An XxY memory device is implemented by one 2-4 decoder and four memory units, as
ID: 2078919 • Letter: A
Question
An XxY memory device is implemented by one 2-4 decoder and four memory units, as show in Fig4 below.
a) What are the values of X and Y?
b) For (A4 A3 A2 A1 A0) = (10010) and R/Wbar is HIGH, what is the decimal value of (Q7 Q6 Q5 Q4 Q3 Q2 Q1 Q0), where Q0 is LSB?
c) For (A4 A3 A2 A1 A0) = (00010) and R/Wbar is HIGH, what is the decimal value of (Q7 Q6 Q5 Q4 Q3 Q2 Q1 Q0), where Q0 is LSB?
d) Consider the burst feature in Fig below, where (A'2 A'1 A'0) are internal addresses for memory. This burst feature allows the memory to read or write up for burst loctions using a single address. Note: the counter in the address burst logic is the counter in Problem 1. For R/Wbar is HIGH, how many bytes can be read out? For this burst reading:
d1) Redo. d2) Suppose (A4 A3 A2 A1 A0) = (10010) , what is the decimal value of (Q7 Q6 Q5 Q4 Q3 Q2 Q1 Q0), according to the internal address sequence?
Schedule
Schedule
A2A, A R/W 2-to-4 e decoder A CS (US CS 010 010 Counter Burst in Problem 1 Lowest-order bits of control internal burst address CLK 6 0 0 0 0 0 0 Az A, A Lowest-order bits of external burst address (b) (a) Fig. 4: (a) An xxY memory device (b) address burst logicExplanation / Answer
a. This is a 32 X 8 memory device which means there are 32 memory locations, each containing 8 bits. Therefore X is 32 and Y is 8.
b. When A4 A3 A2 A1 A0 = 10010 which means
A4 A3 = 10 => Y2 of the decoder will be activated => 3rd memory unit from top will get selected
A2A1A0= 010 => which means 2nd memory location in this unit will get selected
Since R/Wbar is high, we will read from this particular memory location. Therefore
Q7 Q6 Q5 Q4 Q3 Q2 Q1 Q0 = 10111000 = 184 in decimal
c. When A4 A3 A2 A1 A0 = 00010 which means
A4 A3 = 00 => Y0 of the decoder will be activated => 1st memory unit from top will get selected
A2A1A0= 010 => which means 2nd memory location in this unit will get selected
Since R/Wbar is high, we will read from this particular memory location. Therefore
Q7 Q6 Q5 Q4 Q3 Q2 Q1 Q0 = 00110111 = 55 in decimal
d. In burst reading, when R/Wbar is high, the counter will go through from 0 to 7. Hence 8 address locations from one memory unit will be read out
d1. The address mentioned in the initial part is same as that mentioned in d2. Hence refer d2.
d2. A4A3= 10 => 3rd memory unit from top will be selected
A4
A3
A2
A1
A0
Q2
Q1
Q0
A2’
A1’
A0’
Q7
Q6
Q5
Q4
Q3
Q2
Q1
Q0
Decimal
1
0
0
1
0
0
0
0
0
1
0
1
0
1
1
1
0
0
0
184
0
0
1
0
1
1
0
1
0
1
0
1
1
1
87
0
1
0
0
0
0
1
1
0
1
1
0
1
1
219
0
1
1
0
0
1
1
0
1
1
0
1
1
1
183
1
0
0
1
1
0
0
1
1
1
1
0
0
0
120
1
0
1
1
1
1
0
0
1
1
1
1
0
1
61
1
1
0
1
0
0
0
0
0
1
0
1
0
1
21
1
1
1
1
0
1
0
1
1
1
1
1
1
1
127
A4
A3
A2
A1
A0
Q2
Q1
Q0
A2’
A1’
A0’
Q7
Q6
Q5
Q4
Q3
Q2
Q1
Q0
Decimal
1
0
0
1
0
0
0
0
0
1
0
1
0
1
1
1
0
0
0
184
0
0
1
0
1
1
0
1
0
1
0
1
1
1
87
0
1
0
0
0
0
1
1
0
1
1
0
1
1
219
0
1
1
0
0
1
1
0
1
1
0
1
1
1
183
1
0
0
1
1
0
0
1
1
1
1
0
0
0
120
1
0
1
1
1
1
0
0
1
1
1
1
0
1
61
1
1
0
1
0
0
0
0
0
1
0
1
0
1
21
1
1
1
1
0
1
0
1
1
1
1
1
1
1
127
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