Working on a college assignment and I am not sure about a certain part of the question.

In a previous question that required calculations done on an ALU, the next question looks for similar functions, but with ROM.
There is an:
A input, which is 4-bit input number
B input, which is also a 4-bit input number
S input, which is a 3-bit select input
and F, which is a 4-bit output number

The question asks how many addressable locations are required and what is the minimum memory cell size at each location.

In working out the first part I have worked out that there are 2^11 = 2048 addressable locations.

But for this second part I am stumped. Is it a case of working out how many bits can be stored at each location?

Thanks for the help

 

To my old brain the question seems a bit vague.
A diagram would be helpful.
Generally as to how many bits can be stored at a location, well this will depend on the actual devices used and their connection within the circuit.

In the olden days the most common memory devices were either 4 or 8 bit wide for static RAM, 1 bit wide for Dynamic RAM and 8 bit wide for ROM/PROM/EPROM........tho I think I may have seen mention of 4 bit wide ROM\PROM\EPROM devices.

As you can see storage devices come in many different DATA widths, even then 8 bit wide devices (for example) can be used to create memory with greater data widths. If chip select & address pins are connected together but each device connects to different 8 bit groups of data bits you can increase the width of the data at each location.
ie. device one is connected to D0 to D7, device 2 -> D8 to D15, device 3 D16-> D23 and finally D24 to D31. As can be seen each memory location is now 32 bits wide.

Put plainly a single bit device will store 1 bit per memory location.
A 4 bit device will store 4 bits per memory location.
A 8 bit device will store 8 bits per memory location.
A 16 bit device will store 16 bits per memory location.
And so on and so on.

In the example you give one could interpret the inputs as being ADDRESS and the output as being DATA. Or are some inputs DATA or are ALL inputs DATA.

 

Consider what the purpose of the ROM is for. You provide 11 bits of input with the hope of getting an output that goes with those inputs. How many bits are associated with each input combination?

 

Consider what the purpose of the ROM is for. You provide 11 bits of input with the hope of getting an output that goes with those inputs. How many bits are associated with each input combination?

Thanks for your reply. Not sure what you are asking here, is it just the four bits?

 

Thanks for your reply. Not sure what you are asking here, is it just the four bits?

Yep, just four bits. Give some thought and ask if you don't understand why.