• Electromagnetics I
  • Ch 3: Transmission Lines
  • Loc 3.22
  • Electromagnetics I
  • Ch 3
  • Loc 3.22

Single-Reactance Matching

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.

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

The first version is shown in Figure 3.22.1. The purpose of the transmission line is to transform the load impedance

into a new impedance

for which

. This can be done by solving the equation (from Section 3.15)

(3.22.1)

for

, using a numerical search, or using the Smith chart. The characteristic impedance

and phase propagation constant

of the transmission line are independent variables and can be selected for convenience. Normally, the smallest value of

that satisfies Equation 3.22.1 is desired. This value will be

because the real part of

is periodic in

with period

.

After matching the real component of the impedance in this manner, the imaginary component of

may then be transformed to the desired value (

) by attaching a reactance

in series with the transmission line input, yielding

. Therefore, we choose

(3.22.2)

The sign of

determines whether this reactance is a capacitor (

) or inductor (

), and the value of this component is determined from

and the design frequency.

Exercise

#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

to a load impedance of

at 1.5 GHz. The characteristic impedance and phase velocity of the transmission line are

and 0.6c respectively.

Solution:

From the problem statement:

and

are the source and load impedances respectively at

GHz. The characteristic impedance and phase velocity of the transmission line are

and

respectively.

The reflection coefficient

(i.e.,

with respect to the characteristic impedance of the transmission line) is

The length

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

where here

. So a more-specific form of the equation that can be solved for

(as a step toward finding

) is:

By trial and error (or using the Smith chart if you prefer) we find

rad for the primary line, yielding

for the input impedance after attaching the primary line.

We may now solve for

as follows: Since

(Section 3.8), we find

Therefore

.

The impedance of the series reactance should be

to cancel the imaginary part of

. Since the sign of this impedance is negative, it must be a capacitor. The reactance of a capacitor is

, so it must be true that

Thus, we find the series reactance is a capacitor of value

.

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

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.,

where

,

, and

is the discrete parallel susceptance; i.e., the imaginary part of the discrete parallel admittance.

So, the procedure is as follows. The transmission line is used to transform

into a new admittance

for which

. First, we note that

where

is characteristic admittance. Again, the characteristic impedance

and phase propagation constant

of the transmission line are independent variables and can be selected for convenience. In the present problem, we aim to solve the equation

for the smallest value of

, 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

in parallel with the transmission line input. Since we desire

in parallel with

to be

, the desired value is

The sign of

determines whether this is a capacitor (

) or inductor (

), and the value of this component is determined from

and the design frequency.

In the following example, we address the same problem raised in Exercise 1, now using the parallel reactance approach:

Exercise

#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

to a load impedance of

at 1.5 GHz. The characteristic impedance and phase velocity of the transmission line are

and 0.6c respectively.

Solution:

From the problem statement:

and

are the source and load impedances respectively at

GHz. The characteristic impedance and phase velocity of the transmission line are

and

respectively.

The reflection coefficient

(i.e.,

with respect to the characteristic impedance of the transmission line) is

The length

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

where here

mho and

mho. So the equation to be solved for

(as a step toward finding

) is:

By trial and error (or the Smith chart) we find

rad for the primary line, yielding

mho for the input admittance after attaching the primary line.

We may now solve for

as follows: Since

(Section 3.8), we find

Therefore,

.

The admittance of the parallel reactance should be

mho to cancel the imaginary part of

. The associated impedance is

. Since the sign of this impedance is negative, it must be a capacitor. The reactance of a capacitor is

, so it must be true that

Thus, we find the parallel reactance is a capacitor of value

.

Comparing this result to the result from the series reactance method (Exercise 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.

Additional Reading

  • “Smith chart” on Wikipedia.
 

Ellingson, Steven W. (2018) Electromagnetics, Vol. 1. Blacksburg, VA: VT Publishing. https://doi.org/10.21061/electromagnetics-vol-1 CC BY-SA 4.0

 
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