Rockwell Automation 1771-VHSC , D17716.5.74 Very High S - Manual (Page 61)

Application Considerations C–4
5V Differential Line Driver Example
You want to use a 5V differential line driver in your encoder when
you have a long cable run and/or high input frequency or narrow
input pulses (input duty cycle < 50%). The top circuit (NO TAG)
shows a typical 5V differential line driver. The output is connected to
the field wiring arm terminal 1 and is sourcing current and the output
to terminal 2 is sinking current. JPR5 is connected to short out
resistor R1.
Important: Neither output of the differential line driver can be
connected to ground. Damage could occur to your
driving device.
To be sure that your device will drive the 1771-VHSC, you must
know the electrical characteristics of the output driver component
used in your signal source device. The output voltage differential
V
diff
= (V
oh
- V
ol
) is critical, because this is the drive voltage across
the 1771-VHSC input terminals 1 and 2, and the photodiode current
is a function of V
diff
- V
drop
.
The manufacturer of your shaft encoder or other pulse-producing
device can provide information on the specific output device used.
Note: Any signal source which uses a standard TTL output device
driver rated to source 400
µ
A or less in the high logic state is not
compatible with the 1771-VHSC module.
Many popular differential line drivers, such as the 75114,
75ALS192, and the DM8830 have similar characteristics and can
source or sink up to 40mA.
In general, the output voltage V
oh
will be higher both as the supply
voltage and the ambient temperature increase. For example, vendor
data for the 75114 shows V
oh
will be about 3.35V at V
cc
= 5 V, I
oh
=
10mA and 25 o C. Vol will be about 0.075V under the same
conditions. This means V
differential
= V
oh
- V
ol
= 3.27V if the part is
sourcing 10mA. Looking at the curves, if the part were sourcing
5mA you would see V
diff
= 3.425 - 0.05 = 3.37V.
Assuming that you could supply 5mA to the 1771-VHSC input
terminals, how much voltage across the field wiring arm terminals
would be required? V
drop
would be about 1.9V as previously noted.
And 5mA through 150 ohms gives an additional 0.75V drop. Thus,
you would have to apply about (1.9V + 0.75V) = 2.65V across the
terminals to cause a current of 4mA to flow through the photodiode.
The 75114 will give about 3.3V at V
cc
= 5V and 25 o C. Thus you
know that this driver will cause more current to flow than the
minimum required at 5mA.