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



We have learned the impedance matching for Type #1 & Type #2 impedance using formulas & their conversions, and we’ll continue discussing how to match Type #3 & Type #4 impedance in this article.

As we’ve discussed in Part I, 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 the resulted impedance or admittance is a real number 1.
###### Matching Type #3 impedance: r < 1, g < 1, x > 0 or b < 0.

Type #3 impedance is located within the shaded area in the Smith chart as showed in Fig. 1.

Fig. 1  Type #3 impedance in the Smith chart

The process to match a Type #3 impedance into 50Ω:

There are 4 options to match this Type #3 impedance, we’ll discuss them one-by-one below.

A. Matching Type #3 impedance, Option #1 and Option #2

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

Fig. 2  Normalize the impedance

2. Convert 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. 3  Convert impedance to admittance

3. Add a capacitor $$C_p$$ in shunt with the admittance so the real part $$r_1$$ equals 1 after the resultant admittance is converted back to impedance. The real part $$g_1$$ of the admittance $$y_1$$ remains unchanged with this added component.

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

And,

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

Where $$C_p$$ is the admittance of the added capacitor.

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,  $${b_1}=\pm \sqrt {g-g^2}$$

The shunt capacitor $$C_p$$ is either

$$C_p=- \sqrt {g-g^2}-b$$  — Option #1.

or

$$C_p=\sqrt {g-g^2}-b$$  — Option #2.

Also,

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

$$x_1 \ge 0$$ — 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 a capacitor $$C_s$$  in series with impedance of $$-x_1$$.

Option #2: add an inductor $$L_s$$ in series with impedance of $$-x_1$$.

Fig. 5  Add a component in series to tune out the imaginary part of the impedance

Example: match this Type #3 impedance $$z=r+jx=0.6+j1.6$$ using Option #1 & Option #2.

Ans.

This is a resistor in series with an inductor.

Convert the impedance $$z$$ to admittance $$y$$:

$$g={r\over {r^2+x^2}}={0.6\over {0.6^2+1.6^2}}=0.205$$ $$b={-x\over {r^2+x^2}}={-1.6\over {0.6^2+1.6^2}}=-0.548$$

So, $${b_1}=\pm \sqrt {g-g^2}=\pm \sqrt {0.205-0.205^2}=\pm 0.404$$

Option #1, $$b_1=-0.404$$, add a capacitor $$C_p$$ in parallel and then a capacitor $$C_s$$ in series.

Fig. 6  Type #3 impedance with Option #1 matching

$${C_p}={b_1}-b=-0.404-(-0.548)=0.144 .. (admittance)$$

$${x_1}={-{b_1}\over {g^2+{b_1}^2}}={0.404\over {0.205^2+(-0.404)^2}}=1.97$$

$${C_s}=-{x_1}=-1.97 .. (impedance)$$

Option #2, $$b_1=0.404$$, add a capacitor $$C_p$$ in parallel and then an inductor $$L_s$$ in series.

Fig. 7  Type #3 impedance with Option #2 matching

$${C_p}={b_1}-b=0.404-(-0.548)=0.952 .. (admittance)$$

$${x_1}={-{b_1}\over {g^2+{b_1}^2}}={-0.404\over {0.205^2+(-0.404)^2}}=-1.97$$

$${L_s}=-{x_1}=1.97 .. (impedance)$$

B. Matching Type #3 impedance, Option #3 and Option #4

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

Fig. 8  Normalize the impedance

2. Add a capacitor $$C_s$$ in series with the impedance so the real part $$g_1$$ equals 1 after the resultant impedance is converted to admittance. The real part $$r_1$$ of the impedance $$z_1$$ remains unchanged with this added component.

Fig. 9  Add a capacitor in series with the impedance

$$z_1=r_1+jx_1$$

$$r_1=r$$, and $$x_1={x+{C_s}}$$

Where $$C_s$$ is the impedance of the added capacitor in series.

Convert $$z_1$$ 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,  $${x_1}=\pm \sqrt {r-r^2}$$

The capacitor $$C_s$$ in series is either

$$C_s= \sqrt {r-r^2}-x$$  — Option #3.

or

$$C_s=-\sqrt {r-r^2}-x$$  — Option #4.

Also,

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

$$b_1 \le 0$$ — Option #3

$$b_1 > 0$$ — Option #4

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

Option #3: add an in shunt capacitor $$C_p$$ with admittance of $$-b_1$$.

Option #4: add an in shunt inductor with admittance of $$-b_1$$.

Fig. 10  Add a component in shunt to tune out the imaginary part of the admittance

Example: match this Type #3 impedance $$z=r+jx=0.4+j0.6$$ using Option #3 & Option #4.

Ans.

This is a resistor in series with an inductor.

Add a capacitor $$C_s$$ in series to obtain $$g_1=1$$.

$$z_1=r_1+jx_1=r+j(x+C_s)$$

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

Also,

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

$$b_1={-0.49\over {0.4^2+0.49^2}}=-1.225$$ — Option #3

$$b_1={0.49\over {0.4^2+0.49^2}}=1.225$$ — Option #4

Option #3, $$x_1=0.49$$, $$b_1=-1.225$$, add a capacitor $$C_s$$ in series and then a capacitor $$C_p$$ in shunt.

Fig. 11  Type #3 impedance with Option #3 matching

$$C_s=x_1-x=0.49-0.6=-0.11 .. (impedance)$$

$$C_p=-b_1=-(-1.225)=1.225 .. (admittance)$$

Option #4, $$x_1=-0.49$$, $$b_1=1.225$$, add a capacitor $$C_s$$ in series and then an inductor $$L_p$$ in shunt.

Fig. 12  Type #3 impedance with Option #4 matching

$$C_s=x_1-x=-0.49-0.6=-1.09 .. (impedance)$$

$$L_p=-b_1=-1.225 .. (admittance)$$

Question: match this Type #3 impedance $$z=r+jx=0.2+j0.8$$ using Option #1,  Option #2, Option #3, and Option #4.

Ans.

Option #1: $$C_p=0.721, C_s=-1.549$$

Option #2: $$C_p=1.632, L_s=1.549$$

Option #3: $$C_s=-0.4, C_p=2.0$$

Option #4: $$C_s=-1.2, L_p=-2.0$$

###### Matching Type #4 impedance: r < 1, g < 1, x < 0 or b > 0.

Type #4 impedance is located within the shaded area in the Smith chart as showed in Fig. 13.

Fig. 13  Type #4 impedance in the Smith chart

The process to match a Type #4 impedance into 50Ω:

There are 4 options to match this Type #4 impedance, we’ll discuss them one-by-one below.

A. Matching Type #4 impedance, Option #1 and Option #2

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

Fig. 14  Normalize the impedance

2. Convert 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. 15  Convert impedance to admittance

3. Add an inductor $$L_p$$ in shunt with the admittance so the real part $$r_1$$ equals 1 after the resultant admittance is converted back to impedance. The real part $$g_1$$ of the admittance $$y_1$$ remains unchanged with this added component.

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

And,

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

Where $$L_p$$ is the admittance of the added inductor.

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,  $${b_1}=\pm \sqrt {g-g^2}$$

The shunt inductor $$L_p$$ is either

$$L_p=- \sqrt {g-g^2}-b$$  — Option #1.

or

$$L_p=\sqrt {g-g^2}-b$$  — Option #2.

Also,

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

$$x_1 \ge 0$$ — Option #1

$$x_1 < 0$$ — Option #2

4. Simply tune out $$x_1$$ to obtain $$z_2=1$$ by adding a capacitor or inductor in series.

Option #1: add a capacitor $$C_s$$  in series with impedance of $$-x_1$$.

Option #2: add an inductor $$L_s$$ in series with impedance of $$-x_1$$.

Fig. 17  Add a component in series to tune out the imaginary part of the impedance

Example: match this Type #4 impedance $$z=r+jx=0.4-j0.6$$ using Option #1 & Option #2.

Ans.

This is a resistor in series with a capacitor.

Convert the impedance $$z$$ to admittance $$y$$:

$$g={r\over {r^2+x^2}}={0.4\over {0.4^2+(-0.6)^2}}=0.769$$ $$b={-x\over {r^2+x^2}}={-0.6\over {0.4^2+(-0.6)^2}}=1.154$$

So, $${b_1}=\pm \sqrt {g-g^2}=\pm \sqrt {0.769-0.769^2}=\pm 0.421$$

Option #1, $$b_1=-0.421$$, add an inductor $$L_p$$ in shunt and then a capacitor $$C_s$$ in series.

Fig. 18  Type #4 impedance with Option #1 matching

$${L_p}={b_1}-b=-0.421-1.154=-1.575 .. (admittance)$$

$${x_1}={-{b_1}\over {g^2+{b_1}^2}}={0.421\over {0.769^2+(-0.421)^2}}=0.548$$

$${C_s}=-{x_1}=-0.548 .. (impedance)$$

Option #2, $$b_1=0.421$$, add an inductor $$L_p$$ in parallel and then an inductor $$L_s$$ in series.

Fig. 19  Type #4 impedance with Option #2 matching

$${L_p}={b_1}-b=0.421-1.154=-0.733 .. (admittance)$$

$${x_1}={-{b_1}\over {g^2+{b_1}^2}}={-0.421\over {0.769^2+0.421^2}}=-0.548$$

$${L_s}=-{x_1}=0.548 .. (impedance)$$

B. Matching Type #4 impedance, Option #3 and Option #4

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

Fig. 20  Normalize the impedance

2. Add an inductor $$L_s$$ in series with the impedance so the real part $$g_1$$ equals 1 after the resultant impedance is converted to admittance. The real part $$r_1$$ of the impedance $$z_1$$ remains unchanged with this added component.

Fig. 21  Add an inductor in series with the Type #4 impedance

$$z_1=r_1+jx_1$$

$$r_1=r$$, and $$x_1={x+{C_s}}$$

Where $$C_s$$ is the impedance of the added capacitor in series.

Convert $$z_1$$ 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,  $${x_1}=\pm \sqrt {r-r^2}$$

The capacitor $$C_s$$ in series is either

$$C_s= \sqrt {r-r^2}-x$$  — Option #3.

or

$$C_s=-\sqrt {r-r^2}-x$$  — Option #4.

Also,

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

$$b_1 \le 0$$ — Option #3

$$b_1 > 0$$ — Option #4

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

Option #3: add an in shunt capacitor $$C_p$$ with admittance of $$-b_1$$.

Option #4: add an in shunt inductor with admittance of $$-b_1$$.

Fig. 22  Add a component in shunt to tune out the imaginary part of the admittance

Example: match this Type #4 impedance $$z=r+jx=0.6-j1.6$$ using Option #3 & Option #4.

Ans.

This is a resistor in series with a capacitor.

Add an inductor $$L_s$$ in series to obtain $$g_1=1$$.

$$z_1=r_1+jx_1=r+j(x+L_s)$$

$${x_1}=\pm \sqrt {r-r^2}=\pm \sqrt {0.6-0.6^2}=\pm 0.49$$

Also,

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

$$b_1={-0.49\over {0.4^2+0.49^2}}=-0.816$$ — Option #3

$$b_1={0.49\over {0.4^2+0.49^2}}=0.816$$ — Option #4

Option #3, $$x_1=0.49$$, $$b_1=-0.816$$, add an inductor $$L_s$$ in series and then a capacitor $$C_p$$ in shunt.

Fig. 23  Type #4 impedance with Option #3 matching

$$L_s=x_1-x=0.49-(-1.6)=2.09 .. (impedance)$$

$$C_p=-b_1=-(-0.816)=0.816 .. (admittance)$$

Option #4, $$x_1=-0.49$$, $$b_1=0.816$$, add an inductor $$L_s$$ in series and then an inductor $$L_p$$ in shunt.

Fig. 24  Type #4 impedance with Option #4 matching

$$L_s=x_1-x=-0.49-(-1.6)=1.11 .. (impedance)$$

$$L_p=-b_1=-0.816 .. (admittance)$$

Question: match this Type #4 impedance $$z=r+jx=0.2-j0.8$$ using Option #1,  Option #2, Option #3, and Option #4.

Ans.

Option #1: $$L_p=-1.632, C_s=-1.549$$

Option #2: $$L_p=-0.721, L_s=1.549$$

Option #3: $$L_s=1.20, C_p=2.0$$

Option #4: $$L_s=0.40, L_p=-2.0$$

‘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.’