TS634

7/9

Component calculation:

Let us consider the equivalent circuit for a single

ended configuration, Figure 4.

Let us consider the unloaded system. Assuming

the currents through R1, R2 and R3

as respectively:

As Vo° equals Vo without load, the gain in this

case becomes :

The gain, for the loaded system will be (1):

As shown in figure5, this system is an ideal gener-

ator with a synthesized impedance as the internal

impedance of the system. From this, the output

voltage becomes:

with Ro the synthesized impedance and Iout the

output current. On the other hand Vo can be ex-

pressed as:

By identification of both equations (2) and (3), the

synthesized impedance is, with Rs1=Rs2=Rs:

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 negligeable, it comes the following

resistance divider:

After choosing the k factor, Rs will equal to

1/2RL(k-1).

A good impedance matching assumes:

From (4) and (5) it becomes:

By fixing an arbitrary value for R2, (6) gives:

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

R1 from (1), and it comes:

with GL the required gain.

Figure 4 : Single ended equivalent circuit

1/2

R1

R2

R3

+

_

Vi

Vo

Rs1

-1

Vo°

1/2

RL

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

-------

–

----------------------------------- 1

()

,

==

Vo

ViG

()

RoIout

()

–

=

2

()

,

Vo

Vi 1

2R2

R1

-----------

R2

R3

-------

++





1

R2

R3

-------

–

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

Rs1Iout

1

R2

R3

-------

–

--------------------- 3

()

,

–

=

Figure 5 : Equivalent schematic. Ro is the syn-

thesized impedance

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)

R1

2R2/[2(1-R2/R3)GL-1-R2/R3]

R2 (=R4)

Abritrary fixed

R3 (=R5)

R2/(1-Rs/0.5RL)

Rs

0.5RL(k-1)

Ro

Rs

1

R2

R3

-------

–

----------------- 4

()

,

=

Ro

Vi.Gi

Iout

1/2

RL

Ro

kVoRL

RL

2Rs1

+

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

=

Ro

1

2

---RL 5

()

,

=

R2

R3

-------

1

2Rs

RL

---------- 6

()

,

–

=

R3

R2

1

2Rs

RL

----------

–

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

=

R1

2R2

21

R2

R3

-------

–





GL

1

–

R 2

R 3

-------

–

--------------------------------------------------------- 7

()

,

=