8085 program to convert 8 bit BCD number into ASCII Code

Problem – Write an assembly level language program to convert 8 bit BCD number to its respective ASCII Code. 

Assumptions –
Starting address of program: 2000 
Input memory location: 2050 
Output memory location: 3050 and 3051 

ASCII Code for Digits 0 – 9

Example –

Algorithm –

1. Input the content of 2050 in accumulator 
   
2. Move content of Accumulator to register B 
   
3. Separate the least significant digit using AND with 0F and ADD 30 to accumulator 
   
4. Store content of accumulator to memory location 3050 
   
5. Move content of register B to Accumulator 
   
6. Separate the most significant digit using AND with F0 
   
7. Rotate Content of Accumulator 4 times 
   
8. ADD 30 to accumulator 
   
9. Store content of accumulator to memory location 3051 
   

Program –

Address	Mnemonics	Comments
2000	LDA 2050	A <- [2050]
2003	MOV B, A	B <- A
2004	ANI 0F	A <- A & 0F
2006	ADI 30	A <- A + 30
2008	STA 3050	[3050]<-A
200B	MOV A, B	A <- B
200C	ANI F0	A <- A & F0
200E	RLC	Rotate A left
200F	RLC	Rotate A left
2010	RLC	Rotate A left
2011	RLC	Rotate A left
2012	ADI 30	A <- A + 30
2014	STA 3051	[3051]<-A
2017	HLT	Stop Execution

Explanation –

1. LDA 2050 load the content of memory location 2050 to accumulator 
   
2. MOV B, A copy the content of accumulator to register B 
   
3. ANI 0F AND the content of accumulator with immediate data 0F 
   
4. ADI 30 ADD 30 to accumulator 
   
5. STA 3050 store the content of accumulator to memory location 3050 
   
6. MOV A, B copy the content of register B to accumulator 
   
7. ANI F0 AND the content of accumulator with immediate data F0 
   
8. RLC rotate the content of accumulator left without carry 
   
9. RLC rotate the content of accumulator left without carry 
   
10. RLC rotate the content of accumulator left without carry 
    
11. RLC rotate the content of accumulator left without carry 
    
12. ADI 30 ADD 30 to accumulator 
    
13. STA 3051 store the content of accumulator to memory location 3051 
    
14. HLT stops the execution of program  

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