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I am using the CNY70 for a project, but i need information from the datasheet. In Figure 11, which i have added below, the relative collector current, \$I_{Crel}\$, is shown versus the displacement, s, for six different distances, d.

As i understand, d is a measure for the distance between the reflective sensors emitter and the reflective surface, as shown in the image in the upper right corner of fig. 11. But i can't seem to find a description of the displacement s. From the image, it seems as if the displacement s, is the distance from the center of the CNY70 to some surface perpendicular to the reflective surface. But that does not make any sense at all.

Fig. 11 in CNY70 Datasheet

I hope I have provided enough information. Thanks in advance.
Sorry if i left something out, i am new here and in the field of electrical engineering.

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  • \$\begingroup\$ I agree that it is not well explained. To me it looks like they could be specifying the relative output as you get closer to the edge of the reflecting surface. If that is the case, the graph indicates that as long as you are 10mm from an edge, you have no degradation. The only problem is that I would think that the orientation of the sensor would make a difference. Perhaps that is what the 1.5 in the top picture is trying to indicate. \$\endgroup\$ – crj11 Apr 1 '18 at 12:41
  • \$\begingroup\$ I think you are right, saying that it shows the CNY70's displacement from being fully covered to being uncovered. I agree, the orientation of the sensor would make a difference. Thanks for the help. \$\endgroup\$ – J. Fuglsang Apr 1 '18 at 13:42
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enter image description here

Figure 1. The graph is scaled in an unusual manner.

If I was doing this graph I probably would have used 0 as "edge on centre of target" and then used +/-5 mm for the adjustment. These guys are saying that 0 is covered with 5 mm beyond centreline.

As we might expect we get 50% current when the target edge is on the centreline as half the maximum light will be reflected.

The various curves are a function of the beam angle. The closer the target, the narrower the beam and so full or zero light is reflected with a smaller deviation in 's' from the centreline.

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  • \$\begingroup\$ Ah, okay. That makes a whole lot more sense. So at 5 mm displacement, they are saying that only half of the emitters beam is hitting the target, and therefore only half of the beam is reflected back? Thanks you for the quick answer. \$\endgroup\$ – J. Fuglsang Apr 1 '18 at 13:34
  • \$\begingroup\$ You've have understood correctly. \$\endgroup\$ – Transistor Apr 1 '18 at 13:41

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