Smith Chart - Using The Smith Chart To Solve Conjugate Matching Problems With Distributed Components

Using The Smith Chart To Solve Conjugate Matching Problems With Distributed Components

Usually distributed matching is only feasible at microwave frequencies since, for most components operating at these frequencies, appreciable transmission line dimensions are available in terms of wavelengths. Also the electrical behavior of many lumped components becomes rather unpredictable at these frequencies.

For distributed components the effects on reflection coefficient and impedance of moving along the transmission line must be allowed for using the outer circumferential scale of the Smith chart which is calibrated in wavelengths.

The following example shows how a transmission line, terminated with an arbitrary load, may be matched at one frequency either with a series or parallel reactive component in each case connected at precise positions.

Supposing a loss-free air-spaced transmission line of characteristic impedance, operating at a frequency of 800 MHz, is terminated with a circuit comprising a 17.5 resistor in series with a 6.5 nanohenry (6.5 nH) inductor. How may the line be matched?

From the table above, the reactance of the inductor forming part of the termination at 800 MHz is

so the impedance of the combination is given by

and the normalised impedance is

This is plotted on the Z Smith chart at point P₂₀. The line OP₂₀ is extended through to the wavelength scale where it intersects at the point . As the transmission line is loss free, a circle centred at the centre of the Smith chart is drawn through the point P₂₀ to represent the path of the constant magnitude reflection coefficient due to the termination. At point P₂₁ the circle intersects with the unity circle of constant normalised resistance at

.

The extension of the line OP₂₁ intersects the wavelength scale at, therefore the distance from the termination to this point on the line is given by

Since the transmission line is air-spaced, the wavelength at 800 MHz in the line is the same as that in free space and is given by

where is the velocity of electromagnetic radiation in free space and is the frequency in hertz. The result gives, making the position of the matching component 29.6 mm from the load.

The conjugate match for the impedance at P₂₁ is

As the Smith chart is still in the normalised impedance plane, from the table above a series capacitor is required where

Rearranging, we obtain

.

Substitution of known values gives

To match the termination at 800 MHz, a series capacitor of 2.6 pF must be placed in series with the transmission line at a distance of 29.6 mm from the termination.

An alternative shunt match could be calculated after performing a Smith chart transformation from normalised impedance to normalised admittance. Point Q₂₀ is the equivalent of P₂₀ but expressed as a normalised admittance. Reading from the Smith chart scaling, remembering that this is now a normalised admittance gives

(In fact this value is not actually used). However, the extension of the line OQ₂₀ through to the wavelength scale gives . The earliest point at which a shunt conjugate match could be introduced, moving towards the generator, would be at Q₂₁, the same position as the previous P₂₁, but this time representing a normalised admittance given by

.

The distance along the transmission line is in this case

which converts to 123 mm.

The conjugate matching component is required to have a normalised admittance of

.

From the table it can be seen that a negative admittance would require an inductor, connected in parallel with the transmission line. If its value is, then

This gives the result

A suitable inductive shunt matching would therefore be a 6.5 nH inductor in parallel with the line positioned at 123 mm from the load.