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Chp. 4: Series to Parallel Conversion using Quality Factor

The Underrated Magic Trick

A series impedance can be transformed to equivalent parallel impedance and vice versa. This is the most underrated trick yet it is fundamental to matching network design. Math behind series to parallel conversion is simple. You want to know when a series impedance is equal to parallel impedance, so you equate them.

$$ R_s + jX_s = R_p \,||\, jX_p $$
$$ R_s + jX_s = \frac{jR_pX_p}{R_p+jX_p}$$
$$ R_s + jX_s = \frac{R_pX_p^2}{R_p^2+X_p^2}+j\frac{R_p^2X_p}{R_p^2+X_p^2}$$
$$ R_s + jX_s = \frac{R_p}{\frac{R_p^2}{X_p^2}+1}
+j\frac{X_p}{1+\frac{X_p^2}{R_p^2}}$$
$$ R_s + jX_s = \frac{R_p}{Q^2+1}+j\frac{X_p}{1+\frac{1}{Q^2}}$$
$${\large \textcolor{#40CE7F}{\implies R_p = R_s\,(1+Q^2) \;\;\;\text{&}\;\;\; X_p = X_s \left(1+\frac{1}{Q^2}\right)}} $$

This says a series R_{s} is boosted by factor (1+Q^{2}) when it is converted to a parallel resistance, and a series reactance X_{s} is boosted by factor (\(1+\frac{1}{Q^2}\)) when it is converted to a parallel reactance. For Q>>1 (or practically Q>4) this translates to

$${\large \textcolor{#40CE7F}{\implies R_p = Q^2R_s\;\;\;\text{&}\;\;\; X_p = X_s \;\;\; (1)}} $$

This means a resistor got boosted but reactance stayed same which is strange. Let’s explore the physical explanation of why is it so.

Excite a series RL circuit with voltage V_{S} as shown in image below.

$$ V_S = V_R + j QV_R $$
$$ V_S = V_R(1+jQ)$$
$$|V_S| = |V_R|\sqrt{1+Q^2} \;\;\;\; (2) $$
$$\implies |V_R| = \frac{|V_S|}{\sqrt{1+Q^2}} $$

Power dissipated by resistor can be given as:

$$ P = \frac{1}{2}\,\frac{V_R^2}{R} $$
$$ P = \frac{1}{2}\,\frac{V_S^2}{R\,(1+Q^2)} \implies \text{as if } V_S \text{ is delvering power to a resistor }R(1+Q^2)$$

Now this is interesting. Think about it. It says if you were to apply the whole voltage V_{s} across the resistor, the resistor would need to go up by 1+Q^{2} to keep the power same. If you don’t increase the resistor, the power drop across resistor would be huge because it sees the whole V_{S} now instead of small V_{R} as it saw in series case. This is why a series resistor gets a boosting factor of 1+Q^{2} in its parallel equivalent counterpart.

Physically it means voltage across R increased by factor Q. Remember in a series RL circuit, if a resistor has V_{R} voltage, the inductor would have QV_{R} voltage. Now when you move this resistor in parallel to inductor, resistor also taps a bigger voltage QV_{R}. So if voltage goes up by Q, power goes by up Q^{2}, therefore R also has to go up by Q^{2} to keep overall power same. Or you can say if voltage goes up by Q, current has to go down by Q to keep the power same (just like a transformer where if voltage goes up by turn ratio N, current goes down by N). So ratio \(\frac{QV}{\frac{I}{Q}}\) gives you \(Q^2\frac{V}{I}\) i.e., Q^{2} times higher impedance. Ok wait, you just said that it gets boosted by 1+Q^{2}, and now you say Q^{2}. Where did 1 go? To give you intuition behind boosting (i.e., resistor tapping a bigger voltage now), we said this bigger voltage is QV_{R} but it is actually V_{S} and that is almost equal to QV_{R} for Q>>1 as revealed by Eq. (2). So, the factor 1 is there if you want to be very precise, but it wouldn’t be unfair to say Q is usually much greater than one, and to keep things simple (and memorable), we can just say impedance is boosted by Q^{2}.

Now that we argued that it makes sense for a series resistor to get boosted by factor Q^{2} when we place the resistor in parallel, you may ask why didn’t it happen to inductor because Eq. (1) tells us reactance stays almost same.

Let’s go back to series RL circuit. Inductor already had big voltage across it QV_{R}. Going from series to parallel did not get it much bigger of voltage boost. It was QV_{R} in series, and it is V_{S} now in parallel. And as stated before, QV_{R} is almost equal to V_{S} for Q>>1. Therefore inductor value only has to go up a little to keep reactive power same. If inductor also gets a boost of Q^{2}, it will draw huge reactive power, then we can’t say this parallel impedance is equal to series impedance because series inductor would have been drawing relatively small reactive power. And when Q is small, then yes boost in inductance is also visible. For Q>>1, you just don’t see much boost, so it is fair to say series and parallel reactance stays same.

Let’s run quick math to see this through. Reactive power across inductor in series RL circuit can be given as:

$$ Q = \frac{1}{2}\,\frac{(QV_R)^2}{X} $$
$$ Q = \frac{1}{2}\,\frac{Q^2V_S^2}{X\,(1+Q^2)} \;\;\; \text{because}\; V_R = \frac{V_S}{\sqrt{1+Q^2}}$$
$$ Q = \frac{1}{2}\,\frac{V_S^2}{X\,(1+\frac{1}{Q^2})}\implies \text{as if } V_S \text{ is delvering reactive power to a reactance }X\left(1+\frac{1}{Q^2}\right)$$

Note that a common misconception that exists is that this series to parallel impedance transformation is narrowband. The transformation itself is not narrow band, it is Q. Quality factor changes over frequency, and you need to update R and X values based on frequency. If you can do that, it is as wideband as you want. But you won’t be able to do that. Your typical case of use would be: you figured out what is parallel equivalent of your series circuit at one frequency, you went ahead and replaced that series circuit with parallel one. So that’s that. At that frequency, your series and parallel circuit would behave exactly same, at frequencies nearby their response would still be similar because Q wouldn’t have changed much, but once you start looking at far off frequencies, all bets are off because Q would have changed a lot.

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Published: 18 Feb 2023

Last Edit: 18 Feb 2023