电磁干扰抑制设计

P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 Chapter 3 EMI Filter Design Nearly all power circuits contain an input electromagnetic interference (EMI) filter. The main purpose of the EMI filter is to limit the inter- ference that is conducted or radiated from the power circuit. Excessive conducted or radiated interference can cause erratic behavior in other systems that are in close proximity of, or that share an input source with, the power circuit. If this interference affects the power circuit, it can cause erratic operation, excessive ripple, or degraded regulation, which can lead to system level problems. Input EMI filters may also be used to limit inrush current, reduce conducted susceptibility, and suppress spikes. The specifications for the allowable interference are generally driven by the power circuit specification. The most common specifications include MIL-STD-461 for military applications and FCC for commercial applications. Many other EMI specifications also exist. This chapter will deal with the design and analysis of EMI filters that will reduce conducted interference and conducted susceptibility and limit inrush current. The design of the input filter is slightly more critical when the power circuit is a regulated switching circuit, rather than a linear circuit, because a negative input resistance is created by the regulated switching circuit. Although it is possible to simulate the radiated interference of a power circuit, it is beyond the scope of this book. Basic Requirements The design of an input EMI filter begins with the definition of two basic requirements: � The filter must provide the power converter with lower output impe- dance than the negative input resistance of the power circuit. 63 P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 64 Chapter Three � The input filter attenuation must be sufficient to limit the resulting interference to a level that is below the imposed specification. The following flowchart provides a step-by-step approach that may be used to design an input filter. EMI filter design flowchart Define Impedance YES YES YES NO Attenuation Defined? Harmonic Content Known NO Waveform Known? Estimate Waveform Calculate Fourier components Calculate Attenuation Calculate component Values Defining the Negative Resistance The negative resistance of the power circuit can be defined by looking at the following conditions Pin = Poutefficiency Iin = PinVin Rin = VinIin = V2in Pin = V 2 in ∗efficiency Pout P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 EMI Filter Design 65 The input resistance is negative because as the input voltage in- creases, the input current decreases. As a simple example, we can use PSpice to analyze the input resistance of the power circuit. PSpice can analyze the input resistance in a number of ways. The simplest method is the transfer function (.TF) analysis, which calculates the DC gain and the small signal input and output impedance. The following exam- ple uses the PSpice.TF analysis to measure the input resistance of a switching power circuit. Example 1—Input resistance analysis Input File RIN: INPUT RESISTANCE .TF V(5) V1 V1 5 0 20 G1 5 0 Value = { 100/V(5) } .END Output File RIN: INPUT RESISTANCE .TF V(5) V1 V1 5 0 20 G1 5 0 Value = { 100/V(5) } .END .END ∗∗∗SMALL-SIGNAL CHARACTERISTICS V(5)/V1 = 1.000E+00 INPUT RESISTANCE AT V1 = −4.004E+00 OUTPUT RESISTANCE AT V(5) = 0.000E+00 The G1 source simulates a power circuit, which has an input power of 100 W. V1 applies 20 VDC to the power circuit, and the .TF measures the input impedance at node 5 and the output impedance at V1. The results are placed in the output file. Note that PSpice calculated the input impedance as a negative resistance of 4 �, which is in agreement with the above derivation. Defining the Harmonic Content The next step in designing an input EMI filter is to determine the har- monic content of the power circuit input current. If the input current waveform is known, a Fourier analysis can be performed in order to es- tablish the harmonic content of the waveform; however, even if the exact waveform is not known, we can estimate the waveform with reasonable accuracy. The design can be optimized later, if necessary. Consider the pulsating waveform in Fig. 3.1. With a peak amplitude of 1 and a base amplitude of 0, we can compute the Fourier series of P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 66 Chapter Three Input Current 1 0 t T Duty Cycle = t/T Figure 3.1 Pulsating waveform used in the Fourier series computation. harmonic n as follows: An = 2T t∫ 0 sin(nt) Bn = 2T t∫ 0 cos(nt) Cn = √ A2n + B2n If we assume that the input ripple current is pulsating and if we know the duty cycle, we can proceed to the Fourier analysis. If the duty cycle is not known, we will assume a value of 50%. This assump- tion is the worst case, because the Fourier analysis of a pulsed wave- form has a maxima at a value of 50%. In the next example, we will use SPICE to calculate the Fourier coefficients of a 50% duty cycle pulse. Example 2—.FOUR analysis The following example demonstrates the use of the .FOUR analysis. V1 is a pulsed voltage source, which has a 50% duty cycle and a 100-kHz frequency. The .FOUR statement calculates the magnitude and phase of the DC value and the first nine harmonics. The result is placed in the output file as shown below. EX2: DEMONSTRATING THE USE OF THE .FOUR ANALYSIS .OPTIONS NUMDGT=3 .TRAN .01U 20U .FOUR 100KHZ V(1) V1 1 0 PULSE 0 1 0 0 0 5U 10U .END P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 EMI Filter Design 67 FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 5.010000E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (Hz) COMPONENT COMPONENT (DEG) PHASE(DEG) 1 1.000E+05 6.366E-01 1.000E+00 -3.600E-01 0.000E+00 2 2.000E+05 2.000E-03 3.142E-03 8.928E+01 9.000E+01 3 3.000E+05 2.122E-01 3.333E-01 -1.080E+00 4.088E-09 4 4.000E+05 2.000E-03 3.142E-03 8.856E+01 9.000E+01 5 5.000E+05 1.273E-01 2.000E-01 -1.800E+00 2.044E-08 6 6.000E+05 2.000E-03 3.142E-03 8.784E+01 9.000E+01 7 7.000E+05 9.093E-02 1.428E-01 -2.520E+00 5.723E-08 8 8.000E+05 2.000E-03 3.142E-03 8.712E+01 9.000E+01 9 9.000E+05 7.072E-02 1.111E-01 -3.240E+00 1.226E-07 TOTAL HARMONIC DISTORTION = 4.288115E+01 PERCENT As you can see from the output file, the fundamental harmonic has a peak value that is 63.6% of the peak pulse amplitude. Although this does provide the required information, it is far from elegant. A better solution is to calculate the harmonics in Probe. The resulting plot is shown in Fig. 3.2. This is the worst case for a pulsed waveform and could be conserva- tively used for the design of the input filter. Example 3—Using the .STEP command to calculate harmonics The next example uses the PSpice .STEP command to sweep the duty cycle from 5% to 95% and look at the fundamental amplitude of the resulting square wave. As in the previous example, V1 is a pulsed Frequency 0Hz 0.5MHz 1.0MHz 1.5MHz 2.0MHz 2.5MHz 3.0MHz 3.5MHz 4.0MHz 4.5MHz 5.0MHz V(1) 0V 400mV 800mV SEL>> Time 0s 2us 4us 6us 8us 10us 12us 14us 16us 18us 20us V(1) 0V 0.5V 1.0V Figure 3.2 The FFT feature of the Probe graphical waveform postprocessor is used to calculate the harmonics of a square waveform. P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 68 Chapter Three Frequency 0Hz 40KHz 80KHz 120KHz 160KHz 200KHz 240KHz 280KHz 320KHz 360KHz 400KHz 440KHz ... V(1)-V(0) 0V 0.2V 0.4V 0.6V 0.8V 1.0V Figure 3.3 FFT of the .STEP analysis. The waveform with the largest amplitude at 100 kHz corresponds to the 50% duty cycle (TON= 5 µs). voltage source. In this case, the pulse has an initial amplitude of 1 V and switches to 0 V after delay “TON.” “TON” is swept from 0.5 to 9.5 µs in 0.5-µs steps. When the simulation is finished, you can use Probe to display the X-Y data, or you may view the output file in a text editor. You will have a graph of the fundamental harmonic versus “TON.” This confirms the previous statement that the 50% duty cycle was the maxima and provides a reference you may find helpful in the future. X3: .STEP ANALYSIS .PROBE .PARAM TON=0.5u .STEP PARAM TON 0.5u 9.5u 0.5u .TRAN .1U 10U .PRINT TRAN V(1) V1 1 0 PULSE 1 0 {TON} .END The FFT results of the .STEP analysis are shown in the graphs of Figs. 3.3 and 3.4. Example 4 – EMI filter design In order to design the EMI filter, we need to define a converter that will operate with it. For the purpose of this example, let us assume that we have a power converter that will operate with an input voltage of 18 to 32 V DC. The converter output power will be 75 W and will have an operating efficiency of 75%. The converter will have a switching frequency of 100 kHz. The conducted emissions requirement allows the 1-mA peak to be reflected back to the input lines. A second-order filter will be used. P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 EMI Filter Design 69 1 1.00U 3.00U 5.00U 7.00U 9.00U Time in Secs 700M 500M 300M 100M -100.0M Fu nd am en ta l A m pl itu de in V ol ts Fundamental Harmonic vs Ton for 10uSec Pulse Train Figure 3.4 .STEP analysis result shows the 50% duty cycle as the maxima. Let us follow the procedures that were defined in the EMI design flowchart. Step 1 is to calculate the input impedance. Calculating the input impedance. The input impedance was defined ear- lier in this chapter as V2in∗efficiency Pout It is obvious that the lowest impedance will occur at the minimum input voltage. This value can be calculated as 182 × 0.75 75 = 3.24 � Calculating the harmonic content. Because no detail is provided regard- ing the pulse current waveforms, we will assume that the duty cycle is 50%. The average input current is Iavg = PoutVin∗efficiency = 75 18 × 0.75 = 5.56A At a duty cycle of 50%, the peak amplitude will be 11.12 A. In the previous harmonic analysis, we defined the fundamental harmonic to be 0.636Ipk=7.08 A. P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 70 Chapter Three Calculating the required attenuation. With a maximum reflected ripple current of 1-mA peak, we can define the attenuation required as Attenuation = 7.08 0.001 = 7080 = 77 dB Calculating the component values. The attenuation for a second-order filter can be defined as Attenuation = ( fswitch ffilter )2 We can compute the filter frequency as 100 kHz√ Attenuation = 100 kHz 84.14 = 1188 Hz. The values of L and C can be defined by setting their impedances to the input converter input impedance at the filter resonant frequency, as defined above. C = 1 2π ( 1188 ) ( 3.24 ) = 41.35 µF L = 3.24 2π ( 1188 )434 µH Note that the characteristic impedance of the filter is defined by Zo = √ L C = √ 434 µH 41.35 µF = 3.24 � which is equal to the converter input impedance. In an actual design, it is a good practice to provide a 6-dB margin for these characteristics. Damping Elements While this filter provides the proper impedance matching and the re- quired attenuation, the impedance will be quite high at the resonant frequency of the filter. The only damping elements in the circuit are the DC resistance (DCR) of the inductor and the equivalent series resis- tance (ESR) of the capacitor (which we have not defined). It is normally necessary to provide damping of the L-C filter in order to restrict the impedance of the filter at the resonant frequency. A shunt series R-C network is used for this purpose. The value of the damping capacitor is generally 3 to 5 times greater than that of the filter capacitor, and P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 EMI Filter Design 71 C1 41.35U C2 CDAMP I1 AC V(1) R1 RDAMP 1 2 Frequency 100Hz 300Hz 1.0KHz 3.0KHz 10KHz 30KHz 100KHz 300KHz 1.0MHz V(1) 0V 1.0V 2.0V 3.0V 4.0V Figure 3.5 Schematic of the test circuit used to measure the impedance of the filter. The waveform V(1) is equivalent to the impedance because the input is a current (I1 1 0 AC 1). The case for CDAMP = 120µ and RDAMP = 1.6 is shown. the value of the damping resistor is generally close to the characteristic impedance of the filter. The PSpice .Step command is ideal for defining these elements. The following circuit is designed to measure the impedance of the fil- ter, while sweeping the damping capacitor from 120 to 200 µF in 40-µF increments. For each value of the damping capacitor, the damping re- sistor will be swept from 0.5 to 2 times the characteristic impedance (1.6 to 6.4 �) in 0.6-� increments. The PSpice listing and schematic of the test circuit (Fig. 3.5) are shown below. The results are shown below. EX4: TO MEASURE THE IMPEDANCE OF A FILTER .AC DEC 10 100HZ 1MEGHZ .PARAM CDAMP=120u .PARAM RDAMP=1.6 .STEP PARAM CDAMP 120U 200U 40U .STEP PARAM RDAMP 1.6 6.4 .6 .PROBE C1 1 0 41.35U P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 72 Chapter Three C2 1 2 {CDAMP} R1 2 0 {RDAMP} I1 0 1 AC 1 L1 0 1 434U .END The results are provided in the output file and are shown below. Sweep Analysis of EX4.ckt Count CDAMPRDAMPMaximum 1 1.20000e-004 1.60000e+000 3.891 2 1.20000e-004 2.20000e+000 3.440 3 1.20000e-004 2.80000e+000 3.557 4 1.20000e-004 3.40000e+000 3.916 5 1.20000e-004 4.00000e+000 4.395 6 1.20000e-004 4.60000e+000 4.840 7 1.20000e-004 5.20000e+000 5.248 8 1.20000e-004 5.80000e+000 5.619 9 1.20000e-004 6.40000e+000 6.104 10 1.60000e-004 1.60000e+000 2.994 11 1.60000e-004 2.20000e+000 2.869 12 1.60000e-004 2.80000e+000 3.153 13 1.60000e-004 3.40000e+000 3.672 14 1.60000e-004 4.00000e+000 4.161 15 1.60000e-004 4.60000e+000 4.614 16 1.60000e-004 5.20000e+000 5.033 17 1.60000e-004 5.80000e+000 5.580 18 1.60000e-004 6.40000e+000 6.121 19 2.00000e-004 1.60000e+000 2.489 20 2.00000e-004 2.20000e+000 2.593 21 2.00000e-004 2.80000e+000 3.024 22 2.00000e-004 3.40000e+000 3.547 23 2.00000e-004 4.00000e+000 4.038 24 2.00000e-004 4.60000e+000 4.494 25 2.00000e-004 5.20000e+000 5.040 26 2.00000e-004 5.80000e+000 5.591 27 2.00000e-004 6.40000e+000 6.137 The impedance was exceeded with the 120-µF damping capacitor (Fig. 3.6). If we use a 160-µF capacitor, the impedance will be minimized with a 2.2-� damping resistor. A lower impedance could be achieved with a 200-µF damping capacitor and a 1.6-� damping resistor. We will select the 160-µF capacitor and the 2.2-� resistor. The following simulation shows the impedance characteristics and the reflected ripple of the filter (see also Figs. 3.7 and 3.8). EMI2: TO SHOW THE REFLECTED RIPPLE OF THE FILTER .AC DEC 10 100HZ 100KHZ .TRAN 1U 10M 9980U .1u UIC .PROBE P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 EMI Filter Design 73 1 2.10 3.10 4.10 5.10 6.10 Rdamp 6.00 5.00 4.00 3.00 2.00 M ax im um in pe da nc e in O hm s C=120 µF C=200 µF C=160 µF Figure 3.6 Family of curves showing the maximum impedance versus the damping resis- tor value. Each curve represents a different capacitor value. C1 2 0 41.35U C2 2 1 160U R1 1 0 2.2 I1 0 2 AC 1 PULSE 0 11 0.1U 0.1U 0.1U 5U 10U L1 0 2 434U IC=-5.5 .END L1 434U C1 41.35U C2 160U R1 2.2 I1 AC V(2) I(V1) L1[I]4 2 1 Figure 3.7 Circuit used to show the impedance and the reflected ripple of the filter. P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 74 Chapter Three 9.988M -5.611< x > 9.993M -5.609< x > 1 9.982M 9.986M 9.990M 9.994M 9.998M Time in Secs -5.608 -5.609 -5.610 -5.611 -5.612 In du ct or C ur re nt in Am ps ∆x = 5.050U ∆y = 1.908M 1 200 500 1K 2K 5K 10K 20K 50K Frequency in Hz 3.50 2.50 1.50 500M -500M Fi lte r I m pe da nc e V( 2) in A m ps (a) (b) Figure 3.8 Current in the inductor (a) due to a current pulse input, and impedance characteristics over frequency (b) for the filter circuit in Fig. 3.7. Fourth-Order Filters Because the physical size of power converters is continually shrinking, higher order filters are being used more often than not. The filter is P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 EMI Filter Design 75 designed in much the same way as the second-order filter. The following example demonstrates the design of a fourth-order filter using the same design parameters as those that we used for the previous filter. The “octave” rule basically states that resonances should be at least an octave apart. In an effort to be conservative, let us use a factor of 2.5. The attenuation of the filter can be defined as Attenuation = ( fswitch f1 )2 ∗ ( fswitch 2.5 f1 )2 = f 4 switch 6.25 f 41 If we set the attenuation at 7080, as in the previous example, and solve for f1 we obtain f1=6.895 kHz. The second pole is then at 2.5 f1 = 17.237 kHz. The impedance of each section should be designed to be lower than the impedance of the converter, which we had determined to be 3.24 � in the previous example. The filter is loaded by the negative resistance of the converter and produces a combined impedance of Zloaded = Zin ∗ ZoZin + Zo The loaded filter Q is defined as Q = Zloaded Zo where Zo is the filter characteristic impedance defined by Zo = √ L C If we combine the above equations, we have Q = Zin ∗ Zo (Zin + Zo) Zo Zo = − ( Q − 1 Q ) Zin The filter Q is generally maintained below a value of 2. If we set Q = 2 and solve for Zo we obtain Zo = − ( 2 − 1 2 ) (−3.24) = 1.62 � P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 76 Chapter Three If we use this impedance and the calculated resonant frequencies, we can define both inductors and both capacitors. L1 = 1.62 2π ( 6895 ) = 37 µH C1 = 1 2π ( 6895 ) ( 1.62 ) = 14 µF L2 = 1.62 2π ( 17, 237 ) = 15 µH C2 = 1 2π ( 17, 237 ) ( 1.62 ) = 5.7 µF As shown in the previous example, we can use the .Step command to sweep the values of the damping capacitor and the damping resistor. If we use a range of 3 to 5 times the value of the real capacitor, we will sweep the damper capacitor from 42 to 70 µF in steps of 14 µF. We will sweep the damper resistor from one-half to twice the Zo of the filter, i.e., from 0.8 to 3.2 � in 0.2-� steps. The schematic for the fourth-order filter and its impedance response are shown in Fig. 3.9. Note that two 10-M� resistors have been added. To aid circuit con- vergence, the resistors were added to the nodes that are purely reactive. The circuit listing and output file are shown below. A sweep of the maxi- mum impedance as a function of the damping resistor and the damping capacitor was also performed. The results of the sweep are shown in Fig. 3.10. Each curve is for a different value of damping capacitor. 4THORD: A 4TH ORDER FILTER .AC DEC 10 100HZ 1MEGHZ .PROBE .PARAM CDAMP=42u .PARAM RDAMP=0.8 .STEP PARAM CDAMP 42U 70U 14U ∗.STEP PARAMRDAMP .8 3.2 .2 .PRINT AC V(4) VP(4) 1 1 0 5.7U C2 4 2 {CDAMP} R1 2 0 {RDAMP} I1 0 4 AC 1 L2 1 4 37U C3 4 0 14U R2 1 0 10MEG R3 4 0 10MEG L1 0 1 15U .END P1: IML/OVY P2: IML/OVY QC: IML/OVY T1: IML MHBD017-03 Sandler MHBD017-Sandler-v4.cls October 6, 2005 18:53 EMI Filter Design 77 L1 15U C1 5.7U C2 CDAMP R1 RDAMP I1 AC V(4)