Passive Matching Networks

Controlling impedances in RF circuits are essential to maximize power transfer between blocks and to reduce reflections caused by impedance discontinuities along a signal path. In this module, we will define the quality factor, Q, for a device or a circuit, and use this Q to build circuits that modify the impedance seen across a port.

Device Quality Factor

The quality factor, Q, is a measure of how good a device is at storing energy. Thus, the lower the losses, the higher the Q. In general, if we can express the impedance or admittance of a device as

$Y\,\mathrm {or} \,Z={\frac {1}{A+jB}}$

(1)

The quality factor can then be expressed as

$Q={\frac {\left|B\right|}{A}}$

(2)

The imaginary component, $B$ represents the energy storage element, and the real component, $A$, represents the loss (resistive) component.

Series RC Circuit

A lossy capacitor can be modeled as a series RC circuit with the series resistance, $R_{S}$, could represent the series loss due to interconnect resistances. Thus, the admittance of the lossy capacitor can be expressed as

$Y_{S}={\frac {1}{R_{S}+{\frac {1}{j\omega C}}}}$

(3)

We can therefore express the quality factor of the series circuit as:

$Q_{S}={\frac {\frac {1}{\omega C}}{R_{S}}}={\frac {1}{\omega R_{S}C}}={\frac {X_{S}}{R_{S}}}$

(4)

Note that for the lossless case, $R_{S}=0$, and consequently $Q_{S,\mathrm {ideal} }\rightarrow \infty$.

Parallel RC Circuit

A lossy capacitor can also be modeled as a parallel RC circuit, with the parallel resistance, $R_{P}$ could represent the energy loss due to the dielectric leakage of the capacitor. In this case, we can write the impedance of the circuit as:

$Z_{P}={\frac {1}{{\frac {1}{R_{P}}}+j\omega C}}$

(5)

Thus, the quality factor is:

$Q_{P}={\frac {\omega C}{\frac {1}{R_{P}}}}=\omega R_{P}C={\frac {R_{P}}{X_{P}}}$

(6)

We can see that in the lossless case, $R_{P}\rightarrow \infty$, corresponding to $Q_{P,\mathrm {ideal} }\rightarrow \infty$.

RL Circuits

Applying the ideas above to RL circuits, we can get the admittance of a series RL circuit as:

$Y_{S}={\frac {1}{R_{S}+j\omega L}}$

(7)

with quality factor:

$Q_{S}={\frac {\omega L}{R_{S}}}={\frac {X_{S}}{R_{S}}}$

(8)

Similarly, for parallel RL circuits, the impedance and quality factor can be expressed as:

$Z_{P}={\frac {1}{{\frac {1}{R_{P}}}+{\frac {1}{j\omega L}}}}$

(9)

$Q_{P}={\frac {\frac {1}{\omega L}}{\frac {1}{R_{P}}}}={\frac {R_{P}}{\omega L}}={\frac {R_{P}}{X_{P}}}$

(10)

Note that the quality factor is frequency dependent, and for $R_{S}=0$ or $R_{P}\rightarrow \infty$, we get the ideal case where $Q_{S,\mathrm {ideal} }\rightarrow \infty$ or $Q_{P,\mathrm {ideal} }\rightarrow \infty$ respectively.

General Case

In general, for a reactance $X_{S}$ in series with a resistance $R_{S}$,

$Y_{S}={\frac {1}{R_{S}+jX_{S}}}$

(11)

$Q_{S}={\frac {X_{S}}{R_{S}}}$

(12)

and for a reactance $X_{P}={\tfrac {1}{B_{P}}}$ in parallel with a resistance $R_{P}={\tfrac {1}{G_{P}}}$,

$Z_{P}={\frac {1}{G_{P}+jB_{P}}}$

(13)

$Q_{P}={\frac {B_{P}}{G_{P}}}={\frac {R_{P}}{X_{P}}}$

(14)

Series-Parallel Conversions

It is very convenient to be able to convert the series RC or RL circuit to its parallel equivalent or vice-versa, especially in the context of matching circuits. If we equate the parallel and series impedances, $Z_{S}=Z_{P}$, we get

$R_{S}+jX_{S}={\frac {1}{{\frac {1}{R_{P}}}+{\frac {1}{jX_{P}}}}}={\frac {jR_{P}X_{P}}{R_{P}+jX_{P}}}$

(15)

$\left(R_{S}+jX_{S}\right)\left(R_{P}+jX_{P}\right)=jR_{P}X_{P}$

(16)

$R_{S}R_{P}+jR_{P}X_{S}+jR_{S}X_{P}-X_{S}X_{P}=jR_{P}X_{P}$

(17)

We can look at the imaginary and real components separately. For the real components:

$R_{S}R_{P}-X_{S}X_{P}=0$

(18)

${\frac {R_{P}}{X_{P}}}=Q_{P}={\frac {X_{S}}{R_{S}}}=Q_{S}=Q$

(19)

And for the imaginary components:

$R_{P}X_{S}+R_{S}X_{P}=R_{P}X_{P}$

(20)

Using eq. 18 and eq. 20, we get:

${\frac {X_{S}X_{P}}{R_{S}}}X_{S}+R_{S}X_{P}=R_{P}X_{P}$

(21)

${\frac {X_{S}^{2}}{R_{S}^{2}}}R_{S}+R_{S}=R_{S}\left(1+Q^{2}\right)=R_{P}$

(22)

We can use eqs. 18 and 22 to get the relationship between the series and parallel reactances:

$R_{S}R_{P}-X_{S}X_{P}=R_{S}^{2}\left(1+Q^{2}\right)-X_{S}X_{P}=0$

(23)

$R_{S}^{2}\left(1+Q^{2}\right)-X_{S}X_{P}=R_{S}^{2}\left(1+Q^{2}\right){\frac {X_{S}^{2}}{X_{S}^{2}}}-X_{S}X_{P}={\frac {1+Q^{2}}{Q^{2}}}X_{S}-X_{P}=0$

(24)

Rewriting eq. 24 and for $Q^{2}\gg 1$:

$X_{P}={\frac {1+Q^{2}}{Q^{2}}}X_{S}=\left(1+{\frac {1}{Q^{2}}}\right)X_{S}\approx X_{S}$

(25)

Key Results:

• The quality factor is the same for the parallel and series circuits when the impedances are the same.

${\frac {R_{P}}{X_{P}}}=Q_{P}={\frac {X_{S}}{R_{S}}}=Q_{S}=Q$

(26)

• The parallel resistance is larger than the series resistance.

$R_{P}=R_{S}\left(1+Q^{2}\right)$

(27)

• The parallel reactance is approximately equal to the series reactance for $Q^{2}\gg 1$.

$X_{P}=\left(1+{\frac {1}{Q^{2}}}\right)X_{S}\approx X_{S}$

(28)

Basic Matching Networks

We can use our series-parallel conversions to increase or decrease resistances based on the quality factor. Consider the circuit below, where we want to match the source resistance, $R_{S}$ to a load resistance, $R_{L}$ using a lossless matching network:

L-Match Circuits

Consider the circuit below, and assume $R_{\mathrm {in} }<R_{L}$. Since we know that the parallel resistance, $R_{L}$ gets reduced when converted to a series resistance, $R_{\mathrm {in} }$:

$R_{\mathrm {in} }={\frac {R_{L}}{1+Q^{2}}}$

(29)

Where $Q={\frac {R_{L}}{X_{1}}}<math>.Thereactance<math>X_{1}$ also gets transformed into:

$X_{1}^{\prime }=X_{1}{\frac {1}{\left(1+{\frac {1}{Q^{2}}}\right)}}\approx X_{1}$

(30)

Thus, to keep the impedance seen by the source purely resistive, we can cancel out the reactance $X_{1}^{\prime }$ by the reactance $X_{2}=-X_{1}^{\prime }\approx -X_{1}$.

If we use a capacitor for $X_{1}$, then

$L_{1}={\frac {1}{\omega ^{2}C_{1}^{\prime }}}$

(35)

Note that matching circuit is narrowband since the perfect cancellation occurs at only one frequency, $\omega$. If we set $R_{S}=R_{L}^{\prime }$ and define $m={\tfrac {R_{L}}{R_{S}}}$ for $R_{S}<R_{L}$, we can write eq. 33 as

$Q={\sqrt {m-1}}=\omega R_{L}C_{1}$

(36)

We can then solve for the values of $C_{1}$ and $L_{1}$ for a certain frequency $\omega _{0}$ as

$C_{1}={\frac {\sqrt {m-1}}{\omega _{0}R_{L}}}$

(37)

$L_{1}={\frac {1}{\omega _{0}^{2}C_{1}^{\prime }}}\approx {\frac {1}{\omega _{0}^{2}C_{1}}}$

(38)

For the case when $R_{S}>R_{L}$, we just flip the circuit and the equations would still hold.

In general, if we want to convert a resistance from a higher ressitance, $R_{\mathrm {HI} }$, to a lower resistance, $R_{\mathrm {LO} }$, we place the parallel capacitance across $R_{\mathrm {HI} }$. Thus for a certain frequency $\omega _{0}$:

$m={\frac {R_{\mathrm {HI} }}{R_{\mathrm {LO} }}}$

(39)

$Q={\sqrt {m-1}}=\omega _{0}R_{\mathrm {HI} }C_{1}$

(40)

$C_{1}={\frac {\sqrt {m-1}}{\omega _{0}R_{\mathrm {HI} }}}$

(41)

$L_{1}\approx {\frac {1}{\omega _{0}^{2}C_{1}}}$

(42)

Instead of a capacitor, we can place an inductor in parallel with the larger resistance, and use a capacitor to cancel out the series equivalent inductance. However, in this case, the capacitor serves as a DC and low-frequency block.