Skip to main content

Features of 8085 Microprocessor


  • 8085 MPU is an 8-bit microprocessor i.e. it can accept, process and provide 8-bit data.
  • It operates on +5V power supply connected at Vcc.
  • Power supply ground is connected to Vss.
  • It has 8 data lines (D0 - D7).
  • It has 16 address lines (A0 - A15). It can access 2^16 = 64K of memory.
  • It has 8 bit I/O address. It can access 2^8 = 256 I/O ports.
  • It operates on clock cycle with 50% duty cycle.
  • It operates with a 3MHz clock frequency.
  • The lower 8-bit address bus (A0 - A7) and data bus (D0 - D7) are multiplexed to reduce number of external pins.
  • It has 8-bit accumulator, flag register, instruction register, six 8-bit general purpose registers (B, C, D, E, H and L) and two 16-bit registers (SP and PC).
  • It has five hardware interrupts. They are TRAP, RST 7.5, RST 6.5, RST 5.5 and INTR.
  • It has serial I/O control which allows serial communication.
  • It has control signals (IO/M, RD, WR) to control bus cycles.
  • It has five addressing modes. They are: immediate, register, direct, indirect and implied.
  • It has powerful instruction set that can do various arithmetic operations and some logical operations.

Comments

Popular posts from this blog

Program to pack the two unpacked BCD numbers

Statement: Pack the two unpacked BCD numbers stored in memory locations 2000H and 2001H and store the result in memory location 2002H. The least significant digit is stored at 2000H. Example: (2000H) = 04H (2001H) = 09H Result = (2002H) = 94H Program: LDA 2001H                        ; Obtain the most significant BCD digit RLC                                    ; Rotate left  RLC                                    ; Rotate left  RLC                                    ; Rotate left  RLC                                    ; Rotate left...

Program to exchange the contents of two memory locations

Statement: Exchange the contents of memory locations 2000H and 2001H. Example:  Initially, (2000H) = 34H (2001H) = 20H After exchanging, (2000H) = 20H (2001H) = 34H Program 1 using direct addressing instructions: LDA 2000H               ; Get the contents of location 2000H into accumulator MOV B, A                  ; Move the contents of accumulator into register B LDA 2001H                ; Get the contents of location 2001H into accumulator STA 2000H                 ; Store the contents of accumulator into location 2000H MOV A, B                  ; Move the contents of register B into accumulator STA 2001H                 ; Store the contents of accumulator into location 2001H HLT  ...

Program to add the contents of two memory locations

Statement: Add the contents of memory locations 2000H and 2001H and place the result in the memory locations 2002H and 2003H. Example: (2000H) = C3H (2001H) = 29H Result = C3H + A9H = 016CH (2002H) = 6CH (2003H) = 01H Program: LXI H, 2000H                   ; HL points to location 2000H MOV A, M                        ; Move the contents of 2000H into accumulator INX H                                ; HL points to location 2001H ADD M                              ; Add the contents of 2001H with the contents of accumulator INX H                                ; HL points to location 2002H MOV M, A      ...