Memory compilation and chip Splicing

Let's talk about the memory Compilation Method in the computer today. For the computer's internal chip, People split it by bit. Let's take the following chip for example:

Memory preparation

There are 8x4 = 32 squares in the picture, that is, 32 bits. The following four lines are data lines used to retrieve data from the chip. When the chip address is edited, each of the four digits is edited. How many address lines are required for addressing? Is it because there are eight-line addresses that require eight-bit address lines?

Because the chip is an 8x4 Chip, you only need 3 addresses to address it. This is because the eight addresses only need three random combinations of 0 and 1 binary numbers.

PS: The address line determines the number of digits or bytes in the instruction after each addressing.

In, because it is a four-digit addressing, each instruction is 4 bits.

Next we will further understand the memory compilation through CHIP splicing.

Chip Splicing

If we have two chips in the figure above, splice them. There are two scenarios:

Case 1:

Because the chip is 16x4 after splicing, each command requires 4 bits, that is, 4 bits for addressing. In this case, it requires at least four address lines (because the power of 2 is 16), and the scope of the compilation is from 0000 ~~~ 1111.

Case 2:

In this case, an 8x8 chip is spliced. Each address requires 8 bits, that is, one byte for addressing. At this time, the number of digits of its address line does not need to be changed. It can still use three digits, and the compilation range is from 000 ~~~ 111.

Example Analysis

Example 1. How many bytes does the memory address range from a4000h to cbfffh?

Example 2: How many bytes does the memory address range from a4000h to cbfffh?

The above example shows that the question is almost the same. The difference is that one is byte addressing and the other is 4bit addressing. First, we can calculate the total number of K from a4000h to cbfffh (calculated by ourselves), and we can calculate it as 160 K.

If each line of address is a byte (8 bits), we can calculate K bytes;

Similarly, if 4 is used as the address, we can calculate KB of 4 bits, that is, 80 KB of data.

Summary

The memory compilation is actually very simple. As long as you understand the above 38 decoding principles, this estimation will be okay.