Rockwell Automation 1756-HSC ControlLogix High Speed Counter Module - Manual (Page 107)

Rockwell Automation Publication 1756-UM007C-EN-P - November 2011
Application Considerations 107
The previous calculation is necessary because the driving device must cause a
minimum of 5 mA to flow through the photodiode.
The optical isolator manufacturer recommends a maximum of 8 mA to flow
through the photodiode. This current could be exceeded in the 24V position.
To obtain this limit, a DC shunt circuit is included, consisting of D1, Q1, R5
and R6. If the photodiode current exceeds about 8 mA, the drop across R5-R6
will be sufficient to turn Q1 on, and any excess current will be shunted
through D1 and Q1 instead of through the photodiode.
If the driving device is a standard 5V differential line driver, D2 and D3
provide a path for reverse current when the field wiring arm terminal 1 is logic
low and terminal 2 is logic high. The combined drop is about the same at the
photodiode (about 1.4V). The circuit appears more symmetrical, or balanced,
to the driver as opposed to just one diode.
Detailed Circuit Analysis
In the previous example, we used a constant 2.0V drop across the photodiode
and R5-R6. To calculate the true photodiode current, consider the
photodiode, D1, Q1, R5 and R6 as one circuit. The voltage drop across D1
and Q1 is always equal to the drop across the photodiode and R5-R6. We will
call this V
drop
.
First, consider the minimum requirement of I
f
= 4 mA. The V
f
curves for this
photodiode typically has a 1.21…1.29V drop as the junction temperature
varies from 70…25 °C. Let’s call it 1.25V. With 4 mA current, R5 and R6 will
drop (80.4  x 4 mA) = 0.32V. Thus, at 4 mA:
V
drop
= (1.25V + 0.32V) = 1.57V.
Consider when I
f
= 8 mA or above. With the temperature about halfway
between 25…70 °C, V
f
becomes about 1.25V. R5-R6 will now drop 0.64V
(80.4  x 8 mA). That means:
V
drop
= 1.25V + 0.64V = 1.89V.
The V
be
of Q1 is now sufficient to start to turn Q1 on. If the current through
the photodiode increases to 9 mA, V
be
becomes 0.72V and Q1 is fully on. Any
additional current (supplied by a 24V applied input) is shunted away from the
photodiode and dissipated in Q1 and D1.