Increasing the Line Level Using Active Impedance Matching
TS615
First let’s consider the unloaded system. We can assume that the currents through R1, R2 and R3 are
respectively:
2-R---V--1---i, -(--V----i---R–-----V2----o----°---)and (---V----i--R-+----3-V-----o---)-
As Vo° equals Vo without load, the gain in this case becomes:
G = -V----o----(--n---V-o----il-o----a----d----) = -1----+-----12----R----R--–----1----2--RR----------+23-------RR----------23----
The gain, for the loaded system is given by Equation 6:
GL = V-----o----(--w-----i-V-t--h-i---l-o----a----d----) = 12-- 1-----+-----12-----R---R--–----1----2-RR-----------+23-------RR----------23----
Equation 6
The system shown in Figure 74 is an ideal generator with a synthesized impedance acting as the internal
impedance of the system. From this, the output voltage becomes:
Vo = (ViG) – (Ro ⋅ Iout)
where Ro is the synthesized impedance and Iout the output current.
On the other hand Vo can be expressed as:
Equation 7
Vo
=
V
i
1
+
2---R-R---1--2--
+
RR-----23--
----------------1-----–----RR----------23--------------------
–
R---1--s---–-1---I-RR----o------23-u-----t
Equation 8
By identification of both Equation 7 and Equation 8, the synthesized impedance is, with Rs1=Rs2=Rs:
Ro = 1-----–R-----RR-s---------23----
Equation 9
Figure 75: Equivalent schematic. Ro is the synthesized impedance
Ro
Iout
Vi.Gi
1/2RL
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