# Power Tip 24: Convert parallel impedances to series impedances

Robert Kollman, Texas Instruments -June 21, 2010

(Editor's note: there is a linked list of all entries in this series here.)

This Power Tip shows you how to do a quick conversion of parallel-to-series complex impedances (and vice versa). It also shows that a graphical representation of this transformation as a function of frequency looks much like a Smith Chart (Editor's note: the Smith Chart is a graphical technique developed in the 1920s for analysis of transmission lines, impedance matching, antennas and more, and still used extensively in RF-related design).

This tip is useful in simplifying transformer equivalent circuits or filter networks down to two terminal devices. Figure 1 shows the transformation equations for converting a parallel circuit into a series circuit (see Appendix 1 for the derivation).

Figure 1: These circuits are equivalent at one frequency.
(Click on image to enlarge)

Interestingly, these expressions create circles in the Rs/Xs series plane, if one of the paralleled components is held fixed and the other is varied from an open to a short. The variation can be either from the component value changing or from the impedance of components changing with frequency. Figure 2 shows examples of these variations. The x-axis represents the series resistance and the y-axis represents the series reactance.

Figure 2: Constant parallel resistance maps as a circle.
(Click on image to enlarge)

There are two circles here, one for a constant parallel resistance and a second for a constant reactance. The constant resistance line is symmetric around the x-axis. When the reactance is a near open circuit, the impedance is equal to the parallel resistance. As the reactance is reduced, the path of the curve follows a circle toward the origin. It will go positive with an inductive component and negative with a capacitive component. As the reactive impedance is reduced, the curve tends toward zero. The circle is centered on the x-axis at a distance of half of the parallel resistance with a radius of the same.

Also, note that the slope of a line drawn from the origin and a point on the circle is the Q of the circuit. This means that the lowest Q occurs with the larger values of parallel reactive impedance and that highest Q occurs with low parallel reactance.

Another interesting thing about this circle is that it can represent the impedance of a parallel resonant L-C-R circuit. Referring to the constant parallel R curve, at low frequency, the inductor impedance is small and you start at the origin. As the frequency increases, the impedance tracks positive in the first quadrant until the capacitance reactance equals the inductive reaction at resonance (1 on the x-axis). You then shift into the second quadrant and continue around the circle.

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

So how do you use this? This can be useful when you need to figure out how an inductor DC resistance (DCR) and capacitor equivalent series resistance (ESR) will affect the output impedance of a power supply filter. This is shown in Figure 3.

Figure 3: Series-to-parallel conversion simplifies circuit analysis.
(Click on image to enlarge)

The output impedance is highest at resonance, so the filter resonance frequency must first be calculated. Next, do a series-to-parallel conversion on the inductor-DCR combination and the capacitor-ESR combination. Finally, simply combine the three now parallel resistors that are now in parallel. For instance, if you had a 47 μF ceramic capacitor with essentially 0 Ω ESR and a 10 μH output inductor with 50 mΩ DCR. The resonant frequency is 7 kHz. At this frequency, the inductor has 0.4 Ω of reactance, giving a Q of eight and a parallel resistance of 3 Ω. An even quicker method is use the characteristic impedance ((L/C)½) for the inductor’s reactance at resonance.

Appendix 1: Converting the parallel circuit to series.
At one frequency, the two circuits of Figure 1 (above) are equivalent. Calculating the series equivalent from the parallel components:

(Click on equation to enlarge)

Setting real and imaginary terms equal, dividing numerator and denominator by Xp2, and substituting Q = Rp/Xp:

(Click on equation to enlarge)

Similarly, solving for Xs:

(Click on equation to enlarge)