8085 is a 8 bit microprocessor and so A register which is also known as accumulator is also 8 bit.
A register pair typically consists of two registers, each of which can contain a specific number of bits, commonly 8, 16, or 32 bits, depending on the architecture. Therefore, the total number of bits in a register pair is double the bit width of a single register. For example, if each register is 16 bits, the register pair would contain 32 bits.
The 8085 has a 16 bit address bus.
The 8085 is an 8 bit processor, so its word length is 8 bits.
8 bits.
There are seven 8 bit registers in the 8085. A, B, C, D, E, H, and L
24 bits are needed for the program counter. Assuming the instructions are 32 bits, then 32 bits are needed for the instruction register.
Generally, 8 bits at a time. Some instructions deal with 16 bit numbers.
The 8085 is an 8-bit microprocessor. Even though there are some 16-bit registers (BC, DE, HL, SP, PC), with some 16-bit operations that can be performed on them, and a 16-bit address bus, the accumulator (A), the arithmetic logic unit (ALU), and the data bus are 8-bits in size, making the 8085 an 8-bit computer.
8086 is a 16bit processor.
32 bit
We have only one flag register of 8 bits. Bits description is as follows (Assuming D0=LSB & D7=MSB) D7=Sign Bit. D6= Zero Flag D4= Auxiliary Carry Flag D2 = Parity Flag D0= Carry Flag.
The 8085 microprocessor has a 16-bit address bus and an 8-bit data bus. This means it can address up to 2^16 (or 65,536) memory locations, while it can transfer 8 bits of data at a time. The combination of these buses allows the 8085 to efficiently access and process data from memory.