If a 256K bit SRAM chip has 8 data lines, what are the number of address lines it has?

The 256K-bit SRAM chip has 8 data lines.

There are 256 1024 storage units.

There are 8 data lines that indicate that 8 address units can be connected at the same time.

Note also requires 256 1024 8 addresses to represent every 8 addresses.

That's 32 1024 different values.

How many bits of an address can represent that many numbers? That is, how many powers of 2 is greater than or equal to this number?

So the answer is 15

Note: Yes".bits", noStorage unit

8 data lines, indicating that each memory cell is 8 bits.

256k 8 = 32k memory cells.

32k --2 15, so there are 15 address lines.

Storage capacity = 2 m n, where m is the address bus bit and n is the data bus bit.

1k=2∧10

b=8b 1kb=2∧10×8b

The storage capacity mentioned in the title is 256k bits, and obviously 256k bits and 256kb bits are not equal).

256k=2 8 2 10=2 18 while 256kb=2 21256k=2 18=2 m 8

m=15

256k 'bits' is a bit of binary code, and the smallest storage unit in a computer is bytes, which is 8 bits, and 256 8 is the real capacity?

The storage capacity of the SRAM chip is 64K*16 bits, and the chip has 16 address lines and 16 data lines.

The storage capacity is calculated using 2 n, where n represents the number of address lines. 2 16 = 65536, in the computer it is said that its storage capacity can be expanded to a maximum of 64k.

Memory chip capacity = number of cells Number of data line bits, so the data line of 64k*16-bit chip is 16.

SRAM means static random storage hail memory in Chinese, which is a type of random access memory. The so-called "static" means that as long as the memory is kept energized, the data stored in it can be maintained constantly.

In contrast, the data stored in Dynamic Random Access Memory (DRAM) needs to be updated periodically. However, when the power goes out**, the data stored by the SRAM disappears (known as volatilememory), unlike the ROM or flash memory that can store data after a power outage.

The 22nd power of 2 is 4m, and the bit width is 8, so it is 4MB. For example:

It means that there are 8 bits of data, that is, there are 8 data lines;

1024=2 10, so there are 10 address lines;

The storage capacity is 16kb, no number of digits are specified. If it is 8 bits, it is 14 address lines and 8 data lines.

The ROM chip has a capacity of 8K 8 bits, this address line is 13 bits, and the data line is 8 bits.

The chip has 10 address lines and 8 data lines.

Since the DRAM chip has a storage capacity of 512K 8 bits, it is the smallest unit of data storage.

is 8 bits, or 1 byte. Therefore, the data line requires a total of 8 bits, i.e., 8 data lines, usually d(0) d(7).

At the same time, it is possible to know the 512k bits of memory, from 2 19 524, 288 512k, so we can represent the DRAM address in the order of 19 addresses.

However, the internal memory units of DRAM mostly adopt a line-and-line structure, i.e., in time-division multiplexing, the address lines transmit lines and line signals, so the number of address lines should be reduced to 10 address lines. Address values are redundant.

The structure of the internal storage unit of DRAN is shown in the following figure:

If the storage capacity is 512 8, there are 8 data lines; Because Wheel 2 19 = 512K, there are 19 ZD address lines. 512 indicates that 512 1024=2 19-bit data can be stored and 8-bit data can be read and written at the same time.

Because its storage capacity is 512K x 8 bits, its data line is 8, and the number of address lines is 219 = capacity = (total number of SDRAM units) * (bit width.

2^24) *32 =2^29 bit。

Note: I'm counting the TQ2440 development board here.

The total capacity of the entire SDRAM, not just the total capacity of each SDRAM individually. If the capacity of each slice of SDRAM alone = the capacity of the entire SDRAM of Moxiao 2.

Note: The unit of calculation here is bit, if we convert it to bytes, it is 2 29 bit 8 = 2 26 bytes = 2 6 MB = 64 MB.

If the storage capacity is 512 8, there are 8 data lines; Since 2 19 = 512k, there are 19 address lines.

512 indicates that 512 1024 = 2 19-bit data can be stored, and 8-digit data can be read and written at the same time.

DRAM chip is dynamic random access memory, DRAM can only hold data for a short period of time and needs to be refreshed regularly, DRAM is much more complex than SRAM, because in the data storage process of DRAM, the stored information needs to be constantly refreshed, which is the biggest difference between them. Kaichai is bored.

The minimum number of leads for this chip is 19, excluding the power supply and ground terminals. The capacity is 512, 512 is the 9th power of Shenkai 2, 9 address lines are needed, 8 bits means that there are 8 data lines, and the chip also needs 1 chip selection line, 1 read and write line, a total of 19.

However, in practice, the reading and writing line rarely uses one, and generally uses one to read and one to write. In the exam question, it is believed that one can be used for the reading and writing line (theoretically one is possible), so the answer to the Sakura blind jujube case is 19. If it is in actual use, it is best to use a chip with two wires (a chip with a read/write line is not common), so that the efficiency and stability will be better.

In addition to the power supply and ground terminal, the minimum number of chip pins is 20, mainly 10 address lines plus 8 data lines plus 2 read and write lines, so 12 + 8 + 2 = 20, so the minimum number of chip pins is 20.

The address line is used to transmit address information. As a simple example, when the CPU is looking for a piece of data in memory or hard disk, it first finds the address through the address line, and then retrieves the data through the data line.

1024*8 bits = 2 10b = 1k, so 10 address lines are needed.

The data cable is used to transmit data. Because the chip is 8 bits, it needs 8 data lines. In addition to the normal address line and data line, the read and write line is also required, so in addition to the power supply and ground terminal, 20 wires are also needed, so the minimum number of pins is 20.

SRAM chip pins consist:

1. Power supply, ground.

2. Address line.

3. Data cable.

4. a control line.

In addition to the power supply and grounding wire, because it is 1024 * 8 bits, that is, the capacity is 1024byte (bytes), so there are a total of 10 address lines (2 10 = 1024), from the capacity of 1028 * 8, it can be seen that it is the SRAM of the 8bit data bus, so its data line has 8, and the control line is fixed, and the capacity has nothing to do with it, the control line has WE (write control line), RE (read control line), CS (chip selection line) a total of 3. So the minimum number of pins is 10 + 8 + 3 = 21, and your answer is 20 who calculated this, teacher? Don't you want CS (slice selection line) this one?

Anyway, the least SRAM control line I've ever seen is 3.

Most RAM ICs are controlled and operated by a read/write control line, and a few RAM ICs are controlled by two inputs for read and write respectively!

Most RAM integrated circuits use a read/write control line.

The chip has 10 address lines and 8 data lines.

Since the DRAM chip has a storage capacity of 512K 8 bits, the smallest unit of data storage is 8 bits, that is, one byte, so its data line needs a total of 8 bits of data lines, that is, 8 data lines, usually bits d(0) d(7).

At the same time, it can be seen that the word size of the memory is 512K, which is 2 19 = 524 and 288 = 512K, so the address of DRAM can be represented in 19 address sequences.

However, the DRAM internal memory unit mostly adopts the row and column structure, that is, the address line is multiplexed to transmit the row and column signals, so the number of address lines should be reduced to 10 address lines. In this case, there is redundancy in the value of the address.

The structure of the DRAN internal storage unit is as follows:

The address line of the memory is related to the capacity of the memory, similar to 10,000 people have **, ** number has at least 5 digits, but the difference is that the computer uses binary instead of decimal internally. How much is the capacity of the memory, is represented by how many binary numbers, then the number of address lines is how many, for example, the capacity is 4 bits, expressed by two binary numbers, then the address line is 2, 8 bits, expressed by three binary numbers, the address line should be 3, so push down, the content capacity can be expressed by how many binary numbers, which is equivalent to how many times 2 of 1 binary number, then the number of address lines is how much. 512k should refer to 512kb, which is equivalent to 4MB (converted according to 1 to 8), and needs to be represented by 22-bit binary numbers, which is equivalent to 2 times 22 times, so 22 address lines are used.

The data line refers to the width of the data transmitted at one time, and the width of 8 bits should be 8 data lines.

Because its storage capacity is 512k*8 bits, its data line is 8 and the number of address lines is 19. Because 2 19 = 512k.