivalent resistance as seen by the source e It is 400 By evaluating
65. ource that is to say the voltage in the first node minus the voltage in the second node is stored in a variable called v and the name of the source For example for a source called j5 the voltage drop is stored in vj5 e Thecurrent through the source flowing from the first node towards the second node is stored in a variable called i and the name of the source For example the current flowing through a source called jx from its first node towards its second node is stored in ijx This may seem redundant but it is actually rather handy e The power consumed by the source is stored in a variable called p and the name of the source For example for a source called j12 the power consumed is stored in pj12 the power delivered is the negative of that and can be found via pj12 e The equivalent resistance of the rest of the circuit as seen by a source is stored in a variable called r and the name of the source For example the equivalent resistance of a circuit as seen from a source called j2 is stored in variable rj2 123 Illustrative examples A simple example J and R B11 s Example 8 1 Given the circuit below determine the current and voltage drop in Ri true2s si sidc j 0 1 10m r1 1 0 20k irld vr1 In the line above we are doing three things the first is setting the SI flag to true the second is asking Symbulator to run a DC simulation of the circuit described between quotations and the third is asking for t
66. part of the question In order to find the current through R we cannot use as we did before the variable irL because now the load is not the only thing connected to the terminals of the circuit equivalent there is also a current source The expressions stored in irL vrL and prL are only valid when a load RL is the only thing connected to the circuit equivalent So we have no other option than to run a simulation The fastest way to do this is to use the contents of the variable eqcir Inside this variable we find the following circuit description jN 0 n iNo rE n 0 rEq rL n 0 L The values for iNo and rEq are 163 164 already in the calculator s memory So we change the value of the load to 1680 and add the 180mA source flowing from node 0 to node n the nodes used by eqcir s dc jN 0 n iNo rE n 0 rEq rL n 0 168 j 0 n 180m irL The answer 06 is correct there is a current of 60mA flowing through R from O to n Using eqcir is meant to save you time If you find it confusing to use just don t use it Element o Ideal Op Amp Describing an ideal operational amplifier There are many types of operational amplifiers Symbulator can simulate the ideal This is how linear type In Symbulator an ideal Op Amp is described as follows first the name of ideal Op Amps the Op Amp which must start with the letter o and a coma second the name of an are described in input node another coma third the name of the oth
67. rox v1 v2 irt irt irl ir2 va vb The answer we get indicates that the equivalent resistance given by v1 v2 It is 7 2 KO and that 1121 11 mA h 133 mA l2 444 mA and V35 7 0 8 V This is correct 120 Symbolic problems with solve HK5 s Figure 1 24a solve Determine ix and v in the following circuit named the nodes thus bottom is O top left is 1 top right is 2 named the elements according to their value this facilitates remembering who s who in the circuit also defined the resistors nodes in the direction of the current indicated in the diagram s dc e18 1 0 18 ra 1 0 ra r6 1 0 6 r5 2 1 5 evx 2 0 vx Explore the answers Since ir5 which we know is 12A is in terms of vx we can find vx solve ir5 12 vx We get that v is 78 V which is correct Evaluating ir6 we find that i is 3 A The fact that we can find numerical answers in this problem can be quite puzzling until one realizes that ignoring the value of Ra doesn t matter due to the circuit s structure it is not needed it to answer the two questions we have been asked Symbolic problems with Expert B11 s Example 6 19 Expert a Determine the source current b Find the source voltage E This problem having three unknown element values and three known answers is a perfect candidate for the expert mode Below is my circuit description true gt s si s ex e1 1 0 e r1 1 0 2k r2 1 0 rr2 r3 1 0 rr3 Select DC Add these t
68. s Practice Problem 2 10 Find the equivalent resistance of the circuit below 20 Q s er ra0 a 0 8 rb0 6 0 0 18 9 rb1 6 1 2 r10 1 0 0 20 1 p 20 5 a b Choose DC When Done evaluate req and find the equivalent resistance is 11 O 146 AS2 s Example 2 11 Find the equivalent conductance of the circuit below 5S 2 s er r12 1 2 1 5 r10 1 0 1 6 r20 2 0 p 1 8 1 12 1 0 Choose DC Wait for Done Evaluate 1 req to find the equivalent conductance is 10 S AS2 s Practice Problem 2 11 Find the equivalent conductance of the circuit below s er r12 1 2 p 1 8 1 4 r20 2 0 p 1 2 1 12 1 6 1 0 Choose DC Wait for Done Evaluate 1 req to find the equivalent conductance is 4 S Examples with dependent sources The er script can be applied to a second type of passive circuits circuit that include resistors and dependent sources but no independent sources A circuit with no independent sources can only be reduced to an equivalent resistance not to a Th venin or Norton equivalent so the proper script to use is er The script is used in exactly the same manner as for resistive circuits making sure the dependent sources are described properly according to the conventions of Symbulator notation Below is a series of examples of passive circuits with dependent sources Bo2 s Drill Problem 3 14 Find the equivalent resistance of the circuit below 6 4 4 b s er r4 a 0 4 ri a 0 6 ji a 0 iri 2 a 0
69. start with the letter j and a coma second the name of the first This is how node another coma third the name of the second node another coma and fourth current sources the value of the source in amperes No coma at the end The value should be given in i are described in terms of the current flowing through the source from the first node towards the second Symbulator node This means that the value of the source is how much current leaves the source out of the second node and also how much current enters the source s first node When s si is An ideal current source called j1 connected between nodes 3 and O with a current of true 5A through it from node 3 towards node 0 would be described thus j1 3 0 5 Since Symbulator the value of current sources is often given in milli amps Symbulator will interpret any interprets any m in the value of a current source as an indication that the values are in mA as long as m in the value of the current source to mean milli e g s si is true For example a value of 2mA can be entered in many ways 2 1000 8m and 2E 3 The only difference is that the first will be treated as an exact value Simulation answers For each current source in a circuit Symbulator will store a series of answers in the current folder of the calculator e The voltage of each of its nodes with reference to the ground is stored in a variable called v and the name of the node e The voltage drop in the s
70. t symbolic if one or more of the element has an unknown value It is the ability to simulate symbolic circuits that gives Symbulator its name and sets it apart from other programs distinguish between two types of symbolic simulations Purely symbolic problems The first type is when the desired answers are also symbolic e g given as algebraic functions in terms of unknown values what I call a purely symbolic problem Numerical from symbolic problems The second type is when numerical answers are expected from a symbolic circuit what I call a numerical from symbolic problem This solution is possible when they 109 Symbulator has an Expert mode for faster and better results in symbolic to numeric problems Unknown values can be entered as a variable Just make sure the variable is empty 110 give us additional information about the circuit Numerical from symbolic problems can be solved in two ways in Symbulator e f they are simple e g if there s just one or two unknown values in the circuit and we are only asked for one or two numerical answers we can simulate the circuit symbolically and then solve for the numerical answers using solve e fthey are not simple e g if there are many unknown values in the circuit or we are asked for many numerical answers it is easier and more time efficient to use the Expert mode of Symbulator whose gate is s ex cir Let s see an example of each approach to solving numeri
71. traint But anyway here it goes true gt s si s dc e 1 0 vs r1 1 2 10kK r2 2 0 33k 0 0 2 0 vo vs 2 vo vs 4 vo vs 6 The answer 6 6 13 2 19 8 is correct within the linear realm but since the output is constrained to no more than 15V or less than 15V the answer is Vo 15V for vs 6V AS2 s Practice Problem 5 3 Inverting Find the output voltage of the op amp e g vo and calculate the current through the feedback resistor e g the 15kQ resistor 15 KQ 40 mV My answer below Notice we used the m for milli in the value of the voltage source true gt s si s dc e 2 0 40m r1 2 1 5k rf 1 0 15k 0 0 1 0 approx vo irf The answer 12 8 E 6 is correct 168 AS2 s Example 5 4 Inverting Determine Vo true gt s si s dc e6 1 0 6 r1 1 a 20k e2 60 0 2 rf a o 40k 0 a b 0 vo The answer 6 is correct TR5 s Example 4 14 Inverting Find the input output relationship of the circuit below a ees 9 m L gy The formal way to symbulate this circuit would be as described below s dc e 1 0 vs r1 1 b r1 r2 6 0 r2 r3 b a r3 r4 a 0 r4 0 0 a o rl o 0 rl However since we are only interested in the ratio of the input to the output a smarter and faster as in 49 seconds instead of 55 seconds way to symbulate it is this sidc e 1 0 1 r1 1 b r1 r2 b 0 r2 r3 b a r3 r4 a 0 r4 0 0 a 0 rl1 0 0 1 When we evaluate vo v1 both approaches get the same answer shown left below r
72. uced to a Th venin or Norton equivalent One way to find the Th venin or Norton equivalent using Symbulator is to run a first simulation of the circuit to find out the voltage between the two nodes where we want to find the equivalent this is the Th venin voltage or Vr and then to run a second simulation with a short circuit connected between these two nodes to find out the circuit running through it this is the Norton current or Ino We can then get the equivalent resistance or Rea by dividing Vru Ino Notice that Rea is often called the Th venin resistance An easier way to find the Th venin and Norton equivalent of a circuit in Symbulator is to run the th script which does exactly the same thing described above automatically Script th The th script finds the Th venin and Norton equivalents of an active circuit It takes The th script three arguments the circuit description the first node and the second node With th finds the Thevenin and e the value of the Th venin voltage Vra is stored in vth Norton e the value of the Norton current Ino is stored in ino and equivalents of a e the value of equivalent resistance Req is stored in req circuit It stores the answers in variables as shown left th also stores other useful expressions into variables which we will discuss later RM3 s Practice Problem 9 4 Find the Th venin and Norton equivalents of the circuit below 149 150 s th e 1 0 3
73. ymbulator has a way to help you reduce two resistors connected in parallel to their equivalent It works as part of circuit descriptions and is typed thus p When you do not need to know the current through or power consumed in each individual resistor you can reduce parallel resistors using p inside your circuit description Examples As a first example here is a simulation in which it makes sense to use p In B11 s Example 7 4 shown in page 116 it makes sense to reduce R and Rs to an equivalent resistor since we do not need to know their individual currents or power use s dc e 1 0 16 8 r1 1 2 9 r2 1 2 6 r3 2 3 4 re 3 0 p 6 3 r6 2 0 3 131 For Symbulator dependent sources are just sources They require no special notation 132 provides the right answers e g ls is ir3 1 A and voltage drop in area B is v2 6V The second example is of a simulation in which using p makes no sense In B11 s Example 8 3 shown in page 130 it makes no sense to reduce Ri and Rz because you need to know the value of the current through R1 As third example here is a simulation where you can reduce part of the resistors using p In B11 s Example 6 22 shown in page 130 we must leave R alone because we need to know the current through it but we can reduce R2 and Rs to their equivalent using p since we do not need to know their individual currents or power use true gt s si s dc jt 0 1 12m r1 1 0 1k re 1

Part 1: DC Analysis

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