## Using the previous delay loop, what count is required to implement a 0.14 millisecond delay? Assume a 16-MHz crystal.

## First determine how many clock cycles are needed.

## Answer: (0.14e-3)/(125e-9) = 1120.

## 2 + 4x + 2 = 1120 ==> x = 279.

## Loop starts by decrementing the count so need LDX #280.

1 * (2) + (Loop Count - 1) * (1 + 3) + 1 * (1 + 1) = # Clock Cycles