Radio Frequency Integrated Circuit Design - Webs

Impedance Matching
Much as in the previous section, the analysis of either Figure 4.16(a) or
Figure 4.16(b) begins by finding the equivalent impedance of the network. In
the case of Figure 4.16(b), the impedance is given by
Z in = j�L 1R + j�L 2R − � 2 L 1L 2
R + j�L 2
Equivalently, the admittance can be found:
Yin = j�R 2 (L 1 + L 2) − � 2 L 2 2R + j� 3 L 1L 2 2
−� 2 R 2 (L 1 + L 2) 2 − � 4 L 2 1L 2 2
Thus, the inverse of the real part of this equation gives R eq:
R eq = −R 2 (L 1 + L 2) 2 − � 2 L 2 1L 2 2
− RL 2 2
77
(4.11)
(4.12)
R� = (L 1 + L 2) 2 + L 2 1
Q 2 2
L 2 � (4.13)
2
where Q 2 is the quality factor of L 2 and R in parallel. As long as Q 2 is large,
then a simplification is possible. This is equivalent to stating that the resistance
of R is large compared to the impedance of L 2, and the two inductors form a
voltage divider.
R eq ≈ R�L 1 + L 2
L � 2
2
(4.14)
The equivalent inductance of the network can be found as well. Again,
the inverse of the imaginary part divided by j� is equal to the equivalent
inductance:
L eq = �R 2 (L 1 + L 2) 2 − � 2 L 2 1L 2 2�
R 2 (L 1 + L 2) 2 + � 2 L 1L 2 2
=
(L 1 + L 2) 2 − L 2 1
Q 2 2
L 1 + L 2 + L 1
Q 2 2
Making the same approximation as before, this simplifies to
L eq ≈ L 1 + L 2
(4.15)
(4.16)
which is just the series combination of the two inductors if the resistor is absent.