Electronic Circuits and Diagram-Electronics Projects and Design

7 Responses to “Multiplication of Two 8 bit numbers in 8085”

• Neel Terdal says:

  November 17, 2011 at 11:17 am

  MVI C,00
  MVI A,02
  STA 00A0
  LDA 00A0
  MOV B,A
  MVI A,04
  STA 00A1
  LDA 00A1
  MOV D,A
  MVI A,00
  LABEL:ADD B
  DCR D
  JNZ LABEL
  STA 00A2
  HLT
• Neel Terdal says:

  November 17, 2011 at 10:59 am

  5th line should be
  MOV D,A
• Qatada says:

  March 18, 2011 at 6:59 am

  What are the two numbers being multiplied and in what base?
• pruthviraj mane says:

  March 11, 2011 at 2:18 am

  pls give simple ckt based on 8085
• SADDY says:

  January 15, 2011 at 11:09 pm

  THANKS BRO..
• arindam says:

  August 31, 2009 at 2:30 am

  LXI H, 2200 : Initialize the memory pointer
  MOV E, M : Get multiplicand
  MVI D, 00H : Extend to 16-bits
  INX H : Increment memory pointer
  MOV A, M : Get multiplier
  LXI H, 0000 : Product = 0
  MVI B, 08H : Initialize counter with count 8
  MULT: DAD H : Product = product x 2
  RAL
  JNC SKIP : Is carry from multiplier 1 ?
  DAD D : Yes, Product =Product + Multiplicand
  SKIP: DCR B : Is counter = zero
  JNZ MULT : no, repeat
  SHLD 2300H : Store the result
  HLT : End of program
• arindam says:

  August 31, 2009 at 2:26 am

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