About the problem that the memory is composed of RAM chips

foreword

Conduct microcomputer principle "preview" and prepare for the exam. See this topic about composing a memory from multiple RAM chips. Although it is very simple, it is still recorded.

basic concept

Because I am afraid that everyone's foundation is as weak as mine, if I directly talk about the topic of RAM forming memory, it may be confusing. So I will take a few questions as practice first.

topic one

<1> A certain ROM has a storage capacity of 16K*4 bits, and the chip has ( ) address lines and ( ) data lines.
<2> This topic is very simple at the beginning, you only need to know that the first 16K indicates the number of storage units, and the latter 4 bits indicate the number of storage units.
<3> At this time, someone said, what is the number of storage units, and what is the number of bits?
<4> To put it bluntly, the number of storage units is related to the address lines. For example, here is 16K = 2^14. So here are 14 address lines.
<5>The number of digits is related to the data lines, and the number of digits is the number of data lines. For example, 4 bits here are 4 data lines.

Topic two

<1> The storage capacity of a certain ROM is 16K 4 bits, what is the storage capacity of the chip?
<2> Let's look at the data line first, and we find that it is 4 bits. And 1 byte is 8 bits. So the end result is 16K (4/8)=8KB.

Problems with memory made up of RAM

A memory with 14-bit address and 8-bit data, ask:
<1> How many bytes of data can this memory store?
<2> If the memory consists of 8K*4-bit RAM chips, how many chips are needed?
<3> How many address chip selects are needed?

topic one

With the above foundation, this is very simple.
2^14 * (8/8) = 16K * 1B = 16KB

Topic two

This topic is also very simple, just divide the address line by the address line, and divide the data line by the data line. According to the formula below, the answer is 4 pieces.

$$\frac{16K\ast 8}{8K\_\ast 4}=\frac{16K}{8K\_}\ast \frac{8}{4}=2\ast 2=4$$

Topic three

Because the second question is calculated, 4 pieces of RAM are needed for combination. So, 2^2=4, so the answer is a 2-bit address.