The primary oscillator frequency on explorer 16 is 14.535 MHZ. PLL is used. What is the instruction cycle frequency (Fcy)?
Fosc (Processor/Peripheral clock Frequency) = 14.535 x 4 = 58.14/2 = 29.07 MHz.
Tcy (Instruction Cycle) = 1/29.07 = 0.0344uSec.
is this correct?
Yes, your calculations are correct. The instruction cycle frequency (Fcy) is calculated by dividing the processor/peripheral clock frequency (Fosc) by 2, which in this case is 29.07 MHz. The period of each instruction cycle (Tcy) is then calculated by taking the reciprocal of Fcy, which gives a value of 0.0344 microseconds (or 34.4 nanoseconds).
Yes, your calculations are correct. The primary oscillator frequency on the Explorer 16 board is 14.535 MHz. Since a PLL (Phase-Locked Loop) is used, the Fosc (Processor/Peripheral Clock Frequency) is calculated by multiplying the oscillator frequency by 4, resulting in 58.14 MHz. Since the processor operates on a 2:1 Tcy (Instruction Cycle to Clock Cycle) ratio, the instruction cycle frequency (Fcy) is half of the Fosc, which is 29.07 MHz. The corresponding Tcy (Instruction Cycle Time) can be calculated as 1 divided by Fcy, which results in 0.0344 microseconds (or 34.4 nanoseconds).
Yes, your calculations are correct. To find the instruction cycle frequency (Fcy), you need to know the processor or peripheral clock frequency (Fosc). In this case, the primary oscillator frequency on the Explorer 16 board is given as 14.535 MHz.
The PLL (Phase-Locked Loop) is used to multiply the oscillator frequency by a factor of 4, bringing the Fosc to 14.535 MHz x 4 = 58.14 MHz.
However, since most instructions on the microcontroller require 4 clock cycles to complete, the instruction cycle frequency (Fcy) is half of the Fosc. So, Fcy = 58.14 MHz / 2 = 29.07 MHz.
To calculate the duration of each instruction cycle (Tcy), you can take the inverse of the Fcy. So, Tcy = 1 / 29.07 MHz = 0.0344 microseconds (or 34.4 nanoseconds).
Therefore, your calculations for Fcy and Tcy are correct.