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Memory chips of four different sizes as below are available
1. 32 K × 4 2. 32 K × 163. 8 K × 8 4. 16 K × 4
All the memory chips as mentioned in the above list are Read/Write memory. What minimal combination of chips or chip alone can map full address space of 8085 microprocessor ?
- a)1 and 2
- b)1 Only
- c)2 Only
- d)4 Only
Correct answer is option 'C'. Can you explain this answer?
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Memory chips of four different sizes as below are available1. 32 K ...
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Memory chips of four different sizes as below are available1. 32 K ...
To map the full address space of the 8085 microprocessor, we need to consider the memory chips available and their sizes.
Given memory chips:
1. 32 K (32 kilobytes) with 4-bit data width
2. 32 K with 16-bit data width
3. 8 K (8 kilobytes) with 8-bit data width
4. 16 K with 4-bit data width
The 8085 microprocessor has a 16-bit address bus, which means it can address up to 2^16 = 64 kilobytes of memory.
To determine the minimal combination of chips needed, we need to find a combination that can provide a total memory size of 64 kilobytes while utilizing the available chips efficiently.
Let's analyze each memory chip and see how they can be combined:
1. 32 K (4-bit data width):
This chip alone cannot provide the required 64 kilobytes of memory.
2. 32 K (16-bit data width):
This chip alone can provide the required 64 kilobytes of memory. Since this chip has a 16-bit data width, it can directly interface with the 8085 microprocessor.
3. 8 K (8-bit data width):
This chip alone cannot provide the required 64 kilobytes of memory.
4. 16 K (4-bit data width):
This chip alone cannot provide the required 64 kilobytes of memory.
From the analysis above, we can see that the only chip that can provide the required memory size is chip 2 (32 K with 16-bit data width). Therefore, the minimal combination of chips or chip alone that can map the full address space of the 8085 microprocessor is option 'C' (2 Only).
Note: It's important to consider both the memory size and the data width of the chips when determining the minimal combination. In this case, chip 2 is the only one that satisfies both requirements.
Given memory chips:
1. 32 K (32 kilobytes) with 4-bit data width
2. 32 K with 16-bit data width
3. 8 K (8 kilobytes) with 8-bit data width
4. 16 K with 4-bit data width
The 8085 microprocessor has a 16-bit address bus, which means it can address up to 2^16 = 64 kilobytes of memory.
To determine the minimal combination of chips needed, we need to find a combination that can provide a total memory size of 64 kilobytes while utilizing the available chips efficiently.
Let's analyze each memory chip and see how they can be combined:
1. 32 K (4-bit data width):
This chip alone cannot provide the required 64 kilobytes of memory.
2. 32 K (16-bit data width):
This chip alone can provide the required 64 kilobytes of memory. Since this chip has a 16-bit data width, it can directly interface with the 8085 microprocessor.
3. 8 K (8-bit data width):
This chip alone cannot provide the required 64 kilobytes of memory.
4. 16 K (4-bit data width):
This chip alone cannot provide the required 64 kilobytes of memory.
From the analysis above, we can see that the only chip that can provide the required memory size is chip 2 (32 K with 16-bit data width). Therefore, the minimal combination of chips or chip alone that can map the full address space of the 8085 microprocessor is option 'C' (2 Only).
Note: It's important to consider both the memory size and the data width of the chips when determining the minimal combination. In this case, chip 2 is the only one that satisfies both requirements.
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Memory chips of four different sizes as below are available1. 32 K × 4 2. 32 K × 163. 8 K × 8 4. 16 K × 4All the memory chips as mentioned in the above list are Read/Write memory. What minimal combination of chips or chip alone can map full address space of 8085 microprocessor ?a)1 and 2b)1 Onlyc)2 Onlyd)4 OnlyCorrect answer is option 'C'. Can you explain this answer?
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