Delay loops

Single loop

RETARDO
    MOVLW   .N
    MOVWF   CONTADOR
BUCLE
    DECFSZ  CONTADOR,F
    BRA     BUCLE
    RETURN

With CALL: Σ CI ≈ 3N + 4. Body only: ≈ 3N + 2.

PDF example: N=200 @ 20 MHz

Σ CI = 605 → delay = 121 µs.

Two nested loops (PDF structure)

Course PDF slide: two- and three-loop nested delays with cycle counts and formulas — PDF slide 8: line-by-line cycle count and Σ CI = 3NM + 4M + 5 (two loops) / 3NMP + 4MP + 4P + 5 (three loops). — Clase Tema 4 y 5 — Delays · UNEXPO · Prof. Ing. Yoel Pire

    MOVLW   .M
    MOVWF   CONTADOR_2
BUCLE_2
    MOVLW   .N
    MOVWF   CONTADOR_1
BUCLE_1
    DECFSZ  CONTADOR_1,F
    BRA     BUCLE_1
    DECFSZ  CONTADOR_2,F
    BRA     BUCLE_2
    RETURN

Σ CI ≈ 3NM + 4M + 5 (no CALL).

Three nested loops

Σ CI ≈ 3NMP + 4MP + 4P + 5

Maximum @ 20 MHz (N=M=P=255)

This table is the class exercise from the PDF («Create a comparative table of maximum delay for 1, 2, and 3 loops @ 20 MHz»). The 2-loop row (Σ CI = 196,100, 39.22 ms) comes from that slide — not from the partial exam.

PDF exercise: comparative table of max delays for 1, 2, and 3 loops at 20 MHz — PDF slide 9 — class exercise: N=M=P=255 and table with 770 / 196,100 / 50,005,250 cycles. — Clase Tema 4 y 5 — Delays · UNEXPO · Prof. Ing. Yoel Pire

Loops	Σ CI	Max delay
1	770	0.154 ms
2	196,100	39.22 ms
3	50,005,250	~10 s

Ejercicio · Parcial

Comparative table (PDF exercise)

Verify row 1: N=255, Σ CI=770, delay = 770 × 0.2 µs = 0.154 ms.

Ejercicio · Parcial

How much with 2 loops? (same slide)

N=M=255 → Σ CI = 3×255×255 + 4×255 + 5 = 196,100 → 39.22 ms.

Next: Calculation and applications.