In 8085 microprocessor the address bus is of _______ bits.
16
The question asks about the size of the address bus in the 8085 microprocessor. Understanding the different buses in a microprocessor is fundamental to understanding its operation and capabilities.
Understanding the 8085 Microprocessor Address Bus
The 8085 is an 8-bit general-purpose microprocessor developed by Intel. Like other microprocessors, it communicates with memory and input/output (I/O) devices using buses. The three main types of buses are:
- Address Bus: Carries the addresses of memory locations or I/O ports that the CPU wants to access. It is unidirectional, meaning information flows only from the CPU to the memory/I/O devices.
- Data Bus: Carries the actual data being transferred between the CPU and memory/I/O devices. It is bidirectional, allowing data to flow in both directions.
- Control Bus: Carries control signals (like read, write, I/O request, memory request) from the CPU to devices and status signals from devices to the CPU. It is also bidirectional.
8085 Address Bus Size and Addressing Capacity
The size of the address bus determines the maximum amount of memory that a microprocessor can address. The 8085 microprocessor has:
- An 8-bit data bus.
- A 16-bit address bus.
A 16-bit address bus means it has 16 address lines. Let's call these lines $\text{A}_0$ through $\text{A}_{15}$. Each address line can carry either a 0 or a 1. The total number of unique addresses that can be generated is given by $2^{\text{number of address lines}}$.
For the 8085 with its 16-bit address bus, the total number of unique addresses is $\text{2}^{16}$.
Calculating $\text{2}^{16}$:
\( \text{2}^{16} = \text{2}^{10} \times \text{2}^{6} \)
\( \text{2}^{10} = 1024 \text{ (approximately 1K)} \)
\( \text{2}^{6} = 64 \)
\( \text{2}^{16} = 1024 \times 64 = 65536 \)
Therefore, the 8085 microprocessor can address up to 65,536 unique memory locations or I/O ports. Since memory capacity is typically measured in kilobytes (KB), where 1 KB = 1024 bytes, 65536 bytes is equal to $\text{65536 / 1024} = 64$ KB.
Conclusion
The address bus in the 8085 microprocessor is 16 bits wide. This allows it to address $2^{16} = 65536$ memory locations, which is equal to 64 KB of memory space.
Based on the options provided and the technical specifications of the 8085 microprocessor, the correct answer is 16 bits.
| Bus Type | Size (bits) | Function | Direction |
|---|---|---|---|
| Address Bus | 16 | Specifies Memory/I/O addresses | Unidirectional (CPU to Memory/I/O) |
| Data Bus | 8 | Transfers data | Bidirectional |
| Control Bus | Varies | Carries control/status signals | Bidirectional |
Revision Table: Key 8085 Facts
| Feature | Value |
|---|---|
| Data Bus Size | 8 bits |
| Address Bus Size | 16 bits |
| Maximum Memory Capacity | 64 KB |
| Clock Speed | 3 MHz (typical) |
| Number of Pins | 40 |
| Architecture | 8-bit microprocessor |
Additional Information on 8085 Addressing
The 8085 microprocessor uses a multiplexed address/data bus. This means that some pins are used for both address and data information at different times during an instruction cycle.
- Pins $\text{AD}_0$ to $\text{AD}_7$ are used for the lower 8 bits of the address ($\text{A}_0$ to $\text{A}_7$) during the first clock cycle (T1 state) and for 8 bits of data ($\text{D}_0$ to $\text{D}_7$) during later clock cycles.
- Pins $\text{A}_8$ to $\text{A}_{15}$ are used only for the higher 8 bits of the address and are not multiplexed.
An external latch (like the 74LS373) is needed to hold the lower 8 bits of the address stable while the pins transition to carrying data. The Address Latch Enable (ALE) signal from the 8085 indicates when the address is available on the multiplexed bus.
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2. The Commission did record the statements of ryots, sahukars and eye-witnesses.
Select the correct answer using the code given below:
