TS615 datasheet(26/27 Pages) STMICROELECTRONICS

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TS615

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Unlike the level Vo° required for a passive imped-

ance, Vo° will be smaller than 2Vo in our case. Let

us write Vo°=kVo with k the matching factor vary-

ing between 1 and 2. Assuming that the current

through R3 is negligible, it comes the following re-

sistance divider:

After choosing the k factor, Rs will equal to

1/2RL(k-1).

A good impedance matching assumes:

From (eq4) and (eq5) it becomes:

By fixing an arbitrary value of R2, (eq6) gives:

Finally, the values of R2 and R3 allow us to extract

R1 from (eq1), and it comes:

with GL the required gain.

Figure 77 : Components Calculation for

Impedance Matching Implementation

kV oRL

2R s1

------------------------------

--- RL eq5

()

--------1

2Rs

----------- eq6

()

–

GL (gain for the

loaded system)

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]

R2 (=R4)

Abritrary fixed

R3 (=R5)

R2/(1-Rs/0.5RL)

0.5RL(k-1)

Load viewed by

each driver

kRL/2

2Rs

-----------

–

---------------------

2R2

--------

–



 GL 1

–

--------

–

----------------------------------------------------------- eq7

()

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