TS615
INCREASING THE LINE LEVEL BY USING AN
ACTIVE IMPEDANCE MATCHING
With a passive matching, the output signal ampli-
tude of the driver must be twice the amplitude on
the load. To go beyond this limitation an active
matching impedance can be used. With this tech-
nique, it is possible to keep a good impedance
matching with an amplitude on the load higher
than the half of the output driver amplitude. This
concept is shown in figure 74 for a differential line.
Figure 74 : TS615 as a differential line driver with
an active impedance matching
100n
Vcc+
1k
Vi
+ Vcc+
_
GND
R2
1/2 R1 R3
Vcc/2
1/2 R1 R5
Vi
10µ
100n
1k GND
R4
+ Vcc+
100n
_
GND
Rs1
Vo°
Vo°
Rs2
1µ
10n
Vo
RL
1:n
Hybrid
& 100Ω
Transformer
Vo
2-R---V--1---i, (---V----i---R–-----V2----o----°---) and(---V----i--R-+----3-V-----o---)-
As Vo° equals Vo without load, the gain in this
case becomes :
G = V-----o----(--n---V-o----il-o----a----d----) = 1-----+-----12----R----R--–----1----2--RR----------+23-------RR----------23----
The gain, for the loaded system will be (eq1):
GL = V-----o----(--w-----i-V-t--h-i---l-o----a----d----) = 12-- 1-----+-----1-2---R----R--–----1----2-RR-----------+23-------RR----------23---- ,(eq1)
As shown in figure76, this system is an ideal gen-
erator with a synthesized impedance as the inter-
nal impedance of the system. From this, the out-
put voltage becomes:
Vo = (ViG) – (RoIout),(eq2)
with Ro the synthesized impedance and Iout the
output current. On the other hand Vo can be ex-
pressed as:
Component Calculation
Let us consider the equivalent circuit for a single
ended configuration, Figure75.
Figure 75 : Single ended equivalent circuit
+
Vi
_
1/2R1
R2
R3
Rs1
Vo°
-1
Vo
1/2RL
Vo
=
V
i
1
+
2----R-----2--
R1
+
RR-----23--
----------------1-----–----RR----------23--------------------
–
R---1--s---–-1---I-RR----o-----23--u-----t ,(
eq3)
By identification of both equations (eq2) and
(eq3), the synthesized impedance is, with
Rs1=Rs2=Rs:
Ro = 1-----–R-----RR-s---------23---- ,(eq4)
Figure 76 : Equivalent schematic. Ro is the
synthesized impedance
Ro
Iout
Vi.Gi
Let us consider the unloaded system. Assuming
the currents through R1, R2 and R3
1/2RL
as respectively:
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