written 8.4 years ago by | • modified 2.9 years ago |
written 8.4 years ago by |
Step 1: Total EPROM required = 64 KB
Chip size available = 16 KB
$∴ \text{Number of chips required} = \frac{64 KB}{16 KB} = 4$
$∴ \text{Number of sets required} = \frac{Number of chips}{Number of banks} = \frac{4}{2} = 2$
SET 1: Ending address of SET 1 = FFFFFH
SET size = Chip size x 2 = 16 KB x 2 = 32 KB
i.e. $\frac{0000}{0} \frac{0111}{7} \frac{1111}{F} \frac{1111}{F} \frac{1111}{F}$
$\text{Starting address = Ending address – SET size} \\ \hspace{2.8cm} = FFFFFH – 07FFFH \\ \hspace{2.8cm} = F8000H$
SET 2: Ending address of SET 2 = F7FFFH (previous ending - 1)
SET size = Chip size x 2 = 16 KB x 2 = 32 KB
i.e. $\frac{0000}{0} \frac{0111}{7} \frac{1111}{F} \frac{1111}{F} \frac{1111}{F}$
$\text{Starting address = Ending address – SET size} \\ \hspace{2.8cm} = F7FFFH – 07FFFH \\ \hspace{2.8cm} = F0000H$
Step 2: Total RAM required = 32 KB
Chip size available = 8 KB
$∴ \text{Number of chips required} = \frac{32 KB}{8 KB} = 4$
$∴ \text{Number of sets required} = \frac{Number of chips}{Number of banks} = \frac{4}{2} = 2$
SET 1: Starting address = 00000H
SET size = Chip size x 2 = 8 KB x 2 = 32 KB
i.e. $\frac{0000}{0} \frac{0011}{3} \frac{1111}{F} \frac{1111}{F} \frac{1111}{F}$
$\text{Ending address = Starting address – SET size} \\ \hspace{2.8cm} = 04000H + 03FFFH \\ \hspace{2.8cm} = 07FFFH$
SET 2: Starting address = 04000H (previous ending - 1)
SET size = Chip size x 2 = 8 KB x 2 = 32 KB
i.e. $\frac{0000}{0} \frac{0011}{3} \frac{1111}{F} \frac{1111}{F} \frac{1111}{F}$
$\text{Ending address = Starting address – SET size} \\ \hspace{2.8cm} = 04000H – 03FFFH \\ \hspace{2.8cm} = 07FFFH$
Step 4: Final Implementation: