TS615
Increasing the Line Level Using Active Impedance Matching
Let us write Vo°=kVo, where k is the matching factor varying between 1 and 2. If we assume that the
current through R3 is negligible, we can calculate the output resistance, Ro:
Ro = R-----L-k---V-+----o-2---R-R---L--s---1--
After choosing the k factor, Rs will equal to 1/2RL(k-1).
For a good impedance matching we assume that:
Ro = 12-- RL
From Equation 9 and Equation 10, we derive:
Equation 10
RR-----23-- = 1 – 2--R--R---L--s--
By fixing an arbitrary value of R2, Equation 11 becomes:
R3 = 1-----–--R---2----2R----R-----L----s----
Finally, the values of R2 and R3 allow us to extract R1 from Equation 6, so that:
Equation 11
with GL the required gain.
R1 = ------------------------2----R-----2--------------------------
21 – RR-----23-- GL – 1 – RR-----23--
Table 7. Components calculation for impedance matching implementation
GL (gain for the loaded
system)
R1
R2 (=R4)
R3 (=R5)
Rs
Load viewed by each
driver
GL is fixed for the application requirements
GL=Vo/Vi=0.5(1+2R2/R1+R2/R3)/(1-R2/R3)
2R2/[2(1-R2/R3)GL-1-R2/R3]
Arbitrarily fixed
R2/(1-Rs/0.5RL)
0.5RL(k-1)
kRL/2
Equation 12
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