You cannot address 1GB memory with the 8085 or the 8086/8088 without some kind of external demultiplexor that is software controlled. The address bus on the 8085 is 16 bits, giving addressibility of 64KB; while the address bus on the 8086/8088 is 20 bits, giving addressibility of 1MB.
To address 1GB, you need a 30 bit address bus.
In a 1GB memory space, the total number of bytes is (1 \times 2^{30} = 1,073,741,824) bytes. Since memory addresses typically start at 0, the memory address of the last byte would be (1,073,741,824 - 1 = 1,073,741,823). Therefore, the memory address of the last byte of a 1GB memory is 1,073,741,823.
It requires 30 address bits to address 1GB of RAM.230 = 1,073,741,824
1GB has way more memory. 1GB is 1000 MB. so 512 MB is about half of 1GB
You need 30 address lines to access 1G of memory. 230 = 1,073,741,824. log2 (1,073,741,824) = 30.
I think the 1gb 2rx16 is best!
4BG is not a term used for memory, as memory is stored in bytes. So the answer to your question is 1GB.
32 bit address line can access 4GB of memory. As 2^10 -> 1KB; 2^20 -> 2MB; 2^30 -> 1GB and so on.... 32 bit gives (2^30) * (2^2) = 1GB * 4 = 4GB;
1GB
You get 1GB of memory.
1gb128mb
1000
memory 1GB RAM