Timers

Timers

In the real time applications, such as traffic light control, digital clock, process control, serial communication, it is important to keep a track with time. For example in traffic light control application, it is necessary to give time delays between two transitions. These time delays are in few seconds and can be generated with the help of executing group of instructions number of times. This software timers are also called time delays or software delays.

To execute instructions, microprocessor takes fix time as per the instruction, since it is driven by constant frequency clock. This makes it possible to introduce delay for specific time between two events. In the following section we will see different delay implementation techniques.

 

1. Timer Delay using NOP Instruction

NOP instruction does nothing but takes 4T states of processor time to execute. So by executing NOP instruction in between two instructions we can get delay of 4 T-state

1 T state = 1/Operating frequency of 8085

 

2. Timer Delay using Counters

Counting can create time delays. Since the execution times of the instructions used in a counting routine are known, the initial value of the counter, required to get specific time delay can be determined.

Using 8-bit counter :

In this program, the instructions DCR C and JNZ BACK execute number of times equal to count stored in the C register. The time taken by this program for execution can be calculated with the help of T-states. The column to the right of the comments indicates the number of T-states in the instruction cycle of each instruction. Two values are specified for the number of T-states for the JNZ instruction. The smaller value is applied when the condition is not met, and the larger value applied when it is met. The first instruction MVI C, count executed only once and it requires 7 T-states. There are count - 1 passes through the loop where the condition is met and control is transferred back to the first instruction in the loop (DCR C). The number of T-states that elapse while C is not zero are (count -1) × (4+10). On the last pass through the loop, the condition is not met and the loop is terminated. The number of T states that elapse in this pass are 4 + 7.

Total T-states required to execute the given program

= 7 + (count -1) × (4 + 10 ) +  (4 + 7)

    MVI C    Loops        Last loop

For count = 5

Number of T-state = X + (5 -1) × (14) + (11) = 7 + 56 + 11 = 74

Assuming operating frequency of 8085A is 2 MHz,

Time required for 1 T-state = 1/2 MHz = 0.5 µsec

Total time required to execute the given program = 74 × 0.5 µsec. = 37 µsec.

Maximum delay possible with 8-bit count

The maximum count that can be loaded in the 8 bit register is FFH (255) so the maximum delay possible with 8 bit count, assuming operating frequency 2 MHz

= (7 + (255 - 1) ×  (14) + (11)) x 0.5 µsec. = 1787 µsec.

With these calculations, it can be noticed that delay with 8 bit count suitable for small delays and not for large delays.

Using 16-bit counter :

In this program, the instructions DCX B, MOV A, C, ORA B and JNZ BACK execute number of times equal to count stored in BC register pair. The instruction LXI B, count is executed only once. It requires 10 T-states. The number of T-states required for one loop = 6 + 4 + 4 + 10 = 24 T-states. The number of T-states required for last loop = 6 + 4 + 4 + 7 = 21 T-states. So total T-states required for execution of given program are

= 10  + (count-1) × 24 + 21

  LXI B      Loops    Last loop

for count = 03FFH(1023₁₀)

Number of T-states = 10 + (1022) × 24 + 21

= 24559

Assuming operating frequency of 8085A as 2 MHz, the time required for,

T state = 0.5 µsec

Total time required to execute the given program

= 24559 × 0.5 µsec

= 12279.5 , µsec

= 12.2795 msec

Maximum delay possible with 16-bit count

The maximum count that can be loaded in the 16 bit register pair is  (65535 H).

So the maximum delay possible with 8 bit count, assuming operating frequency 2 MHz.

= 10(10 + (65535 - 1)  × (24) + (21)) x 0.5 µsec

= 0.786425 sec

If the application requires the delays more than this, then the nested loop technique is used to implement the delays.

 

3. Timer Delay using Nested Loops

In this, there are more than one loops. The innermost loop is same as explained above. The outer loop sets the multiplying count to the delays provided by the innermost loop.

T-states required for execution of inner loop

T_inner = 7 + (Delay count - 1 ) × 14 + 11

T-states required for execution of the given program

= (Multiplier count - 1) × (T _inner +14) + 11

For delay count = 65H (101) and multiplier count

= 51H (81)

T_inner = 7 + (101-l) × 14 + 11

= 1418

Total time required to execute the given program is

(Operating frequency is X MHz) = [(81-1) × (1418 + 14) +11] × 0.5 µsec.

= 57.2855 msec.

 

Lab Experiment 3.2.1 : Generate a delay of 0.4 seconds.

Statement : Write a program to generate a delay of 0.4 sec if the crystal frequency is 5 MHz.

Solution : In 8085, the operating frequency is half of the crystal frequency,

Operating frequency = 5/2 = 2.5 MHz

Time for one T-state = 1/2.5 MHz = 0.4 µsec

Number of T-states required = Required Time / Time for 1T – state =  0.4 sec / 0.4 µsec

= 1 × 10⁶

Delay program :

LXI B, count ; 16-bit count

BACK : DCX B ; Decrement count

MOV A, C

ORA B ; Logically OR B and C

JNZ BACK ; If result is not zero repeat

1 × 10⁶ = 10 + (count - 1) × 24 + 21

count = (1 × 10⁶ – 31 / 24) + 1 ≈ 41666₁₀

count = 41666₁₀ = A2C2H

 

Lab Experiment 3.2.2 : Generate and display binary up counter.

Statement : Write a program for displaying binary up counter. Counter should count numbers from 00 to FFH and it should increment after every 0.5 sec.

Assume operating frequency of 8085 equal to 2 MHz. Display routine is available.

Solution :

LXI SP, 27FFH ; Initialize stack pointer

MVI C, OOH ; Initialize counter

BACK : CALL Display ; Call display subroutine

CALL Delay ; Call delay subroutine

INR C ; Increment counter

MOV A, C

CPI 00 ; Check counter is > FFH

JNZ BACK ; If not, repeat

HLT ; Stop

Delay subroutine :       

Delay : LXI D, count ; Initialize count

BACK : DCXD ; Decrement count

MOV A, E

ORA D ; Logically OR D and E

JNZ BACK; If result is not 0 repeat

RET ; Return to main program

Flowchart :

 

Lab Experiment 3.2.3 : Generate and display BCD up counter with frequency 1 Hz.

Statement : Write a program for displaying BCD up counter. Counter should count numbers from 00 to 99H and it should increment after every 1 sec. Assume operating frequency of 8085 equal to 3 MHz. Display routine is available.

Solution :

LXI SP, 27 FFH ; Initialize stack pointer

MVO C, 00H ; Initialize counter

BACK : CALL Display ; Call display subroutine

CALL Delay ; Call delay subroutine

MOV A, C ;

ADI , 01 ; Increment counter

DAA ; Adjust it for decimal

MOV C, A ; Store count

CPI, 00 ; Check count is > 99

JNZ BACK ; If not, repeat

HLT ; Stop

Delay subroutine :

Delay : MVI B, Multiplier-count; Initialize multiplier count

BACK1: LXI D, Initialize Count

BACK : DCX D ; Decrement count

MOV A, E ;

ORA D ; Locally OR D and E

JNZ BACK ; If result is not 0, repeat

DCR B ; Decrement multiplier count

JNZ BACK1 ; If not zero, repeat

RET ; Return to main program.

Operating frequency : 3 MHz

Time for one T-state = 1/ 3MHz = 0.333 µsec

Flowchart :

 

Lab Experiment 3.2.4 : Generate and display BCD down counter with frequency 1 Hz

Statement : Write a program for displaying BCD down counter. Counter should count numbers from 99 to 00 and it should decrement after every 1 sec. Assume display and delay routines are available.

Solution : Flowchart :

Source program wioth logic 1 :

LXI SP, 27 FFH ; Initialize stack pointer

MVI C, 99H         ; Initialize counter

BACK : CALL Display ; Call display subroutine

CALL Dealy ; Call delay subroutine

ADI 99H ; * (Explain later)

DAA ; Adjust for decimal

CPI 99H ; Compare with last count

JNZ BACK ; If no, repeat

*Addition :

Program with logic 2 :

LXI SP, 27FFH ; Initialize stack pointer

MVI C, 99H ; Initialize counter = 99

BACK : Call Display ; Call display subroutine

Call Delay ; Call Delay subroutine

MOV A, C ; Get the count

ANI 0FH ; Check for lower nibble

JNZ SKIP ; If it is not 0FH go to skip

MOV A, C  ; Else get the count

SBI 06 ; and subtract 6

MOV C, A ; Store the count

SKIP : DCR C ; Decrement count

MOV A, C ; Get the count

CPI 99H ; Check it for last count

JNZ BACK ; If not, repeat

HLT ; Stop

Review Questions

1. Write a program to count from 0 to 9 with one second delay between each count. At the count of 9, the counter should reset itself to 0 and repeat the sequence continuously. Assume the clock

AU : Dec.-11, Marks 8

2. Explain the need of software timers.

3. Explain how software delays can be implemented using counters.