The BIU of 8086 contains 16 bit segment registers. What is the width of address sent out by BIU of 8086 miocroprocessor?
20 bit
The given question is about the BIU (Bus Interface Unit) of the 8086 microprocessor, and it asks about the width of the address sent out by the BIU.
The 8086 microprocessor is a 16-bit microprocessor with a segmented memory architecture. This architecture involves using segment registers, which are 16-bit registers. These segment registers are combined with a 16-bit offset to form a 20-bit physical address. Let's understand how this is done:
- The 8086 has four segment registers: Code Segment (CS), Data Segment (DS), Stack Segment (SS), and Extra Segment (ES).
- Each segment register holds the base address of a segment, and it is a 16-bit register.
- The 20-bit physical address is calculated using the formula: \(\text{Physical Address} = \text{(Segment Register Value)} \times 10_{hex} + \text{Offset}\)
- This involves shifting the 16-bit segment register left by 4 bits (multiplying by \(2^4\) or 16), and then adding an offset address, which is also 16-bit, resulting in a 20-bit physical address.
To clarify, here's an example:
- If the segment register contains: \(1234_{hex}\),
- The segment address becomes: \(12340_{hex}\)
- If the offset is: \(0020_{hex}\),
- The physical address is: \(12360_{hex}\)
The key point here is that this method allows the 8086 to address up to 1 MB of memory using a 20-bit address despite having 16-bit segment registers and offsets.
Therefore, the correct answer is that the width of the address sent out by the BIU of 8086 is 20 bit.
Conclusion: The BIU of the 8086 microprocessor uses a 20-bit address bus, allowing it to address 1 MB of memory, which is a significant feature of this microprocessor.
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