TS634ID 查看數據表（PDF） - STMicroelectronics
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TS634ID Datasheet PDF : 9 Pages
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TS634
Component calculation:
Let us consider the equivalent circuit for a single
ended configuration, Figure 4.
Figure 4 : Single ended equivalent circuit
+
Vi
_
1/2R1
R2
R3
Rs1
Vo°
Vo
-1
1/2RL
By identification of both equations (2) and (3), the
synthesized impedance is, with Rs1=Rs2=Rs:
Ro
=
------R----s------
1
–
R-----2-
R3
,(
4)
Figure 5 : Equivalent schematic. Ro is the syn-
thesized impedance
Vi.Gi
Ro
Iout
1/2RL
Let us consider the unloaded system. Assuming
the currents through R1, R2 and R3
as respectively:
-2R---V--1---i,
(---V----i---–-----V----o---°---)-
R2
a
nd
(---V----i---+-----V----o----)
R3
As Vo° equals Vo without load, the gain in this
case becomes :
G
=
V-----o---(--n----o---l---o---a---d----)
Vi
=
-1----+-----1-2---R----R-–----1----2-RR----------+23------RR---------23---
The gain, for the loaded system will be (1):
GL
=
V-----o---(---w----i--t--h----l--o---a----d---)-
Vi
=
-1-
2
-1----+------2---R----R-----1----2------+-----RR---------23---
1
–
R-----2-
R3
,(
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:
Vo = (ViG) – (RoIout),(2)
with Ro the synthesized impedance and Iout the
output current. On the other hand Vo can be ex-
pressed as:
Vo
=
V
i
1
+
-2---R----2--
R1
+
RR-----23-
----------------------------------------------
1
–
R-----2-
R3
–
R-----s---1---I---o---u----t
1
–
R-----2-
R3
,(
3)
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:
Ro
=
-----k---V-----o---R----L-------
RL + 2Rs1
After choosing the k factor, Rs will equal to
1/2RL(k-1).
A good impedance matching assumes:
Ro
=
-1-
2
R
L
,(
5
)
From (4) and (5) it becomes:
R-----2-
R3
=
1
–
-2---R----s-
RL
,(6
)
By fixing an arbitrary value for R2, (6) gives:
R3
=
-------R----2--------
1 – -2---R----s-
RL
Finally, the values of R2 and R3 allow us to extract
R1 from (1), and it comes:
R1 = ------------------------2----R----2------------------------- ,(7)
2
1
–
RR-----23-
GL
–
1
–
R-----2-
R3
with GL the required gain.
GL (gain for the
loaded system)
R1
R2 (=R4)
R3 (=R5)
Rs
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]
Abritrary fixed
R2/(1-Rs/0.5RL)
0.5RL(k-1)
7/9
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