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RS vAGvA {V$% u-G ("#$% N*S .S) uNNe$R-m u (";$* cC5 KY )*+ 9d &@5 7Y- =_- )@) ./$* ("w$A a[5 (J )4s [...]... #200 R3, HERE 1 2 1 Do vậy tổng thời gian giữ chậm là [(200 2) + 1 + 1] 1.085 = 436 .17ms Thông thường ta tính thời gian giữ chậm dựa trên các lệnh bên trong vòng lặp và bỏ qua các chu kỳ đồng hồ liên quan với các lệnh ở ngoài vòng lặp Trong ví dụ 3. 15 giá trị lớn nhất mà R3 có thể chứa là 255, do vậy một cách tăng độ trễ là sử dụng lệnh UOP (không làm gì) trong vòng lặp để tiêu tốn thời gian một cách... dụ 3. 15 Ví dụ 3. 15: Hãy tìm kích thước của thời gian giữ chậm trong chương trình sau, nếu tần số giao động thach anh là 11.0592MHz A, #55H P1, A ACALL DELAY CPL A SJMP AGAIN ; Time delay DELAY: MOV R3, #200 HERE : DJNZ R3, HERE RET AGAIN: MOV MOV Lời giải: Từ bảng A-1 của phụ lục Appendix A ta có các chu kỳ máy sao cho các lệnh của chương trình con giữ chậm là: DELAY: HERE : MOV DJNZ RET R3, #200... 1.085ms = 1629.67ms 3. 3 .3 Độ trễ thời gian của vòng lặp trong vòng lặp Một cách khác để nhận được giá trị từ độ trễ lớn là sử dụng một vòng lặp bên trong vòng lặp và cũng được gọi là vòng lặp lồng nhau Xem ví dụ 3. 17 dưới đây Ví dụ 3. 17: Đối với một chu kỳ máy 1.085ms hãy tính thời gian giữ chậm trong chương trình con sau: DELAY: AGAIN: HERE: MOV MOV NOP NOP DJNZ DJNZ RET R2, #200 R3, #250 R3, HERE R2, AGAIN... một cách đơn giản Điều này được chỉ ra trong ví dụ 3. 16 dưới đây Ví dụ 3. 16: Hãy tìm độ trễ thời gian cho chương trình con sau Giả thiết tần số dao động thạch anh là 11.0592MHz DELAY: MOV HERE : NOP NOP NOP NOP DJNZ R3, #250 R3, HERE RET Số chu kỳ máy 1 1 1 1 1 2 1 Lời giải: Thời gian trễ bên trong vòng lặp HERE là [250 (1 + 1 + 1 + 1 + 1 + 2)] 1.0851ms = 1627.5ms Cộng thêm hai lệnh ngoài vòng lặp.. .Lệnh (a) MOV (b) DEC (c) DJNZ (d) LJMP (e) SJMP (f) NOP (g) MUL Chu kỳ máy 1 1 2 2 2 1 4 R3, #55 R3 R2, target AB Thời gian thực hiện 1 1.085 ms = 1.085 ms 1 1.085 ms = 1.085 ms 2 1.085 ms = 2.17 ms 2 1.085 ms = 2.17 ms 2 1.085 ms = 2.17 ms 1 1.085 ms = 1.085 ms 4 1.085 ms = 4 .34 ms 3. 3.2 Tính toán độ trễ Như đã trình bày ở trên đây, một chương trình con giữ chậm gồm... nhiên, các lệnh MOV R3, #250 và DJNZ R2, AGAIN ở đầu và cuối vòng lặp AGAIN cộng (3 200 1.085ms) = 651ms vào thời gian trễ và kết quả ta có 217000 + 651 = 217651ms = 217.651 miligiây cho tổng thời gian trễ liên quan đến chương trình con giữ chậm DELAY nói trên Lưu ý rằng, trong trường hợp vòng lặp lồng nhau cũng như trong mọi vòng lặp giữ chậm khác thời gian xấp xỉ gần dúng vì ta bỏ qua các lệnh đầu... đến chương trình con giữ chậm DELAY nói trên Lưu ý rằng, trong trường hợp vòng lặp lồng nhau cũng như trong mọi vòng lặp giữ chậm khác thời gian xấp xỉ gần dúng vì ta bỏ qua các lệnh đầu và cuối trong chương trình con . )*o (2:$% 7Y- ("#$% N*^& =- EE - FF• 72M) )*-J (*C$* )@) ./$* $*V5 ( _- =C $*V5 .Z-s i%*ÜJ .C -GzD 7<$ +Gz7 c5(e )PJ c' $*_ (2:$% 7Y- K# = _- 7>J )*o *-/ $ (* 3- )PJ c'. q2M$% (* 3- %-J$ ("ễ ("#$% =? 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