主题：计算机小白问题： -- 桃儿
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家园 计算机小白问题： how many 128*8 RAM chips are needed to provide a memory capacity of 2048 bytes?
128是什么意思？ 第二个8呢？
how many lines of the address must be used to access 2048 bytes?how many of these lines are connected to the address inputs of all chips?
我自己有答案， 但是不知道自己想的是否对，以前学过，现在手里没书
- 复 计算机小白问题：
家园 that is all about binary 128=2^7, 8bit = 1 BYTE, 1024=2^10, 128*8=2^10 bit;
a) 128 x 8 RAM => 128 Bytes (8bits) / chip => Nbr of chips = 2048 / 128 = 16
b) 2048 = 2^11, We have 16 chips => 4 bits for chip select Number of common address lines is 11 - 4 = 7 Summary: 4 bits (MSB) to select the correct chip and 7 bits (LSB) to select the memory location inside the selected chip.
c) We need 4 x 16 line decoder.
BTW, this is something very popular on the net. try to google for yourself.
- 复 计算机小白问题：
家园 自己顶