EE-SX1088-datasheet question

Hi there i was looking at this part EE-SX1088 and when look at the data sheet it keeps reffering to a light current i was wondering what does that mean exactly. does that mean if i put a higher voltage into the diode will i get a higher light current or will that only increase the Foward current

The “Light Current” specification characterizes the amount of current that the detection transistor will allow to pass under the stated test conditions; with 10V applied across the collector and emitter, and 20mA of drive current to the diode on the input side, the transistor will pass between 0.5 and 14mA at a device temperature of 25°C.


Increasing the drive current to the illuminating LED will tend to increase the light current, just as increasing the current supplied to the base of a standard bipolar transistor will increase the flow of current from collector to emitter. There are of course limits here, most notably the absolute maximum forward current values specified in the datasheet. Operating at or near these limits will tend to increase the rate at which the device degrades.

Thank you for the response
I just have a bit of a follow-up question in the electrical and optical characteristics chart for the collector-emitter saturated voltage it has the values" 0.15 V typ., 0.4 V max"
does that mean that the device will start to degrade if I go over 0.4 V because, my design has a Vcc of 5V connected to the collector side with an output also on the collector side connected to an ic.

The saturation voltage refers to the potential that appears between collector and emitter when the light received from the detector is significantly more than required to pass the amount of current that is allowed to flow from collector to emitter, again subject to the stated test conditions. In this case, with 20mA drive current and 0.1mA flowing through the output, the output transistor will drop 150mV typically, 400mV maximum.

Depending on the input thresholds of the input device you’re connecting to, this may or may not present an issue from a logic level perspective; check the datasheet for that device. Also, consider using a resistor to ensure an appropriate logic level when the output is in an “off” state; if that device does not have a weak pull-down, it would be possible for parasitic capacitances to retain sufficient charge to falsely provide a logic high signal with the output device off, if I interpret the mention of your intended use correctly.

So, looking at the diagram on page 2 of the datasheet,
image
if you select R-load such that the output current is 0.1mA when the phototransistor is fully on (saturated) and apply 20mA to the input LED, the output voltage will typically be 0.15V lower than Vcc (4.85V with Vcc = 5V) and no more than 0.4V lower than Vcc.

I am little confused about this diagram as well just because i don’t understand why IL(light current) is going into Vcc or is it coming out of Vcc i was thinking of light current as an equivalent to Ib as an npn transistor

The light current is the collector current (Ic), not the base current (Ib), induced by the light from the LED. It will be drawn from Vcc through the collector to the emitter of the transistor. The light itself would be analogous to Ib.