1, chip 2732, that is, 4K x 8bit=32Kb address line 12. Data line 8 chips 21141K x 4bit address line 10. 4 data cables, 16K×8b RAM, 14 address cables. Memory address range: 2K, by 2^11=2048=2K. That's why we need 11The on-chip address line, the chip selection address line only needs one, which is divided into two groups by the inverter to control the chip selection signal of the two groups (two pieces of RAM are a group) RAM.
2. The capacity is 1Mbit. The SRAM chip of model 628128 is defined as 131072word×8bit, that is 128K×8bit, that is, 1M bit static RAM. SRAM6264 chip memory capacity is 8K* 8,6264. The metric unit of capacity is the liter. Capacity also refers to the number of units of objects that an object or space can hold. Now widely used in computer hard disk,ong>2732 Chip capacity , so the word length is 16. The system program area and user program area are 8kWord2732 chip capacity , so if the ROM part uses 2732, 8k×16÷ (4k×8) = 4 2732 chip capacity , The RAM part requires 8k×16÷ (8k×8) = 2 pieces. There are only 20 address buses in total, and the addressing space is a total of 1MWord, so the system program area can be allocated to 00Low address from 000h, assign the user program area to high address from 80000h.
2, bank and chip selection is mainly used for address decoding bank can be understood as a memory chip with a capacity of X chip selection is the enabling chip of the chip, 0 indicates that the chip is not selected, 1 indicates that the chip is selected. For example, the system has 8M memory, divided into 8 banks (0~7), each capacity is 1M, then the on-chip address uses 20 bit coding. Slice selection addresses use 3-bit encoding.
3, the previous figure: the capacity of each piece of 27128 is 128kbit, that is8 data cables, 14 address cables. Four pieces of 27128 split the entire external addressing space of 51. Therefore, the top two bits of port P2 connect the four pieces of 27128 through the 2-4 decoder, and allocate the 2732 chip capacity mapping space. Slice 1: The minimum 128kbit space is allocated, and when P2[7:6] is 00, the strobe is [0000H: 3FFFH].
4, the 8 RAM chips are connected in parallel through the PCB line, which can form an 8Kx8bit memory. A minimum of 13 address lines are required to complete all addressing. Three address lines are required to complete PBANK slice selection. 128 chipset select address lines, in which two 512*4 RAM chips share one chipset select address line. The chipset selection address line is the shared chip selection line.
is generally like this, we mention an address such as 0xa000 actually defaults to 1 byte (the next address is 0xa001), but 1 byte = 8 bit so 160K means 160K Bytes, When using chips to form byte, consider using bits to form Byte.
K*1 bit chip means that this chip has 1024 units, each unit a bit of binary, because 2^10=1024, so in order to distinguish between these 1024 units you need 10 address lines. One chip provides 1K of space, 32K is 32 pieces. The composition of this system requires a total of 32*8=256 chips, a group of 8 pieces, a total of 32 groups.
capacity is 4K*8bit, 6264 capacity is 8K*8bit (3) Determine the address range of each group of memory chips.
1, 【 Answer 】 : The commonly used EPROM chip has 2712732762712272527512, where 27 is the code name of the EPROM chip, and the last two digits represent the storage capacity of the EPROM. For example, the 64 of 2764 stands for 64 KB (bit). In terms of bytes, each Byte is 8 bits, the storage capacity of 2764 is 8K x 8 bits, that is, 8KB (Byte, byte).
2, EPROM chip: The storage capacity of the 2764 is 8K (8K*8bit). The storage capacity of the 27128 is 16K (16K*8bit). The storage capacity of the 27256 is 32K (32K*8bit). EPROM was invented by Israeli engineer Dov Frohman. It is a computer memory chip that retains data after a power outage - that is, non-volatile.
3, if it is necessary to burn the software code is long, the chip program memory space is limited, should expand the external ROM (program memory), the maximum can be extended to 64k bytes, commonly used chips have 2764,27128,27256,27512, etc., their storage space is 8k, 16k, 32k, 64k bytes.
4, ROM chip: (27-EPROM) 2716 (2K×8), 2732 (4K×8), 2764 (8K×8), 27128 (16K×8), 27256 (32K)×8), 27512 (64K×8), etc. (28-EEPROM) : 2816 (2K×8), 2864 (8K×8). RAM chip: 6116 (2K×8 bit), 6264 (8K×8 bit), 62256 (32K×8 bit).
【 Answer 】 : The commonly used EPROM chip has 2712732762712272527512, of which 27 isThe last two digits represent the memory capacity of the EPROM chip. For example, the 64 of 2764 stands for 64 KB (bit). In terms of bytes, each Byte is 8 bits, the storage capacity of 2764 is 8K x 8 bits, that is, 8KB (Byte, byte).
is a memory chip used to store data A0 to A12 for 13 address signal input lines, indicating that the chip capacity is 2 to the 13th power, that is, 8K D0 to D7 for the data line, indicating that each memory unit of the chip stores a byte (8-bit binary number). When reading on the chip, as outputLine, used as input line when programming the chip. CE is the input signal and the low level is valid.
AT27 series is not a single chip microcomputer, is EPROM, such as AT273AT276AT27BV010, different models represent different storage capacity. Earlier, it was used to expand the program space of the 8051 series MCU (8051 ROM), the early 8051 series MCU did not have FLASH, so the program needed to be placed in the EPROM. It is still used in the field of single-chip computers, but it is generally used when the code space exceeds 64K.
MAX232 chip is a single power level conversion chip designed by MAXIM company for RS-232 standard serial port, using 5v single power supply. The MAX220-MAX249 series line drivers/receivers are designed for EIA/TIA-232E and V.28/V.24 communication interfaces, especially in applications where ±12V power is not available. These devices are particularly suitable for battery-powered systems due to their low-power shutdown mode, which reduces power consumption to less than 5uW.