TS634
7/9
Compon e nt calc ulation:
Let us consider the equivalent c ircuit 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 generator with a synthesized impedance as the i nterna l
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 expressed a s:
By identification of both e quat ions (2) a nd (3), the
synthesized impedance is, with Rs1=Rs2=Rs:
Unlike the level Vo° required for a passive impedance, Vo° will be smaller than 2Vo in our case. Let
us write Vo°=kVo with k t he matching factor varying between 1 and 2. Assum ing that the current
through R3 i s negligeable, it c omes the fo llowing
resistance divider:
After choosing the k factor, Rs will equal to
1/2RL(k-1).
A good impedance matching assume s:
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
2
Vi
R
1
---------
Vi Vo°
–
()
R
2
--------------------------
and
Vi Vo
+
()
R
3
----------------------- -
,
G
Vo noload()
Vi
-------------------------------
1
2R2
R
1
---------- -
R
2
R
3
-------
++
1
R
2
R
3
------ -
–
-----------------------------------
==
GL
Vo withload()
Vi
------------------------------------
1
2
-- -
1
2R2
R
1
---------- -
R
2
R
3
-------
++
1
R
2
R
3
------ -
–
-----------------------------------
1(),==
Vo ViG()RoIout()
–= 2
()
,
Vo
Vi
1
2R2
R
1
---------- -
R
2
R
3
------ -
++
1
R
2
R
3
------ -
–
---------------------------------------------- -
Rs1Iout
1
R
2
R
3
------ -
–
---------------------
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) Abritra ry fixed
R3 (=R5) R2/(1-Rs/0.5RL)
Rs 0.5RL(k-1)
Ro
Rs
1
R
2
R
3
-------
–
---------------- -
4(),=
Ro
Vi.Gi
Iout
1/2
RL
Ro
kVoRL
RL2Rs
1+
---------------------------
=
Ro
1
2
-- -
RL5()
,=
R
2
R
3
-------
1
2
Rs
RL
----------
6(),–=
R
3
R
2
1
2
Rs
RL
--------- -
–
-------------------
=
R
1
2R2
21
R
2
R
3
-------
–
GL
1–
R
2
R
3
-------
–
---------------------------------------------------------
7(),=