Square waveform of 1ms port

[anon10457]

Hi,
I've written the code below to generate square waveform of 1ms on a port; Polll a port pin ;count the no. of times it goes up

ORG 0H
JMP MAIN
SUBRT: CPL P2.3
MOV c, P2.3
Jc OVER ;if c=1 jump to over
OVER: INC R0 ; count the no. of times pin went up
RET
MAIN: CLR P2.3
MOV TMOD,#01
MOV TL0 ,#66
MOV TH0 ,#0FCH
SETB P2.3
SETB TR0
HERE: LCALL SUBRT
AGAIN: JNB TF0, HERE ; polling the port pin i.e if overflow occurred or not
CLR TR0
CLR TF0
CLR P2.3
END

I've used timers in the code to generate the waveform of 1ms but unable to reach 1ms ; somehow managed to reach 982micro sec. Is the code alright? ?
Please guide.
I've attached the 2 screenshot(one with simulator tab and other with port tab displayed).

Thanks
Anon10457

[basant]

Anon10457,

Reaching exact value may not be possible all the time. Hence 982microsec is fine. Code looks ok at the high level. I hope that you have tested it properly.

-- Basant

[anon10457]

Hi,
I've tried to write the code below to generate 1 machine cycle whenever 1ms is passed.

ORG 0H
JMP MAIN
SUBRT1:
INC R0 ;stores seconds
MOV A,R0
SUBB A,R1
JZ OVER ; if seconds reached 60
RET
SUBRT2: ;produce 1 machine cycle
CLR TR0 ;Stop the timer 0
CLR TF0 ;Clear timer 0 flag
MOV TL0,#0FAH
MOV TH0,#0FFH
CLR P2.7
SETB P2.7 ;SET high P2.3
SETB TR0 ;Start timer 0
LOOP: JNB TF0, LOOP
RET
MAIN:
CLR P0.3
MOV R1,#60
CLR TR0
MOV TMOD,#01
MOV TL0 ,#026H
MOV TH0 ,#0FCH ;1ms 2us
SETB P2.3
SETB TR0
L1: LCALL SUBRT1
AGAIN: JNB TF0, L1
OVER: LCALL SUBRT2
END

But in the exercise 11.7 it was asked to generate 1machine cycle (I am considering here that crystal frequency is 11.0592Mhz) whenever 1sec(or 10^6 micro sec) has passed.
so what value to store in TMOD so that it runs for around 10^6 micro sec.? ?

Also as per my calculation the value to be stored in timer0 to count till 1sec is going beyond the range of it( even if set to 16bit mode; its max range is 65536) i.e it is coming to 856122.
Calculation:

To calculate the values to be loaded
into the TL and TH registers, look at
the following example
􀂾 Assume XTAL = 11.0592 MHz, we can
use the following steps for finding the TH,
TL registers’ values
1. Divide the desired time delay by 1.085 us
2. Perform 65536 – n, where n is the decimal
value we got in Step1
3. Convert the result of Step2 to hex, where
yyxx is the initial hex value to be loaded into
the timer’s register
4. Set TL = xx and TH = yy

Confused. So cant even think of going to 1min.

Thanks,
Anon10457

[basant]

Anon10457,

You also need to put an appropriate outer loop around timer possibly a nested loop to reach the correct delay. For example, if you are able to generate delay of 1ms through basic timer operation, then this part needs to be run for 60,000 times to reach 1min. This 60,000 times needs to be implemented through some loops. I would suggest that first try to do it for 1sec then extend your code for 1 min.

-- Basant

[anon10457]

Hi,
In the code below the as the value of R0 and R1 finishes to zero it comes to RETI instruction. What does this RETI instruction really does> ? Does it signals to jump back to MAIN??

ORG 0H ;send pulse for 1machine cycle whenever 1sec has passed.
JMP MAIN
ORG 000BH ;Timer 0 int. vector table
LJMP ISR_T0 ;jump to ISR
;--The main program for initialization
;ORG 0030H ;after vector table space
SUBRT1:
CPL P2.3
DEC R1
MOV A,R1
JZ SUBRT2
JNZ AGAIN
RET
SUBRT2:
DJNZ R0,L1
;DEC R0
;SJMP L1
RET
MAIN:
CLR TR0
MOV R4,#1
MOV TMOD,#01
MOV IE,#82H ;IE=10000010 (bin) enable;130 in decimal
MOV TL0 ,#026H ;FC26 value kept in register to
MOV TH0 ,#0FCH ; make it run for 1msec.
SETB P2.3
SETB TR0
;MOV R2,#2 ;as 60sec will take too long so counting till 2sec.
L2:MOV R0,#2 ;10
L1: MOV R1,#10 ;100
AGAIN: JNB TF0,AGAIN ;Monitor timer flag 0
ISR_T0:
CLR TR0
CLR TF0
MOV TL0 ,#026H ;FC26 value kept in register to
MOV TH0 ,#0FCH ;make it run for 1msec.
SETB TR0
LCALL SUBRT1
;if R1 is zero then decrement R0 and jump to L1
;DJNZ R1,AGAIN
;DJNZ R0,L1
RETI ------------------------>>>>>line 42
CLR TR0 ;Stop the timer 0
CLR TF0 ;Clear timer 0 flag
CLR P2.3
END

I am unable to make out how to end the program bcos after encountering the RETI instruction it displays...''Invalid return from Interrupt" line 42.
Please guide.
Also where can I find in the simulator that what crystal frequency it is using/making?

Thanks,
Anon10457

[anup]

Anon10457,

RETI instruction is for returning from an interrupt service routine. Does your code fall through to the RETI instruction while not executing an interrupt service routine? In case yes, the simulator will barf (like above).

Thanks,

Anup

[anon10457]

Hi,
No, code executes the service routine completely then comes across RETI. By the way where should it return to after executing RETI. ?

In the slide explaining interrupt configuration code is something like this:

ORG 00H
LJMP MAIN
ORG 0BH
LJMP ISR_T0

ORG 30H
MAIN: MOV IE,#82H
MOV TMOD,#01H
MOV TL0,#42H
MOV TH0,#03H
CPL P2.3
SETB TR0
AGAIN: SJMP AGAIN

ISR_T0: CLR TF0
MOV TL0,#42H
MOV TH0,#03H
CPL P2.3
SETB TR0
RETI

there is no END. where should it come to? If it comes just after RETI then what are its chances of getting executed ? Moreover where will it lead to after RETI.?

When will this code end.?

Thanks,
Anon10457

[basant]

Anon10457,

RETI will result in jumping back to the place wherefrom interrupt originated. In this case, it should jump back to the next instruction where timer got over and it resulted in interrupt i.e. loop with the label AGAIN.

END should come right at the end. So it can come after RETI insturction as there is nothing after that.

-- Basant

[anon10457]

Hi,
Then the above code is a case of Infinite Loop. Right?
How to end the loop then in such case of interrupt?

Thanks,
Anon10457

[basant]

Anon10457,

As you are writing the program to practice timer and interrupt together, infinite loop is ok. If you have some other tasks to do, you could done the same in parallel without putting any infinite loop.

-- Basant

[anon10457]

Hi,

ORG 00H
LJMP MAIN

ORG 000BH ;Timer 0 int. vector table
LJMP ISR_T0 ;jump to ISR


ISR_T0: CLR TF0
MOV TL0,#26H
MOV TH0,#0FCH
CPL P2.3
SETB TR0
RETI

ORG 30H
MAIN: MOV IE,#82H
MOV TMOD,#01H
MOV TL0,#26H
MOV TH0,#0FCH
CPL P2.3
SETB TR0
AGAIN: SJMP AGAIN

CLR TR0 ;Stop the timer 0
CLR TF0 ;Clear timer 0 flag
END

What I want to it to do is to get out of the interrupt. So wish to know how to end the interrupt execution?
Where should I make changes so that it stops executing the interrupt section again and again?

Thanks,
Anon10457

[basant]

Anon10457,

You are getting interrupt again because timer has been switched on again in ISR. Since you want to generate a square waveform on P2.3, this strategy looks alright. One possibility is that you can decide to terminate this, once waveform has been generated say for 10 times. In that case, you can maintain the count in one of the registers/memory location and do not set timer again in ISR. I guess this is what you are looking for.

-- Basant

[anup]

Anon10457,

You return from the interrupt handler using the RETI instruction. However, if the interrupt comes immediately after you come out of the handler, you will again go back to the handler. You must make sure that the interrupt frequency is low enough in order to do some other useful work.

Thanks,

Anup

[anon10457]

Hi,

ORG 00H
LJMP MAIN

ORG 000BH ;Timer 0 int. vector table
LJMP ISR_T0 ;jump to ISR


ISR_T0: CLR TF0

MOV TL0,#78H
MOV TH0,#0FFH
CPL P2.3
SETB TR0
RETI

ORG 30H
MAIN: MOV IE,#82H
MOV R0,#03
MOV TMOD,#01H
MOV TL0,#26H
MOV TH0,#0FCH
CPL P2.3
SETB TR0
AGAIN: SJMP AGAIN

MOV A,#05----------------------------------------------------------->>>>>>>>>>>>>>>>>>>>>>>>>>>>value in accumulator
L1: CLR TR0 ;Stop the timer 0
CLR TF0 ;Clear timer 0 flag
END

I've reduced the frequency in ISR(value ff78). Still the value in the accumulator does not gets updated to 5.


Please Guide.

Thanks,
Anon10457

[basant]

Anon10457,

Reducing the frequency is not going to help because infinite loop in main is still there. What I had suggested was that you emit 0/1 on P2.3 fixed number of times and keep checking the same in ISR. Once the count is over, don't set the timer again. You also need to find a way to terminate infinite loop in main. Please pay closer attention to the program logic.

-- Basant

[anon10457]

Hi,

Modified code:

ORG 00H
LJMP MAIN

ORG 000BH ;Timer 0 int. vector table
LJMP ISR_T0 ;jump to ISR


ISR_T0:
CLR TF0
DEC R0
MOV A,R0
JZ L1

CPL P2.3

RETI

ORG 30H
MAIN:
MOV IE,#82H
MOV R0,#01
MOV TMOD,#01H
MOV TL0,#78H
MOV TH0,#0FFH

SETB TR0
AGAIN: SJMP AGAIN

MOV A,#05
L1: CLR TR0 ;Stop the timer 0
MOV A,#05
CLR TF0 ;Clear timer 0 flag
END

I am still unable to make out. If the above code is alright then what is the point of having interrupt. ? Same could be achieved with a subroutine. ?Here interrupt is acting more or less as subroutine .

Thanks,
Anon10457

[basant]

Anon10457,

Your logic is still incorrect. Your program will produce only one instance of 0/1 on P2. Further, you should come out of subroutine including ISR only through RET/RETI. Otherwise your stack will get corrupted. Currently you are jumping out of ISR into main code directly.

Benefit of interrupt comes when you are having multiple tasks e.g. serial communication and timer are active at the same time. You can't handle multiple events without interrupt. Further, unlike ISR which can be triggered asynchronously, a normal subroutine can be triggered only through sequential execution of the code.

Regarding your program logic, why don't you properly define what you want to do, write code accordingly and if it doesn't work, you have great debugger in the simulator to localize the problem.

-- Basant

[anon10457]

Hi,
Actually I want to make this loop finite by making it execute the RETI instruction as it finishes executing ISR.
But after RETI it jumps back to AGAIN loop.
How to make the changes so that MAIN is finite?.

ORG 00H
LJMP MAIN
ORG 000BH ;Timer 0 int. vector table
LJMP ISR_T0 ;jump to ISR
ISR_T0:
;CLR TF0
CPL P2.3
RETI
ORG 30H
MAIN:
MOV IE,#82H
MOV R0,#01
MOV TMOD,#01H
MOV TL0,#78H
MOV TH0,#0FFH
SETB TR0
AGAIN: SJMP AGAIN
MOV A,#05
CLR TR0 ;Stop the timer 0
MOV A,#05
CLR TF0 ;Clear timer 0 flag
END

In the code above I've avoided jumping from ISR. Still unable to make it finite.

Thanks,
Anon10457

[anon10457]

Hi,

ORG 0H ; GENERATE 1MACHINE CYCLE WHENVER 1SEC HAS PASSED VIA INTERRUPT
JMP MAIN
ORG 000BH ;Timer 0 int. vector table
LJMP ISR_T0 ;jump to ISR
ISR_T0:
CLR TF0
CPL P3.2
RETI ;
SUBRT:
CLR TR0
CLR TF0
MOV TL0,#026H ; VALUES FITTED TO RUN FOR
MOV TH0,#0FCH ; 1 MACHINE CYCLE
SETB TR0
RET; FROM HERE IT JUMPS TO LINE2 I.E JMP MAIN, WHY SO? ?
MAIN:
MOV IE,#82H
MOV TMOD,#01H
CLR TR0
CLR TF0
MOV TL0,#026H ; VALUES FITTED TO RUN FOR
MOV TH0,#0FCH ; 1 MACHINE CYCLE
SETB TR0
L2:MOV R0,#2 ;10
L1:MOV R1,#10 ;100
AGAIN: JNB TF0 , AGAIN
DJNZ R1,SUBRT
DJNZ R0,L1
CLR TR0
CLR TF0
END

In the code above why it jumps to line 2 as it finishes the subroutine. ? ?

Please Guide.

Thanks,
Anon10457

[basant]

Anon10457

We have already answered this in post #12. We can't give more pointed answer than this.

Thanks
Basant

[basant]

Anon10457,

You are jumping to SUBRT like regular jump and then while returning back, you are using RET instruction. RET will look at return address on the stack. Since SUBRT was not arrived from ACALL, what return address is there on the stack is unpredictable. Most likely this is the reason why you end up at arbitraty location after RET from SUBRT.

Note that all subroutine calls must be in pair i.e. call through ACALL and return through RET, otherwise stack will get corrupted.

-- Basant

[anon10457]

Hi,

I've written the code below to implement digital clock via interrupt.

ORG 0H ; GENERATE 1MACHINE CYCLE WHENVER 1SEC HAS PASSED VIA INTERRUPT
JMP MAIN
ORG 000BH ;Timer 0 int. vector table
LJMP ISR_T0 ;jump to ISR
ISR_T0:
CLR TF0
CPL P3.2
RETI
SUBRT:
CLR TR0
CLR TF0
MOV TL0,#026H ; VALUES FITTED TO RUN FOR
MOV TH0,#0FCH ; 1 MACHINE CYCLE
SETB TR0
LOOP: JNB TF0, LOOP
RET
SUBRT2:
INC R5 ;holds no. of seconds passed.
INC R3
MOV A,R3
SUBB A,R4
JZ SUBRT3
RET
SUBRT3: ;generate pulse of 1machinecyle if 1sec has passed.
CLR TR0
CLR TF0
CLR P2.3
MOV TL0,#0FFH ; keep 65535 to genrate pulse of 1machhine cycle i.e 1.085micro sec
MOV TH0,#0FFH
SETB TR0
SETB P2.3
RET
MAIN:
MOV R4,#1
MOV IE,#82H
MOV TMOD,#01H
MOV R2,#1 ; keep the no. of seconds want to run i.e 60 cos 60 sec =1min
L2:MOV R0,#10 ;keep 10
L1:MOV R1,#100 ;keep 100 cos 10*100=1000 i.e 1sec
L3: LCALL SUBRT
DJNZ R1, L3
LCALL SUBRT
DJNZ R0,L1
LCALL SUBRT2 ;keep count of seconds passed.
DJNZ R2,L2
;LCALL SUBRT

CLR TR0
CLR TF0
END

Issue is when I do single stepping I can see the pulse generated of 1machine cycle but if run the program completely at the end no pulse is seen.? ?

Thanks
Anon10457

[anon10457]

Hi,

The code below to implement event counter through INT0.

ORG 00H
LJMP MAIN
ORG 000BH
LJMP ISR_INT0
ISR_INT0:
CLR TF0
CPL P3.2
;SETB P2.3
RETI

SUBRT:
MOV TL0,#26H
MOV TH0,#0FFH
SETB TR0
LOOP: JNB TF0, LOOP ; after overflow will jump to ISR_INT0
RET

MAIN:
MOV TMOD,#51H ; timer 1 as counter, timer 0 as timer
MOV IE, #81H
MOV TH1,#00H
MOV TL1,#00H
SETB TR1
ACALL SUBRT

END

Is the code alright? No pulse is generated at the end via interrupt. Why so?

Please give hint.

Thanks,
Anon10457

[anon10457]

Hi,
In exercise 12.3 digital clock using timer interrupt is to be combined with external interrup INT0 and snippet of the code is shown below:

ORG 0H ; GENERATE 1MACHINE CYCLE WHENVER 1SEC HAS PASSED VIA INTERRUPT
JMP MAIN
ORG 000BH ;Timer 0 int. vector table
LJMP ISR_T0 ;jump to ISR---> LINE 4
ORG 0003H
LJMP ISR_INT0
ISR_T0:
CLR TF0
CPL P3.2
RETI ; PROBLEM LIES HERE I.E IT JUMPS TO 'AGAIN' LOOP , AS OVERFLOW IS SET NEXT, JUMPS TO ISRT ,THEN ; TO 'AGAIN'LOOP...KEEPS ON DOING IT. NEVER FINISHES? ?

ISR_INT0:
CLR TF0
CPL P3.2
;SETB P2.3
RETI
SUBRT:
CLR TR0
CLR TF0
MOV TL0,#026H ; VALUES FITTED TO RUN FOR
MOV TH0,#0FCH ; 1 MACHINE CYCLE
SETB TR0
LOOP: JNB TF0, LOOP
RET
SUBRT2:
INC R5 ;holds no. of seconds passed.
INC R3
MOV A,R3
SUBB A,R4
JZ SUBRT3
RET
SUBRT3: ;generate pulse of 1machinecyle if 1sec has passed.
CLR TR0
CLR TF0
CLR P2.3
MOV TL0,#0FFH ; keep 65535 to genrate pulse of 1machhine cycle i.e 1.085micro sec
MOV TH0,#0FFH
SETB TR0
SETB P2.3
RET
SUBRT4:
MOV TL0,#026H
MOV TH0,#0FFH
SETB TR0
LOOP: JNB TF0, LOOP ; after overflow will jump to ISR_INT0
RET
MAIN:
MOV R4,#1
;MOV IE,#82H
SETB IE.0
SETB IE.1
SETB IE.7
MOV TH1,#00H
MOV TL1,#00H
SETB TR1
MOV TMOD,#051H
MOV R2,#1 ; keep the no. of seconds want to run i.e 60 cos 60 sec =1min
L2:MOV R0,#2 ;keep 10
L1:MOV R1,#2 ;keep 100 cos 10*100=1000 i.e 1sec
L3: LCALL SUBRT
DJNZ R1, L3
LCALL SUBRT
SETB TR1 ; starting timer1 as counter and
ACALL SUBRT4 ; calling subroutine.
DJNZ R0,L1
LCALL SUBRT2 ;keep count of seconds passed.
DJNZ R2,L2
;LCALL SUBRT
CLR TR0
CLR TF0
END

I receive the following error while compilation:

Compiling file: s10.asm
Initializing pre-processor ...
Warning at 5 in s10.asm: This ORG has lower value than the previous one
Syntax error at 45 in s10.asm: Label was already defined: `loop'
Compilation error at 4 in s10.asm: Unable to overwrite already reserved program memory at address 0xB -- compilation failed
Compilation error at 4 in s10.asm: Unable to overwrite already reserved program memory at address 0xC -- compilation failed
Compilation error at 4 in s10.asm: Unable to overwrite already reserved program memory at address 0xD -- compilation failed
Compilation error at 4 in s10.asm: Unable to overwrite already reserved program memory at address 0xE -- compilation failed
Pre-processing FAILED !
Creating code listing file ... -> "s10.lst"
5 errors, 1 warning

Unable to understand this error . As INT0 in interrupt vector table starts at 0003H and TF0 stored at 000BH and the same is initialized in the beginning. Why still the error comes up?

Thanks
Anon10457

[basant]

Anon10457,

I think that the warning at line 5 and other errors are related. First fix the warning and remove the error for label loop. The program should compile fine after that.

Regarding your posts #22 and #23, what I can suggest is that use debugger properly by setting breakpoints and single stepping into the program. We will also give it a shot sometime towards evening. I can't figure out what is wrong just by looking at the code.

-- Basant

[basant]

Anon10457,

In the code below, the problem is that you have not enabled timer 0 interrupt correctly. You should check the structure of IE register again to see what should be the correct value to enable timer0.

Hi,

The code below to implement event counter through INT0.

MOV TMOD,#51H ; timer 1 as counter, timer 0 as timer
MOV IE, #81H
MOV TH1,#00H

Is the code alright? No pulse is generated at the end via interrupt. Why so?

-- Basant

[basant]

Anon10457,

If you can see that pulse is getting generated in single stepping, then I don't see any reason why it will not be there when you normally execute the program. One guess is that there may be some issue in displaying the waveform in the simulator. In any case, I would suggest you to double check via breakpoint, singlestepping etc. that you are indeed getting expected pulses.

-- Basant

Hi,

I've written the code below to implement digital clock via interrupt.


Issue is when I do single stepping I can see the pulse generated of 1machine cycle but if run the program completely at the end no pulse is seen.? ?

Thanks
Anon10457