Processing math: 100%
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Physics LibreTexts

3.22: Single-Reactance Matching

( \newcommand{\kernel}{\mathrm{null}\,}\)

An impedance matching structure can be designed using a section of transmission line combined with a discrete reactance, such as a capacitor or an inductor. In the strategy presented here, the transmission line is used to transform the real part of the load impedance or admittance to the desired value, and then the reactance is used to modify the imaginary part to the desired value. (Note the difference between this approach and the quarter-wave technique described in Section 3.19. In that approach, the first transmission line is used to zero the imaginary part.) There are two versions of this strategy, which we will now consider separately.

The first version is shown in Figure 3.22.1. The purpose of the transmission line is to transform the load impedance ZL into a new impedance Z1 for which Re{Z1} = Re{Zin}. This can be done by solving the equation (from Section 3.15) Re{Z1}=Re{Z01+Γej2βl1Γej2βl} for l, using a numerical search, or using the Smith chart.1 The characteristic impedance Z0 and phase propagation constant β of the transmission line are independent variables and can be selected for convenience. Normally, the smallest value of l that satisfies Equation ??? is desired. This value will be λ/4 because the real part of Z1 spans all possible values every λ/4.

m0093_fSRMs.png Figure 3.22.1: Single-reactance matching with a series reactance.

After matching the real component of the impedance in this manner, the imaginary component of Z1 may then be transformed to the desired value (Im{Zin}) by attaching a reactance Xs in series with the transmission line input, yielding Zin=Z1+jXS. Therefore, we choose Xs=Im{ZinZ1} The sign of Xs determines whether this reactance is a capacitor (Xs<0) or inductor (Xs>0), and the value of this component is determined from Xs and the design frequency.

Example 3.22.1: Single reactance in series

Design a match consisting of a transmission line in series with a single capacitor or inductor that matches a source impedance of 50Ω to a load impedance of 33.9+j17.6 Ω at 1.5 GHz. The characteristic impedance and phase velocity of the transmission line are 50Ω and 0.6c respectively.

Solution

From the problem statement: ZinZS=50 Ω and ZL=33.9+j17.6 Ω are the source and load impedances respectively at f=1.5 GHz. The characteristic impedance and phase velocity of the transmission line are Z0=50 Ω and vp=0.6c respectively.

The reflection coefficient Γ (i.e., ZL with respect to the characteristic impedance of the transmission line) is

ΓZLZ0ZL+Z00.142+j0.239

The length l of the primary line (that is, the one that connects the two ports of the matching structure) is determined using the equation:

Re{Z1}=Re{Z01+Γej2βl1Γej2βl}

where here Re{Z1}=Re{ZS}=50 Ω. So a more-specific form of the equation that can be solved for βl (as a step toward finding l) is:

1=Re{1+Γej2βl1Γej2βl}

By trial and error (or using the Smith chart if you prefer) we find βl0.408 rad for the primary line, yielding Z150.0+j29.0 Ω for the input impedance after attaching the primary line.

We may now solve for l as follows: Since vp=ω/β (Section 3.8), we find

β=ωvp=2πf0.6c52.360 rad/m

Therefore l=(βl)/β 7.8 mm.

The impedance of the series reactance should be jXsj29.0 Ω to cancel the imaginary part of Z1. Since the sign of this impedance is negative, it must be a capacitor. The reactance of a capacitor is 1/ωC, so it must be true that

12πfC29.0 Ω

Thus, we find the series reactance is a capacitor of value C3.7pF.

The second version of the single-reactance strategy is shown in Figure 3.22.2. The difference in this scheme is that the reactance is attached in parallel. In this case, it is easier to work the problem using admittance (i.e., reciprocal impedance) as opposed to impedance; this is because the admittance of parallel reactances is simply the sum of the associated admittances; i.e., Yin=Y1+jBp where Yin=1/Zin, Y1=1/Z1, and Bp is the discrete parallel susceptance; i.e., the imaginary part of the discrete parallel admittance.

m0093_fSRMp.png Figure 3.22.2: Single-reactance matching with a parallel reactance

So, the procedure is as follows. The transmission line is used to transform YL into a new admittance Y1 for which Re{Y1} = Re{Yin}. First, we note that Y11Z1=Y01Γej2βl1+Γej2βl where Y01/Z0 is characteristic admittance. Again, the characteristic impedance Z0 and phase propagation constant β of the transmission line are independent variables and can be selected for convenience. In the present problem, we aim to solve the equation Re{Y1}=Re{Y01Γej2βl1+Γej2βl} for the smallest value of l, using a numerical search or using the Smith chart. After matching the real component of the admittances in this manner, the imaginary component of the resulting admittance may then be transformed to the desired value by attaching the susceptance Bp in parallel with the transmission line input. Since we desire jBp in parallel with Y1 to be Yin, the desired value is Bp=Im{YinY1} The sign of Bp determines whether this is a capacitor (Bp>0) or inductor (Bp<0), and the value of this component is determined from Bp and the design frequency.

In the following example, we address the same problem raised in Example 3.22.1, now using the parallel reactance approach:

Example 3.22.2: Single reactance in parallel

Design a match consisting of a transmission line in parallel with a single capacitor or inductor that matches a source impedance of 50Ω to a load impedance of 33.9+j17.6 Ω at 1.5 GHz. The characteristic impedance and phase velocity of the transmission line are 50Ω and 0.6c respectively.

Solution

From the problem statement: ZinZS=50 Ω and ZL=33.9+j17.6 Ω are the source and load impedances respectively at f=1.5 GHz. The characteristic impedance and phase velocity of the transmission line are Z0=50 Ω and vp=0.6c respectively.

The reflection coefficient Γ (i.e., ZL with respect to the characteristic impedance of the transmission line) is

ΓZLZ0ZL+Z00.142+j0.239

The length l of the primary line (that is, the one that connects the two ports of the matching structure) is the solution to:

Re{Y1}=Re{Y01Γej2βl1+Γej2βl}

where here Re{Y1}=Re{1/ZS}=0.02 mho and Y0=1/Z0=0.02 mho. So the equation to be solved for βl (as a step toward finding l) is:

1=Re{1Γej2βl1+Γej2βl}

By trial and error (or the Smith chart) we find βl0.126 rad for the primary line, yielding Y10.0200j0.0116 mho for the input admittance after attaching the primary line.

We may now solve for l as follows: Since vp=ω/β (Section 3.8), we find

β=ωvp=2πf0.6c52.360 rad/m

Therefore, l=(βl)/β 2.4 mm.

The admittance of the parallel reactance should be jBp+j0.0116 mho to cancel the imaginary part of Y1. The associated impedance is 1/jBpj86.3 Ω. Since the sign of this impedance is negative, it must be a capacitor. The reactance of a capacitor is 1/ωC, so it must be true that

12πfC86.3 Ω

Thus, we find the parallel reactance is a capacitor of value C1.2pF.

Comparing this result to the result from the series reactance method (Example 3.22.1), we see that the necessary length of transmission line is much shorter, which is normally a compelling advantage. The tradeoff is that the parallel capacitance is much smaller and an accurate value may be more difficult to achieve.


  1. For more about the Smith chart, see “Additional Reading” at the end of this section.↩

This page titled 3.22: Single-Reactance Matching is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Steven W. Ellingson (Virginia Tech Libraries' Open Education Initiative) .

Support Center

How can we help?