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RS-485: Transmitting Full-Duplex Data over Single Twisted-Pair Cable Application Note
Equation 6
shows that to eliminate V
making R
/R
equals k:
B
D
R
B
------- -
(EQ. 7)
=
k
R
D
With the V
component eliminated, the receiver input voltage is purely dependent of V
OD1
V
----------------------------- -
V
=
AB1
1
-- -
1
+
k
Then, substituting k with
V
---------------------------------
V
=
AB1
(EQ. 8)
-----------------------
1
+
R
1.6
Deriving the R
Figure 11
shows that there are two leakage current paths parallel to the termination resistor. Their combined
current value, I
+ I
LK1
LK2
loading of the main signal path. This requires that the sum of the resistor values (R
the termination resistor:
R
+
R
20R
(EQ. 9)
B
D
At the same time, the gain requirement of
R
=
R
k
(EQ. 10)
B
D
Setting
Equation 10
=
Equation 9
20R
T
R
--------------
=
D
k
+
1
Substituting k with
Equation 7
(EQ. 11)
R
=
20
R
D
R15AN0009EU0100 Rev.1.00
Mar 4, 2022
, its bracket, k - R
OD1
OD2
R
D
--------- -
+
R
IN
Equation 7
gives
Equation
V
OD2
OD2
------------------------------------------------ -
=
R
1
D
------- -
1
+
R
+
D
R
R
B
B
IN
and R
Resistor Values
B
D
, should be about 10% of I
T
Equation 7
A
R
B
R
D
Z
D
Y
Figure 11. Leakage Currents must be ≤ 10% of Termination Current
and solving for R
gives
Equation
2 R
+
2R
=
20
T
T
S
/R
, must become zero. This is accomplished by
B
D
8:
1
--------- -
R
IN
, the current through the termination resistor, to minimize
T
demands that
I
LK1
I
LK2
R
R
R
D
B
R
S
R
S
gives the first resistor value with:
D
11:
R
2 R
2R
+
T
T
S
------------------------------------------------
R
+
2 R
+
2R
T
T
S
OD2
, R
) is about 20 times that of
B
D
B
P
I
R
V
T
T
Bus
N
:
Page 6
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