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V(5)/V1 = 1.000E+00 INPUT RESISTANCE AT V1 = 4.004E+00 OUTPUT RESISTANCE AT V(5) = 0.000E+00
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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 le. Note that PSpice calculated the input impedance as a negative resistance of 4 , which is in agreement with the above derivation. De ning the Harmonic Content The next step in designing an input EMI lter is to determine the harmonic content of the power circuit input current. If the input current waveform is known, a Fourier analysis can be performed in order to establish 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
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Input Current 1 0 t T
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Duty Cycle = t/T
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Pulsating waveform used in the Fourier series computation.
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harmonic n as follows: 2 An = T
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sin(nt)
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2 Bn = T
cos(nt)
Cn =
2 A2 + Bn n
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 assumption is the worst case, because the Fourier analysis of a pulsed waveform has a maxima at a value of 50%. In the next example, we will use SPICE to calculate the Fourier coef cients 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 rst nine harmonics. The result is placed in the output le 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
EMI Filter Design
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 5.010000E-01 HARMONIC FREQUENCY FOURIER NORMALIZED NO (Hz) COMPONENT COMPONENT 1 1.000E+05 6.366E-01 1.000E+00 2 2.000E+05 2.000E-03 3.142E-03 3 3.000E+05 2.122E-01 3.333E-01 4 4.000E+05 2.000E-03 3.142E-03 5 5.000E+05 1.273E-01 2.000E-01 6 6.000E+05 2.000E-03 3.142E-03 7 7.000E+05 9.093E-02 1.428E-01 8 8.000E+05 2.000E-03 3.142E-03 9 9.000E+05 7.072E-02 1.111E-01 PHASE NORMALIZED (DEG) PHASE(DEG) -3.600E-01 0.000E+00 8.928E+01 9.000E+01 -1.080E+00 4.088E-09 8.856E+01 9.000E+01 -1.800E+00 2.044E-08 8.784E+01 9.000E+01 -2.520E+00 5.723E-08 8.712E+01 9.000E+01 -3.240E+00 1.226E-07
TOTAL HARMONIC DISTORTION = 4.288115E+01 PERCENT
As you can see from the output le, 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 conservatively used for the design of the input lter.
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
1.0V
0.5V
0V 0s V(1) Time 800mV 2us 4us 6us 8us 10us 12us 14us 16us 18us 20us
400mV
SEL>> 0V 0Hz V(1)
0.5MHz
1.0MHz
1.5MHz
2.0MHz
2.5MHz Frequency
3.0MHz
3.5MHz
4.0MHz
4.5MHz
5.0MHz
The FFT feature of the Probe graphical waveform postprocessor is used to calculate the harmonics of a square waveform.
Three
1.0V
0.8V
0.6V
0.4V
0.2V
0V 0Hz
40KHz
80KHz 120KHz ... V(1)-V(0)
160KHz
200KHz
240KHz Frequency
280KHz
320KHz
360KHz
400KHz
440KHz
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 s).
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 s in 0.5- s steps. When the simulation is nished, you can use Probe to display the X-Y data, or you may view the output le in a text editor. You will have a graph of the fundamental harmonic versus TON. This con rms the previous statement that the 50% duty cycle was the maxima and provides a reference you may nd 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
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