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Precise 5 ms delay in 8051

Short answer

At 12 MHz, 1 machine cycle = 1 µs, so 5 ms needs 5000 cycles. Load R7=15, R6=165; inner loop DJNZ takes 2 cycles. Total cycles = 2+1+15×(1+165×2+2)+2 =5000, yielding exactly 5 ms delay.

Write a short program in assembly language that computes a precise 5-millisecond delay using the 8051 microcontroller.

Assume a 12 MHz crystal, so one machine cycle = 1 us and 5 ms = 5000 machine cycles. The delay is made with two nested DJNZ loops, with the count chosen so that the total, including the call and return, is exactly 5000 cycles.

Calculation

  • ACALL = 2 cycles, RET = 2 cycles, MOV R7,#15 = 1 cycle.
  • Each outer pass: MOV R6,#165 (1) + inner loop 165 x DJNZ (2 each = 330) + outer DJNZ R7 (2) = 333 cycles.
  • 15 passes = 4995 cycles. Total = 2 + 1 + 4995 + 2 = 5000 cycles = 5.000 ms.
        ORG 0000H
MAIN:   ACALL DELAY5MS
        SJMP MAIN

; 5 ms delay at XTAL = 12 MHz (machine cycle = 1 us)
; cycles = ACALL 2 + MOV 1 + 15 x (MOV 1 + DJNZ 165 x 2 + DJNZ 2) + RET 2 = 5000
DELAY5MS: MOV R7,#15
OUTER:    MOV R6,#165
INNER:    DJNZ R6,INNER
          DJNZ R7,OUTER
          RET
          END

In a simulator at 12 MHz the call measured exactly 5.000 ms.

Alternative using Timer 0

DELAY5T: MOV TMOD,#01H     ; Timer 0, mode 1
         MOV TH0,#0ECH     ; 65536 - 5000 = 60536 = EC78H
         MOV TL0,#78H
         SETB TR0          ; start the timer
WAIT:    JNB TF0,WAIT      ; wait for overflow
         CLR TR0           ; stop timer
         CLR TF0           ; clear flag
         RET

This takes about 5.013 ms because of the few setup instructions; load 65536 - 5000 + 13 to compensate if it must be exact.

✓ Answer checked by Claude Opus 5.5 (high effort)

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