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Most shift register tutorials I've seen have the output pins connected to ground (for instance: http://www.arduino.cc/en/Tutorial/ShiftOut), however I've seen at least one circuit that connects them to VCC instead (https://learn.adafruit.com/assets/7766).

These circuits both use the same chip (74hc595), and aside from reversing the LEDs look otherwise identical. is there something I'm missing?

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Not really. High pins source current and low pins sink current, which means that outputs that lead to ground through LEDs are active high and outputs that lead to VCC are active low.

It may be an issue with logic families whose drive strength is highly asymmetrical, e.g. they can source much more current than they can sink or vice versa, but HCMOS has mostly symmetrical drive strength (within 20% or so).

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  • \$\begingroup\$ Actually, HCMOS does not have symmetrical drive strength, since hole mobility is less than electron mobility, and the makers of HCMOS did not make the effort to completely compensate. The asymmetry, however, is much less than for the old bipolar (TTL, LSTTL, etc) logic families. \$\endgroup\$ – WhatRoughBeast Feb 15 '15 at 22:31
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    \$\begingroup\$ Right, I was thinking of Advanced CMOS. \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 15 '15 at 22:32
  • \$\begingroup\$ Even AC/ACT is not perfectly symmetrical. For a 74HC244, for instance, 24 mA / 4.5 volt outputs have different (Vcc - Voh) and Vol. It's true that both will put out 24 mA at usable levels, but that's not quite the same as symmetrical. \$\endgroup\$ – WhatRoughBeast Feb 15 '15 at 22:46

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