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I'm drawing the footprint for APDS-9253 and in the datasheet I'm looking at package drawings. As far as I understand, the top-left drawing is a view from component side, the top-right one from solder side.

It seems the pin 1 is placed at the opposite side of the standard pinout. Is it correct or am I wrong?

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  • \$\begingroup\$ Well, (1) Page 23 of datasheet shows that the pin/pad with cut corner is pin 1. (2) If you compare the two drawings, the pin numbering is mirrored, ie, for one drawing, pin 1 is on left, for another drawing, pin 1 is on right. (3) If we use the British standard technical/machine drawing convention, we are looking at the front (Front View of First Angle Projection), we are looking the IC's marking face, and the pads are hidden from our eyes), ie, the pads should be drawn as "hidden" lines. \$\endgroup\$
    – tlfong01
    Feb 4 at 11:35
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    \$\begingroup\$ @tlfong01, this is clear to me. Sorry, perhaps my question wasn't clear enough. My doubt is usually the pin 1 is on the top-left when you see the IC from above. I've never seen an IC with this mirrored (pin 1 on top-right and then in CW) pinout \$\endgroup\$
    – Mark
    Feb 5 at 3:54
  • \$\begingroup\$ Ha, I agree you saying "pin 1 is on the top-left when you see the IC from above". Perhaps there is some ambiguity of the phrase "see IC from above". I always think that if my eyes are looking at the surface with the "IC marking", then Pin 1 is top left. But for SMD IC, the pins are not showing, because "too short", therefore hidden "under the marking surface" (That is why, I said should be shown as "hidden lines) I am only 80% sure though. (Ages ago I got a pass with distinction in GCE A Level in technical/mechanical drawing, and earned my living for 3 three years as a draughtsman. :) \$\endgroup\$
    – tlfong01
    Feb 5 at 4:35
  • \$\begingroup\$ One more thing. Let us look at the middle figure. Using the BS (British Standard of First Angle Projection) This middle guy is what you see if your eyes are on the right side of the IC, and "project" (push) what you see to the left (vertical plane/wall). Now (a) you see the tall, thin rectangle, which is the "side view" of the IC, (a) The top (two dimensional) surface with marking.now appears to be a vertical line on the left of the rectangle, (b) The right vertical line is a thicker rectangle. / to continue, ... \$\endgroup\$
    – tlfong01
    Feb 5 at 6:10
  • \$\begingroup\$ (c) If we zoom into this tall thin rectangle, you will notice the right part has three small rectangles, at top, middle, and bottom. From this "side view" picture, the pins/pads 1, 6 overlap, similarly pins 2, 5 and pins 3, 4 overlap into one. (d) This confirms (Note 1) that the 6 pins/pads are at the bottom surface of the IC. Note 1 - I am not surprised it you are still confused. You need to wiki MultiView Projection:en.wikipedia.org/wiki/Multiview_projection, and carefully read the First Angle Projection section at the middle of the wiki article. Happy learning. Cheers. \$\endgroup\$
    – tlfong01
    Feb 5 at 6:17
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In looking at their pinout:

enter image description here

You can see the left image shows pin 1 (I've circled it) from the TOP. In the right image they have flipped to the BOTTOM view and you can see pin 1 is now on the other side. I've circled that one also.

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    \$\begingroup\$ Thanks. But that was exactly my question. Usually the standard pinout is the one on the right: pin 1 on the top-left and then go ahead in CCW). This is the first time I see an IC with such an inverted numeration! \$\endgroup\$
    – Mark
    Feb 5 at 3:52

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