Increasing the Line Level Using Active Impedance Matching

TS615

33/36

First let’s consider the unloaded system. We can assume that the currents through R1, R2 and R3 are

respectively:

The gain, for the loaded system is given by Equation 6:

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:

Equation 7

where Ro is the synthesized impedance and Iout the output current.

On the other hand Vo can be expressed as:

Equation 8

By identification of both Equation 7 and Equation 8, the synthesized impedance is, with Rs1=Rs2=Rs:

Equation 9

Figure 75: Equivalent schematic. Ro is the synthesized impedance

2Vi

R1

---------

Vi

Vo

°

–

()

R2

---------------------------and

Vi

Vo

+

()

R3

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

,

G

Vo noload

()

Vi

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

1

2R2

R1

-----------

R2

R3

--------

++

1

R2

R3

--------

–

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

==

GL

Vo withload

()

Vi

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

1

2

---

1

2R2

R1

-----------

R2

R3

--------

++

1

R2

R3

--------

–

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

==

Vo

Vi G

()

Ro Iout

⋅

()

–

=

Vo

Vi 1

2R2

R1

-----------

R2

R3

--------

++





1

R2

R3

--------

–

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

Rs1Iout

1

R2

R3

--------

–

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

–

=

Ro

Rs

1

R2

R3

--------

–

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

=

Ro

Vi.Gi

Iout

1/2

RL