Rockwell Automation 1756-HSC ControlLogix High Speed Counter Module - Manual (Page 109)

Rockwell Automation Publication 1756-UM007C-EN-P - November 2011
Application Considerations 109
Assuming that you could supply 5 mA to the 1756-HSC input terminals, how
much voltage across the field-wiring arm terminals would be required? V
drop
would be about 1.6V as previously noted. And 4 mA through 150  gives an
additional 0.60V drop. Thus, you would have to apply about (1.6V + 0.60V) =
2.20V across the terminals to cause a current of 4 mA to flow through the
photodiode. The 75114 gives about 3.3V at V
cc
= 5V and 25 °C (77 °F). Thus
you know that this driver causes more current to flow than the minimum
required at 4 mA.
To determine how much current flows, use the following equation:
V
drive
- V
drop
= V
resistor
3.3V - 1.6V = 1.7V
1.5V/150  = 11.3 mA
As you can see, 1.6V
drop
is too low.
Remember that V
drop
varies linearly from about 1.6V to 2.0V as I
f
varies from
4…8 mA. Recalculate assuming V
drop
= 2.0V.
V
drive
- V
drop
= V
resistor
3.3V - 2.0V = 1.3V
1.3V/150  = 8.7 mA
The resulting 8.7 mA is consistent with our assumption of V
drop
= 2.0V
at I
f
= 8 mA. This shows that driver 75114 causes about 8 mA to flow through
the photodiode.
+12 to +24V Single-ended Driver
Some European-made encoders use a circuit similar to the lower circuit in the
figure below. The current capable of being sourced is limited only by the 22 
resistor in the driver output circuit (R).