Question

Requires building a 128K x 24 bit memory bank using 32KByte memories

1. Identify address and data bus width for the 128K x 24 bank
2. Identify address and data bus width for 32KBytes memories
3. Build the corresponding memory bank
4. Expresses the functions of the Address Encoder using the most significant bits of the bank's address bus.

Answer 1

Given memory Bank size 128K x 24 bit 32 KB memories - Size of Address Bus 8 Data Bus 128k x 24 2 x 20 x 24 bd subit 24 bit 24 bit २ 217 x 24 bit - - = 17 bits Address bus size Data Bus size = su bili zu bet 2 Address 8 Date Bus = 32 KB 2x [IB & bit X2 x 2bit 15 X 8 bet 2 8 bit 2 bit 2 8 bit Address Bub z 15 bits Date bus = & bit play Ebit 128 K x 24 #f memories required 3 32 K x 8 2x 3 12

2 32 Kx8 328x3 32KX3 32kX8 32k18 32 KX3 132kg 22 KXY 2 32kx8 32k88 321x2 32 2X 8 3 memory organisation Decoder size 니 2x22 De corden
address bus is use to carry address or location of the data to read or write, while data bus is use to carry data from/ to the location.