The most significant difference between the Intel 8085 and 8086 microprocessors is that the 8085 is an 8-bit system and the 8086 is a 16-bit system. This difference allows the 8086 system to have a much larger set of operational instructions and can make calculations to more significant places. Note: the 8085 processor does have two 16-bit registers. The pointer and the program counter.
The 8085 is an 8-bit computer, with only limited capability to do 16 bit arithmetic. In order to add two 16-bit numbers, NUM1 and NUM2, together, and store the result at NUM3, you can use the code... LHLD NUM1 XCHG LHLD NUM2 DAD D SHLD NUM3 If you want to subtract NUM1 from NUM2, you need to take the two's complement first, by inverting it and adding one... LHLD NUM1 MOV A,H CMA MOV H,A MOV A,L CMA MOV L,A INX H ... and then continue with adding NUM2... XCHG LHLD NUM2 DAD D SHLD NUM3
As a product of its prime factors: 2*2*2*2 = 16 or as 2^4 = 16
2 x 2 x 2 x 2 = 16
multiplication is commutative. interpret % as the fraction 1/100. 16 percent of 40 = 16 x 1/100 x 40 40 percent of 16 = 40 x 1/100 x 16 same numbers in any order and you get the same answer when the only operation is multiplication.
The 8085 was replaced with the 8086/8088. As such, there is no 16 bit version of the 8085.
Because Intel designed it that way. The 8085 was designed as a 8 bit computer in a 16 bit address space. This means that the PC (Program Counter) and SP (Stack Pointer) should be 16 bits in size.
The most significant difference between the Intel 8085 and 8086 microprocessors is that the 8085 is an 8-bit system and the 8086 is a 16-bit system. This difference allows the 8086 system to have a much larger set of operational instructions and can make calculations to more significant places. Note: the 8085 processor does have two 16-bit registers. The pointer and the program counter.
The 8085 has a 16 bit address bus.
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.
No. The 8086 has instructions not present in the 8085. The 8086 was marketed as "source compatible" with the 8085, meaning that there was a translator program which could convert assembly language code for the 8085 into assembly language code for the 8086. However, this does not mean that the compiled 8086 assembly code would then run on an 8085; among other things, the 8086 was a true 16-bit processor, as opposed to the 8085 which was an 8-bit processor that supported a few 16-bit operations.
The stack pointer is 16 bits in size on the 8085 because that is how Intel designed it. The address bus is also 16 bits, so it made sense for the program and stack to be located anywhere in that address space.
Because it is an 8-bit microprocessor.
Because that's how Intel designed it. The 8085 is an 8-bit computer operating on a 16-bit address space.
The 8086/8088 microprocessor family is a 16 bit microprocessor. The 8086 implementation also has a 16 bit data bus, but the 8088 implementation has an 8 bit data bus, comparable to the 8085. The 8088 implementation was intended as a logical upgrade from the 8085, while keeping the complexity of the system on an equal footing as the 8085.
16 bit addition
The program counter (PC) and stack pointer (SP) registers are 16-bit registers in the 8085 and in the 8086/8088 because that is how Intel designed the processors.