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:
- Add a lossless element, capacitor or inductor, to get the real part of either impedance or admittance to be 1.
- 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.
Fig. 4 Add a capacitor in shunt with the admittance
$$g_1=g={r\over {r^2+x^2}}$$
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.
Fig. 16 Add an inductor in shunt with the admittance
$$g_1=g={r\over {r^2+x^2}}$$
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.’