Topic: Microprocessors and Microcontrollers
For the below mentioned 8051 assembly code... Calculate time required for execution of Part 1
- A. 2495600 $\mu s$
- B. 2496300 $\mu s$
- C. 2495300 $\mu s$
- D. 2496600 $\mu s$
Answer: C. 2495300 $\mu s$
Explanation: With a 12 MHz clock and 1 machine cycle taking 12 clock cycles, 1 machine cycle = 1 $\mu s$. The code forms a triple nested loop: R0 (100) -> R1 (50) -> R2 (248). The innermost loop (Part 3) takes $248 \times 2 = 496$ cycles. The middle loop executes Part 2 (1 cycle), Part 3 (496 cycles), and the DJNZ R1 (2 cycles), totaling $(1 + 496 + 2) \times 50 = 24950$ cycles. The outer loop executes Part 1 (1 cycle), the middle loop (24950 cycles), and the DJNZ R0 (2 cycles), totaling $(1 + 24950 + 2) \times 100 = 2495300$ machine cycles. This translates to 2495300 $\mu s$.