I created a circuit in which all the address inputs of an AT28C256 are connected to outputs of shift registers (74HC164).
In the future I will do many of these circuits, however I haven't yet decided if I need the full 32KB of memory.
Let's just assume that I later decide I need only 8K of memory for my application (for example, AT28C64).
After looking at the datasheets, I notice the difference in pin-out between the two chips is that on the AT28C64, A13 and A14 are replaced with NC (no connects).
Is it OK anyway to connect such pins to outputs of shift registers (and be able to make a AT28C64 like a drop-in replacement for AT28C256) or are the NC pins internally connected to something?
All chips in my question are DIP ICs.
Best Answer
The datasheet of the AT28C64 makes it pretty clear: There are "No Connect" (NC) pins, those are not connected in the chip so you can hook them up to a voltage. There are also "Don't connect" (DC) pins which you shouldn't connect but the DIP package doesn't have any of those. That's the obvious part.
The not so obvious part is that there are multiple versions of the AT28C64 and some of them have a READY output instead of a NC on pin 1. To avoid shorting things out, you should put a 330 Ohm (or so) resistor in series with pin 1 of the EEPROM so that if you plug in an AT28C64 with a READY output, the current gets limited.
With that resistor in series with pin 1, swapping in an 8K ROM instead of the 32K one should work just fine.