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V b = E Rx Rx+Ry V b = 15V 2.222kΩ 2.222kΩ+1 kΩ V b ≈10.34V The second sub-circuit will use the 6 volt source. Therefore, the 15 volt source will be replaced with its ideal internal resistance, a short. This new circuit is shown in Figure 6.13. The current directions are as follows: current exits the source and travels through the 4 kΩ producing a voltage drop + to − from right to left. At node b the current splits. Now the 1 kΩ and 5 kΩ are in parallel so they both see the same voltage. Some of the current flows down through the 5 kΩ resistor producing a voltage drop + to − from top to bottom, just as it did in the first sub-circuit. The remainder of the current flows to the left, through the 1 kΩ, producing a voltage drop + to − from right to left. Once again, Vb can be determined via voltage divider, this time between the 4 kΩ and the parallel combo of the 5 kΩ and 1 kΩ is approximately 833.3 Ω. V b = E Rx Rx+Ry V b = 6V 833.3Ω 833.3Ω+4 kΩ V b ≈1.034V Both sub-circuits show polarities of + to − from top to bottom for Vb, therefore these two voltages simply add, for a total of approximately 11.37 volts. This agrees nicely with the result obtained using source conversions. The slight deviation in the final digit is no doubt due to the rounding of intermediate values, such as resistor combinations. It is instructive to pursue this example further and investigate a current. In the source conversion version it was discovered that the current through the 1 kΩ resistor is 3.62 mA and flowing left to right. Using superposition, the corresponding current in the first sub-circuit can be found using KVL and Ohm's law, (15 V − 10.34 V)/ 1 kΩ, or 4.66 mA flowing left to right. In the second sub-circuit, this current can be obtained using Ohm's law alone, 1.034 V / 1 kΩ, or 1.034 mA flowing right to left. As this current opposes the current from the first sub-circuit, the two values are subtracting leaving approximately 3.63 mA flowing left to right, agreeing with the original answer within rounding error. 179 Figure 6.13 Second sub-circuit for the circuit of Figure 6.11. | DCElectricalCircuitAnalysis_Page_179_Chunk1101 |
Computer Simulation Computer Simulation For further verification, the circuit of Example 6.4 is entered into a simulator as shown in Figure 6.14. A DC operating point analysis is performed, as shown in Figure 6.15. The voltage at node 2, which is Vb in the original circuit, agrees nicely with the values computed previously. This exploration shows that the superposition technique offers a fairly straightforward method of solving a variety of multi-source series-parallel circuits. It is not without its limitations, though. First, if a circuit has a large number of sources, superposition can become a bit tedious as it requires as many sub-circuits as there are sources, and every one of these sub-circuits requires separate solutions. Second, there are some series-parallel configurations that superposition along with series- parallel analysis techniques will not solve. We will investigate methods of dealing with these issues later in this chapter and in the following chapter. 180 Figure 6.14 The circuit of Example 6.4 in a simulator. Figure 6.15 Simulation results for the circuit of Example 6.4. | DCElectricalCircuitAnalysis_Page_180_Chunk1102 |
6.4 Thévenin's Theorem 6.4 Thévenin's Theorem Thévenin's theorem, named after Léon Charles Thévenin, is a powerful analysis tool. For DC, it states: Any single port linear network can be reduced to a simple voltage source, Eth, in series with an internal resistance, Rth. An example is shown in Figure 6.16. The phrase “single port network” means that the circuit is cut in such a way that only two connections exist to the remainder of the circuit (two connection points makes up one port). The remainder of the circuit may be a single component or a large multi-component sub-circuit. As there are many ways to cut a typical circuit, there are many possible Thévenin equivalents. The important thing is that there are only two connection points between the two portions of the circuit and neither point has to be ground. Consider the circuit shown in Figure 6.17. Suppose we cut the circuit immediately to the left of R4. That is, we will find the Thévenin equivalent that drives R4. The first step is to make the cut, removing the remainder of the circuit (in this case, just R4). We then determine the open circuit output voltage. This is the maximum voltage that could appear between the cut points and is called the Thévenin voltage, Eth. This is shown in Figure 6.18. In a circuit such as this, basic series-parallel analysis may be used to find Eth. This process turns out to be quite straightforward in this particular circuit. Due to the open, no current flows through R3, thus no voltage is developed across R3, and therefore Eth must equal the voltage developed across R2 which may be obtained via a voltage divider with resistor R1 and source E. The second part is finding the Thévenin resistance, Rth. Beginning with the “cut” circuit, replace all sources with their ideal internal resistance (thus shorting voltage sources and opening current sources). From the perspective of the cut point, look back into the circuit and simplify by making appropriate series and parallel combinations to determine its equivalent resistance. This is shown in Figure 6.19. Looking in from where the cut was made (right-to-left here), we find that R1 and R2 are in parallel, and this combination is then in series with R3. Thus, Rth = R3 + (R1 || R2). A common error is to determine the equivalent resistance that the source drives. Remember, everything is determined from the vantage point of the cut or port. Thévenin equivalents are not limited to single source circuits. It is possible to find the equivalent of a network with several sources. Finding the open circuit output voltage will undoubtedly require some extra work, for example the use of superposition of source conversions. Finding the Thévenin resistance is unchanged, just remember to replace every source with its ideal internal resistance before simplifying the network. 181 Figure 6.16 Thévenin equivalent circuit. Figure 6.17 Example circuit. Figure 6.18 Finding Eth.. | DCElectricalCircuitAnalysis_Page_181_Chunk1103 |
As noted earlier, the original circuit could be cut in a number of different ways. We might, for example, want to determine the Thévenin equivalent that drives R2 in the circuit of Figure 6.17. This cut appears in Figure 6.20. Clearly, this will result in different values for both Eth and Rth. For example, Rth is now R1 || (R3 + R4). Finding Thévenin Equivalents in Lab Finding Thévenin Equivalents in Lab The procedure for experimentally determining an equivalent in the laboratory mimics the analytical approach. The first step is to figuratively cut the circuit and isolate the section that is to be converted. At this point we have the open circuit version of the circuit and all we need do is connect a multimeter to the open port to determine Eth. There are two methods for determining Rth. The first method works well for simple circuits that use only resistors and current and/or voltage sources. The second method is more generally applicable and will work for circuits with active devices such as transistors. To use the first method, the sources are physically removed from the circuit and replaced with their ideal internal resistance. Thus voltage sources are replaced with a shorting wire and current sources are left as opens. Do not simply place shorting wires across the terminals of voltage sources as doing so will cause an overload and potentially damage the sources. Once this is completed, a multimeter is placed at the open port and set to read resistance. The indicated value is Rth. The second method uses a variable resistance, namely either a rheostat or a decade box, and exploits the voltage divider rule. In this version, the sources are left active (powered up) and are not replaced with opens or shorts. Once Eth is measured, the variable resistor is placed across the port connections. This resistance is adjusted until the port voltage drops to exactly half of Eth. In the equivalent circuit, there are only two resistances of concern, Rth and this variable load resistance, and they are in series. Consequently, if the load voltage is now half of the open circuit voltage (Eth), then the other half of the voltage must be dropping across the equivalent internal resistance (Rth). For this to be true in a series circuit, the two resistances must have the same value. Thus, we simply remove the rheostat from the circuit and use a multimeter to determine its precise value. If a decade box is used instead, the value is determined directly by reading the knob settings. 182 Figure 6.19 Finding Rth.. Figure 6.20 An alternate cut or port location. | DCElectricalCircuitAnalysis_Page_182_Chunk1104 |
Whether determined analytically or empirically, the Thévenin equivalent circuit can replace the original single port network regardless of what the original was connected to. The same voltages and currents will be seen in this other portion, and it won't matter if the other portion is comprised of a single resistor, multiple resistors, multiple resistors and multiple sources, or even multiple resistors and sources wired to a selection of piquant cheeses. The Thévenin equivalent is a true functional equivalent and can be used on any linear bilateral network. Example 6.5 Determine the Thévenin equivalent of the circuit driving the 1 kΩ in Figure 6.21. Verify that the equivalent produces the same voltage across this resistor as the original circuit. First, we'll redraw the circuit showing the portion to be Théveninized, as shown in Figure 6.22. The open circuit output voltage will be the voltage across the 800 Ω resistor as there will be no voltage drop across the 200 Ω resistor (as no current flows through it into the open). This is found via a simple voltage divider. Eth = E Rx Rx+R y Eth = 10 V 800Ω 800Ω+100Ω Eth ≈8.889V The equivalent resistance is found by shorting the voltage source and then simplifying the circuit. The result is 200 Ω in series with the parallel combination of the 100 Ω and 800 Ω. 200 + 100 || 800 ≈ 288.89 Ω. To determine Vc, we can use a voltage divider between the 1 kΩ and the 288.89 Ω along with the equivalent source voltage of 8.889 volts. V c = Eth RL RL+Rth V c = 8.889V 1k Ω 1 kΩ+288.89Ω V c ≈6.897V Now let's determine Vc in the original circuit using series-parallel analysis techniques. Perhaps the quickest approach is a pair of voltage dividers. Vb is found via a divider between the parallel combination of the 1 kΩ + 200 Ω and the 800 Ω, against the 100 Ω. Vc is is then found using that voltage with a divider between the 1 kΩ and 200 Ω. Using this approach, Vb is approximately 8.276 volts and Vc is approximately 6.897 volts, providing an excellent match. 183 Figure 6.21 Circuit for Example 6.5. Figure 6.22 Finding Eth. | DCElectricalCircuitAnalysis_Page_183_Chunk1105 |
It might seem that the Thévenin method is the “long way home” in this example, and it is, but it has the advantage of being more efficient if several different loads are being considered. For example, suppose we decided to determine the output voltage not just for the 1 kΩ, but for a group of a half dozen different resistors. The double voltage divider would have to be determined for each load resistor using the straight series-parallel method but only a single divider needs to be computed for each load when using the Thévenin method. The Thévenin method is also of great use when determining maximum power transfer, as we shall see a later in this chapter. Computer Simulation Computer Simulation To verify the results of Example 6.5, the original circuit is recreated in a simulator along its Thévenin equivalent, as shown in Figure 6.23. A DC operating point analysis is performed and the results are shown in Figure 6.24. Note that the voltages across the identical 1 kΩ loads (nodes 3 and 5) are virtually the same, indicating functional equivalence between the two. The slight deviation is due to the rounding of the Thévenin voltage and resistance. 184 Figure 6.23 The original circuit of Example 6.5 in a simulator along with the equivalent. | DCElectricalCircuitAnalysis_Page_184_Chunk1106 |
As a further check, the simulation is run again, but this time using an alternate load resistor value. The value chosen is the Thévenin resistance value. By matching the resistances, this should produce a 50/50 voltage divider and the load voltage should equal half of the Thévenin source voltage, or approximately 4.444 volts. That is precisely what we find, as seen in Figure 6.25. As mentioned previously, Thévenin's theorem can be applied to much more complex multi-source circuits, and the item being driven need not be just a single resistor. This is illustrated in the next example. 185 Figure 6.24 Results of the simulation showing equivalence. Figure 6.25 Results of the simulation for matched resistance. | DCElectricalCircuitAnalysis_Page_185_Chunk1107 |
Example 6.6 Determine the Thévenin equivalent of the circuit driving the resistor/voltage source combo in Figure 6.26. Verify that the equivalent produces the same voltage across this resistor as the original circuit. Determining Rth is not particularly difficult here. After shorting the 10 volt source and open the current source, it can be seen that the 1 kΩ and 3 kΩ are in parallel, yielding 750 Ω. This is in series with the 1250 Ω for a total of 2 kΩ, which is in turn in parallel with the 6 kΩ. The final result is 1.5 kΩ. Finding Eth is a little more involved. One possibility is to use a source conversion on the 10 volt source as that will leave us with two parallel current sources that may be combined directly. The converted source will be 10 V/ 1kΩ, or 10 mA, in parallel with 1 kΩ. This results in a total of 25 mA feeding upwards with 1 kΩ || 3 kΩ, or 750 Ω. We can use the current divider rule to determine the current flowing through the 1250 Ω plus 6 kΩ branch, and then use Ohm's law to find the open circuit voltage (i.e., the voltage across the 6 kΩ). I 6k = I S Rx Rx+Ry I 6k = 25 mA 750Ω 750Ω+7250Ω I 6k = 2.34375 mA Eth = I 6k×R6k Eth = 2.34375 mA×6 k Ω Eth = 14.0625V Another approach to determine this value would be to use superposition. This is left as an additional exercise. Turning our attention to the voltage produced across the 6 volt with 500 Ω sub-circuit, we have a simple series loop with Eth opposing this 6 volt source, leaving 8.0625 volts to produce a clockwise current through Rth in series with the 500 Ω (a total of 2 kΩ). That current will be 8.0625 V / 2 kΩ, or 4.03125 mA. This will produce a drop across the 500 Ω of 2.015625 volts with a polarity of + to − from top to bottom. This will add to the 6 volt source resulting in a final potential of 8.015625 volts. The results are verified via the simulation technique used previously. The node voltages for both the original and equivalent circuits are shown in Figure 6.27. 186 Figure 6.26 Circuit for Example 6.6. | DCElectricalCircuitAnalysis_Page_186_Chunk1108 |
6.5 Norton's Theorem 6.5 Norton's Theorem Norton's theorem is credited to Edward Lawry Norton. In a nutshell, Norton's theorem is the current source version of Thévenin's theorem. That is, a single port DC network can be reduced to a single current source, IN, with parallel internal resistance, RN. Indeed, once you understand one of them, the other is but a minor extension, so we will not need to spend a great deal of time here once the basics are outlined sufficiently. The process of determining the Norton equivalent current and resistance is very similar to that employed for the Thévenin equivalent. First, the Norton resistance is found the same way as is the Thévenin resistance: replace all sources with their ideal internal resistance and then perform appropriate series and parallel combinations to reduce this to a single resistance value. Consequently, the Norton and Thévenin resistance values are identical, RN = Rth. Second, instead of finding the open circuit output voltage, we find the short circuit output current. This is the Norton current. Instead of thinking in terms of a voltmeter at the opened load, we think in terms of connecting a shorting ammeter across the load. Either way, we're looking for the maximum value that can be obtained. Perhaps the most useful thing to remember here is that if we can create a Thévenin equivalent for a network then it must be possible to create a Norton equivalent. Indeed, once a Thévenin equivalent is found, a source conversion can be performed on it to yield the Norton equivalent! The opposite, of course, is also true. 187 Figure 6.27 Results of the simulation using the original and equivalent circuits. | DCElectricalCircuitAnalysis_Page_187_Chunk1109 |
6.6 Maximum Power Transfer Theorem 6.6 Maximum Power Transfer Theorem Given a simple voltage source with internal resistance, a useful question to ask is “What value of load resistance will yield the maximum amount of power in the load?” While it is not true that maximizing load power is a goal of all circuit designs, it is a goal of a portion of them and thus worth a closer look. Consider the basic circuit depicted in Figure 6.28 with source E, source internal resistance Ri and load resistance R. We would like to describe the load power in terms of the load resistance. To make the job easier, we may normalize the voltage source E to 1 volt and the source resistance Ri to 1 Ohm. By doing this, R also becomes a normalized value, that is, it no longer represents a simple resistance value but rather represents a ratio in comparison to Ri. In this way the analysis will work for any set of source values. Note that the value of E will equally scale the power in both Ri and R, so a precise value is not needed, and thus, we may as well chose 1 volt for convenience. The power in the load can be determined by using I2R where I = E / (Ri+R). Using our normalized values I = 1 / (1+R) and thus the load power is: P =( 1 1+R) 2 R or after expanding, P = R R 2+2 R+1 We now have an equation that describes the load power in terms of the load resistance. Before we go any further, take a look at what this equation tells you, in general. It is obvious that maximum power will not occur at the extremes. If R = 0 or R = ∞ (i.e., shorted or opened load) the load power is zero. While shorting the load will yield the maximum load current, it will also yield zero load voltage and thus no power. Similarly, opening the load will yield maximum load voltage but it will also yield zero load current, and again, no load power. To find the precise value that produces the maximum load power, the proof can be divided into two major portions. The first involves graphing the function and the second requires differential calculus to solve for a precise value. We shall proceed with the graphing portion which will lead us to the answer. The more rigorous proof of the second method is detailed in Appendix C. Graphing the Power Function Graphing the Power Function P = R / P = R / (R2+2R+1 R+1) This may be done in parts, looking at the contribution of each term, and then combining to form the final result. First, consider the denominator R2 + 2R + 1. It consists of three segments. The most simple is the horizontal line at +1. The 2R term creates a straight line with slope 2 starting at the origin. The R2 term creates a simple curve with increasing slope that starts at the origin, crosses the horizontal +1 line at R = 1 and also crosses the 2R line at R = 2. These items are drawn individually and 188 Figure 6.28 Defining maximum power transfer. | DCElectricalCircuitAnalysis_Page_188_Chunk1110 |
then added together as illustrated in Figure 6.29. Of course, we must remember that we really want 1 / ( R2 + 2R + 1) and so we plot the reciprocal of the combo as shown in Figure 6.30 (red curve). We also include the numerator term, R. This is shown as a straight line (blue) with a slope of 1. Finally, these two curves are multiplied together to produce the load power equation (black). 189 Normalized Components and Total 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Normalized Load Resistance 0 0.5 1 1.5 2 2R R2+2R+1 R2 1 Figure 6.29 The three terms of the power equation plotted individually and then summed. Figure 6.30 The power equation and components plotted. Normalized Components 0.0 0.5 1.0 1.5 2.0 2.5 Normalized Load Power 0 0.05 0.1 0.15 0.2 0.25 Normalized Load Resistance 0 0.5 1 1.5 2 2.5 R 1/(R2+2R+1) R/(R2+2R+1) (right axis) | DCElectricalCircuitAnalysis_Page_189_Chunk1111 |
A close examination of the power curve will show that the peak occurs at R = 1. This is easier to see if we plot the completed power curve using a logarithmic horizontal axis and also scale the vertical axis to 100%, as shown in Figure 6.31. The peak is more apparent and the curve is symmetrical in shape rather than lopsided. This reinforces the idea that the ratio of the resistances is what matters. Finally, we can state: Maximum load power will be achieved when the load resistance is equal to the internal resistance of the driving source. No other value of load resistance will produce a higher load power. For the circuit of Figure 6.28, this means that R must equal Ri. As an exercise, we can try substituting a few values around the peak to verify this. For example, given E = 1 V and Ri = 1 Ω, we calculate P for R = 0.5 Ω, 1 Ω, and 2 Ω: P0.5 = 0.5/(0.52+2∙0.5+1) = 2/9 P1 = 1/(12+2∙1+1) = 2/8 = ¼ P2 = 2/(22+2∙2+1) = 2/9 You can also try this with extremely small variations such as R = 0.9999 Ω along with R = 1.0001 Ω and you will not achieve a value equal to or greater than ¼ Watt. While matching the resistance produces the maximum load power, it does not produce maximum load current or maximum load voltage. In fact, this condition produces a load voltage and a load current that are half of their maximums. Their 190 Load Power as a Percentage of Maximum 0 20 40 60 80 100 Normalized Load Resistance 0.01 0.1 1 10 100 Figure 6.31 The load power curve with logarithmic axis showing symmetry. | DCElectricalCircuitAnalysis_Page_190_Chunk1112 |
product, however, is at the maximum. Further, efficiency at maximum load power is only 50% (i.e., only half of all generated power goes to the load with the other half being wasted internally). Values of R greater than Ri will achieve higher efficiency but at reduced load power. Sometimes we favor efficiency over maximal load power. As any linear two point network can be reduced to something like Figure 6.28 by using Thévenin's theorem, combining the two theorems allows us to determine maximum power conditions for any resistor in a complex circuit. 6.7 Delta-Y Conversions 6.7 Delta-Y Conversions Certain component configurations, such as bridged networks, cannot be reduced to a single resistance using basic series-parallel conversion techniques. One method for simplification involves converting sections into more convenient forms. The configurations in question are three-point networks containing three resistors. Due to the manner in which they drawn, they are referred to as delta networks and Y networks17. These configurations are shown in Figures 6.32. In this particular case the delta version is drawn upside down so that its terminal designations match those of the Y configuration. Alternately, if they are slightly redrawn they are known as pi (also called “π”) networks and T (also called “tee”) networks. These configurations are shown in Figure 6.33. 17 In some sources the capital Greek letter delta (Δ) is used instead of spelling out “delta” and the letter Y is spelled out as “wye”. Thus, you may come across discussion of “Δ-Y ”, “Δ-wye” or “delta-wye” networks. It's all the same stuff. 191 Figure 6.32 Delta and Y (Δ-Y) networks. Figure 6.33 Alternate form: Pi and T(π-T) networks. | DCElectricalCircuitAnalysis_Page_191_Chunk1113 |
It is possible to convert back and forth between delta and Y networks. That is, for every delta network, there exists a Y network such that the resistances seen between the X, Y and Z terminals are identical, and vice versa. Consequently, one configuration can replace another in order to simplify a larger circuit. Δ-Y Conversion Δ-Y Conversion A true equivalent circuit would present the same resistance between any two terminals as the original circuit. Consider the unloaded case for the circuits of Figure 6.32 (i.e., just these networks with nothing else connected to them). The equivalent resistances seen between each pair of terminals for the delta and Y respectively are: RXY = Ra || (Rb+Rc) = Rd + Re (6.1) RXZ = Rb || (Ra+Rc) = Rd + Rf (6.2) RZY = Rc || (Rb+Ra) = Re + Rf (6.3) Assuming we have the delta and are looking for the Y equivalent, note that we have three equations with three unknowns (Rd, Re and Rf). Thus, they can be solved using a term elimination process. If we subtract Equation 6.3 from Equation 6.1 we eliminate the second resistance (Re) and arrive at a difference between the first and third unknown resistances (Rd − Rf). This quantity can then be added to Equation 6.2 to eliminate the third resistance (Rf), leaving just the first unknown resistance (Rd). (Rd + Re) − (Re + Rf) = (Rd − Rf) = Ra || (Rb+Rc) − Rb || (Ra+Rc) (Rd + Rf) + (Rd − Rf) = 2Rd = 2( Rb || (Ra+Rc) + Ra || (Rb+Rc) − Rc || (Ra+Rb) ) Therefore, Rd = Rb || (Ra+Rc) + Ra || (Rb+Rc) − Rc || (Ra+Rb) which, after simplifying, is: Rd = Ra Rb Ra+Rb+Rc (6.4) Similarly, we can show that Re = Ra Rc Ra+Rb+Rc (6.5) R f = Rb Rc Ra+Rb+Rc (6.6) Note that if three identical resistors are used, the values of the Y equivalent will all be one-third of that value. 192 | DCElectricalCircuitAnalysis_Page_192_Chunk1114 |
Y-Δ Conversion Y-Δ Conversion For the reverse process of converting Y to delta, start by noting the similarities of the expressions for Rd, Re and Rf. If two of these expressions are divided, a single equation for Ra, Rb or Rc will result. For example, using Equations 6.4 and 6.5: Rd Re = Ra Rb Ra+Rb+Rc Ra Rc Ra+Rb+Rc Rd Re = Ra Rb Ra Rc Rd Re = Rb Rc Therefore, Rb Rc = Rd Re Rb = Rc Rd Re This process can be repeated for Equations 6.4 and 6.6 to obtain an expression for Ra. The two expressions for Ra and Rb can then be substituted into Equation 6.4 to obtain an expression for Rc that utilizes only Rd, Re and Rf. A similar process is followed for Ra and Rb resulting in: Ra = Rd Re+ReR f +Rd R f R f (6.7) Rb = Rd Re+Re R f +Rd R f Re (6.8) Rc = Rd Re+Re R f +Rd R f Rd (6.9) If three identical resistors are used, the values of the delta equivalent will all be three times that value, the inverse of the situation when converting from delta to Y. Thus, equations 6.4, 6.5 and 6.6 can be used to convert a delta network into a Y network, and equations 6.7, 6.8 and 6.9 can be used to convert a Y network into a delta network. An example of how this will tame an otherwise incorrigible series- parallel network is next. 193 | DCElectricalCircuitAnalysis_Page_193_Chunk1115 |
Example 6.7 Determine the equivalent of the bridge circuit shown in Figure 6.34. This circuit uses a five resistor bridge which cannot be further simplified using basic series-parallel combinations. As a result, the techniques presented in Chapter 5 will not be sufficient to obtain a solution. A delta-Y conversion is proposed to simplify the circuit instead. We begin by defining a delta configuration using the top three resistors (shown in blue). When replaced with a Y configuration, we will have resistors in series with the 2 kΩ and 4 kΩ, and a third resistor in series with the source. This new network can solved using basic series-parallel techniques. Note that a Y-delta is not as good of a choice. For example, using the 1 kΩ, 2 kΩ and the 5kΩ as a sideways Y, the delta version leaves resistors in parallel with the 3 kΩ and the 4 kΩ, and another in parallel with the source. This is straightforward to solve, but unfortunately, node b disappears in the process. We shall use Equations 6.4, 6.5 and 6.6 to perform the delta-Y conversion. The orientation in Figure 6.34 is skewed compared to the reference of Figure 6.32. To make them match, imagine rotating the networks of Figure 6.32 slightly clockwise so that point X is at the very top (matching node a), leaving Y and Z at the bottom (matching c and b, respectively). Given this new orientation, Ra is the 3 kΩ, Rb is the 1 kΩ and Rc is the 5 kΩ. Rd = Ra Rb Ra+Rb+Rc Rd = 3k Ω1 kΩ 3 kΩ+1 kΩ+5kΩ Rd ≈333.333Ω Re = Ra Rc Ra+Rb+Rc Re = 3 kΩ5k Ω 3k Ω+1k Ω+5k Ω Re ≈1.666667 k Ω R f = Rb Rc Ra+Rb+Rc R f = 1 kΩ5k Ω 3 kΩ+1 kΩ+5 kΩ R f ≈555.556Ω 194 Figure 6.35 Equivalent for the circuit of Figure 6.34. Figure 6.34 Circuit for Example 6.7. | DCElectricalCircuitAnalysis_Page_194_Chunk1116 |
The equivalent network is placed back into the original circuit as shown in Figure 6.35 (red replaces blue). We now have a 333.333 Ω in series with 2.555556 kΩ || 5.666667 kΩ. To determine Vb and Vc the total equivalent resistance can be used to find the source current. From there, a current divider may be employed along with Ohm's law to find the node voltages. RTotal = 333.333Ω+2.555556 kΩ||5.666667k Ω RTotal ≈2.0946 kΩ I s = E RTotal I s ≈ 12 V 2.0946 kΩ I s ≈5.729 mA I b = I s Rright Rleft+Rright I b ≈5.729mA 5.666667kΩ 2.555556k Ω+5.666667kΩ I b ≈3.9484 mA V b = I b×R V b ≈3.948mA×2k Ω V b ≈7.8968V By similar process, Vc is approximately 7.1226 volts. To verify the results of Example 6.7, we can take the two node voltages, apply them back to the original circuit, and determine whether or not they satisfy KVL or KCL. For example, examining node b, the current flowing through the 1 kΩ resistor has to equal the sum of the currents flowing through the 2 kΩ and 5 kΩ resistors, via KCL. I 1k = V a−V b R I 1k ≈12 V−7.8968V 1kΩ I 1k ≈4.1032 mA I 5k = V b−V c R I 5k ≈7.8968 V−7.1226 V 5 kΩ I 5k ≈0.1548mA 195 | DCElectricalCircuitAnalysis_Page_195_Chunk1117 |
I 2k = I 1k −I5k I 2k ≈4.1032 mA −0.1548mA I 2k ≈3.9484 mA This result matches the current Ib calculated previously. For completeness sake, the process can be replicated for node c. Computer Simulation Computer Simulation For further verification, both the original circuit of Figure 6.34 and the converted circuit of Figure 6.35 are entered into a simulator, as shown in Figure 6.36. Both resistive networks are connected to a common power supply. Nodes 2 and 4 correspond to Vb, while nodes 3 and 5 correspond to Vc. The results of a DC operating point simulation are shown in Figure 6.37. 196 Figure 6.36 Original and converted bridge circuits in simulator. | DCElectricalCircuitAnalysis_Page_196_Chunk1118 |
The voltages match each other perfectly, along with the manual calculation. In closing, we see that it is possible to swap delta networks with Y networks, and achieve identical results. 6.8 Summary 6.8 Summary In this chapter we have examined several new techniques and theorems to assist with the analysis of DC electrical circuits. Beginning with more practical models for voltage and current sources, we added an internal resistance which sets limits on the source's maximum output. For a voltage source, this resistance is in series, its ideal value being a short. For current sources, the resistance is in parallel, its ideal value being an open. Source conversions allow us to create an equivalent current source for any practical voltage source and vice versa. An equivalent source is one that will create the same voltage across (and current into) whatever the new source is connected to as did the original source. In some cases, this swap allows differing sources to be combined into a single source, simplifying analysis. The superposition theorem states that, for any multi-source linear bilateral network, the contributions of each source may be determined independent of all other sources, the final result being the summation of the contributions, cognizant of current directions and voltage polarities. Thus, the original circuit of N sources generates N new circuits, one for each source under consideration and with all other sources replaced by their ideal internal resistance. 197 Figure 6.37 Simulation results for original and converted bridges. | DCElectricalCircuitAnalysis_Page_197_Chunk1119 |
Thévenin's and Norton's theorems allow the simplification of complex linear single port (i.e., two connecting points) networks. The Thévenin equivalent consists of a voltage source with series a resistance while the Norton equivalent consists of a current source with a parallel resistance. These equivalents, when replacing the original sub-circuit, will create the same voltage across the remainder of the circuit with the same current draw. That is, the remainder of the circuit will see no difference between being driven by the original sub-circuit or by either the Thévenin or Norton equivalents. The maximum power transfer theorem states that for a simple voltage source with an internal resistance driving a single resistor load, the maximum load power will be achieved when the load resistance equals the internal resistance. At this point, efficiency will be 50%. If the load resistance is higher than the internal resistance, the load power will not be as great, however, the system efficiency will increase. Delta-Y conversions allow the generation of equivalent “three connection point” resistor networks. Resistor networks with three elements in the shape of a triangle or delta (with one connection point at each corner) can be converted into a three element network in the shape of a Y or T, or vice versa. The two versions will behave identically to the remainder of the circuit. This allows the simplification of some circuits and eases analysis. Review Questions Review Questions 1. Define the term linear bilateral network. 2. What are the ideal internal resistances of voltage sources and current sources? 3. Outline the process of converting a voltage source into a current source, and vice versa. 4. In general, describe the process of using superposition to analyze a multi- source circuit. 5. What do Thévenin's and Norton's theorems state? How are they related? 6. What does the maximum power transfer theorem state? How might it be used with Thévenin's or Norton's theorems? 7. What are delta and Y configurations? How are they related? 198 | DCElectricalCircuitAnalysis_Page_198_Chunk1120 |
6.9 Exercises 6.9 Exercises Analysis Analysis 1. For the circuit shown in Figure 6.38, determine the equivalent current source. 2. Given the circuit shown in Figure 6.39, determine the equivalent current source. 3. Determine the equivalent current source for the circuit shown in Figure 6.40. 4. For the circuit shown in Figure 6.41, determine the equivalent current source. 199 Figure 6.38 Figure 6.39 Figure 6.41 Figure 6.40 | DCElectricalCircuitAnalysis_Page_199_Chunk1121 |
5. For the circuit shown in Figure 6.42, determine the equivalent voltage source. 6. Given the circuit shown in Figure 6.43, determine the equivalent voltage source. 7. Determine the equivalent voltage source for the circuit shown in Figure 6.44. 8. For the circuit shown in Figure 6.45, determine the equivalent voltage source. 200 Figure 6.42 Figure 6.43 Figure 6.44 Figure 6.45 | DCElectricalCircuitAnalysis_Page_200_Chunk1122 |
9. Given the circuit shown in Figure 6.46, determine the equivalent voltage source. 10. Using source conversion, find Vb for the circuit shown in Figure 6.47. 11. Using source conversion, find the current through the 3 kΩ resistor in the circuit of Figure 6.48. 12. Using source conversion, find Vb for the circuit shown in Figure 6.49. 201 Figure 6.46 Figure 6.47 Figure 6.48 Figure 6.49 | DCElectricalCircuitAnalysis_Page_201_Chunk1123 |
13. Using source conversion, find Va for the circuit shown in Figure 6.50. 14. Using source conversion, find Vb for the circuit shown in Figure 6.51. 15. Using superposition, determine Vb for the circuit shown in Figure 6.47. 16. Using superposition, find the current through the 3 kΩ resistor for the circuit of Figure 6.47. 17. Using superposition, find the current through the 3 kΩ resistor for the circuit of Figure 6.48. 18. Using superposition, determine Vab for the circuit shown in Figure 6.48. 19. Using superposition, determine Vb for the circuit shown in Figure 6.49. 20. Using superposition, find the current through the 4 kΩ resistor for the circuit of Figure 6.49. 21. Using superposition, find the current through the 30 kΩ resistor for the circuit of Figure 6.50. 22. Using superposition, determine Va for the circuit shown in Figure 6.50. 23. Using superposition, determine Vba for the circuit shown in Figure 6.51. 24. Using superposition, find the current through the 10 kΩ resistor for the circuit of Figure 6.51. 202 Figure 6.50 Figure 6.51 | DCElectricalCircuitAnalysis_Page_202_Chunk1124 |
25. Using superposition, find the current through the 1.5 kΩ resistor for the circuit of Figure 6.52. 26. Using superposition, determine Vab for the circuit shown in Figure 6.52. 27. Using superposition, determine Vb for the circuit shown in Figure 6.53. 28. Using superposition, find the current through the 200 Ω resistor for the circuit of Figure 6.53. 29. Using superposition, find the current through the 4 kΩ resistor for the circuit of Figure 6.54. 30. Using superposition, determine Vb for the circuit shown in Figure 6.54. 31. Is it possible to determine Vb in Figure 6.54 by using source conversions instead of superposition? Why/why not? 203 Figure 6.52 Figure 6.53 Figure 6.54 | DCElectricalCircuitAnalysis_Page_203_Chunk1125 |
32. Using superposition, determine Vbc for the circuit shown in Figure 6.55. 33. Using superposition, find the current through the 10 kΩ resistor for the circuit of Figure 6.55. 34. Using superposition, find the currents through the 100 Ω and 700 Ω resistors for the circuit shown in Figure 6.56. 35. Using superposition, determine Vbd for the circuit shown in Figure 6.56. 36. Is it possible to determine Vbd in Figure 6.56 by using source conversions instead of superposition? Why/why not? 37. Using superposition, determine Vad for the circuit shown in Figure 6.57. 204 Figure 6.55 Figure 6.56 Figure 6.57 | DCElectricalCircuitAnalysis_Page_204_Chunk1126 |
38. Using superposition, find the current through the 20 kΩ resistor for the circuit shown in Figure 6.57. 39. Using superposition, find the current through the 12 kΩ resistor for the circuit shown in Figure 6.58. 40. Using superposition, determine Vb for the circuit shown in Figure 6.58. 41. Using superposition, determine Vb for the circuit shown in Figure 6.59. 42. Using superposition, find the current through the 100 Ω resistor for the circuit of Figure 6.59. 43. Using superposition, find the current through the 5 kΩ resistor for the circuit of Figure 6.60. 44. Using superposition, determine Vc for the circuit shown in Figure 6.60. 205 Figure 6.58 Figure 6.59 Figure 6.60 | DCElectricalCircuitAnalysis_Page_205_Chunk1127 |
45. Given the circuit shown in Figure 6.61, determine the Thévenin equivalent circuit that is driving the 4 kΩ resistor. 46. Given the circuit shown in Figure 6.61, determine the Norton equivalent circuit driving the 4 kΩ resistor. 47. Given the circuit shown in Figure 6.61, determine the Norton equivalent circuit driving the 12 kΩ resistor. 48. Determine the Thévenin equivalent circuit driving the 12 kΩ resistor for the circuit shown in Figure 6.62. 49. Given the circuit shown in Figure 6.63, determine the Thévenin equivalent circuit that is driving the 20 kΩ resistor. 50. For the circuit shown in Figure 6.63, determine the Norton equivalent circuit driving the 4 kΩ resistor. 206 Figure 6.61 Figure 6.62 Figure 6.63 | DCElectricalCircuitAnalysis_Page_206_Chunk1128 |
51. Given the circuit shown in Figure 6.64, determine the Norton equivalent circuit driving the 40 Ω resistor. 52. Determine the Thévenin equivalent circuit driving the 10 Ω resistor for the circuit shown in Figure 6.64. 53. Given the circuit shown in Figure 6.65, determine the Norton equivalent circuit driving the 12 kΩ resistor. 54. Given the circuit of Figure 6.61, determine the power in the 4 kΩ resistor. If this resistor can be replaced with any other value, is it possible to achieve a higher power? Why/why not? 55. Given the circuit of Figure 6.62, determine the power in the 12 kΩ resistor. If this resistor can be replaced with any other value, is it possible to achieve a higher power? Why/why not? 56. Given the circuit of Figure 6.64, determine the power in the 40 Ω resistor. If this resistor can be replaced with any other value, is it possible to achieve a higher power? Why/why not? 57. Given the circuit of Figure 6.65, determine the power in the 6 kΩ resistor. If this resistor can be replaced with any other value, is it possible to achieve a higher power? Why/why not? 207 Figure 6.64 Figure 6.65 | DCElectricalCircuitAnalysis_Page_207_Chunk1129 |
Design Design 58. Consider the 4 kΩ resistor to be the load in Figure 6.61. Determine a new value for the load in order to achieve maximum load power. Also determine the maximum load power. 59. Consider the 12 kΩ resistor to be the load in Figure 6.62. Determine a new value for the load in order to achieve maximum load power. Also determine the maximum load power. 60. Consider the 40 Ω resistor to be the load in Figure 6.64. Determine a new value for the load in order to achieve maximum load power. Also determine the maximum load power. 61. Consider the 6 kΩ resistor to be the load in Figure 6.65. Determine a new value for the load in order to achieve maximum load power. Also determine the maximum load power. 62. Redesign the circuit of Figure 6.52 so that it uses only current sources and produces the same component currents and voltages as the original circuit. 63. Redesign the circuit of Figure 6.54 so that it uses only current sources and produces the same component currents and voltages as the original circuit. 64. Redesign the circuit of Figure 6.58 so that it uses only voltage sources and produces the same component currents and voltages as the original circuit. 65. Redesign the circuit of Figure 6.60 so that it uses only voltage sources and produces the same component currents and voltages as the original circuit. 66. Convert the delta network of Figure 6.66 into a Y network. 208 Figure 6.66 | DCElectricalCircuitAnalysis_Page_208_Chunk1130 |
67. Convert the pi network of Figure 6.67 into a T network. 68. Convert the Y network of Figure 6.68 into a delta network. 69. Convert the T network of Figure 6.69 into a pi network. 209 Figure 6.67 Figure 6.68 Figure 6.69 | DCElectricalCircuitAnalysis_Page_209_Chunk1131 |
Challenge Challenge 70. Redesign the circuit of Figure 6.70 so that it uses only current sources and produces the same node voltages as the original circuit. 71. Using any combination of techniques, find the current through the 9 kΩ resistor for the circuit of Figure 6.70. 72. Using any combination of techniques, determine Vbc for the circuit shown in Figure 6.70. 73. Is it possible to determine Vbc in Figure 6.70 by using just source conversions? Why/why not? 74. Using superposition, find the current through the 25 Ω resistor for the circuit of Figure 6.71. 210 Figure 6.70 Figure 6.71 | DCElectricalCircuitAnalysis_Page_210_Chunk1132 |
75. Using superposition, find Vab in the circuit of Figure 6.71. 76. Redesign the circuit of Figure 6.71 using only voltage sources so that it achieves the same node voltages as the original. 77. Is it possible to determine Vbc in Figure 6.72 by using just source conversions or just superposition? Why/why not? 78. Using any combination of techniques, find the current through the 100 Ω resistor for the circuit shown in Figure 6.72. 79. Redesign the circuit of Figure 6.72 so that it uses only voltage sources and produces the same node voltages as the original circuit. 80. Determine the Thévenin and Norton equivalents driving the 40 Ω resistor for the circuit shown in Figure 6.53. 81. Determine the Thévenin and Norton equivalents driving the 12 kΩ resistor for the circuit shown in Figure 6.54. 82. Given the circuit of Figure 6.63, determine if the 4 kΩ resistor is the optimal value to achieve maximum power dissipation in that resistor. If it is not, determine the value that will produce maximum power in the resistor along with the resulting power. 83. For the circuit of Figure 6.73, determine an equivalent circuit using just a single voltage source. 211 Figure 6.72 Figure 6.73 | DCElectricalCircuitAnalysis_Page_211_Chunk1133 |
Simulation Simulation 84. Using DC bias simulations, compare the original circuit of problem 1 to its converted equivalent. Do this by connecting a resistor to the output terminals, trying several different resistance values and checking to see if the two circuits always produce the same voltage across this resistor. 85. Using DC bias simulations, compare the original circuit of problem 5 to its converted equivalent. Do this by connecting a resistor to the output terminals, trying several different resistance values and checking to see if the two circuits always produce the same voltage across this resistor. 86. Perform a DC bias simulation on the circuit of problem 11 to verify the node voltages. 87. Perform a DC bias simulation on the circuit of problem 13 to verify the node voltages. 88. Perform a DC bias simulation on the circuit of problem 19 to verify the node voltages. 89. Perform a DC bias simulation on the circuit of problem 21 to verify the resistor current. 90. Create DC bias simulations of the original and equivalent circuits generated in problem 45 to determine if the equivalent circuit is truly equivalent. Do this by substituting several different values for the 4 kΩ resistor in both circuits to see if the same load voltage is obtained for both circuits. 91. Create DC bias simulations of the original and equivalent circuits generated in problem 49 to determine if the equivalent circuit is truly equivalent. Do this by substituting several different values for the 20 kΩ resistor in both circuits to see if the same load voltage is obtained for both circuits. 92. Create DC bias simulations of the original and equivalent circuits generated in problem 53 to determine if the equivalent circuit is truly equivalent. Do this by substituting several different values for the 12 kΩ resistor in both circuits to see if the same load voltage is obtained for both circuits. 93. Perform a DC bias simulation on the circuit of problem 63 to verify that the node voltages of the new design match those of the original. 94. Perform a DC bias simulation on the circuit of problem 65 to verify that the node voltages of the new design match those of the original. 95. Using either Monte Carlo simulation or multiple DC bias simulations, verify that the resistance calculated in problem 55 achieves maximum power. Do this by trying several resistor values near the calculated value, and determining the power for each based on the squared load voltages. 212 | DCElectricalCircuitAnalysis_Page_212_Chunk1134 |
96. Using either Monte Carlo simulation or multiple DC bias simulations, verify that the resistance calculated in problem 56 achieves maximum power. Do this by trying several resistor values near the calculated value, and determining the power for each based on the squared load voltages. This XKCD comic is too large to fit on this page. Just go to this link instead: https://xkcd.com/1732/ 213 | DCElectricalCircuitAnalysis_Page_213_Chunk1135 |
7 7 Nodal & Mesh Analysis Nodal & Mesh Analysis, Dependent Sources , Dependent Sources 7.0 Chapter Learning Objectives 7.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Utilize nodal analysis techniques to solve for voltages in multi-source series-parallel networks. • Utilize mesh analysis techniques to solve for currents in multi-source series-parallel networks. • Analyze networks using dependent voltage and current sources. 7.1 Introduction 7.1 Introduction This chapter presents the culmination of the analysis of DC circuits that employ resistors along with any number of voltage and/or current sources. Here we shall focus on nodal analysis and mesh analysis. Both nodal analysis and mesh analysis generate a system of simultaneous linear equations that are used to solve the circuit for various voltages or currents. There are several methods that can be used to solve the simultaneous equations including substitution, Gauss-Jordan elimination and expansion by minors. These methods are reviewed in Appendix B and are not covered in this chapter. Instead, to keep clutter to a minimum and focus on the circuit analysis aspects, the explanations and examples will simply detail the process of examining the circuit and applying basic circuit laws to create the system of equations. From there, the specific technique used to solve these simultaneous equations is up to you based on your personal preferences. It is particularly efficient to obtain a more advanced scientific calculator that can solve these equations directly rather than plowing through the solution manually. By doing so, you can spend your precious time mastering the process of circuit analysis and creating the equations. Manual solution techniques, though not necessarily difficult, can be tedious, time consuming and error prone. If you plan on continuing your study into AC electrical circuits, you should consider obtaining a calculator that can solve simultaneous equations with complex coefficients. Such calculators can be expensive when purchased new, such as the Texas Instruments TI-89 and Nspire models. On the used market, perfectly satisfactory older models such as the TI-85 and TI-86 can be found at considerable discount. Another model to consider is the Casio FX-9750GII, although it is not quite as powerful as some of the other units mentioned. Along with nodal and mesh, we shall also introduce the concept of dependent sources. These are current and voltage sources whose value is not fixed to some particular value, but rather is dependent on some other current or voltage in the circuit. What makes this interesting is that this controlling current or voltage may itself be affected by the value produced by the dependent source. Dependent sources are often used to model the behavior of active electronic devices such as bipolar and field effect transistors. 214 | DCElectricalCircuitAnalysis_Page_214_Chunk1136 |
7.2 Nodal Analysis 7.2 Nodal Analysis Nodal analysis is a technique that can be applied to virtually any circuit. In general use, it might be considered a universal solution technique as there are no practical circuit configurations that it cannot handle. Nodal analysis relies on the application of Kirchhoff's current law to create a series of node equations that can be solved for node voltages. These equations are based on Ohm's law and will be of the form I = V/R, or more generally, I = (1/RX)∙VA + (1/RY)∙VB... Once the node voltages are obtained, finding any branch currents or component powers becomes an almost trivial exercise. We will examine two variations; a general version that can be used with both voltage and current sources, and a second somewhat quicker version that can be used with circuits only driven by current sources. General Method General Method Consider the circuit shown in Figure 7.1. We begin by labeling connection nodes and assigning current directions. We are particularly interested in current junctions, that is, places where currents can combine or split. These are also known as summing nodes and are circled in green on the figure. The current directions are chosen arbitrarily and may be the opposite of reality. This is not a problem. If we assign directions that are wrong, the resulting directions will ultimately show up reversed but the computed node voltages will be just fine. One node is chosen as the reference. This is the point to which all other node voltages will be measured against. Typically, the reference node is ground, although it does not have to be. We now write a current summation equation for each summing node, except for the reference node. In this circuit there is only one node where currents combine (other than ground) and that's node b. Points a and c are places where components connect, but they are not summing nodes, so we can ignore them for now. Using KCL on node b we can say: I1 + I2 = I3 215 Figure 7.1 Basic dual voltage source circuit with currents and nodes defined. | DCElectricalCircuitAnalysis_Page_215_Chunk1137 |
Next, we describe these currents in terms of the node voltages and associated components via Ohm's law. For example, I3 is the node b voltage divided by R3 while I1 is the voltage across R1 divided by R1. This voltage is Va − Vb. Therefore, V a −V b R1 +V c −V b R2 = V b R3 Noting that Va = E1 and Vc = E2, with a little algebra this can be reduced to a series of products of conductances and voltages: ( 1 R1)E1 +( 1 R2)E2 =( 1 R1 + 1 R2 + 1 R3)V b All quantities are known except for Vb and thus it is easily found with a little more algebra. As we shall see, this conductance-voltage product format turns out to be a convenient way of writing these equations. Also, note that the first two terms on the left reduce to fixed current values. If there had been more nodes, there would have been more equations, one for each node. For current sources, a more direct approach is possible. Consider the circuit of Figure 7.2. We start as before, identifying nodes and labeling currents. This is shown in Figure 7.3. We then write current summation equations at each node (except for ground). We consider currents entering a node as positive, and exiting as negative. Node a: I1 = I3 + I4 Node b: I3 = I2 + I5, and rearranging in terms of the fixed source, Node b: −I2 = −I3 + I5 216 Figure 7.2 Basic dual current source circuit. Figure 7.3 Basic dual current source circuit with currents and nodes defined. | DCElectricalCircuitAnalysis_Page_216_Chunk1138 |
The currents are then described by their Ohm's law equivalents: Node a: I 1 = V a −V b R3 +V a R1 Node b: −I 2 =−V a −V b R3 +V b R2 Expanding and collecting terms yields: Node a: I 1 =( 1 R1 + 1 R3)V a −( 1 R3)V b Node b: −I 2 =−( 1 R3)V a +( 1 R3 + 1 R2)V b As the resistor values and currents are known, simultaneous equation solution techniques may be used to solve for the node voltages. Once again, there will be as many equations as node voltages. It is very important that the terms “line up” when the final system of equations is written out. That is, there should be a column for the Va terms, a column for the Vb terms, and so on. They should not be written out in random order. This format will make it much it easier to enter the coefficients into a calculator or solve manually. Example 7.1 Determine Vb in the circuit of Figure 7.4. This circuit has two current summing nodes, ground and node b. Let us assume that the currents from the two sources flow into node b and the exiting current flows down through the 5 kΩ resistor. It will be convenient if we find the conductance equivalents of the resistors before continuing. Taking the reciprocals we find: 1 kΩ = 1 mS, 4 kΩ = 0.250 mS, and 5 kΩ = 0.2 mS. Through KCL we can say: I1k + I4k = I5k Replacing these currents with their Ohm's law equivalents yields, 15 V −V b 1k Ω + 6V −V b 4 kΩ = V b 5 kΩ collecting terms, 15mA+1.5mA =( 1 1kΩ + 1 4k Ω + 1 5k Ω)V b 217 Figure 7.4 Circuit for Example 7.1. | DCElectricalCircuitAnalysis_Page_217_Chunk1139 |
and solving for Vb, V b = 16.5 mA 1mS +0.25mS +0.2mS V b = 11.379volts As Vb is higher than the 6 volt source, our assumed current direction for the 4 kΩ resistor was incorrect; we assumed right to left but it is in fact left to right, flowing from 11.379 volts to 6 volts. As you can see, this did not present a problem. For comparison sake, this circuit was solved in Chapter 6 using source conversions and also using superposition. Inspection Method Inspection Method The system of equations can be obtained directly through inspection if the circuit contains no voltage sources. For the node under inspection, sum all of the current sources connected to it to obtain the current constant. The conductance term for that node will be the sum of all of the conductances connected to that node. For the other node conductances, determine the conductances between the node under inspection and these other nodes. These terms will all be negative. As a crosscheck, the set of equations produced must exhibit diagonal symmetry, that is, if a diagonal is drawn from the upper left to the lower right through the conductance-voltage pairs, then the coefficients found above the diagonal will have to match those found below the diagonal as we move perpendicular from the diagonal. This is illustrated in Figure 7.5. Diagonal symmetry can be seen in Example 7.1 by noting the matching “−1/R3” coefficients in the final pair of equations. As an example of the inspection method, focusing on node a in Figure 7.2, we find the fixed current source I1 feeding it (entering, therefore positive). The conductances directly connected to node a are 1/R1 and 1/R3, yielding the coefficient for Va. The only conductance common between nodes a and b is 1/R3, yielding the Vb coefficient. The inspection method is summarized as follows: 1. Verify that the circuit uses only current sources with resistors and no voltage sources. If voltage sources exist, they must be converted to current sources before proceeding. 2. Find all of the current summing nodes and number them. Also decide on the reference node (usually ground). 3. To generate an equation, locate the first node. This is the node of interest and the next few steps will be associated with it. 218 1 V a -2 Va -6 Va -9 Va -2 Vb 5 V b -4 Vb -7 Vb -6 Vc -4 Vc 8 V c 0 V c -9 Vd -7 Vd 0 V d 3 V d Figure 7.5 Diagonal symmetry. | DCElectricalCircuitAnalysis_Page_218_Chunk1140 |
4. Sum the current sources feeding the node of interest. Entering is deemed positive while exiting is deemed negative. The sum is placed on one side of the equals sign. 5. Next, find all of the resistors connected to the node of interest and write them as a sum of conductances on the other side of the equals sign, the group being multiplied by this node's voltage (e.g., V1). 6. Now find all of the resistors that are connected to the node of interest and to other nodes (except the ground reference). For each of these other nodes, multiply the sum of the conductances between the node of interest and this other node by this other node's voltage, and then subtract that product from the equation built so far. Once all other nodes are considered, this equation is finished. 7. Find the next node and treat this as the new node of interest. 8. Repeat steps 4 through 7 until all nodes have been treated as the node of interest. Each iteration creates a new equation. There will be as many equations as there are nodes, less the reference node. Check for diagonal symmetry and solve. The inspection method is best observed in action; as in the following example. Example 7.2 Determine Va and Vb in the circuit of Figure 7.6. Also determine the current flowing through the 100 Ω resistor. This circuit has three current summing nodes; ground, node a and node b. Also, it uses only current sources and no voltage sources. Therefore, we are able to use the inspection method to obtain the system of equations (in this case, two equations). We begin by focusing on node a, our first node of interest. We will build the first expression piece by piece. First we find the current sources feeding this node: 800mA −2A = ... Next, we find all of the resistors connected to node a and write them as conductances, the group being multiplied by node voltage a: 800 mA −2A =( 1 10 Ω + 1 50 Ω + 1 100Ω)V a ... Now find all of the resistors that are connected to this node and to the other nodes. Multiply those resistors (expressed as conductances) by the other associated node voltages and subtract the products from the expression built 219 Figure 7.6 Circuit for Example 7.2. | DCElectricalCircuitAnalysis_Page_219_Chunk1141 |
so far. Repeat for all remaining nodes except the ground reference. In this example there is only one other node, node b, and thus only one iteration. 800 mA −2 A =( 1 10Ω + 1 50Ω + 1 100Ω)V a −( 1 50Ω + 1 100 Ω)V b Finally, simplify the constants and coefficients, and the first expression is complete: −1.2 A = 130mSV a −30 mSV b Now we repeat the entire process for the next equation. Node b is our new node of interest. The fixed current sources are: −800mA −300 mA = ... Next, we find all of the resistors connected to node b and write them as conductances, the group being multiplied by node voltage b: −800mA −300 mA =( 1 25Ω + 1 50Ω + 1 100Ω)V b ... Now we find all of the resistors that are connected to this node and to the other nodes. Multiply those resistors (expressed as conductances) by these other node voltages and subtract those products from the expression built so far. Make sure the terms align vertically based on the node voltages. −800 mA −300 mA = −( 1 50Ω + 1 100Ω)V a+( 1 25Ω + 1 50Ω + 1 100Ω)V b Finally, simplify the constants and coefficients, and this expression is complete: −1.1 A =−30 mSV a +70 mSV b We now have two equations with two unknowns. Let's check for diagonal symmetry: −1.2 A = 130mSV a −30 mSV b −1.1 A =−30 mSV a +70 mSV b The perpendicular coefficient is −30 mS on each side. We can now solve the system. The results are Va = −14.27 volts and Vb = −21.83 volts. The current flowing through the 100 resistor is (−14.27 V − (−21.83 V))/100 Ω, or approximately 75.6 mA flowing left to right. 220 | DCElectricalCircuitAnalysis_Page_220_Chunk1142 |
Let's verify that these values are correct. We can perform a KCL summation at node a and see if it balances. We already know that 2 amps and 75.6 milliamps exit while 800 milliamps enters. We only need to find the currents through the 10 Ω and 50 Ω resistors. First, notice that the 50 Ω sees the same voltage as the 100 Ω resistor. As it is half the resistance, it must produce twice the current, or 151.2 milliamps exiting. The current through the 10 Ω is found via Ohm's law, or −14.27 V/10 Ω, which is 1.427 amps entering. Entering: 0.8 A + 1.427 A = 2.227 A Exiting: 2 A + 0.0756 A + 0.1512 A ≈ 2.227 A KCL is satisfied. To complete the verification we perform the same summation on node b. This is left as an exercise. Example 7.3 Write the node equations for the circuit of Figure 7.7. This circuit has four current summing nodes; ground and nodes a, b, and c. Also, it uses only current sources and no voltage sources. Therefore, we are able to use the inspection method to obtain the system of equations (three equations). As we have already seen, we will wind up with conductance values in the final equations. We can save some work later by finding the conductance value of each resistor right now. The result is shown in Figure 7.8. We begin by focusing on node a, our first node of interest. We will build the first expression piece by piece. First we find the current sources feeding this node. That's just the 1.5 A source, entering is positive. 1.5A = ... Next, we find all of the conductances connected to node a and multiply by node voltage a: 1.5A = (0.2S +0.25S)V a ... Find all of the conductances that are connected to this node and to the other nodes. Multiply those conductances by the other associated node voltages and subtract those products from the expression built so far. Repeat for all remaining nodes except the ground reference. 221 Figure 7.7 Circuit for Example 7.3. Figure 7.8 Circuit of Figure 7.7 shown with conductances instead of resistances. | DCElectricalCircuitAnalysis_Page_221_Chunk1143 |
1.5A = (0.2S +0.25S)V a −(0.25S)V b −(0)V c Finally, simplify the constants and coefficients, and the first expression is complete: 1.5A = 0.45SV a −0.25SV b −0V c We leave the zero term in just for alignment. Now we repeat the entire process for the next equation. The node of interest is now node b. Here is the result: 0 =−0.25 SV a +0.8SV b −0.5V c And finally for node c we have: 1A = 0V a −0.5SV b +0.6V c Our three equations are (with some padding to see the columns better): 1.5A = 0.45SV a −0.25SV b −0V c 0 =−0.25 SV a +0.8SV b −0.5V c 1A = −0V a −0.5SV b +0.6V c Check for diagonal symmetry: we have pairs of −0.25, 0, and −0.5 straddling the diagonal. Now the system of equations can be solved for the three node voltages and verified using KCL summations at each node. This is left as an exercise. Converting Sources and Other Simplifications Converting Sources and Other Simplifications Given circuits with voltage sources, it may be easier to convert them to current sources and then apply the inspection technique rather than using the general approach outlined initially. There is one trap to watch out for when using source conversions: the voltage across or current through a converted component will most likely not be the same as the voltage or current in the original circuit. This is because the location of the converted component will have changed. For example, the circuit in Figure 7.9 could be solved using nodal analysis by converting the voltage source and the associated resistance into a current source. That is, E/R1 would be converted into a source I3 with a parallel resistor R1. This is shown in Figure 7.10. 222 Figure 7.9 Circuit with both current and voltage sources. | DCElectricalCircuitAnalysis_Page_222_Chunk1144 |
Here is the trap: in the converted circuit, although R1 still connects to node a, the other end no longer connects to the voltage source. Rather, the right side now connects to node c. Therefore, the voltage drop across R1 in the converted circuit is not likely to equal the voltage drop seen across R1 in the original circuit (the only way they would be equal is if E was 0 volts). In the converted circuit, nodes a and c have not changed from the original, so the original voltage across R1 can be determined via Va, Vc and E in the original circuit. If a circuit uses voltage sources exclusively, or even a very large proportion of voltage sources, an alternate technique called mesh analysis may be a better choice than using numerous conversions. Mesh analysis will be detailed later in this chapter. A final item to consider is to simplify resistor networks in order to reduce the number of nodes. Fewer nodes means fewer equations and a quicker solution. For example, a portion of a network might consist of two parallel resistors which are in series with a third resistor. Ordinarily, the common junction of these three resistors would constitute a current summing node and an equation would need to be derived for it. It may be easier to simply replace the trio with a single resistor that equals the series-parallel combination value. This would remove the node and simplify creating the system of equations, but not alter the remainder of the circuit. Supernode Supernode On occasion you may come across a circuit like the one shown in Figure 7.11 that has a voltage source without a series resistance associated with it. Without that resistance, it becomes impossible to create an expression for the current passing through the source using the general method, and impossible to convert the voltage source into a current source in order to use the inspection method. One possible way out of this quandary is to simply add a very small resistor in series with it so that a source conversion is possible. The resistor in question would have to be much smaller than any surrounding resistors in order to have minimal impact on the results. A reduction by two orders of magnitude would generally yield a variation smaller than that produced by resistor tolerances in all but high precision circuits. The other way out is to use a supernode. A supernode is, in effect, the combination of two nodes. It relies on a simple observation. If we examine the circuit of Figure 7.11, the path of the voltage source produces identical currents flowing into and out of nodes a and b. As a consequence, if we treat the two nodes as one big node, then when we write a KCL summation, these two terms will cancel. To see just how this works, refer to Figure 7.12. 223 Figure 7.11 Circuit for supernode analysis. Figure 7.10 Circuit converted to current sources only. | DCElectricalCircuitAnalysis_Page_223_Chunk1145 |
In this version we have replaced the voltage source with its ideal internal resistance, a short. We have also labeled the two nodes of interest, a and b, and labeled the currents, drawn with convenient directions. Due to the shorted voltage source, nodes a and b are now the same node. Consider the currents entering and exiting this combined or “super” node. On the left side (formerly node a) we see a constant current Ix entering while I1, I2 and I3 are exiting. On the right side (formerly node b) we see Iy exiting along with I4, and entering we see I1 and I2. Now let's set up the entering currents on the left side of the equals sign with the exiting currents on the right: Σ Iin = Σ I out I x+I 1+I 2 = I y+I 1+I 2+I3+I 4 This can be simplified to: I x −I y = I 3 +I 4 Writing this in terms of Ohm's law we have: I x −I y = 1 R1 V a + 1 R3 V b We also know that Va − Vb = E from the original circuit. Assuming all sources and resistors are known, that makes two equations with two unknowns, solvable using simultaneous equation techniques. This is illustrated in the following example. 224 Figure 7.12 Circuit modified for supernode analysis. | DCElectricalCircuitAnalysis_Page_224_Chunk1146 |
Example 7.4 Find Va and Vb for the circuit of Figure 7.13. As shown in Figure 7.14, we short the 60 volt source and write a current summation at the a b supernode: Σ Iin = Σ I out 1A+I 1+I 2 = 2.5 A+I 1+I 2+I 3+I 4 This can be simplified to: −1.5 A = I 3 +I 4 Writing this in terms of Ohm's law we have: −1.5A = 1 4ΩV a + 1 10ΩV b −1.5 A = 0.25SV a +0.1SV b We also know that Va − Vb = 60 volts. Therefore Vb = Va − 60 volts. Substituting this into the prior equation yields: −1.5 A = 0.25SV a +0.1S(V a −60 V) −1.5 A = 0.25SV a +0.1SV a −6A 4.5 A = 0.35SV a V a = 12.857 V We know that Vb is 60 volts below Va, so Vb = −47.143 volts. To verify, we will perform a KCL summation at each node. For node a, assuming I1 exits as drawn: I 1 = 1A −V a 4Ω −V a−V b 20Ω I 1 = 1 A−12.857V 4Ω −12.857V−(−47.143 V) 20Ω I 1 =−5.2143A (negative exit means it's entering) 225 Figure 7.14 Circuit modified for supernode analysis. Figure 7.13 Circuit for Example 7.4. | DCElectricalCircuitAnalysis_Page_225_Chunk1147 |
Doing likewise for node b, and assuming I1 enters as drawn: I 1 = 2.5 A+ V b 10Ω −V a−V b 20Ω I 1 = 2.5 A+−47.143 V 10Ω −12.857V−(−47.143V) 20Ω I 1 =−5.2143A (negative enter means it's exiting) These currents match meaning that the current through the voltage source is verified to be the same at both terminals, as it must be. An alternative to the basic supernode technique is to simply describe one node voltage in terms of another at the outset. This is illustrated in the example following. Example 7.5 Find Va, Vb and Vc for the circuit of Figure 7.15. Once again we have a situation of a voltage source without a series resistance. Without having to short it and thus treating nodes a and c as an explicit supernode, we can instead note by observation that the currents entering and exiting the voltage source must be identical. The circuit is redrawn in Figure 7.16 with currents labeled and using equivalent conductances in place of the resistances. This time the voltage source is left in. We start with the observation that Vc = Va − 8 V. In other words, Vc is locked to Va and if we find one of them, we can determine the other. Therefore, instead of writing equations using three nodes, we shall instead refer to node c in reference to node a, that is we shall write Va − 8 V wherever we need Vc. Thus, this three node circuit will only need two equations. We begin at node a and apply KCL as usual. Σ Iin = Σ I out I 1+I 3 = I 2 This is expanded using Ohm's law and we solve for I1: I 1 = I 2 −I 3 I 1 = 0.2SV a −0.5S(V b −V a) I 1 = 0.7SV a −0.5SV b 226 Figure 7.15 Circuit for Example 7.5. | DCElectricalCircuitAnalysis_Page_226_Chunk1148 |
On to node b: 1A = I 3+I 4 1A = 0.5S(V b −V a)+0.25S(V b −V c) 1A = 0.5S(V b −V a)+0.25S(V b −(V a −8V)) 1A = 0.5S(V b −V a)+0.25S(V b −V a+8 V) −1 A =−0.75V a+0.75V b And finally node c: I 4 = I1+I5 I 1 = I 4 −I 5 I 1 = 0.25S(V b −V c)−0.1SV c I 1 = 0.25S(V b −(V a−8V))−0.1S(V a −8 V) I 1 = 0.25S(V b −V a+8 V)−0.1S(V a −8V) I 1 =−0.35V a+0.25V b+2.8A The final equations for nodes a and c both equal I1, meaning they equal each other. Thus, 0.7SV a −0.5SV b = −0.35SV a+0.25SV b+2.8 A 2.8 A = 1.05SV a −0.75SV b The final equations are: 2.8A = 1.05SV a −0.75SV b −1 A =−0.75SV a +0.75SV b The solution is Va = 6 volts and Vb ≈ 4.6667 volts. As Vc is 8 volts less than Va, then Vc = −2 volts. 227 Figure 7.16 Circuit of Example 7.5 with currents labeled and using conductances. | DCElectricalCircuitAnalysis_Page_227_Chunk1149 |
Computer Simulation Computer Simulation In order to verify the result of Example 7.5, the circuit is entered into a simulator as shown in Figure 7.17. A DC operating point simulation is run. The results are shown in Figure 7.18 and match the calculated values perfectly. Node 1 corresponds to Vb, node 2 corresponds to Va and node 3 corresponds to Vc. 228 Figure 7.17 Circuit of Example 7.5 in the simulator. Figure 7.18 Simulation results for the circuit of Example 7.5. | DCElectricalCircuitAnalysis_Page_228_Chunk1150 |
7.3 Mesh Analysis 7.3 Mesh Analysis In some respects mesh analysis is a mirror of nodal analysis. While nodal analysis leverages KCL to create a series of node equations that are used to solve for node voltages, mesh analysis uses KVL to create a series of loop equations that can be solved for mesh currents. A mesh current should not be confused with a branch current. While branch currents represent the current flowing through a particular component, mesh currents combine to create branch currents. That is, the current through any particular component may be an individual mesh current or a combination of two mesh currents. Circuits using complex series-parallel arrangement with multiple voltage and/or current sources may be solved using this technique. It is, however, limited to planar circuits. Planar circuits are those that can be drawn on a flat plane without having any of their wires cross. In essence, these circuits can be drawn so that they appear like a series of window panes. Non-planar circuits may be 3D in appearance and in such a circuit it becomes impossible to define the loop equations as any given component might have more than two meshing currents. General Method General Method Consider the circuit of Figure 7.19. We begin by designating a series of loops. These loops should be minimal in size and cover all components at least once. By convention, the loops are drawn clockwise. There is nothing magic about them being clockwise, it is just a matter of consistency. The current path along a loop is referred to as a mesh current. In the circuit of Figure 7.19 we have two loops, and thus, two mesh currents, I1 and I2. Note that all components exist in at least one loop (and sometimes in more than one loop, like R3). Depending on circuit values, one or more of these loop's directions may in fact be opposite of reality. This is not a problem. If this is the case, the currents will show up as negative values, and thus we know that they're really flowing counter-clockwise. We begin by writing KVL equations for each loop. Loop 1: E1 = voltage across R1 + voltage across R3 Loop 2: −E2 = voltage across R2 + voltage across R3 (E2 is negative as I2 is drawn flowing out of its negative terminal.) 229 Figure 7.19 Circuit for mesh analysis. | DCElectricalCircuitAnalysis_Page_229_Chunk1151 |
Expand the voltage terms using Ohm's law. Loop 1: E1 = I1 ∙ R1 + (I1 − I2) R3 Loop 2: −E2 = I2 ∙ R2 + (I2 − I1) R3 Expanding and collecting terms yields: Loop 1: E1 = (R1 + R3) I1 − R3 ∙ I2 Loop 2: −E2 = R3 ∙ I1 + (R2 + R3) I2 Assuming that the resistor values and source voltages are known, we have two equations with two unknowns. These can be solved for I1 and I2 using simultaneous equation solution techniques such as determinants or Gauss-Jordan elimination. Example 7.6 Find Vb and Vbc for the circuit of Figure 7.20. The first step is to define a set of clockwise loops that are of minimum size and which cover all components. These loops are drawn in Figure 7.21. Also, the polarities of the voltage drops produced by these currents are drawn next to the components in the same color for easy identification. Note that some components see only one current, such as the 3 kΩ resistor that sees only I2; and some see two currents in opposing directions, such the 5 kΩ resistor that sees both I2 and I3. With multiple current directions flowing through a single component, obeying the resulting voltage polarities becomes very important. We now write KVL summations around each loop. If the current passing through a component sees a polarity of + to − then this is counted as a voltage drop while the reverse counts as a voltage rise. This is true even for voltage sources (i.e., a voltage source may appear negative in one loop and positive in another). ΣV rises = ΣV drops Loop 1: 12 V = V 1k+V 2k Loop 2: 0V = V 3k+V 5k+V 1k Loop 3: 0V = V 5k+V 4k+V 2k These are expanded using Ohm's law. The current for the loop under consideration is deemed positive while any opposing (i.e., meshing) current is deemed negative: 230 Figure 7.20 Circuit for Example 7.6. | DCElectricalCircuitAnalysis_Page_230_Chunk1152 |
Loop 1: 12 V = 1k(I 1 −I2)+2 k(I1 −I 2) Loop 2: 0V = 3 k(I2)+5k(I 2 −I 3)+1k(I 2 −I 1) Loop 3: 0V = 5 k(I 3 −I 2)+4k(I 3)+2k(I 3 −I1) The terms are expanded: Loop 1: 12 V = 1k I 1 −1 k I 2+2 k I 1 −2 k I 3 Loop 2: 0V = 3 k I 2+5k I 2 −5k I3+1k I 2 −1 k I 1 Loop 3: 0V = 5 k I 3 −5k I 2+4k I 3+2k I 3 −2k I1 Like terms are grouped, once again creating aligned columns for the current coefficients: Loop 1: 12 V = (1k+2k)I 1 −1 k I 2 −2 k I3 Loop 2: 0V = −1k I 1+(3 k+5k+1 k) I2 −5k I 3 Loop 3: 0V = −2k I1 −5k I 2+(5k+4k+2k)I 3 Which simplifies to: Loop 1: 12 V = 3k I1 −1k I2 −2 k I3 Loop 2: 0V = −1k I 1 +9 k I 2 −5k I 3 Loop 3: 0V = −2k I1 −5k I2+11 k I 3 Note that we have diagonal symmetry. The solution is I1 = 5.729 mA, I2 = 1.6258 mA and I3 = 1.7806 mA. To find Vb we need to find the net current through the 2 kΩ resistor. That's I1 − I3, or 5.729 mA − 1.7806 mA, which is 3.9484 mA. When multiplied by 2 kΩ we find Vb = 7.8968 volts. Similarly, Vbc = (I3 − I2) 5 kΩ = (1.7806 mA − 1.6258 mA ) 5 kΩ, or 774 mV. Computer Simulation Computer Simulation In order to verify the results of the analysis, the circuit is entered into a simulator and a virtual voltmeter is placed across the 5 kΩ resistor. This is shown in Figure 7.22. The results agree nicely with the original analysis. As nice as this is, in a practical circuit we need to be concerned about the effects of component tolerance. It would be extraordinarily odd if during production each resistor was equal to precisely the nominal value specified. In reality, each resistor will have a stated tolerance. Thus, with all five resistors changing slightly from unit to unit, we should expect the currents and voltages to vary as well. The question is, what would be a typical spread? This can be addressed via a Monte Carlo analysis. This analysis is, in fact, a series of simulations. Each simulation uses randomized component values that fall within the stated tolerance spread. For example, if we specify that the 1 kΩ resistor has a 5% tolerance, then the actual resistor used for a 231 Figure 7.21 Circuit of Example 7.6 with current loops and polarities added. | DCElectricalCircuitAnalysis_Page_231_Chunk1153 |
simulation will be a randomly generated value between 950 Ω and 1050 Ω. We then state how many trials we'd like, each one using their own unique randomized values for each component. The results of a Monte Carlo analysis using the DC operating point simulation is shown in Figure 7.23. Ten randomized trials were generated for the value of Vb. Each of the five resistors was assigned a tolerance of 5%. As can be seen, voltage values both above and below the nominal simulation can result. 232 Figure 7.22 Circuit of Figure 7.20 in a simulator with a virtual voltmeter. | DCElectricalCircuitAnalysis_Page_232_Chunk1154 |
Inspection Method Inspection Method Like nodal analysis, there is a method to generate the equations by inspection. Simply focus on one loop, which we'll call the loop under inspection, and ask the following questions: what is the total source voltage in this loop? This yields the voltage constant for the equation. Then sum all of the resistance values in the loop under inspection. This yields the coefficient for that current term. For the remaining current coefficients in this equation, sum the resistances that are in common between the loop under inspection and the other loops, respectively. These values will always be negative. Repeat this process for all loops, with each loop in turn becoming the loop under inspection. As was the case with nodal analysis, the set of equations produced must exhibit diagonal symmetry, that is, if a diagonal is drawn from the upper left to the lower right through the IR pairs, then the coefficients found above the diagonal will have to match those found below the diagonal. While it is possible to extend this technique to include current sources, it is often easier and less error-prone to convert the current sources into voltage sources and continue with the direct inspection method outlined above. Either way, it is important to remember that the number of loops determines the number of equations to be solved. 233 Figure 7.23 Monte Carlo analysis results for the circuit of Figure 7.20. | DCElectricalCircuitAnalysis_Page_233_Chunk1155 |
Example 7.7 Find Vc for the circuit of Figure 7.24. First, we'll label the loops, as shown in Figure 7.25. Now, beginning with loop 1, sum all of the voltage sources in this loop. That's +50 volts. Secondly, sum all of the resistors in this loop. That's 100 Ω + 200 Ω, or 300 Ω. This is the coefficient for the I1 term. The coefficient for the I2 term is the sum of all resistances that are in common between loops 1 and 2. That's just the 200 Ω resistor. The coefficient for the I3 term would be the sum of all resistances that are in common between loop 1 and loop 3. In this circuit, there aren't any so the coefficient is zero. The first equation is: Loop 1: 50 V = 300I1 −200I2 −0I3 Repeat the process for loops 2 and 3: Loop 2: 0V = −200I1 +900I2 −400I3 Loop 3: 20V = −0I1 −400I2 +900I3 Note that the 20 volt source shows up positive because of the direction of I3 (into the negative terminal and out of the positive terminal). As required, we have diagonal symmetry. The solution is I1 = 214.47 mA, I2 = 71.698 mA and I3 = 54.088 mA. Vc is the net current through the 400 Ω resistor times 400 Ω. The net current is I2 − I3, or 71.698 mA − 54.088 mA, which is 17.61 mA. When multiplied by 400 Ω we find Vb = 7.044 volts. Note that we use I2 − I3 and not I3 − I2. The reason is because we are trying to find the voltage from node c to ground, and I2 is the current flowing in that direction. In contrast, using I3 − I2 would yield the voltage from ground to node c; the same magnitude but opposite sign. As a crosscheck, KVL states that Vc + V500 has to equal the −20 volt source. The drop across the 500 Ω is 54.088 mA times 500 Ω, or 27.044 volts from left to right. Starting at node c of 7.044 volts and then dropping 27.044 volts yields − 20 volts as expected. 234 Figure 7.24 Circuit for Example 7.7. Figure 7.25 Circuit of Example 7.7 with current loops identified. | DCElectricalCircuitAnalysis_Page_234_Chunk1156 |
Supermesh Supermesh Sometimes you may run across a current source which has no associated internal resistance, such as found in the circuit of Figure 7.26. This is similar to the situation seen previously under nodal analysis where a voltage source does not have a specified internal resistance. There are two ways of solving this predicament. The first is add a very large resistance in parallel with the current source and then perform a source conversion on the pair so that the inspection method of mesh can be followed. The larger the value of this resistor, the more accuracy that is obtained. As a general rule it should be, at minimum, at least a couple of orders of magnitude larger than any surrounding resistor, and preferably larger. The second method is to use supermesh. A supermesh is a larger mesh loop than contains other mesh loops inside of it. Consider the circuit shown in Figure 7.26. In the center we have a current source, Is, which lacks an associated internal resistance. Two traditional mesh loops, I1 and I2, are labeled as usual. The problem here is that we cannot use an Ohm's law based IR voltage drop for Vb. We have no way to express this as the voltage across Is is an unknown. On the other hand, what we do know is that Is must equal the combination of the traditional mesh currents I1 and I2. That is, from the perspective of the first loop, Is = I1 − I2. Remember, one or both of the mesh currents could be negative, and thus rotating counterclockwise. At this point we invoke the idea of a supermesh loop. First, we replace the offending current source with its ideal internal resistance (an open). The supermesh loop is drawn which encompasses the original two loops. This is shown in Figure 7.27. The supermesh loop is drawn in red and labeled. 235 Figure 7.26 Circuit for supermesh. Figure 7.27 Supermesh labeled. | DCElectricalCircuitAnalysis_Page_235_Chunk1157 |
We now perform a KVL summation around the supermesh loop, similar to what we have done previously. The difference this time is that we need to recognize that components each see one of the original mesh currents; namely I1 or I2 in this case. We do not solve for a supermesh current, we simply use the supermesh to define the loop for the KVL summation. The summation follows: ΣV rises = ΣV drops E1 = V R1+V R2+E2 The voltage drops across the resistors can be expanded using Ohm's law, using the original mesh current associated with each resistor. E1 −E 2 = I 1 R1+I2 R2 By inspection, I s = I 1 −I 2 or I 2 = I 1 −I s We now have two equations with two unknowns and can solve for I1 and I2. This procedure is illustrated in the following example. Example 7.8 Find Vb for the circuit of Figure 7.28. First, we'll label the loops, as shown in Figure 7.29. Next, we'll perform a KVL summation around the supermesh loop. ΣV rises = ΣV drops 20 V = V R1+V R2+12V Expand using Ohm's law and rearrange: 20 V −12 V = 5ΩI1+8ΩI 2 8V = 5ΩI 1+8ΩI 2 236 Figure 7.28 Circuit for Example 7.8. | DCElectricalCircuitAnalysis_Page_236_Chunk1158 |
By inspection we know 3A = I2 −I 1 or I 2 = I 1+3A We can substitute this expression into the prior supermesh expression and solve for I1: 8 V = 5ΩI 1+8ΩI 2 8 V = 5ΩI 1+8Ω(I1+3A) 8 V = 5ΩI 1+8ΩI1+24 V −16 V = 13Ω I 1 I 1 ≈−1.231A Thus, I2 = −1.231 A + 3 A, or 1.769 A. To determine Vb we simply subtract the drop across the 5 Ω resistor from the 20 volt source. V b = 20 V −I 15Ω V b = 20 V −(−1.231A)5Ω V b ≈26.15 V As a crosscheck, we could also add the voltage across the 8 Ω resistor to the 12 volt source: V b = 12V +I28Ω V b = 12V +1.769 A 8Ω V b ≈26.15 V Comparison of Nodal and Mesh Comparison of Nodal and Mesh Compared to nodal analysis, mesh analysis has the advantage of dealing with resistances rather than conductances when writing the system of equations. Further, the mesh inspection method works with voltage sources which tends to be convenient for many circuits, while the nodal inspection method requires current sources. On the down side, the resulting set of mesh currents requires further processing in order to find either branch currents or node voltages, while nodal analysis produces node voltages directly with no further processing. Mesh also has the disadvantage of being limited to planar circuits while there are no such limits to nodal. 237 Figure 7.29 Circuit of Figure 7.28 with supermesh labeled. | DCElectricalCircuitAnalysis_Page_237_Chunk1159 |
7.4 Dependent Sources 7.4 Dependent Sources A dependent source is a current or voltage source whose value is not fixed (i.e., independent) but rather which depends on some other circuit current or voltage. The general form for the value of a dependent source is Y = kX where X and Y are currents and/or voltages and k is the proportionality factor. For example, the value of a dependent voltage source may be a function of a current, so instead of the source being equal to, say, 10 volts, it could be equal to twenty times the current passing through a particular resistor, or V=20I. There are four possible dependent sources: the voltage-controlled voltage source (VCVS), the current-controlled voltage source (CCVS), the voltage-controlled current source (VCCS), and the current-controlled current source (CCCS). The source and control parameters are the same for both the VCVS and the CCCS so k is unitless (although it may be given as volts/volt and amps/amp, respectively). For the VCCS and CCVS, k has units of amps/volt and volts/amp, respectively. These are referred to as the transresistance and transconductance of the sources with units of ohms and siemens. The schematic symbols for dependent or controlled sources are usually drawn using a diamond. Also, there may be a secondary connection for the controlling current or voltage. Examples of a voltage-controlled voltage source, current-controlled voltage source, voltage-controlled current source and a current-controlled current source are shown in Figure 7.30, left-to-right. On each of these symbols, the control element is shown to the left of the source. This portion is not always drawn on a schematic. Instead, the source simply may be labeled as a function, for example, V = 0.02 IX where IX is the controlling current. Dependent sources are not “off-the-shelf” items in the same way that a battery is. Rather, dependent sources usually are used to model the behavior of more complex devices. For example, a bipolar junction transistor commonly is modeled as a CCCS while a field effect transistor may be modeled as a VCCS18. Similarly, many op amp circuits are modeled as VCVS systems. Solutions for circuits using dependent sources follow along the lines of those established for independent sources (i.e., the application of Ohm's law, KVL, KCL, etc.), however, the sources are now dependent on the remainder of the circuit which tends to complicate the analysis. 18 For details, see Fiore, J.M., Semiconductor Devices: Theory and Application, and Operational Amplifiers and Linear Integrated Circuits: Theory and Application, both free OER titles. 238 Figure 7.30 Dependent sources (L to R): VCVS, CCVS, VCCS, CCCS. | DCElectricalCircuitAnalysis_Page_238_Chunk1160 |
In general, there are two possible configurations: isolated and coupled. A example of the isolated form is shown in Figure 7.31. In this configuration, the dependent source (center) does not interact with the sub- circuit on the left driven by the independent source, thus it can be analyzed as two separate circuits. Solutions for this form are relatively straightforward in that the control value for the dependent source can be computed directly. This value is then substituted into the dependent source and the analysis continues as is usual. Sometimes it is convenient if the solution for a particular voltage or current is defined in terms of the control parameter rather than as a specific value (e.g., the voltage across a particular resistor might be expressed as 8 VA instead of 12 volts, where VA is 1.5 volts). The second type of circuit (coupled) is somewhat more complex in that the dependent source can affect the parameter that controls the dependent source. In other words, the dependent source(s) will contribute terms that include the controlling parameter(s), thus partly controlling itself. Some additional effort will be in order to solve these circuits. To illustrate, consider the circuit of Figure 7.32. In this example it should be obvious that the current from the dependent source can affect the voltage at node a, and it is this very voltage that in turn sets up the value of the current source. Circuits of this type can be analyzed using mesh or nodal analysis. Nodal analysis works well here and is illustrated following. 239 Figure 7.31 Dependent source: isolated configuration. Figure 7.32 Dependent source: coupled configuration. | DCElectricalCircuitAnalysis_Page_239_Chunk1161 |
We begin by defining current directions. Assume that the currents through R1 and R3 are flowing into node a, the current through R2 is flowing out of node a, and the current through R4 is flowing out of node b. We shall number the branch currents to reflect the associated resistor. The resulting KCL equations are: Σ I in = Σ I out Node a: I 1+I3 = I 2 Nodeb: k V a = I 3+I 4 The currents are then described by their Ohm's law equivalents: Node a: E−V a R1 + V b−V a R3 = V a R2 Nodeb: k V a = V b−V a R3 + V b R4 Expanding terms yields: Node a: E R1 −V a R1 +V b R3 −V a R3 = V a R2 Nodeb: k V a = V b R3 −V a R3 +V b R4 Collecting terms and simplifying yields: Node a : E R1 =( 1 R1 + 1 R2 + 1 R3)V a −1 R3 V b Nodeb: 0 =−(k+ 1 R3)V a +( 1 R3 + 1 R4)V b Values for the resistors, k, and E normally are known, so the analysis proceeds directly. Also, it is worth remembering that it is possible to perform source conversions on dependent sources, within limits. The same procedure is followed as for independent sources. The new source will remain a dependent source (e.g., converting VCVS to VCCS). This process is not applicable if the control parameter directly involves the internal impedance (i.e., is its voltage or current). Time for an example, this one utilizing a simplified model of a transistor amplifier. 240 | DCElectricalCircuitAnalysis_Page_240_Chunk1162 |
Example 7.9 Find Vb and Vc for the circuit of Figure 7.33. This CCCS would be typical of a simple model of a bipolar junction transistor (nodes a, b, and c). Ideally, the output voltage, Vc, would be equal to the input voltage (1 V) times the resistor ratio of 15 kΩ / 2 kΩ and inverted, or approximately −7.5 volts. In reality, it usually comes up a little short of this. Let's see how well this works out. This circuit is a good candidate for nodal analysis using the general method. Note that the points labeled a and b are the same node so we can write just two KCL summations. Using the current directions as draw on the schematic, for node b we have: Σ I in = Σ I out I x+100I x = V b 2 kΩ 1011V−V b 10 k Ω = V b 2 kΩ 10.1 mA =( 1 2 kΩ + 101 10 kΩ)V b 10.1 mA = 10.6 mSV b V b = 0.95283 V For node c we have: Σ I in = Σ Iout −V c 15 kΩ = 100 I x − V c 15 kΩ = 100 1V−V b 10 k Ω V c = −7.0755 V Alternately, instead of writing the second KCL summation we could have used Vb to determine Ix, i.e., (1 − Vb)/10 kΩ. As the current through the 15 k Ω resistor is 100Ix, we could then use Ohm's law to find Vc. In any case, we do see that Vc is inverted and just shy of the 7.5 volt estimate. 241 Figure 7.33 Circuit for Example 7.9. | DCElectricalCircuitAnalysis_Page_241_Chunk1163 |
Computer Simulation Computer Simulation For verification, the dependent source circuit of Example 7.9 is entered into a simulator as shown in Figure 7.34. A DC operating point analysis is run, the results of which are shown in Figure 7.35. The output voltage shows roughly −7.08 volts and a Vb of just under 1 volt, thus achieving excellent agreement with the manual calculation. While this simulation is effective in this case, just using a dependent source to stand in for a transistor is rather limited. There are many other, perhaps more subtle, elements to a proper transistor model that will ensure both greater accuracy and proper results under a wide range of operating conditions. Any quality simulator will include a parts library using simulation models that are fine-tuned to particular manufacturer part numbers. 242 Figure 7.34 Circuit of Figure 7.33 in a simulator. Figure 7.35 Simulation results for the circuit of Figure 7.33. | DCElectricalCircuitAnalysis_Page_242_Chunk1164 |
7.5 Summary 7.5 Summary Nodal analysis can be used to solve virtually any complex multi-source DC electrical circuit. It is based on KCL, writing expressions involving each node in the circuit. A system of equations results, there being as many equations as there are nodes in the circuit, minus the reference node (which usually is taken as ground). The set of equations will exhibit diagonal symmetry, which can be used as a crosscheck before setting out to solve them. The solution will be a complete set of node voltages. From these, any branch current may be determined as needed. There are two different methods of creating the system of equations. The first method is deemed the general method and will work for a mix of current sources and voltage sources. Individual currents are defined based on the node voltages and any known current sources. KCL is then applied at each node, followed by simplification and combination of terms to arrive at the end equations. The second approach uses the inspection method. If the circuit contains only current sources (or if the voltage sources are converted), this method allows direct generation of the system of equations without the need for simplification and thus is less prone to error. Mesh analysis can be used to solve any planar complex multi-source DC electrical circuit. It is based on KVL, writing expressions involving each closed loop in the circuit. The loops are minimally sized and the set of loops must cover every component in the circuit. A system of equations results, there being as many equations as there are loops. As with nodal analysis, the set of equations will exhibit diagonal symmetry. The solution will be a complete set of mesh currents. From these, any node voltage may be determined. Like nodal, mesh offers two different methods of creating the system of equations. The general method will work for a mix of current sources and voltage sources. Individual loops are defined based on the meshing currents passing through each component. KVL is then applied around each loop, followed by simplification and combination of terms to arrive at the end equations. In contrast, if the circuit contains only voltage sources (or if the current sources are converted), then the inspection method may be used. This method allows direct generation of the system of equations and is faster and less error prone. Dependent sources are current or voltage sources whose value depends on the current or voltage developed in some other part of the circuit. There are four types: current controlled current source, current controlled voltage source, voltage controlled current source and voltage controlled voltage source. These sources are used commonly to model the characteristics of active devices such bipolar and field effect transistors. Techniques for solution tend to be a bit more involved than when using constant sources, however, nodal analysis in particular tends to work well. 243 | DCElectricalCircuitAnalysis_Page_243_Chunk1165 |
Review Questions Review Questions 1. Describe the practical differences between nodal analysis and mesh analysis. 2. What is diagonal symmetry? Of what use is it? 3. What are the differences between the general method and the inspection method of nodal analysis? 4. What are the differences between the general method and the inspection method of mesh analysis? 5. What is a supernode? 6. What is a supermesh? 7. Describe the concept of dependent sources and how they differ from independent or constant sources. 7.6 Exercises 7.6 Exercises Analysis Analysis 1. Given the circuit in Figure 7.36, write the mesh loop equations. 2. Using mesh analysis, determine the value of Vb for the circuit shown in Figure 7.36. 3. For the circuit shown in Figure 7.36, use mesh analysis to determine the current through the 1 kΩ resistor. 4. Given the circuit in Figure 7.37, write the mesh loop equations and the associated determinants. 244 Figure 7.36 Figure 7.37 | DCElectricalCircuitAnalysis_Page_244_Chunk1166 |
5. Using mesh analysis, determine the value of Vb for the circuit shown in Figure 7.37. 6. For the circuit shown in Figure 7.37, use mesh analysis to determine the current through the 500 Ω resistor. 7. Given the circuit in Figure 7.38, write the mesh loop equations. 8. Using mesh analysis, determine the value of Vb for the circuit shown in Figure 7.38. 9. For the circuit shown in Figure 7.38, use mesh analysis to determine the current through the 75 Ω resistor. 10. Given the circuit in Figure 7.39, write the mesh loop equations and the associated determinants. 11. Using mesh analysis, determine the value of Vbc for the circuit shown in Figure 7.39. 12. For the circuit shown in Figure 7.39, use mesh analysis to determine the current through the 10 kΩ resistor. 245 Figure 7.38 Figure 7.39 | DCElectricalCircuitAnalysis_Page_245_Chunk1167 |
13. Given the circuit in Figure 7.40, write the mesh loop equations. 14. Using mesh analysis, determine the value of Vac for the circuit shown in Figure 7.40. 15. For the circuit shown in Figure 7.40, use mesh analysis to determine the current through the 4 kΩ resistor. 16. Given the circuit in Figure 7.41, write the mesh loop equations and the associated determinants. 17. Using mesh analysis, determine the value of Vc for the circuit shown in Figure 7.41. 18. For the circuit shown in Figure 7.41, use mesh analysis to determine the current through the 10 kΩ resistor. 19. Given the circuit in Figure 7.42, write the mesh loop equations. 246 Figure 7.40 Figure 7.41 Figure 7.42 | DCElectricalCircuitAnalysis_Page_246_Chunk1168 |
20. Using mesh analysis, determine the value of Vbd for the circuit shown in Figure 7.42. 21. For the circuit shown in Figure 7.42, use mesh analysis to determine the current through the 500 Ω resistor. 22. Given the circuit in Figure 7.43, write the mesh loop equations. 23. Using mesh analysis, determine the value of Vad for the circuit shown in Figure 7.43. 24. For the circuit shown in Figure 7.43, use mesh analysis to determine the current through the 3 kΩ resistor. 25. Given the circuit in Figure 7.44, write the mesh loop equations. 247 Figure 7.44 Figure 7.43 | DCElectricalCircuitAnalysis_Page_247_Chunk1169 |
26. Using mesh analysis, determine the value of Ve for the circuit shown in Figure 7.44. 27. For the circuit shown in Figure 7.44, use mesh analysis to determine the current through the 9 kΩ resistor. 28. Given the circuit in Figure 7.45, write the mesh loop equations and the associated determinants. 29. Using mesh analysis, determine the value of Vbc for the circuit shown in Figure 7.45. 30. Given the circuit shown in Figure 7.45, use mesh analysis to determine the current through the 600 Ω resistor. 31. Given the circuit in Figure 7.46, write the mesh loop equations. 32. Using mesh analysis, determine the value of Vbc for the circuit shown in Figure 7.46. 33. Given the circuit shown in Figure 7.46, use mesh analysis to determine the current through the 800 Ω resistor. 248 Figure 7.45 Figure 7.46 | DCElectricalCircuitAnalysis_Page_248_Chunk1170 |
34. For the circuit in Figure 7.47, write the mesh loop equations. 35. Using mesh analysis, determine the value of Va for the circuit shown in Figure 7.47. 36. For the circuit shown in Figure 7.47, use mesh analysis to determine the current through the 3 kΩ resistor. 37. Given the circuit in Figure 7.48, write the mesh loop equations. 38. Using mesh analysis, determine the value of Vc for the circuit shown in Figure 7.48. 39. For the circuit shown in Figure 7.48, use mesh analysis to determine the current passing through the 8.5 kΩ resistor. 40. Given the circuit in Figure 7.49, write the mesh loop equations (consider using source conversion). 249 Figure 7.47 Figure 7.48 Figure 7.49 | DCElectricalCircuitAnalysis_Page_249_Chunk1171 |
41. Using mesh analysis, determine the value of Va for the circuit shown in Figure 7.49. 42. For the circuit shown in Figure 7.49, use mesh analysis to determine the current passing through the 30 kΩ resistor. 43. Given the circuit in Figure 7.50, write the mesh loop equations (consider using source conversion). 44. Using mesh analysis, determine the value of Vb for the circuit shown in Figure 7.50. 45. For the circuit shown in Figure 7.50, use mesh analysis to determine the current through the 5 kΩ resistor. 46. Given the circuit in Figure 7.51, write the node equations. 47. Using nodal analysis, determine the value of Vb for the circuit shown in Figure 7.51. 48. For the circuit shown in Figure 7.51, use nodal analysis to determine the current through the 3 kΩ resistor. 250 Figure 7.50 Figure 7.51 | DCElectricalCircuitAnalysis_Page_250_Chunk1172 |
49. Given the circuit in Figure 7.52, write the node equations. 50. Using nodal analysis, determine the value of Vb for the circuit shown in Figure 7.52. 51. For the circuit shown in Figure 7.52, use nodal analysis to determine the current passing through the 12 kΩ resistor. 52. Given the circuit in Figure 7.53, write the node equations. 53. Using nodal analysis, determine the value of Vba for the circuit shown in Figure 7.53. 54. For the circuit shown in Figure 7.53, use nodal analysis to determine the current passing through the 100 Ω resistor. 55. Given the circuit in Figure 7.54, write the node equations. 251 Figure 7.52 Figure 7.53 Figure 7.54 | DCElectricalCircuitAnalysis_Page_251_Chunk1173 |
56. Using nodal analysis, determine the value of Vac for the circuit shown in Figure 7.54. 57. For the circuit shown in Figure 7.54, use nodal analysis to determine the current passing through the 20 kΩ resistor. 58. Given the circuit in Figure 7.55, write the node equations. 59. Using nodal analysis, determine the value of Vb for the circuit shown in Figure 7.55. 60. For the circuit shown in Figure 7.55, use nodal analysis to determine the current through the 100 Ω resistor. 61. Given the circuit in Figure 7.56, write the node equations. 62. Using nodal analysis, determine the value of Vb for the circuit shown in Figure 7.56. 63. For the circuit shown in Figure 7.56, use nodal analysis to determine the current through the 40 Ω resistor. 252 Figure 7.55 Figure 7.56 | DCElectricalCircuitAnalysis_Page_252_Chunk1174 |
64. Given the circuit in Figure 7.48, write the node equations. 65. Using nodal analysis, determine the value of Vd for the circuit shown in Figure 7.48. 66. For the circuit shown in Figure 7.48, use nodal analysis to determine the current passing through the 20 kΩ resistor. 67. Given the circuit in Figure 7.49, write the node equations using the general approach. Do not use source conversions. 68. Using nodal analysis, determine the value of Vab for the circuit shown in Figure 7.49. 69. For the circuit shown in Figure 7.49, use nodal analysis to determine the current passing through the 30 kΩ resistor. 70. Given the circuit in Figure 7.50, write the node equations. 71. Using nodal analysis, determine the value of Vb for the circuit shown in Figure 7.50. 72. For the circuit shown in Figure 7.50, use nodal analysis to determine the current through the 4 kΩ resistor. 73. For the circuit of Figure 7.57, determine Vb. 74. For the circuit of Figure 7.58, determine Vc. 253 Figure 7.57 Figure 7.58 | DCElectricalCircuitAnalysis_Page_253_Chunk1175 |
75. Given the circuit of Figure 7.59, determine Vb. 76. Find the current through the 10 kΩ resistor given the circuit of Figure 7.60. 77. Given the circuit of Figure 7.61, determine Vc. 78. In the circuit of Figure 7.62, determine Va. 254 Figure 7.59 Figure 7.60 Figure 7.61 Figure 7.62 | DCElectricalCircuitAnalysis_Page_254_Chunk1176 |
79. For the circuit of Figure 7.63, determine Va. 80. For the circuit of Figure 7.64, determine Va. Challenge Challenge 81. Given the circuit in Figure 7.43, write the node equations. 82. Using nodal analysis, determine the value of Vb for the circuit shown in Figure 7.43. 83. For the circuit shown in Figure 7.43, use nodal analysis to determine the current through the 3 kΩ resistor. 84. Given the circuit in Figure 7.41, write the node equations. 85. Using nodal analysis, determine the value of Vc for the circuit shown in Figure 7.41. 86. For the circuit shown in Figure 7.41, use nodal analysis to determine the current through the 8 kΩ resistor. 87. Given the circuit in Figure 7.45, write the node equations and the associated determinants. 88. Using nodal analysis, determine the value of Vbc for the circuit shown in Figure 7.45. 255 Figure 7.63 Figure 7.64 | DCElectricalCircuitAnalysis_Page_255_Chunk1177 |
89. For the circuit shown in Figure 7.45, use nodal analysis to determine the current through the 2 kΩ resistor. 90. Given the circuit of Figure 7.65, determine Vc. 91. Find the current through the 10 kΩ resistor in the circuit of Figure 7.66. 92. Given the circuit of Figure 7.67, determine Vc. 256 Figure 7.65 Figure 7.66 Figure 7.67 | DCElectricalCircuitAnalysis_Page_256_Chunk1178 |
93. Given the circuit of Figure 7.68, determine the current through the 5 kΩ resistor. 94. For the circuit of Figure 7.69, determine Vb. 95. For the circuit of Figure 7.70, determine Vc. 257 Figure 7.68 Figure 7.69 Figure 7.70 | DCElectricalCircuitAnalysis_Page_257_Chunk1179 |
96. For the circuit of Figure 7.71, determine Va. 97. For the circuit of Figure 7.72, determine Vb. Simulation Simulation 98. Perform a DC bias simulation on the circuit depicted in Figure 7.42 to verify the component currents. 99. Perform a DC bias simulation on the circuit depicted in Figure 7.44 to verify the component currents. 100. Perform a DC bias simulation on the circuit depicted in Figure 7.42 to verify the loop currents and node voltages. 101. Perform a DC bias simulation on the circuit depicted in Figure 7.51 to verify the node voltages. 102. Perform a DC bias simulation on the circuit depicted in Figure 7.54 to verify the node voltages. 103. Perform a DC bias simulation on the circuit depicted in Figure 7.55 to verify the node voltages. 258 Figure 7.71 Figure 7.72 | DCElectricalCircuitAnalysis_Page_258_Chunk1180 |
Notes Notes ♫♫ ♫♫ 259 | DCElectricalCircuitAnalysis_Page_259_Chunk1181 |
8 8 Capacitors Capacitors 8.0 Chapter Learning Objectives 8.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Describe the theoretical and practical aspects of capacitor construction. • Describe the current-voltage characteristic behavior of capacitors. • Utilize component data sheets to determine operating characteristics of capacitors. • Determine the initial and steady-state equivalents of resistor-capacitor networks. • Determine the transient response of basic RC networks. 8.1 Introduction 8.1 Introduction This chapter introduces another passive device, the capacitor. Capacitors are fundamentally different from resistors in terms of both their construction and their operation. For starters, when placed in DC circuits, capacitors are not ohmic, unlike resistors. Their current-voltage characteristic does not respond to Ohm's law. Instead, their current-voltage characteristic is dynamic in nature. Further, in the ideal case, capacitors do not dissipate power. Indeed, capacitors are energy storage devices. In a way, you could imagine them to be a little like rechargeable batteries, but unlike batteries these devices could be charged or discharged in tiny fractions of a second and last for billions upon billions of cycles. Capacitors are an integral part of modern electronic systems. They are used in AC-to-DC power supplies to help smooth and stabilize the output voltage. In audio and communications systems they are used in filters, for example to control the high and low frequency response of amplifiers and similar equipment, or for tuning purposes. Essentially, in any application that needs to smooth out a varying voltage, store electric charge or filter a signal; a capacitor is likely to be used. Much of the original work on capacitors was done in the 18th century. One of the earliest examples of a capacitor was the Leyden jar, which consists of a glass bottle or jar lined with metal foil on the inside and the outside. Benjamin Franklin's famous kite experiment made use of a Leyden jar. In the early days, capacitors were known as condensers (not to be confused with the condensers used in air conditioning systems). 260 | DCElectricalCircuitAnalysis_Page_260_Chunk1182 |
8.2 Capacitance and Capacitors 8.2 Capacitance and Capacitors A capacitor is a device that stores energy. Capacitors store energy in the form of an electric field. At its most simple, a capacitor can be little more than a pair of metal plates separated by air. As this constitutes an open circuit, DC current will not flow through a capacitor. If this simple device is connected to a DC voltage source, as shown in Figure 8.1, negative charge will build up on the bottom plate while positive charge builds up on the top plate. This process will continue until the voltage across the capacitor is equal to that of the voltage source. In the process, a certain amount of electric charge will have accumulated on the plates. The ability of this device to store charge with regard to the voltage appearing across it is called capacitance. Its symbol is C and it has units of farads (F), in honor of Michael Faraday, a 19th century English scientist who did early work in electromagnetism. By definition, if a total charge of 1 coulomb is associated with a potential of 1 volt across the plates, then the capacitance is 1 farad. 1 farad ≡ 1 coulomb / 1 volt (8.1) or more generally, C= Q V (8.2) Where C is the capacitance in farads, Q is the charge in coulombs, V is the voltage in volts. From Equation 8.2 we can see that, for any given voltage, the greater the capacitance, the greater the amount of charge that can be stored. We can also see that, given a certain size capacitor, the greater the voltage, the greater the charge that is stored. These observations relate directly to the amount of energy that can be stored in a capacitor. 261 Figure 8.1 Basic capacitor with voltage source. + - + - + - + - + - + - + - electron flow - + | DCElectricalCircuitAnalysis_Page_261_Chunk1183 |
Unsurprisingly, the energy stored in capacitor is proportional to the capacitance. It is also proportional to the square of the voltage across the capacitor. W =1 2 CV 2 (8.3) Where W is the energy in joules, C is the capacitance in farads, V is the voltage in volts. The basic capacitor consists of two conducting plates separated by an insulator, or dielectric. This material can be air or made from a variety of different materials such as plastics and ceramics. This is depicted in Figure 8.2. For practical capacitors, the plates may be stacked alternately or even made of foil and formed into a rolled tube. However it is constructed, the characteristics of the dielectric will play a major role in the performance of the device, as we shall see. In general, capacitance increases directly with plate area, A, and inversely with plate separation distance, d. Further, it is also proportional to a physical characteristic of the dielectric; the permittivity, ε. Thus, capacitance is equal to: C=ε A d (8.4) Where C is the capacitance in farads, A is the plate area in square meters, d is the plate separation distance in meters, ε is permittivity of the dielectric between the plates. It should be noted that the effective plate area is somewhat larger than the precise physical area of the plates. This is due to a phenomenon called fringing. Essentially, the electric field lines bulge outward at the plate edges rather than maintain uniform parallel orientation. This is illustrated in Figure 8.3 262 l w d Plate 1 Plate 2 Dielectric Figure 8.2 Components of a generic capacitor. | DCElectricalCircuitAnalysis_Page_262_Chunk1184 |
From Equation 8.4 it is obvious that the permittivity of the dielectric plays a major role in determining the volumetric efficiency of the capacitor, in other words, the amount of capacitance that can be packed into a given sized component. Some dielectrics are notably more efficient than others. To make comparisons easier, relative permittivity is often used, that is, the ratio of the dielectric's permittivity to that of a vacuum, ε0. A table of relative permittivity for a variety of dielectrics is shown in Figure 8.4. A number of common dielectrics, such as various poly plastic films and mica, exhibit permittivities two to six times that of air, but there are also ceramic dielectrics whose dielectrics are hundreds to thousands of times that of air. Material Relative permittivity, εr = ε/ε0 Vacuum 1 (ε0=8.85E−12 farads/meter) Air 1.00058986 (at STP) PTFE/Teflon 2.1 Polyethylene/XLPE 2.25 Polyimide 3.4 Polypropylene 2.2–2.36 Polystyrene 2.4–2.7 Polyester (Mylar) 3.1 Paper 1.4 Mica 3–6 Silicon dioxide 3.9 Rubber 7 Diamond 5.5–10 Silicon 11.68 Titanium dioxide 86–173 Strontium titanate 310 Calcium copper titanate >250,000 At quick glance, it might seem that choosing the dielectric with the highest permittivity would be the best choice but this is not necessarily the case. There are several other factors that go into this decision including temperature stability, leakage resistance (effective parallel resistance), ESR (equivalent series resistance) and breakdown strength. For an ideal capacitor, leakage resistance would be infinite and ESR would be zero. Unlike resistors, capacitors do not have maximum power dissipation ratings. Instead, they have maximum voltage ratings. The breakdown strength of the dielectric will 263 + + + + + + + + - - - - - - - - Figure 8.3 Capacitor electric field with fringing. Figure 8.4 Relative permittivity of various dielectrics. Data derived from Wikipedia and other sources. | DCElectricalCircuitAnalysis_Page_263_Chunk1185 |
set an upper limit on how large of a voltage may be placed across a capacitor before it is damaged. Breakdown strength is measured in volts per unit distance, thus, the closer the plates, the less voltage the capacitor can withstand. For example, halving the plate distance doubles the capacitance but also halves its voltage rating. Figure 8.5 lists the breakdown strengths of a variety of different dielectrics. Comparing the tables of Figures 8.4 and 8.5 hints at the complexity of the situation. For instance, consider polystyrene versus polypropylene. Polystyrene offers modestly increased permittivity yet polypropylene has a considerable advantage in terms of breakdown strength. As a consequence, the plates can be placed much closer together when using polypropylene while achieving the same voltage rating as a capacitor using polystyrene. Therefore, the polypropylene capacitor will require less volume for the same capacitance. As an added benefit, polypropylene exhibits high temperature stability and low moisture absorption, among other characteristics. Comparing polypropylene to polyester, we find that polyester's improved permittivity along with a similar breakdown strength yields improved volumetric efficiency over polypropylene. Unfortunately, polyester suffers from greater temperature dependence. Substance Breakdown Strength (kV/mm) Air 3.0 Borosilicate glass 20–40 PTFE (Teflon, insulating film) 60–173 Polyethylene 19–160 Polypropylene 650 Polystyrene 19.7 PEEK (Polyether ether ketone) 23 Polyester (Mylar) 580 Neoprene rubber 15.7–26.7 Distilled water 65–70 Waxed paper 40–60 Mica 118 Diamond 2,000 PZT (ceramic) 10–25 Capacitor Styles and Packaging Capacitor Styles and Packaging Capacitors are available in a wide range of capacitance values, from just a few picofarads to well in excess of a farad, a range of over 1012. Unlike resistors, whose physical size relates to their power rating and not their resistance value, the physical size of a capacitor is related to both its capacitance and its voltage rating (a 264 Figure 8.5 Dielectric strength of various dielectrics. Data derived from Wikipedia and other sources.c | DCElectricalCircuitAnalysis_Page_264_Chunk1186 |
consequence of Equation 8.4. Modest surface mount capacitors can be quite small while the power supply filter capacitors commonly used in consumer electronics devices such as an audio amplifier can be considerably larger than a D cell battery. A sampling of capacitors is shown in Figure 8.6. Toward the front and left side of the photo are a variety of plastic film capacitors. The disk-shaped capacitor uses a ceramic dielectric. The small square device toward the front is a surface mount capacitor, and to its right is a teardrop-shaped tantalum capacitor, commonly used for power supply bypass applications in electronic circuits. The medium sized capacitor to the right with folded leads is a paper capacitor, at one time very popular in audio circuitry. A number of capacitors have a crimp ring at one side, including the large device with screw terminals. These are aluminum electrolytic capacitors. These devices tend to exhibit high volumetric efficiency but generally do not offer top performance in other areas such as absolute accuracy and leakage current. They usually are polarized, meaning that the leads must match the polarity of the applied voltage. Inserting them into a circuit backwards can result in catastrophic failure. The polarity is usually identified by a series of minus signs and/or a stripe that indicates the negative lead. Tantalum capacitors are also polarized but are typically denoted with a plus sign next to the positive lead. A variable capacitor used for tuning radios is shown in Figure 8.7. One set of plates is fixed to the frame while an intersecting set of plates is affixed to a shaft. Rotating the shaft changes the amount of plate area that overlaps, and thus changes the capacitance. For large capacitors, the capacitance value and voltage rating are usually printed directly on the case. Some capacitors use “MFD” which stands for “microfarads”. While a capacitor color code exists, rather like the resistor color code, it has generally fallen out of favor. For smaller capacitors a numeric code is used that echoes the color code. Typically it consists of a three digit number such as “152”. 265 Figure 8.6 A variety of capacitor styles and packages. Figure 8.7 A variable capacitor. Figure 8.8 Capacitor schematic symbols (top-bottom): non-polarized, polarized, variable. | DCElectricalCircuitAnalysis_Page_265_Chunk1187 |
The first two digits are the precision portion and the third digit is the power of ten multiplier. The result is in picofarads. Thus, 152 is 1500 pf. The schematic symbols for capacitors are shown in Figure 8.8. There are three symbols in wide use. The first symbol, using two parallel lines to echo the two plates, is for standard non-polarized capacitors. The second symbol represents polarized capacitors. In this variant, the positive lead is drawn with a straight line for that plate and often denoted with a plus sign. The negative terminal is drawn with a curved line. The third symbol is used for variable capacitors and is drawn with an arrow through it, rather like a rheostat. In order to obtain accurate measurements of capacitors, an LCR meter, such as the one shown in Figure 8.9, may be used. These devices are designed to measure the three common passive electrical components: resistors, capacitors and inductors19. Unlike a simple digital multimeter, an LCR meter can also measure the values at various AC frequencies instead of just DC, and also determine secondary characteristics such as equivalent series resistance and effective parallel leakage resistance. Capacitor Data Sheet Capacitor Data Sheet A portion of a typical capacitor data sheet is shown in Figure 8.10. This is for a series of through-hole style metallized film capacitors using polypropylene for the dielectric. First we see a listing of general features. For starters, we find that the capacitors use a flame retardant epoxy coating and are also RoHS compliant. We then move to a set of electrical performance specifications. For example, we see that this series is available in two variants, one rated at 800 volts DC and the other rated at 1600 volts DC. Further, tolerance is available as either ±3% or ±5%. Dissipation factor (tan δ) is a measure of particular importance for AC operation and is proportional to the ESR (equivalent series resistance, ideally 0), smaller being better. The insulation resistance indicates the value of an effective parallel leakage resistance (higher is better), here, some 30,000 MΩ. Finally, we see physical size data, essential for printed circuit board layouts. Capacitors in Series and in Parallel Capacitors in Series and in Parallel Multiple capacitors placed in series and/or parallel do not behave in the same manner as resistors. Placing capacitors in parallel increases overall plate area, and thus increases capacitance, as indicated by Equation 8.4. Therefore capacitors in parallel add in value, behaving like resistors in series. In contrast, when capacitors are placed in series, it is as if the plate distance has increased, thus decreasing capacitance. Therefore capacitors in series behave like resistors in parallel. Their value is found via the reciprocal of summed reciprocals or the product-sum rule. 19 Inductors are the subject of the next chapter. 266 Figure 8.9 An LCR meter, designed to read capacitance, resistance and inductance. | DCElectricalCircuitAnalysis_Page_266_Chunk1188 |
Example 8.1 Find the equivalent capacitance of the network shown in Figure 8.11. These capacitors are all in parallel, and thus, the equivalent value is the sum of the three capacitances: CTotal = C1+C 2+C 3 CTotal = 1μ F+100nF+560nF CTotal = 1.66μ F 267 Figure 8.11 Circuit for Example 8.1. Figure 8.10 Capacitor data sheet. Courtesy of Panasonic | DCElectricalCircuitAnalysis_Page_267_Chunk1189 |
Example 8.2 Find the equivalent capacitance of the network shown in Figure 8.12. In this circuit, we find that the left and middle capacitors are in parallel. This combination is in series with the capacitor to the right: Cleft = C1+C 2 Cleft = 3.3μF+4.7μF Cleft = 8μ F CTotal = Cleft C3 Cleft+C3 CTotal = 8μ F16μ F 8μ F+16μ F CTotal ≈5.33μ F If a circuit contains nothing but a voltage source in parallel with a group of capacitors, the voltage will be the same across all of the capacitors, just as it is in a resistive parallel circuit. If the circuit instead consists of multiple capacitors that are in series with a voltage source, as shown in Figure 8.13, the voltage will divide between them in inverse proportion. In other words, the larger the capacitance, the smaller its share of the applied voltage. The voltages can also be found by first determining the series equivalent capacitance. The total charge may then be determined using the applied voltage. Finally, the individual voltages are computed from Equation 8.2, V = Q/C, where Q is the total charge and C is the capacitance of interest. This is illustrated in the following example. Example 8.3 Find the voltages across the capacitors in Figure 8.14. The first step is to determine the total capacitance. As these are in series, we can use the reciprocal rule: CTotal = 1 1 C1 + 1 C2 + 1 C3 CTotal = 1 1 2μ F+ 1 4μ F+ 1 8μ F CTotal ≈1.143μ F 268 Figure 8.12 Circuit for Example 8.2. Figure 8.13 A simple capacitors-only series circuit. Figure 8.14 Circuit for Example 8.3. | DCElectricalCircuitAnalysis_Page_268_Chunk1190 |
From here we determine the total charge: Q = V C Q = 12 V1.143μ F Q = 13.71μC Charge is constant across all of the series capacitors, therefore: V 2uF = Q C V 2uF = 13.71μC 2μ F V 2uF = 6.855V V 4uF = Q C V 4uF = 13.71μC 4μ F V 4uF = 3.427V V 8uF = Q C V 8uF = 13.71μC 8μ F V 8uF = 1.714V The sum of the three voltages is 12 volts (within rounding error) and verifies KVL as expected. A Practical Tip A Practical Tip While it may be tempting to try, do not attempt to verify the operation of Example 8.3 in the laboratory using a standard DMM. The reason is because the internal resistance of a typical digital voltmeter is many orders of magnitude lower than the leakage resistance of the capacitors. As a result, charge will be transferred to the meter, ruining the measurement. It would be akin to trying to measure the voltages across a string of resistors, each in excess of 100 MΩ, with a meter whose internal resistance is 1 MΩ. The meter's resistance dominates the parallel combination and causes excessive loading which ruins the measurement. A special sort of voltmeter, an electrostatic voltmeter or electrometer, is needed for these types of measurements. These are sometimes referred to as non-charge transfer meters. 269 | DCElectricalCircuitAnalysis_Page_269_Chunk1191 |
Current-Voltage Relationship Current-Voltage Relationship The fundamental current-voltage relationship of a capacitor is not the same as that of resistors. Capacitors do not so much resist current; it is more productive to think in terms of them reacting to it. The current through a capacitor is equal to the capacitance times the rate of change of the capacitor voltage with respect to time (i.e., its slope). That is, the value of the voltage is not important, but rather how quickly the voltage is changing. Given a fixed voltage, the capacitor current is zero and thus the capacitor behaves like an open. If the voltage is changing rapidly, the current will be high and the capacitor behaves more like a short. Expressed as a formula: i=C d v d t (8.5) Where i is the current flowing through the capacitor, C is the capacitance, dv/dt is the rate of change of capacitor voltage with respect to time. A particularly useful form of Equation 8.5 is: d v d t = i C (8.6) An alternate way of looking at Equation 8.5 indicates that if a capacitor is fed by a constant current source, the voltage will rise at a constant rate (dv/dt). It is continuously depositing charge on the plates of the capacitor at a rate of I, which is equivalent to Q/t. As long as the current is present, feeding the capacitor, the voltage across the capacitor will continue to rise. A good analogy is if we had a pipe pouring water into a tank, with the tank's level continuing to rise. This process of depositing charge on the plates is referred to as charging the capacitor. For example, considering the circuit in Figure 8.15, we see a current source feeding a single capacitor. If we were to plot the capacitor's voltage over time, we would see something like the graph of Figure 8.16. 270 Figure 8.15 Capacitor with current source. V t Figure 8.16 Capacitor voltage versus time. | DCElectricalCircuitAnalysis_Page_270_Chunk1192 |
As time progresses, the voltage across the capacitor increases with a positive polarity from top to bottom. With a theoretically perfect capacitor and source, this would continue forever, or until the current source was turned off. In reality, this line would either begin to deflect horizontally as the source reached its limits, or the capacitor would fail once its breakdown voltage was reached. The slope of this line is dictated by the size of the current source and the capacitance. Example 8.4 Determine the rate of change of voltage across the capacitor in the circuit of Figure 8.17. Also determine the capacitor's voltage 10 milliseconds after power is switched on. First, note the direction of the current source. This will produce a negative voltage across the capacitor from top to bottom. The rate of change is: dv dt = i C dv dt =−5μ A 30 nF dv dt ≈−166.7volts per second Thus, for every second, the voltage rises another −166.7 volts. Assuming it is completely uncharged when power is applied, after 10 milliseconds it will have risen to −166.7 V/s times 10 ms, or −1.667 volts. Equation 8.6 provides considerable insight into the behavior of capacitors. As just noted, if a capacitor is driven by a fixed current source, the voltage across it rises at the constant rate of i/C. There is a limit to how quickly the voltage across the capacitor can change. An instantaneous change means that dv/dt is infinite, and thus, the current driving the capacitor would also have to be infinite (an impossibility). This is not an issue with resistors, which obey Ohm's law, but it is a limitation of capacitors. Therefore we can state a particularly important characteristic of capacitors: The voltage across a capacitor cannot change instantaneously. (8.7) This observation will be key to understanding the operation of capacitors in DC circuits. 271 Figure 8.17 Circuit for Example 8.4. | DCElectricalCircuitAnalysis_Page_271_Chunk1193 |
8.3 Initial and Steady-State Analysis of RC Circuits 8.3 Initial and Steady-State Analysis of RC Circuits When analyzing resistor-capacitor circuits, always remember that capacitor voltage cannot change instantaneously. If we assume that a capacitor in a circuit is not initially charged, then its voltage must be zero. The instant the circuit is energized, the capacitor voltage must still be zero. If there is no voltage across the device, then it is behaving like a short circuit. We call this the initial state. Thus, we have our first rule regarding RC circuits: For DC analysis, initially capacitors appear as shorts. (8.8) Consider the circuit of Figure 8.18. Assume that C1 and C2 are initially uncharged and there is no voltage across them. The instant power is applied, the two capacitors appear as short circuits. If we redraw the circuit for this instant in time, we arrive at the equivalent circuit shown in Figure 8.19. Given this equivalent, we can see that shorting C2 places R2 and R3 in parallel, however, they are both shorted out by C1. This leaves only R1 left in the circuit along with the source, E. At this point, currents will begin to flow, and thus begin charging up the capacitors. As the capacitor voltages rise, the current will begin to decrease, and eventually the capacitors will stop charging. At that point no further current will be flowing, and thus the capacitor will behave like an open. We call this the steady- state condition and we can state our second rule: At steady-state, capacitors appear as opens. (8.9) 272 Figure 8.18 A basic resistor-capacitor (RC) circuit. Figure 8.19 A basic RC circuit, initial state. | DCElectricalCircuitAnalysis_Page_272_Chunk1194 |
Continuing with the example, at steady-state both capacitors behave as opens. This is shown in Figure 8.20. This leaves E to drop across R1 and R2. This will create a simple voltage divider. The steady-state voltage across C1 will equal that of R2. As C2 is also open, the voltage across R3 will be zero while the voltage across C2 will be the same as that across R2. In reality, practical capacitors can be thought of as an ideal capacitance in parallel with a very large (leakage) resistance, so there will be a limit to this performance. Example 8.5 Given the circuit of Figure 8.21, find the voltage across the 6 kΩ resistor for both the initial and steady-state conditions assuming the capacitor is initially uncharged. For the initial state the capacitor is treated as a short. The initial state equivalent circuit is drawn below in Figure 8.22. Immediately apparent is the parallel connection between the 6 kΩ and 3 kΩ resistors. This combination is equivalent to 2 kΩ. Therefore, we can perform a voltage divider to find the potential across the 6 kΩ (i.e., the 2 kΩ combo). V 6k = E Rx Rx+Ry V 6k = 24 V 2 kΩ 2 kΩ+1 kΩ V 6k = 16 V For the steady-state condition the capacitor will be fully charged, its current will be zero, and we treat it as an open. The steady-state equivalent circuit is 273 Figure 8.22 Circuit of Figure 8.20, initial state. Figure 8.20 A basic RC circuit, steady-state. Figure 8.21 Circuit for Example 8.4. | DCElectricalCircuitAnalysis_Page_273_Chunk1195 |
drawn below in Figure 8.23. The 3 kΩ resistor is now out of the picture, leaving us with the 6 kΩ in series with the 1 kΩ resistor. Once again, a voltage divider may be used to determine the voltage across the 6 kΩ. V 6k = E Rx Rx+Ry V 6k = 24 V 6k Ω 6k Ω+1 kΩ V 6k = 20.57V 8.4 Transient Response of RC Circuits 8.4 Transient Response of RC Circuits The question remains, “What happens between the time the circuit is powered up and when it reaches steady-state?” This is known as the transient response. Consider the circuit shown in Figure 8.24. Note the use of a voltage source rather than a fixed current source, as examined earlier. The key to the analysis is to remember that capacitor voltage cannot change instantaneously. Assuming the capacitor is uncharged, the instant power is applied, the capacitor voltage must be zero. Therefore all of the source voltage drops across the resistor. This creates the initial current, and this current starts to charge the capacitor (the initial rate being equal to i/C as dictated by Equation 8.6). According to Kirchhoff's voltage law, as the capacitor voltage begins to increase, the resistor voltage must decrease because the sum of the two must equal the fixed source voltage. This means that the circulating current must also decrease. This, in turn, means that the rate of capacitor voltage increase begins to slow. As the capacitor voltage continues to increase, less voltage is available for the resistor, causing further reductions in current, and a further slowing of the rate of capacitor voltage change. Eventually, the capacitor voltage will be nearly equal to the source voltage. This will result in a very small potential across the resistor and an equally small current, slowing subsequent capacitor voltage increases to a near standstill. Theoretically, the capacitor voltage approaches the source voltage but never quite equals it. Similarly, the current drops to near zero, but never completely turns off. This is illustrated in Figure 8.25. 274 Figure 8.23 Circuit of Figure 8.20, steady- state. Figure 8.24 A simple RC circuit. | DCElectricalCircuitAnalysis_Page_274_Chunk1196 |
The dashed red line represents the initial rate of change of capacitor voltage. This trajectory is what would be expected if an ideal current source drove the capacitor, as in Example 8.4. As noted previously, the rate of voltage change versus times is equal to i/C, and therefore in this case, E/RC. If the initial rate of change were to continue unabated, the source voltage, E, would be reached in RC seconds. Consequently, RC is referred to as the charge time constant and is denoted by τ (Greek letter tau). Thus, Time constant, τ = RC (8.10) As noted, once the capacitor begins to charge, the current begins to decrease and the capacitor voltage curve begins to fall away from the initial trajectory. The solid red curve represents the capacitor voltage. Notice that after five time constants the capacitor is nearly fully charged and the circuit is considered to be in steady-state (i.e., the capacitor behaves as an open). Steady-state is reached in approximately five time constants. (8.11) 275 Figure 8.25 Normalized charge and discharge curves. Normalized Charge/Discharge Curves Percent of Maximum 0 20 40 60 80 100 Time Constants 0 1 2 3 4 5 | DCElectricalCircuitAnalysis_Page_275_Chunk1197 |
This sort of recursively dependent operation is characteristic of exponential functions. The equation for the capacitor's voltage charging curve is: V C (t) = E(1 −ϵ −t τ) (8.12) Where VC(t) is the capacitor voltage at time t, E is the source voltage, t is the time of interest, τ is the time constant, ε (also written e) is the base of natural logarithms, approximately 2.718. The dashed blue line shows the initial slope of current change. The solid blue curve shows the circulating current (and by extension of Ohm's law, the resistor voltage). The equation for this curve, which follows the general shape ε−t, is: I (t)= E R ϵ −t τ (8.13) and V R(t) = E ϵ−t τ (8.14) Once power is removed or bypassed, the stored charge on the capacitor will dissipate through any associated resistor(s) creating a discharge current which will end with the capacitor voltage drained back to zero. During the discharge phase, both the capacitor's voltage and current will follow the solid blue curve; Equations 8.13 and 8.14 being appropriate. The discharge time constant may be different from the charge times constant, depending on the associated resistances. A precise derivation of the exponential charge/discharge equations is given in Appendix C. Example 8.6 Given the circuit of Figure 8.26, assume the switch is closed at time t = 0. Determine the charging time constant, the amount of time after the switch is closed before the circuit reaches steady-state, and the capacitor voltage at t = 0, t = 50 milliseconds and t = 1 second. Assume the capacitor is initially uncharged. First, the time constant: τ = RC τ = 50 kΩ2μF τ = 100ms 276 Figure 8.26 Circuit for Example 8.6. | DCElectricalCircuitAnalysis_Page_276_Chunk1198 |
Steady-state will be reached in five time constants, or 500 milliseconds. Therefore we know that VC(0) = 0 volts and VC(1) = 100 volts. To find VC(50 ms) we simply solve Equation 8.12. V C(t) = E(1 −ϵ −t τ) V C (50 ms) = 100V(1 −ϵ −50 ms 100 ms) V C (50 ms) ≈39.35V This value can also be determined graphically from Figure 8.25. The time of 50 milliseconds represents one-half time constant. Find this value on the horizontal axis and then track straight up to the solid red curve that represents the charging capacitor voltage. The point of intersection is at approximately 40% of the maximum value on the vertical axis. The maximum value here is the source voltage of 100 volts. Therefore the capacitor will have reached approximately 40% of 100 volts, or just about 40 volts. Computer Simulation Computer Simulation The circuit of Figure 8.26 is entered into a simulator, as shown in Figure 8.27. In order to reflect the notion of a time-varying circuit with a switch, the 100 volt DC voltage source has been replaced with a rectangular pulse voltage source. This source starts at 0 volts and then immediately steps up to 100 volts. It stays at this level for 500 milliseconds before dropping back to 0 volts. A transient (i.e., time domain) simulation is run, plotting the capacitor voltage over time for the first 500 milliseconds. This is seen in Figure 8.28. There are several items to note here. First, we see that the overall shape perfectly echoes the generic curve presented in Figure 8.25. Second, we see that after the predicted five time constants, or 500 milliseconds, the capacitor voltage has plateaued, indicating 277 Figure 8.27 Circuit of Figure 8.26 in a simulator. | DCElectricalCircuitAnalysis_Page_277_Chunk1199 |
steady-state. Third, at steady-state the capacitor voltage has virtually reached the maximum value set be the source, or 100 volts. Finally, at 50 milliseconds, we see that the capacitor voltage has reached roughly 40 volts, just as predicted. As mentioned previously, it is possible for a circuit to have different charge and discharge time constants. This could occur if a switch introduces or removes component(s) between the charge versus discharge phases. Further, because the capacitor is discharging, its current direction will be opposite to that of the charging current. KVL still must be satisfied, but because the capacitor is now behaving as a source, the magnitude of the discharge resistance's voltage must equal the capacitor voltage magnitude. Therefore, its curve will take the same shape (the solid blue curve of Figure 8.25). These issues will be illustrated in the following example. Example 8.7 For the circuit of Figure 8.29, assume the capacitor is initially uncharged. At time t = 0 the switch contacts position 1. The switch is thrown to position 2 at time t = 50 milliseconds. Determine the charging time constant, the amount of time after the switch is closed before the circuit reaches steady-state, the maximum charging and discharging currents, and the capacitor voltage at t = 0, t = 50 milliseconds, t = 90 milliseconds, and t = 1 second. 278 Figure 8.28 Simulation results for the circuit of Figure 8.25. Figure 8.29 Circuit for Example 8.7. | DCElectricalCircuitAnalysis_Page_278_Chunk1200 |
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