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inductor/capacitor ratio broadens the entire curve. The impedance magnitude at the dip does not change, and thus the current at f0 does not change. In practical terms, the Q for a series circuit, Qseries, may also be defined by the ratio of circuit reactance to the total series resistance at resonance. Qseries = X 0 RT... | ACElectricalCircuitAnalysis_Page_301_Chunk801 |
At resonance XL and XC have the same magnitude, thus we can also say: Q series = √X L X C RT Qseries = 1 RT √X L X C Q series = 1 RT √ 2 π f L 2π f C Which simplifies to: Qseries = 1 RT √ L C (8.13) Effect of Q on Component Voltages Effect of Q on Component Voltages Q will create a multiplying effect on the inductor an... | ACElectricalCircuitAnalysis_Page_302_Chunk802 |
303 Figure 8.8 Series resonance: component voltages for high Q. Series Resonance Voltage Plot: High Q Normalized Voltage 0 2 4 6 8 10 12 Normalized Frequency 0.1 1.0 10.0 0.1 1 10 VR VL VC Series Resonance Voltage Plot: Low Q Normalized Voltage 0 0.5 1 1.5 Normalized Frequency 0.0 0.1 1.0 10.0 100.0 0.01 0.1 1 10 100 V... | ACElectricalCircuitAnalysis_Page_303_Chunk803 |
Example 8.1 Consider the series circuit of Figure 8.10 with the following parameters: the source is 10 volts peak, L = 1 mH, C = 1 nF and R = 50 Ω. Find the resonant frequency, the system Q and bandwidth, and the half-power frequencies f1 and f2. We begin by finding the resonant frequency. f 0 = 1 2π√LC f 0 = 1 2π√1e-3... | ACElectricalCircuitAnalysis_Page_304_Chunk804 |
f 2 = f 0 + BW 2 f 2 = 159kHz + 7.95kHz 2 f 2 ≈163kHz Given the 10 volt peak source, the voltages across the capacitor and inductor at the resonance frequency of 159 kHz would be Q times greater, or 200 volts. At higher or lower frequencies, the increased impedance lowers the current and also lowers the voltages across... | ACElectricalCircuitAnalysis_Page_305_Chunk805 |
For instance, using curve A, Qcoil at 100 kHz is approximately 90. If XL is 450 Ω at this frequency, then Rcoil would be 450 Ω/90, or 5 Ω. Effectively, Qcoil sets the ceiling for the Q of the series resonant circuit, Qseries. That is, the system Q can never be higher than the coil Q. To do so would require less resista... | ACElectricalCircuitAnalysis_Page_306_Chunk806 |
Rcoil = X L Qcoil Rcoil = 663.3Ω 95 Rcoil = 7Ω Combined with the 140 Ω resistor, we are left with 147 Ω, some 5% higher than if we had ignored it. The system Q is: Q series = X L RT Qseries = 663.3Ω 147Ω Qseries = 4.51 The Q is on the low side but not extremely so. Now for the bandwidth: BW = f 0 Q BW = 4.8kHz 4.51 BW ... | ACElectricalCircuitAnalysis_Page_307_Chunk807 |
Also, note that there is a slight spread between the peaks of the capacitor and inductor voltages, with vC slightly below f0 and vL slightly above, again just as expected. At the lowest frequencies, all of the source appears across the capacitor while at the highest frequencies all of the source appears across the indu... | ACElectricalCircuitAnalysis_Page_308_Chunk808 |
Example 8.3 Design a series resonant circuit with a resonant frequency of 100 kHz and a bandwidth of 2 kHz using a 10 mH inductor. Assumes the inductor follows curve B in Figure 8.14. We can find the value for the capacitance by rearranging the resonance frequency equation: f 0 = 1 2π√LC √LC = 1 2π f 0 C = 1 (2π f 0) 2... | ACElectricalCircuitAnalysis_Page_309_Chunk809 |
Rcoil = X L Qcoil Rcoil = 6283Ω 115 Rcoil = 54.6Ω Consequently, we must add 125.7 Ω − 54.6 Ω, or 71.1 Ω, to the series network to achieve the desired system Q. Failure to do so will yield a much higher Q than specified, resulting in a much reduced bandwidth. The completed design is shown in Figure 8.15. 8.3 Parallel Re... | ACElectricalCircuitAnalysis_Page_310_Chunk810 |
an inductor and resistor and finding its series equivalent (i.e., expressing the resulting impedance in rectangular form). After completing the process we should have an equivalent circuit like that shown in Figure 8.18. In this equivalent circuit, R and C are the values from the original circuit while L(p) and Rcoil(p... | ACElectricalCircuitAnalysis_Page_311_Chunk811 |
Substituting this equivalent back into Equation 8.14 yields, Rs Rs 2+X s 2 + −jX s Rs 2+X s 2 = 1 R p + 1 jX p Therefore, 1 Rp = Rs Rs 2+X s 2 1 jX p = −jX s Rs 2+X s 2 Taking the reciprocal results in: Rp = Rs 2+X s 2 Rs (8.15) jX p = j Rs 2+X s 2 X s (8.16) For high Q coils (Qcoil ≥ 10), Xs >> Rs, so we can approxima... | ACElectricalCircuitAnalysis_Page_312_Chunk812 |
Based on Equation 8.19 and the development of Equation 8.13, it can be shown that: Qparallel = RT√ C L (8.20) For higher Q circuits (Qparallel ≥ 10), f0 is found as in the series case (repeating): f 0= 1 2 π√LC (8.2) For lower Q circuits, f0 will be reduced slightly due to the fact that the transformed resistance is fr... | ACElectricalCircuitAnalysis_Page_313_Chunk813 |
If the parallel resonant circuit is driven by a current source, then the voltage produced across the resonant circuit (sometimes referred to as a tank circuit) will echo the shape of the impedance magnitude. In other words, it will effectively discriminate against high and low frequencies and keep only those signals in... | ACElectricalCircuitAnalysis_Page_314_Chunk814 |
Note that the parallel resistor can be used to lower the system Q and thus broaden the bandwidth, however, the system Q can never be higher than the Q of the inductor itself. The inductor sets the upper limit on system Q and therefore, how tight the bandwidth can be. In other words, Qcircuit ≤ Qcoil. This is the same s... | ACElectricalCircuitAnalysis_Page_315_Chunk815 |
Qparallel = RT X L Qparallel = 549.5k Ω 7.41 kΩ Qparallel = 74.1 Our initial assumption of high circuit Q is met. BW = f 0 Qparallel BW = 23.6 kHz 74.1 BW = 318Hz f 1 ≈f 0 −BW 2 f 1 ≈23.6 kHz −318Hz 2 f 1 ≈23.44kHz f 2 ≈f 0 + BW 2 f 2 ≈23.6 kHz +318Hz 2 f 2 ≈23.76kHz To find the circuit voltage at f0, simply multiply t... | ACElectricalCircuitAnalysis_Page_316_Chunk816 |
A frequency domain or AC analysis is run on the circuit, plotting the magnitude of the source voltage (node 1) from 2 kHz to 200 kHz. This will give us roughly a factor of ten on either side of the resonant frequency. The result is shown in Figure 8.23. The plot shows a clear and sharp peak in the low 20 kHz region. No... | ACElectricalCircuitAnalysis_Page_317_Chunk817 |
Example 8.5 For the circuit of Figure 8.25, determine the resonant frequency, the corner frequencies of f1 and f2, the bandwidth and the system Q. Also find the circuit voltage at the resonant frequency. Rcoil = 100 Ω. This circuit is identical to the one in the previous example with the exception of an added 100 k Ω l... | ACElectricalCircuitAnalysis_Page_318_Chunk818 |
Qcoil = 74.1 Rp = 549.5 k Ω We'll assume this is a high Q (≥ 10) circuit. Rp is in parallel with the load resistance of R yielding an effective parallel resistance of 549.5 kΩ || 100 kΩ, or 84.6 kΩ. Qparallel = RT X L Qparallel = 84.6 k Ω 7.41k Ω Qparallel = 11.4 The circuit Q is much reduced but our initial assumption... | ACElectricalCircuitAnalysis_Page_319_Chunk819 |
Example 8.6 Consider the circuit of Figure 8.27 with the following parameters: L=2 mH, C=10 nF, and Qcoil = 25. Determine the resonant frequency and a value for R such that the system bandwidth is 3 kHz. As usual, we'll assume this is a high Q (≥ 10) circuit. This is certainly true of the coil, although we have to dete... | ACElectricalCircuitAnalysis_Page_320_Chunk820 |
We have high Q and can continue18. Ultimately, we need to determine the total parallel resistance required to achieve this Q. Before we can do that we need to determine XL. X L = 2π f L X L = 2π35.59kHz 2mH X L = 447Ω RT = Q parallel×X L RT = 11.86×447Ω RT = 5.3k Ω RT is the parallel combination of R and Rp (the parall... | ACElectricalCircuitAnalysis_Page_321_Chunk821 |
Computer Simulation Figure 8.28 shows the completed design of the previous example captured in a simulator. A 1 mA current source is used for convenience of calculation. Given that RT is 5.3 k Ω, the 1 mA current source should produce 5.3 volts at the resonance frequency of 35.59 kHz. The results of an AC analysis are ... | ACElectricalCircuitAnalysis_Page_322_Chunk822 |
First off, the f0 of approximately 35.6 kHz is verified by both the peak in voltage and the phase angle reaching 0° at this frequency, the latter indicating perfect cancellation between the inductor and capacitor (i.e., the circuit impedance is purely resistive and achieving unity power factor). The cursors are used to... | ACElectricalCircuitAnalysis_Page_323_Chunk823 |
L C = (2π f 0 L) 2+R 2 (2π f 0 L)2 = L C −R2 2π f 0 L =√ L C −R 2 2π f 0 =√ 1 LC −R 2 L 2 2π f 0 = 1 √LC √1 −C R 2 L And finally we come to: f 0 = 1 2π√LC √1 −C R 2 L (8.21) If desired, we can treat the first term as the ordinary series resonant frequency and the second term as a fractional coefficient, as in: f 0 = f ... | ACElectricalCircuitAnalysis_Page_324_Chunk824 |
Assuming we have high circuit Q, the resonant frequency is: f 0= 1 2π√LC f 0= 1 2 π√1 mH100nF f 0=15.92 kHz X L = 2π f L X L = 2π15.92kHz 1mH X L = 100Ω Qcoil = X L Rcoil Qcoil = 100Ω 5Ω Qcoil = 20 There are no other resistances in the circuit, therefore Qcircuit = Qcoil and our initial assumption is correct. The circu... | ACElectricalCircuitAnalysis_Page_325_Chunk825 |
The resonant frequency appears to be just under 16 kHz, as predicted. Cursor-based measurement of the frequency at which the phase crosses 0° yields 15.89 kHz. This turns out to be even closer than it seems. In spite of the high circuit Q, kp was calculated and as expected is very close to unity, namely 0.99875. When m... | ACElectricalCircuitAnalysis_Page_326_Chunk826 |
Alternate Definition for Parallel Resonant Frequency Instead of defining the parallel resonant frequency as the point where the power factor is unity, i.e., where XL and XC have the same magnitude, it can be defined in terms of the frequency where the magnitude of the impedance is maximum. For high Q circuits the two d... | ACElectricalCircuitAnalysis_Page_327_Chunk827 |
This equation will yield a value between the ideal high Q case and the unity power factor case. This can be seen in Figure 8.34 where there is still an impedance peak (as evidenced by the voltage peak) in spite of the fact that a phase angle of 0° is not reached. Furthermore, the frequency of the peak is below that of ... | ACElectricalCircuitAnalysis_Page_328_Chunk828 |
The loudspeaker of Figure 8.36 is a medium-size woofer with a nominal impedance of 8 Ω. First, note the large variation on both the phase and magnitude of the impedance. The parallel items from the model produce an obvious peak in impedance just below 30 Hz. This is referred to as the free air resonance and is denoted ... | ACElectricalCircuitAnalysis_Page_329_Chunk829 |
Parallel resonance is similar to series resonance but in some ways is like its mirror image. In a parallel resonant circuit the inductor will dominate at low frequencies and produce a small net impedance. At high frequencies, the capacitor will dominate and also produce a small net impedance. At resonance, the two effe... | ACElectricalCircuitAnalysis_Page_330_Chunk830 |
8.5 Exercises 8.5 Exercises Inductor Q curves to be used with the exercises below Analysis Analysis 1. A circuit has a resonant frequency of 440 kHz and a system Q of 30. Determine the bandwidth and the approximate values for f1 and f2. 2. A circuit has a resonant frequency of 19 kHz and a bandwidth of 500 Hz. Determin... | ACElectricalCircuitAnalysis_Page_331_Chunk831 |
6. At a certain frequency, an inductor's impedance is 3 + j150 Ω. Determine the parallel resistance and reactance that produces the same value. 7. A certain 75 μH inductor is described by curve B. Determine the equivalent parallel inductor/resistor combination at 1 MHz. 8. A certain 3.3 mH inductor is described by curv... | ACElectricalCircuitAnalysis_Page_332_Chunk832 |
15. Repeat problem 13 but assume instead that the inductor's Rcoil = 15 Ω. 16. Repeat problem 12 but assume instead that the inductor follows curve D. 17. For the circuit shown in Figure 8.39, determine the resonant frequency, system Q and bandwidth. If the source is 20 mA peak, determine the resistor and capacitor vol... | ACElectricalCircuitAnalysis_Page_333_Chunk833 |
20. For the circuit shown in Figure 8.42, determine the resonant frequency, system Q and bandwidth. If the source is 3 volts, determine the inductor and capacitor currents at resonance. Assume the inductor's Q is 30. 21. For the circuit shown in Figure 8.43, determine the resonant frequency, system Q and bandwidth. If ... | ACElectricalCircuitAnalysis_Page_334_Chunk834 |
24. Given the circuit shown in Figure 8.46, determine the resonant frequency, system Q and bandwidth. If the source is 2.5 mA, determine the resistor voltage and the three branch currents at resonance. 25. For the circuit shown in Figure 8.47, determine the resonant frequency, system Q and bandwidth. If the source is 5... | ACElectricalCircuitAnalysis_Page_335_Chunk835 |
29. A parallel resonant circuit consists of a 12 nF capacitor and a 27 μH inductor with a Qcoil of 55. Determine the required additional parallel resistance to achieve a system Q of 40. 30. A series resonant circuit has a design target of f0=200 kHz with a bandwidth of 5 kHz. Which of the inductor curves above (A, B, C... | ACElectricalCircuitAnalysis_Page_336_Chunk836 |
37. Investigate the effects of inductor Q on the system bandwidth of problem 21. Plot the system voltage from 0.01 f0 to 100 f0 three times, the first using the specified coil resistance and then using values ten times larger and ten times smaller. 38. Investigate the effects of component tolerance on the system freque... | ACElectricalCircuitAnalysis_Page_337_Chunk837 |
9 9 Polyphase Power Polyphase Power 9.0 Chapter Learning Objectives 9.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Define the differences between polyphase and single phase systems and detail their advantages. • Determine line voltage, line current, phase voltage and phase curre... | ACElectricalCircuitAnalysis_Page_338_Chunk838 |
9.2 Polyphase Definition 9.2 Polyphase Definition A polyphase system uses multiple current-carrying wires with multiple sub- generators, each with their own unique phase. This allows for considerable delivery of power to the load. The most popular scheme is the three-phase configuration. This can be visualized as three... | ACElectricalCircuitAnalysis_Page_339_Chunk839 |
Of particular importance is the relative phase of each source. As the load will also have three segments or legs (a three-phase load), a consistent delivery of power demands that the three sources be spread equally over time. This means that each source is one-third of a cycle, or 120 degrees, out of phase with the oth... | ACElectricalCircuitAnalysis_Page_340_Chunk840 |
9.2 Three-Phase Connections 9.2 Three-Phase Connections It is possible to configure systems using delta- or Y-connected sources with either delta- or Y-connected loads. One item to note is that delta-connected systems are always three wire systems while Y-connected systems can make use of a fourth neutral wire (the com... | ACElectricalCircuitAnalysis_Page_341_Chunk841 |
each leg of the load must be the same, with the exception of the phase. This is true for both the Y-Y configuration as well as the delta-delta configuration. The tricky bit here is the difference between a source (or load) current or voltage, and the line current or voltage. Line voltage is the voltage magnitude betwee... | ACElectricalCircuitAnalysis_Page_342_Chunk842 |
We begin by focusing on quadrants two and three of the phasor plot. This section is redrawn in Figure 9.8. In reality, any two vectors can be used for the following proof, but this pair turns out to be particularly convenient in its orientation. For ease of use we shall normalize the magnitude of the generator voltage ... | ACElectricalCircuitAnalysis_Page_343_Chunk843 |
degrees, or 60 degrees. As the sum of the interior angles of a triangle must be 180 degrees, this means that the third angle must be 30 degrees. The horizontal leg of the triangle (dark yellow or maybe “spicy mustard”) can be determined because we know both the hypotenuse and the opposite angle. opposite = hypotenuse×s... | ACElectricalCircuitAnalysis_Page_344_Chunk844 |
Example 9.1 A three-phase delta-connected generator feeds a three-phase delta-connected load like the system shown in Figure 9.5. Assume the generator phase voltage is 120 VAC RMS. The load consists of three identical legs of 50 Ω each. Determine the line voltage, load phase voltage, generator phase current, line curre... | ACElectricalCircuitAnalysis_Page_345_Chunk845 |
Example 9.2 A three-phase Y-connected generator feeds a three-phase Y-connected load similar to the system shown in Figure 9.6. Assume the generator phase voltage is 220 VAC RMS. The load consists of three identical legs of 100 Ω each. Determine the line voltage, load phase voltage, generator phase current, line curren... | ACElectricalCircuitAnalysis_Page_346_Chunk846 |
Example 9.3 For the the system shown in Figure 9.9, determine the total apparent and real power delivered to the load. Also find the line voltage. The phase voltage of the source is 240 volts RMS at 60 Hz. Given the fact that the three load legs are all together at one common point (ground), this must be a Y-Y system. ... | ACElectricalCircuitAnalysis_Page_347_Chunk847 |
The real power can be found a few different ways: P = S×cosθ P = 3456VA×cos(−36.87°) P = 2765W P = 3×iload 2×Rload P = 3×4.8A 2×40Ω P = 2765W Computer Simulation Computer Simulation The circuit of Example 9.3 is worthy of a simulation. The first thing to do is to determine an appropriate value of inductance to achieve ... | ACElectricalCircuitAnalysis_Page_348_Chunk848 |
Now we check the line voltage. This was calculated to be 416 volts RMS, or approximately 588 volts peak. The post processor is used to display the result of node voltage 1 minus node voltage 2. This is shown in Figure 9.12. Again, the results are as expected with a peak just under 600 volts. Finally, we will investigat... | ACElectricalCircuitAnalysis_Page_349_Chunk849 |
350 Figure 9.12 One of the line voltages simulated from Figure 9.10. Figure 9.13 Simulated voltage across one of the load resistors in Figure 9.10. | ACElectricalCircuitAnalysis_Page_350_Chunk850 |
Heterogeneous Systems Heterogeneous Systems Systems configured as delta-to-Y and Y-to-delta appear to be a bit more complex than homogeneous systems. We shall refer to these as heterogeneous systems as the structures of the generator and load are of opposite kind. Examples are shown in Figures 9.14 and 9.15, respective... | ACElectricalCircuitAnalysis_Page_351_Chunk851 |
In Figure 9.15, the line voltage equals √3 times the generator phase voltage. The load is delta-connected, so each leg sees the line voltage. Knowing this, each leg of the load current can be computed. Also, the line current equals the generator phase current, and the load phase current will equal the line current divi... | ACElectricalCircuitAnalysis_Page_352_Chunk852 |
Ptotal = 3×iload 2×R Ptotal = 3×(0.664A) 2×200Ω Ptotal = 264W As a crosscheck, the power generated is: Ptotal = 3×igen×v gen Ptotal = 3×0.383A×230V Ptotal = 264 W Power generated equals power dissipated. Example 9.5 A Y-delta system like the one shown in Figure 9.15 has a generator phase voltage of 100 volts RMS at 60 ... | ACElectricalCircuitAnalysis_Page_353_Chunk853 |
The true load power can be found several ways. First, we can use the i2 R form. To do this we need to find the resistive portion of the load. Recall that the power factor is equal to cosine θ. Therefore the impedance angle is: θ = cos −1 PF θ = cos −10.8 θ ≈36.9° The real part is: R = Z cosθ R = 50Ωcos36.9° R = 40 Ω Al... | ACElectricalCircuitAnalysis_Page_354_Chunk854 |
In three-phase systems the situation is potentially complicated by the fact that the load is split into three parts and can be either Y-connected or delta-connected. The process for three phase is essentially the same as it is for single phase, but with a couple slight twists. The first course of action is to determine... | ACElectricalCircuitAnalysis_Page_355_Chunk855 |
First, the power factor is the cosine of the impedance angle. At 60 Hz, the reactance of the 100 mH inductor is −j37.7 Ω. This is in series with the resistance for a load impedance of 100 −j37.7 Ω or 106.920.66° per leg. The cosine of this angle is 0.9357. The voltage across each leg of the load will equal the line vo... | ACElectricalCircuitAnalysis_Page_356_Chunk856 |
C = 1 2π f X C C = 1 2π60 Hz303Ω C ≈8.75μ F These capacitors would be placed directly in parallel with each leg of the load and should result in a reduction of the generator and line currents. Computer Simulation Computer Simulation To see the effect of power factor correction, the circuit used in Example 9.6 is captur... | ACElectricalCircuitAnalysis_Page_357_Chunk857 |
The three power factor correction capacitors are added in parallel with the existing load legs (i.e., from line to line). This is illustrated in Figure 9.19. The transient simulation is repeated. The results are shown in Figure 9.20. The peak current in this version of the circuit is approximately 4.4 amps. Theoretical... | ACElectricalCircuitAnalysis_Page_358_Chunk858 |
Example 9.7 The Y-Y system shown in Figure 9.21 has a generator phase voltage of 230 volts RMS at 50 Hz. The load draws 900 VA with a power factor of 0.85 lagging. Determine the generator phase current. Also determine components to correct the power factor and the new generator phase current once the system is correcte... | ACElectricalCircuitAnalysis_Page_359_Chunk859 |
We're looking for the generator phase current so let's break this down to a single leg, first. The total apparent power, S, is 900 VA. For a single leg that's 300 VA. This is a Y-Y system so the generator phase current and voltage are the same as the load phase current and voltage. The current can be found via the appa... | ACElectricalCircuitAnalysis_Page_360_Chunk860 |
Computer Simulation Computer Simulation In Example 9.7 we computed the generator phase current to be 1.304 amps RMS, which is equivalent to 1.844 amps peak. If the corrected circuit is proper, then the apparent power should fall to the real power, or 255 watts. The resulting generator phase current should be this power... | ACElectricalCircuitAnalysis_Page_361_Chunk861 |
power factor, or cos-1(0.85), which is 31.8 degrees. The fastest way to determine R is to recognize that the real portion is the impedance magnitude times the power factor: R = Z×PF R = 176Ω×0.85 R ≈150Ω The reactive portion can be found via the Pythagorean theorem or by using the power relation. Then we apply the reac... | ACElectricalCircuitAnalysis_Page_362_Chunk862 |
For the comparison, the power factor correction capacitors are added in a delta configuration (across the lines) as shown in Figure 9.24. Another simulation is run, the result shown in Figure 9.25. The peak current has decreased to 1.56 amps. This is just slightly lower than the expected value of 1.568 amps peak. Again... | ACElectricalCircuitAnalysis_Page_363_Chunk863 |
9.4 Summary 9.4 Summary Polyphase systems can be thought of as a group of individual sources of the same magnitude that are synced together and where the load is similarly divided into sections or legs. By spreading out the source currents across the waveform's period, a smooth application of power to the load can be a... | ACElectricalCircuitAnalysis_Page_364_Chunk864 |
9.5 Exercises 9.5 Exercises Unless specified otherwise, assume generator frequencies are 60 Hz for all problems. Analysis Analysis 1. As depicted in Figure 9.26, a three-phase delta-connected generator feeds a delta-connected load. The generator phase voltage is 120 volts and the load consists of three legs of 10 Ω eac... | ACElectricalCircuitAnalysis_Page_365_Chunk865 |
4. Referring to Figure 9.27, if the generator phase voltage is 230 volts and the load is balanced with each leg at 12 Ω, determine the line voltage, line current, generator phase current, load current, load voltage and total load power. 5. As depicted in Figure 9.28, a three-phase delta-connected generator feeds a Y-co... | ACElectricalCircuitAnalysis_Page_366_Chunk866 |
9. A three-phase delta-connected generator feeds a delta-connected load consisting of three legs of 10 Ω in series with j4 Ω of inductive reactance, as shown in Figure 9.30. If the line voltage is 208 volts, find the voltage across each load leg, the current through the wires connecting the load to to the generator, an... | ACElectricalCircuitAnalysis_Page_367_Chunk867 |
13. The three-phase system of Figure 9.7 uses a Y-connected generator feeding a delta-connected load. The load consists of three legs of 40 Ω in series with j30 Ω of inductive reactance, as shown in Figure 9.32. If the generator phase voltage is 230 volts, find the line voltage, the voltage across each load leg, the li... | ACElectricalCircuitAnalysis_Page_368_Chunk868 |
17. A 120 volt three-phase delta-connected generator feeds a delta-connected load consisting of three legs of 75 Ω in series with −j10 Ω of capacitive reactance as shown in Figure 9.34. Find the voltage across each load leg, the current through the wires connecting the load to to the generator, and the apparent and rea... | ACElectricalCircuitAnalysis_Page_369_Chunk869 |
Challenge Challenge 21. Using the Y-Y system of problem 11 and assuming the source frequency is 60 Hz, determine appropriate component values to be added to the load in order to shift the power factor to unity. These new components should be in a delta configuration. Simulation Simulation 22. Use a transient analysis t... | ACElectricalCircuitAnalysis_Page_370_Chunk870 |
Notes Notes ♫♫ ♫♫ 371 | ACElectricalCircuitAnalysis_Page_371_Chunk871 |
10 10 Decibels and Bode Plots Decibels and Bode Plots 10.0 Chapter Learning Objectives 10.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Convert between ordinary and decibel based power and voltage gains. • Utilize decibel-based voltage and power measurements during circuit analys... | ACElectricalCircuitAnalysis_Page_372_Chunk872 |
one, but for purely passive systems it will likely be fractional (i.e., the output quantity is smaller than the input quantity). For example, a simple voltage divider might be said to have a “gain” of 0.2, or some such, meaning that the output signal is only 20% of the input signal. Unlike the ordinary gain measurement... | ACElectricalCircuitAnalysis_Page_373_Chunk873 |
At this point, you may be wondering what the big advantage of the decibel system is. To answer this, recall a few log identities. Normal multiplication becomes addition in the log system, and division becomes subtraction. Likewise, powers and roots become multiplication and division. Because of this, two important thin... | ACElectricalCircuitAnalysis_Page_374_Chunk874 |
Example 10.1 An amplifier has a power gain of 800. What is the decibel power gain? G '=10 log10G G '=10 log10800 G '=10×2.903 G '=29.03dB We could also use our estimation technique: • G = 800 = 8∙102 • 8 is equivalent to 3 factors of 2, or 2∙2∙2, and can be expressed as 3 dB + 3 dB + 3 dB, which is, of course, 9 dB • 1... | ACElectricalCircuitAnalysis_Page_375_Chunk875 |
G '= −10 dB −10dB −10dB −10 dB G '= −40dB Remember, if an increase in signal is produced, the result will be a positive dB value. A decrease in signal will always result in a negative dB value. A signal that is unchanged indicates a gain of unity, or 0 dB. To convert from dB to ordinary form, just invert the steps; tha... | ACElectricalCircuitAnalysis_Page_376_Chunk876 |
G=2×10×10 G=200 While the approximation technique appears to be slower than the calculator, practice will show otherwise. Being able to quickly estimate dB values can prove to be a very handy skill in the electronics field. This is particularly true in larger, multi-stage designs. Example 10.4 A three-stage amplifier h... | ACElectricalCircuitAnalysis_Page_377_Chunk877 |
Note that current gain may be treated in the same manner as voltage gain (although this is less commonly done in practice). Example 10.5 A circuit has an output signal of 2 V for an input of 50 mV. What is A'v? First find the ordinary gain. Av= 2 0.05 =40 Now convert to dB form. A'v=20 log10 40 A'v=20×1.602 A'v=32.04dB... | ACElectricalCircuitAnalysis_Page_378_Chunk878 |
Signal Representation in dBW and dBV Signal Representation in dBW and dBV As you can see from the preceding section, it is possible to spend considerable time converting between decibel gains and ordinary voltages and powers. Because the decibel form does offer advantages for gain measurement, it would make sense to us... | ACElectricalCircuitAnalysis_Page_379_Chunk879 |
P '=10 log10 P 1Watt P'=10 log10 200mW 1W P'=−7dBW For units of dBm, use a 1 milliwatt reference. P '=10 log10 P 1Watt P'=10 log10 200mW 1 mW P'=23 dBm 200 mW, −7 dBW, and 23 dBm are three ways of saying the same thing. Note that the dBW and dBm values are 30 dB apart. This will always be true, because the references a... | ACElectricalCircuitAnalysis_Page_380_Chunk880 |
Example 10.10 A test oscillator produces a 2 volt signal. What is this value in dBV? V '=20log10 V reference V '=20log10 2 V 1V V '= 6.02dB When both circuit gains and signal levels are specified in decibel form, analysis can be very quick. Given an input level, simply add the gain to it in order to find the output lev... | ACElectricalCircuitAnalysis_Page_381_Chunk881 |
Note that the units are dB and not dBW. This is very important! Saying that the gain is “so many” dBW is the same as saying the gain is “so many” watts. Obviously, gains are “pure” numbers and do not carry units such as watts or volts. The usage of a dB-based system is shown graphically in Figure 10.3. Note how the sta... | ACElectricalCircuitAnalysis_Page_382_Chunk882 |
10.3 Bode Plots 10.3 Bode Plots The Bode plot is a graphical response prediction technique that is useful for both circuit design and analysis. It is named after Hendrik Wade Bode, an American engineer known for his work in control systems theory and telecommunications. A Bode plot is, in actuality, a pair of plots: On... | ACElectricalCircuitAnalysis_Page_383_Chunk883 |
Lead Network Gain Response Lead Network Gain Response Reduction in low frequency gain is caused by lead networks. A generic lead network is shown in Figure 10.5. It gets its name from the fact that the output voltage developed across R leads the input. At very high frequencies the circuit will be essentially resistive.... | ACElectricalCircuitAnalysis_Page_384_Chunk884 |
very easy to find the approximate gain at any given frequency as long as fc is known. It is not necessary to go through reactance and phasor calculations. To create a general response equation, start with the voltage divider rule to find the gain: V out V i n = R R−j X c V out V i n = R∠0 √R 2+X c 2∠−arctan X c R The m... | ACElectricalCircuitAnalysis_Page_385_Chunk885 |
To express Av in dB, substitute equation 10.11 into equation 10.5. A'v=20 log10 1 √ 1+ f c 2 f 2 After simplification, the final result is: A'v=−10 log10(1+ f c 2 f 2) (10.12) Where fc is the critical frequency, f is the frequency of interest, A'v is the decibel gain at the frequency of interest. Example 10.13 A circui... | ACElectricalCircuitAnalysis_Page_386_Chunk886 |
V out V i n = R R−j X c V out V i n = R∠0 √R 2+X c 2∠−arctan X c R The phase portion of this is, θ=arctan X c R By using equation 10.6, this simplifies to, θ=arctan f c f (10.13) Where fc is the critical frequency, f is the frequency of interest, θ is the phase angle at the frequency of interest. Often, an approximatio... | ACElectricalCircuitAnalysis_Page_387_Chunk887 |
Example 10.14 A telephone amplifier has a lower break frequency of 120 Hz. What is the phase response one decade below and one decade above? One decade below 120 Hz is 12 Hz, while one decade above is 1.2 kHz. θ=arctan f c f θ=arctan 120Hz 12Hz θ=84.3degrees one decade below f c (i.e, approaching 90 degrees) θ=arctan 1... | ACElectricalCircuitAnalysis_Page_388_Chunk888 |
Note that this equation is almost the same as Equation 10.12. The only difference is that f and fc have been transposed. In a similar vein, we may examine the phase response. At very low frequencies, the circuit is basically capacitive. Because the output is taken across C, Vout will be in phase with Vin. At very high ... | ACElectricalCircuitAnalysis_Page_389_Chunk889 |
Example 10.15 A medical ultra sound transducer feeds a lag network with an upper break frequency of 150 kHz. What are the gain and phase values at 1.6 MHz? Because this represents a little more than a 1 decade increase, the approximate values are −20 dB and −90 degrees, from Figures 10.7 and 10.8, respectively. The exa... | ACElectricalCircuitAnalysis_Page_390_Chunk890 |
Because this circuit involved the use of a single lag network, this is exactly what you would expect. Rise Time versus Bandwidth Rise Time versus Bandwidth For pulse-type signals, the “speed” of a circuit is often expressed in terms of its rise time. If a square pulse such as Figure 10.12a is passed into a simple lag n... | ACElectricalCircuitAnalysis_Page_391_Chunk891 |
0.1V peak = V peak(1−ϵ −t1 RC) 0.1V peak = V peak−V peak ϵ −t1 RC 0.9V peak = V peak ϵ −t1 RC 0.9 = ϵ −t1 RC log 0.9 = −t1 RC t1=0.105 RC (10.17) To find the interval up to the 90% point, follow the same technique using 0.9Vpeak . Doing so yields: t2=2.303 RC (10.18) The rise time, Tr, is the difference between t1 and ... | ACElectricalCircuitAnalysis_Page_392_Chunk892 |
Example 10.16 Determine the rise time for a lag network critical at 100 kHz. f 2=0.35 T r T r=0.35 f 2 T r= 0.35 100kHz T r=3.5μs 10.4 Combining the Elements - Multi-Stage Effects 10.4 Combining the Elements - Multi-Stage Effects A complete gain or phase plot combines three elements: (1) the midband response, (2) the l... | ACElectricalCircuitAnalysis_Page_393_Chunk893 |
Example 10.17 Draw the Bode gain plot for the following amplifier: A'v midband = 26 dB, one lead network critical at 200 Hz, one lag network critical at 10 kHz, and another lag network critical at 30 kHz. The dominant lag network is 10 kHz. There is only one lead network, so it’s dominant by default. • Draw a straight ... | ACElectricalCircuitAnalysis_Page_394_Chunk894 |
There is one item that should be noted before we leave this section, and that is the concept of narrowing. Narrowing occurs when two or more networks share similar critical frequencies, and one of them is a dominant network. The result is that the true −3 dB breakpoints may be altered. Here is an extreme example. Assum... | ACElectricalCircuitAnalysis_Page_395_Chunk895 |
10.7 Exercises 10.7 Exercises Analysis Analysis dB emphasis 1. Convert the following power gains into dB form: a) 10 b) 80 c) 500 d) 1 e) 0.2 f) 0.03. 2. Convert the following dB power gains into ordinary form: a) 0 dB b) 12 dB c) 33.1 dB d) 0.2 dB e) −5.4 dB f) −20 dB. 3. An amplifier has an input signal of 1 mW, and ... | ACElectricalCircuitAnalysis_Page_396_Chunk896 |
18. Which amplifier has the greatest power output? a) 50 watts b) 18 dBW c) 50 dBm. 19. Which amplifier has the greatest power output? a) 200 mW b) −10 dBW c) 22 dBm. 20. A three stage amplifier has voltage gains of 20 dB, 5 dB, and 12 dB respectively. What is the total voltage gain in dB and in ordinary form? 21. If t... | ACElectricalCircuitAnalysis_Page_397_Chunk897 |
38. What are the maximum and minimum phase shifts across the entire frequency spectrum for the circuit of Problem 36? 39. A noninverting DC amplifier has a midband gain of 36 dB, and lag networks at 100 kHz, 750 kHz, and 1.2 MHz. Draw its gain Bode plot. 40. What are the maximum and minimum phase shifts across the enti... | ACElectricalCircuitAnalysis_Page_398_Chunk898 |
Appendix A Appendix A Standard Component Sizes Standard Component Sizes Passive components (resistors, capacitors and inductors) are available in standard sizes. The tables below are for resistors. The same digits are used in subsequent decades up to at least 1 Meg ohm (higher decades are not shown). Capacitors and ind... | ACElectricalCircuitAnalysis_Page_399_Chunk899 |
Appendix B Appendix B Methods of Solution of Linear Simultaneous Equations Methods of Solution of Linear Simultaneous Equations Some circuit analysis methods, such as nodal analysis and mesh analysis, yield a set of linear simultaneous equations. There will be as many equations as there are unknowns. For example, a par... | ACElectricalCircuitAnalysis_Page_400_Chunk900 |
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