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What are Vcc and GND on a shift register?

I thought that I had to provide a positive voltage to the IC via the Vcc, but turns out that I can operate my shift register without powering the Vcc.

Similarly, what does GND on this IC mean?

I am talking about this one (74HC164A):

http://www.onsemi.com/pub_link/Collateral/MC74HC164A-D.PDF

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  • \$\begingroup\$ possible duplicate of What is ground and what does it do?, or maybe electronics.stackexchange.com/questions/17382/… \$\endgroup\$
    – Nick T
    Commented Nov 12, 2014 at 4:24
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    \$\begingroup\$ What makes you think you can operate the chip without powering Vcc? \$\endgroup\$
    – The Photon
    Commented Nov 12, 2014 at 4:26
  • \$\begingroup\$ I checked; couldn't find my answer in the questions you provided the links for. \$\endgroup\$
    – Sidd
    Commented Nov 12, 2014 at 4:27
  • \$\begingroup\$ I am powering it without the Vcc. It is working perfectly. It is somehow drawing power through the RESET pin, which is HIGH. \$\endgroup\$
    – Sidd
    Commented Nov 12, 2014 at 4:29
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    \$\begingroup\$ When you are operating the chip on phantom power via the RESET pin you are outside the normal operating conditions, so there is no guarantee whatsoever about the chips behavior, the chip is allowed to anything. This includes working normal. It also includes working normal for 1 week and then exploding. That it works NOW for YOU in your circuit is no guarantee that it continue to do so. \$\endgroup\$
    – Wouter van Ooijen
    Commented Nov 12, 2014 at 8:15

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You must provide a Vcc between 2.0 and 6.0 V in order to operate the IC you linked to:

enter image description here

If you attempt to operate it with a Vcc outside this range, there is no guarantee what the performance will be or whether it will be functional at all.

It is somehow drawing power through the RESET pin, which is HIGH.

You are powering the chip through the ESD protection diode on the reset pin. This is bad practice --- in a larger circuit, the Vcc of the shift register would be connected to many other devices, and the current required to power them all would be likely to damage either the shift register chip or whatever is driving the reset pin.

If you just have the one IC on a breadboard, the current needed to drive it is so low that you might not damage anything. Even so, it would be wise to use the chip the way it was designed to be used and provide power on the Vcc pin.

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  • \$\begingroup\$ My IC is working without the Vcc even plugged in. It is somehow drawing the power it needs from the RESET pin, which is HIGH. \$\endgroup\$
    – Sidd
    Commented Nov 12, 2014 at 4:30
  • \$\begingroup\$ Moreover, the Vcc on the IC is able to even power up an LED (although doing is disrupting the proper functioning of the IC on other output pins). \$\endgroup\$
    – Sidd
    Commented Nov 12, 2014 at 4:33
  • \$\begingroup\$ Thanks! So, what voltage should I give to the RESET? And is the GND on the chip to be grounded somewhere? \$\endgroup\$
    – Sidd
    Commented Nov 12, 2014 at 4:40
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    \$\begingroup\$ See the datasheet line for the minimum Vih and maximum Vil. They apply to RESET the same as for any other input. And GND should be attached to the negative terminal of your power source. It should also be the same as the ground of whatever other logic is connected to this device (what's producing RESET? Where are the outputs going?) \$\endgroup\$
    – The Photon
    Commented Nov 12, 2014 at 4:42

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