Memory Systems Homework Solutions for ECE 2030 A, B - Prof. David Schimmel, Assignments of Electrical and Electronics Engineering

The solutions to homework 9 of the memory systems course for ece 2030 a and b. It includes true or false questions, filling in the blank exercises, and memory system design tasks. Students can use this document to check their answers and understand the concepts related to dram and sram, memory organization, and addressing.

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Uploaded on 12/01/2009

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Homework 9: Memory Systems
ECE 2030 A, B (Schimmel) - Solutions
1. True or False: Only Statements (b) and (e) are true. The others are false. Reasons were not
required to be given, and are provided here for explanation only.
a. DRAM incorporates an inverter loop to store one bit of data per cell.
This is true of SRAM, not DRAM.
b. SRAM uses more transistors per bit than DRAM.
True.
c. DRAM is slower than SRAM because it uses flip-flops rather than latches.
DRAM is slower, but the reason has nothing to do with flip-flops.
d. SRAM does not require analog circuitry to read data.
A sense amplifier (analog device) is required to read data.
e. DRAM requires periodic “refreshing” due to capacitor discharge and leakage.
True.
f. SRAM stands for “Static Read Access Memory.”
“Static Random Access Memory”
g. For a given chip size, SRAM contains more bits of memory than DRAM.
DRAM uses fewer transistors, hence can fit more memory per unit area.
2. Fill in the blank with a numeral (expressed in base 10):
a. 10 bytes is equivalent to __20___ nibbles.
b. A memory system with 10 address bits has _1024__ unique addresses.
c. A memory system with 1G (or 230) unique addresses requires __30___ address bits.
3. Design a 1024 x 8 memory system using 1024 x 4 chips.
1024
x
4
A9:0
_
R/W
CS
D3:0
1024
x
4
A9:0
_
R/W
CS
D3:0
10
10
4
4
D7:4
D3:0
A9:0
_
R/W
CS
pf3
pf4

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Homework 9 : Memory Systems

ECE 2030 A, B (Schimmel) - Solutions

  1. True or False: Only Statements (b) and (e) are true. The others are false. Reasons were not required to be given, and are provided here for explanation only. a. DRAM incorporates an inverter loop to store one bit of data per cell. This is true of SRAM, not DRAM. b. SRAM uses more transistors per bit than DRAM. True. c. DRAM is slower than SRAM because it uses flip-flops rather than latches. DRAM is slower, but the reason has nothing to do with flip-flops. d. SRAM does not require analog circuitry to read data. A sense amplifier (analog device) is required to read data. e. DRAM requires periodic “refreshing” due to capacitor discharge and leakage. True. f. SRAM stands for “Static Read Access Memory.” “Static Random Access Memory” g. For a given chip size, SRAM contains more bits of memory than DRAM. DRAM uses fewer transistors, hence can fit more memory per unit area.
  2. Fill in the blank with a numeral (expressed in base 10): a. 10 bytes is equivalent to __ 20 ___ nibbles. b. A memory system with 10 address bits has _ 1024 __ unique addresses. c. A memory system with 1G (or 2 30 ) unique addresses requires __ 30 ___ address bits.
  3. Design a 1024 x 8 memory system using 1024 x 4 chips. 1024 x 4 A9: _ R/W CS D3: 1024 x 4 A9: _ R/W D3: 10 10 4 4 D7: D3: A9: _ R/W CS
  1. Design an 8G x 8 memory system using 2G x 8 chips. 2Gx 8 A 30 : _ R/W CS D 7 : 31 8 D7: 0 A 30 : _ R/W CS 2Gx^8 A 30 : _ R/W CS D 7 : 31 8 2Gx 8 A 30 : _ R/W D 7 : 31 8 2Gx 8 A 30 : _ R/W CS D 7 : 31 8 2 - to- 4 I^ DEC 1 EN Y 2 Y 0 Y 3 I 0 Y 1 A 32 A 31
  1. Given the sequence of assembler directives below, give the resulting storage pattern. Show data and addresses in hexadecimal notation. Assume the following:  each address location stores one byte of data  words are 32 bits wide  big endian storage convention  strings are terminated with a null character  addresses increase as you go UP the table .data 0x1111000 1 .byte 65 , 255 .space 2 .asciiz “Yes.” .word 42 Items in gray are not required and may be left blank. The two cells with red question marks must either be left blank, or marked with some indication that those spaces were skipped intentionally or that their contents remain unchanged. Memory Address Memory Contents 0x11110010? 0x1111000f? 0x1111000e? 0x1111000d 0x2a 0x1111000c 0x 00 0x1111000b 0x 00 0x1111000a 0x 00 0x1111000 9 0x 00 0x1111000 8 0x2e 0x1111000 7 0x 73 0x 11110006 0x 65 0x1111000 5 0x 59 0x1111000 4? 0x1111000 3? 0x1111000 2 0xff 0x1111000 1 0x 41 0x1111000 0?