Adding 8 Times a 10bit Number - Electronic Circuits Projects Diagrams Free

Suraj143 · 21st February 2008, 03:41 AM

Hi I want to add a 10 bit number 8 times & place them in two bytes.

Ex:345 X 8 = 2760

Does anybody have any math routine for that I searched in piclist but I couldn't find such a one.

Pommie · 21st February 2008, 03:55 AM

You can use the RLF instruction.

Code:

;put 345 into Acc
	movlw	low .345
	movwf	AccLo
	movlw	high .345
	movwf	AccHi
;now multiply by 8
	bcf	STATUS,C	;gets shifted into bit 0 of AccLo
	rlf	AccLo,f
	rlf	AccHi,f		;*2
	bcf	STATUS,C
	rlf	AccLo,f	
	rlf	AccHi,f		;*4
	bcf	STATUS,C
	rlf	AccLo,f	
	rlf	AccHi,f		;*8
;AccHi/AccLo now contain 2760.

HTH

Mike.

Suraj143 · 21st February 2008, 04:02 AM

Hi Mike Thanks for your input I'm wondering how can I use this for 16F PICs the maximum value it can hold is 255.

Thanks

Pommie · 21st February 2008, 04:12 AM

The code I posted is for 16F Pics. By combining two bytes the maximum value becomes 65535.

Mike.

Suraj143 · 21st February 2008, 04:33 AM

Hi thanks mike !

But i have a small question.

When I multiply 345 X 2 = 690
1010110010 That is correct

But when I double this 690 it will be = 1380
10101100100

Can you tell how this bit0 will be get set in the AccHi register.

Pommie · 21st February 2008, 04:57 AM

An RFL instruction moves all bits left one place. The bottom bit gets the value that was in the carry flag and the top bit gets placed in the carry flag.

Can you answer your question now?

Mike.

Suraj143 · 21st February 2008, 04:59 AM

I have attached the number pattern the green cell is my problem how will it be set?

Pommie · 21st February 2008, 05:21 AM

Via the carry bit.

Mike.

Suraj143 · 21st February 2008, 05:28 AM

Hi mike after reading lot of times now I got understood. I got what I was expected.

What I’m doing with your routine?

I’m going to sample my AD value 8 times & going to divide by 8 to get the average.

Thank you very much mike

Suraj143 · 21st February 2008, 05:30 AM

I just got a point. The bottom bit gets the carry bit value after the instruction completed I believe.
Ex after the RLF completed the bottom bit will get sets.

Now I understood

thanks again mike.

Pommie · 21st February 2008, 05:31 AM

To divide by 8 you need to use RRF.

Mike.

Suraj143 · 21st February 2008, 05:32 AM

Quote:

Originally Posted by Pommie

To divide by 8 you need to use RRF.

Mike.

Oh yes why not

Thanks for that.

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21st February 2008, 03:41 AM	#1
Suraj143	Adding 8 Times a 10bit Number Hi I want to add a 10 bit number 8 times & place them in two bytes. Ex:345 X 8 = 2760 Does anybody have any math routine for that I searched in piclist but I couldn't find such a one.

21st February 2008, 04:02 AM	#3
Suraj143	Hi Mike Thanks for your input I'm wondering how can I use this for 16F PICs the maximum value it can hold is 255. Thanks

21st February 2008, 04:12 AM	#4
Pommie	The code I posted is for 16F Pics. By combining two bytes the maximum value becomes 65535. Mike.

21st February 2008, 04:33 AM	#5
Suraj143	Hi thanks mike ! But i have a small question. When I multiply 345 X 2 = 690 1010110010 That is correct But when I double this 690 it will be = 1380 10101100100 Can you tell how this bit0 will be get set in the AccHi register.

21st February 2008, 04:57 AM	#6
Pommie	An RFL instruction moves all bits left one place. The bottom bit gets the value that was in the carry flag and the top bit gets placed in the carry flag. Can you answer your question now? Mike.

21st February 2008, 04:59 AM	#7
Suraj143	I have attached the number pattern the green cell is my problem how will it be set? Attached Thumbnails

21st February 2008, 05:21 AM	#8
Pommie	Via the carry bit. Mike.

21st February 2008, 05:28 AM	#9
Suraj143	Hi mike after reading lot of times now I got understood. I got what I was expected. What I’m doing with your routine? I’m going to sample my AD value 8 times & going to divide by 8 to get the average. Thank you very much mike

21st February 2008, 05:30 AM	#10
Suraj143	I just got a point. The bottom bit gets the carry bit value after the instruction completed I believe. Ex after the RLF completed the bottom bit will get sets. Now I understood thanks again mike.

21st February 2008, 05:31 AM	#11
Pommie	To divide by 8 you need to use RRF. Mike.