The inputs to the 6527 are optically isolated from the internally circuitry of the board. If you look at figures 3-3 and 3-4 of the User Manual, it shows how to connect a signal.
There is a 3kohm,1/4W resistor and an led in series between the DIG+ and DIG- input lines. When you send a current through this series combination sufficient to light the LED, you get a Logic 1. You should setup your supply and load (external circuitry) to supply at least 1mA of current to the LED, while at the same time making sure not to exceed the power dissipation limit of the resistor and the maximum current of the LED. These specs are given in the manual.
For example, supply a known voltage of 5V to DIG+ and ground DIG-. You will get a 1.5V drop across the LED, and the remainder of
3.5V across the resistor. 3.5V/3kohm = 1.2mA which is sufficient for the LED to light and give a Logic 1. The power dissipated by the resistor is 3.5V^2/3kohm = 4.1mW, which is well below .25W.
