# A Solution for Fast Parallel-Series Impedance Conversion in Texas Instruments Power Supply Design

This article will show you how to quickly convert shunt impedance to series impedance (and vice versa). The article also shows that a graph of this conversion as a function of frequency looks a lot like a Smith chart ” title=”Smith Chart”>Smith Chart. The method described in this article is useful in simplifying the equivalent circuit of a transformer or filter network to two end devices. Figure 1 shows the conversion equation for converting a parallel circuit to a series circuit (see Appendix 1 for the derivation).

Figure 1: These circuits are equivalent circuits at one frequency.

Interestingly, these expressions form some circles in the Rs/Xs series layer if one of the parallel components is fixed and the other varies from open to short. Differences can come from changes in component values or from component impedances that vary with frequency. Figure 2 shows an example of these differences. The X-axis represents series resistance, while the Y-axis represents series reactance.

There are 2 circles here: one for constant parallel resistance and one for constant reactance. The constant resistance line is symmetrical around the X axis. When the reactance is near the open circuit, the impedance is equal to the parallel resistance. Due to the reduced reactance, the curved path follows the circle to the starting point, which is positive for the inductive component and negative for the capacitive component. The curve tends to zero due to the reduced reactance. At 1/2 the parallel resistance distance, the circle is centered on the X axis and has the same radius.

In addition, it should be noted that the slope of the line connecting the starting point and a point on the circle is the Q of the circuit. That is, the value of the shunt reactance is larger when the lowest Q occurs, and the shunt reactance is lower when the highest Q occurs. Another interesting thing about this circle is that it can indicate the impedance of a parallel resonant LCR circuit. Referring to the constant parallel R curve, at low frequencies the Inductor impedance is smaller and you start at the starting point. As the frequency rises, the impedance is positive in the first quarter circle until the capacitive reactance equals the resonant inductive response (1 on the x-axis). After that, you turn into the second quarter circle and continue around the circle.

Figure 2: Constant parallel resistance mapped as a circle.

The second curve shows the impedance circle for fixed reactance and parallel variable resistance. It has the same shape as the constant R-curve, but it is centered on the Y-axis.

So how to use it? It is useful when you need to estimate how much the inductor DC resistance (DCR) and capacitor equivalent series resistance (ESR) contribute to the output impedance of the power filter. Figure 3 illustrates this. The output impedance is highest at resonance, so the filter resonance frequency must first be calculated. Next, perform series-parallel conversion for the inductor-DCR combination and the capacitor-ESR combination. Finally, simply combine the three parallel resistors that are already in parallel. For example, if you have a 47uF ceramic capacitor with essentially 0 Ohm ESR, and a 10µH output inductor with 50 mOhm DCR. The resonant frequency is 7kHz. At this frequency, the inductor has a reactance of 0.4 Ohm, resulting in a Q of 8 and a parallel resistance of 3 Ohm. A faster method is to use the characteristic impedance ((L/C) 0.5) for the inductive reactance at resonance.

Figure 3: Series-parallel conversion simplifies circuit analysis.

Next time, we’ll discuss some methods of isolating power compensation, so stay tuned.

For more details on the content of this article and other power solutions, please visit: www.ti.com.cn/power.

Appendix 1: Series Conversion of Parallel Circuits.

At a certain frequency, the two circuits shown in Figure 1 are equivalent. Compute the series equivalent circuit of the parallel section:

Make the real and imaginary terms equal, divide the numerator and denominator by Xp2, and substitute Q=Rp/Xp.

Similarly, solve for Xs.

Author: Robert Kollman

Texas Instruments (TI)

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