Correct Answer - Option 3 : 9
Concept:
Decoder expansion
n1 × m1 → n2 × m2
D1 D2
Number of D2 decoder required = ∑ K
\(\frac{{{m_2}}}{{{m_1}}} = {K_1}\)
\(\frac{{{K_1}}}{{{m_1}}} = {K_2}\)
\(\frac{{{K_2}}}{{{m_1}}} = {K_3}\)
⋮
Till 0 or 1
Calculation:
Given decoder 1 is 3 × 8 and the second decoder is 6 × 64
\(\frac{{64}}{8} = 8\)
\(\frac{8}{8} = 1\)
Number of 3 × 8 decoders = 8 + 1
Number of 3 × 8 decoders = 9
Important Points
Given
Decoder
|
To be implemented
|
Required
|
2 x 4
|
4 x 16
|
4 + 1 = 5
|
2 x 4
|
3 x 8
|
2 + 0 = 2
|
3 x 8
|
6 x 64
|
8 + 1 = 9
|
4 x 16
|
6 x 64
|
4 + 0 = 4
|
4 x 16
|
8 x 256
|
16 + 1 = 17
|