I just bought 74HC595 shift register IC and I'm having trouble finding the first pin. There is no first pin mark anywhere on the chip as you can see below.
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asked Sep 26, 2013 at 16:02
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2\$\begingroup\$ Probably the lower left. SMT packages of this sort generally tend to be just a physical shrink of the DIP version, though I suppose exceptions are possible, so reading about the package marking in the data sheet wouldn't be a bad idea. Absent that, this part at least is cheap enough that you could determine if that is the case by potentially destructive testing. \$\endgroup\$– Chris StrattonCommented Sep 26, 2013 at 16:03
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1\$\begingroup\$ Lower left was my guess too but heck do I know. \$\endgroup\$– Muhamed KrlićCommented Sep 26, 2013 at 16:04
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\$\begingroup\$ I thought the protection diodes might come in handy if nothing else works, but it turns out that there are couple more body diodes in an arbitrary package (ATtiny85) than I expected ... \$\endgroup\$– jippieCommented Sep 26, 2013 at 17:51
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2 Answers
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You walk to the manufacturer's web site, look the data sheet up: http://www.nxp.com/documents/data_sheet/74HC_HCT595.pdf and there on page 17, you'll find a nice drawing of your package. The drawing will show you that the package is not totally symmetrical, the top edge of one of the longer sides of the package is slanted, and pin 1 is on that side.
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\$\begingroup\$ Yes! That's it! First pin is marked by a slight curve on the chip. \$\endgroup\$ Commented Sep 26, 2013 at 16:16
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Page 17 of the datasheet indicates that Pin 1 is the lower-left pin.
simulate this circuit – Schematic created using CircuitLab
