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If there is an IC with multiple CS (chip select) lines, like the (old) MOS 6540 (for example):
Source: cbmhardware.de
Warning: This pinout may not be 100% correct, please take a look at this website: dasarodesigns.com
Is it correct that you need to enable all of the available CS lines to enable the chip (with enable I mean high for high-active CS and low for low-active _CS inputs)?
Not meant to be a 6540-specific question!
Typically, yes. A chip is not selected unless all its chip select lines are activated. Sometimes the chip selects have different polarity, one is negated and the other is not.
This usually simplifies the chip select logic, and does no harm because if you only need one, you can tie the others to whatever level necessary.
Sometimes the answer is yes. But this is a chip design specific thing and requires that you look at each part's data sheet. There is NO general answer for this.
I can't find a datasheet for the 6540 online anywhere, but I did come across a design for an adaptor that allows you to replace it with a different ROM such as a 2716:
http://www.dasarodesigns.com/product/6540-rom-adapter-rev-1/
In the accompanying text it states:
2) The 6540 has five chip select lines. CS3, CS4, and CS5 are active low meaning that they must be held low (+0v) for the chip to be on (its data lines to be low impedance). CS1 and CS2 are active high, meaning that they are must be high (+5v) for the chip to be on.
As this information appears to have been used to produce a working design it's likely to be correct.
As others have stated, it's important to check the datasheet for the exact answer but certainly in my experience it's usual to have to activate all the chip select lines in order to enable a device.
