Impedance Matching-Using Lump Elements, Formulas, and Conversions-Part I.

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In this article you’ll learn how to match any type of impedance by using formulas and their conversions. The only thing you need to do is enter S11 or to-be-matched impedance and you’ll get the answers by following all steps.

Based on the values of r, g, x, and b, we can roughly categorize the impedance into 4 different types:

  • Type #1: r ≥ 1, x any value.
  • Type #2: g ≥ 1, b any value.
  • Type #3: r < 1, g < 1, x > 0 or b < 0.
  • Type #4: r < 1, g < 1, x < 0 or b > 0.

Fig. 1  Four types of impedance in the Smith chart

 

Theoretically all these 4 types of impedance can be perfectly matched into 50Ω by using only 2 lumped elements, inductors and capacitors, if not considering the limited amount of component values we are able to get as well as their tolerances.

The very basic rules of impedance matching are:

  1. Add a lossless element, capacitor or inductor, to get the real part of either impedance or admittance to be 1.
  2. Add the second lossless element to tune out the remaining imaginary part, reactance or susceptance, so the resultant impedance or admittance is a real number 1.

We’ll discuss how to match these 4 types of impedance one-by-one, including background, process, and results.

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Matching Type #1 impedance: r ≥ 1, x any value.

Type #1 impedance is located within the area of \(r=1\) circle.

 

Fig. 2  Type #1 impedance in the Smith chart 

 

The process to match a Type #1 impedance into 50Ω:

1. Normalize the given impedance. If the impedance is \(Z=R+jX\), then the normalized impedance is \(z=Z/50=r+jx\). 

Fig. 3  Normalize the impedance

 

2. Because \(r ≥ 1\), we need to transform the impedance \(z\) to admittance \(y\) using these 2 equations: $$g={r\over {r^2+x^2}}$$ and, $$b={-x\over {r^2+x^2}}$$

 

Fig. 4  Transform impedance to admittance

 

3. Add either a capacitor, option #1, or an inductor, option #2, in shunt with the admittance so the real part \(r_1\) equals 1 after the resultant admittance is transformed back to impedance. The real part \(g_1\) of the admittance \(y_1\) remains unchanged with this added component. 

Fig. 5  Add a component in shunt with the admittance

 

$$g_1=g={r\over {r^2+x^2}}$$

And, 

$$b_1={{-x\over {r^2+x^2}}+{b_p}}$$

Where \(b_p\) is the admittance of either the added capacitor \(C_p\) or inductor \(L_p\).

Since,

$${r_1}={{g_1}\over {{g_1}^2+{b_1}^2}}={g\over {g^2+{b_1}^2}}=1$$

So, $$g-g^2={b_1}^2$$ 

and,  $${b_1}=\pm \sqrt {g-g^2}$$

If \({b_1}=+ \sqrt {g-g^2} \ge 0\), a shunt capacitor \(C_p\) is added and the admittance is \(C_p={b_1}-b\)  — Option #1.

If \({b_1}=- \sqrt {g-g^2} <0\), a shunt inductor \({L_p}\) is added and the admittance is \(L_p={b_1}-b\)  — Option #2.  

Also,

$${x_1}={-{b_1}\over {g^2+{b_1}^2}}$$

\(x_1 \le 0\) — Option #1

\(x_1 > 0\) — Option #2

 

4. Simply tune out \(x_1\) to obtain \(z_2=1\) by adding an inductor or capacitor in series.

Option #1: add an inductor in series with impedance of \(-x_1\). 

Option #2: add a capacitor in series with impedance of \(-x_1\).

Fig. 6  Add a component in series to tune out the imaginary part of the impedance

 

Example: match this Type #1 impedance \(z=r+jx=2.0+j1.0\).

Ans.

This is a resistor in series with an inductor.

Transform the impedance \(z\) to admittance \(y\):

$$g={r\over {r^2+x^2}}={2.0\over {2.0^2+1.0^2}}=0.4$$ $$b={-x\over {r^2+x^2}}={-1.0\over {2.0^2+1.0^2}}=-0.2$$

So, $${b_1}=\pm \sqrt {g-g^2}=\pm \sqrt {0.4-0.4^2}=\pm 0.49$$

Option #1, add a capacitor \(C_p\) in parallel and then an inductor \(L_s\) in series.

Fig. 7  Type #1 impedance with Option #1 matching

 

If \(b_1=0.49\), then

$${C_p}={b_1}-b=0.49-(-0.2)=0.69 .. (admittance) $$

$${x_1}={-{b_1}\over {g^2+{b_1}^2}}={-0.49\over {0.4^2+0.49^2}}=-1.22$$

$${L_s}=-{x_1}=1.22 .. (impedance) $$

 

Option #2, add an inductor \(L_p\) in parallel and a capacitor \(C_s\) in series.

Fig. 8  Type #1 impedance with Option #2 matching

 

If \(b_1=-0.49\), then

$${L_p}={b_1}-b=-0.49-(-0.2)=-0.29 .. (admittance) $$

$${x_1}={-{b_1}\over {g^2+{b_1}^2}}={0.49\over {0.4^2+0.49^2}}=1.22$$

$${C_s}=-{x_1}=-1.22 .. (impedance) $$

 

Question:

Follow the steps above, match this impedance \(z=1.4-j0.6\).

Ans.   

Option #1: \(C_p=0.23\), \(L_s=0.81\)

Option #2: \(L_p=-0.75\), \(C_s=-0.81\)

 
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Matching Type #2 impedance: g ≥ 1, b any value.

Type #2 impedance is located within the area of \(g=1\) circle.

 

Fig. 9  Type #2 impedance in the Smith chart

 

The process to match a Type #2 impedance into 50Ω:

1. Normalize the given impedance. If the impedance is \(Z=R+jX\), then the normalized impedance is \(z=Z/50=r+jx\). 

Fig. 10  Normalize the impedance

 

2. Add either a capacitor \(C_s\), option #1, or an inductor \(L_s\), option #2, in series with the impedance so the real part \(g_1\) equals 1 after the resultant impedance is transformed to admittance. The real part \(r_1\) of the impedance \(z_1\) remains unchanged with this added component. 

Fig. 11  Add a component in series with the impedance

\(z_1=r_1+jx_1\)

\(r_1=r\), and \(x_1={x+{x_s}}\)

Where \(x_s\) is the impedance of either the added capacitor \(C_s\) or inductor \(L_s\).

 

Transform \(z_1\) to \(g_1\).

$$g_1={{r_1}\over {{r_1}^2+{x_1}^2}}={r\over {r^2+{x_1}^2}}=1$$

So, $$r-r^2={x_1}^2$$

and,  $${x_1}=\pm \sqrt {r-r^2}$$

If \({x_1}=- \sqrt {r-r^2} <0\), a capacitor \({C_s}\) is added in series and the impedance is \(C_s={x_1}-x\)  — Option #1.  

If \({x_1}=+\sqrt {r-r^2} >0\), an inductor \({L_s}\) is added in series and the impedance is \(L_s={x_1}-x\)  — Option #2.  

 

Also, $$b_1={-{x_1}\over {r^2+{x_1}^2}}$$

\({b_1}>0\) — Option #1

\({b_1}<0\) — Option #2

 

4. Simply tune out \(b_1\) to obtain \(y_2=1\) by adding an inductor or capacitor in shunt.

Option #1: add an in shunt inductor with admittance of \(-b_1\). 

Option #2: add an in shunt capacitor with admittance of \(-b_1\).

 

Fig. 12  Add a component in shunt with the admittance \(y_1\)

 

Example: match this Type #2 impedance \(z=r+jx=0.4-j0.2\).

Ans.

\(z\) is a resistor in series with a capacitor.

$${x_1}=\pm \sqrt {r-r^2}=\pm \sqrt {0.4-0.4^2}=\pm 0.49$$

Option #1, add a capacitor \(C_s\) in series and then an inductor \(L_p\) in shunt.

Fig. 13  Type #2 impedance with Option #1 matching

 

If \(x_1=-0.49<0\), then

$$C_s={{x_1}-x}=-0.49-(-0.2)=-0.29 .. (impedance) $$

Also, $$b_1={-{x_1}\over {r^2+{x_1}^2}}={-(-0.49)\over {0.4^2+{(-0.49)}^2}}=1.22$$

$$L_p={-b_1}=-1.22 .. (admittance) $$

 

Option #2, add a inductor \(L_s\) in series and then an capacitor \(C_p\) in shunt.

If \(x_1=0.49>0\), then

$$L_s={{x_1}-x}=0.49-(-0.2)=0.69 .. (impedance) $$

Also, $$b_1={-{x_1}\over {r^2+{x_1}^2}}={-0.49\over {0.4^2+0.49^2}}=-1.22$$

$$C_p={-b_1}=1.22 .. (admittance) $$

 

Question:

Follow the steps above, match this impedance \(z=0.6+j0.4\).

Ans.   

Option #1: \(C_s=-0.89\), \(L_p=-0.82\)

Option #2: \(L_s=0.09\), \(C_p=0.82\)

 

Continue to the next article to learn Type #3 & Type #4 impedance matching.

 

Fig. 14  Type #3 and Type #4 impedance

 

 

 

‘Note: This is an article written by an RF engineer who has worked in this field for over 40 years. Visit ABOUT to see what you can learn from this blog.’

 

 

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