Microchip PIC period register PR2

Question

From Microchip sample code

PR2 = 2083u;     /* Timer2 Period, 19.2 kHz */ 

How does 2083u correspond to 19.2 kHz, which is

1 / 19.2E03 = 52.083u

They don't correspond at all. Mistake by Microchip?

Answer 1

PR2 = 2083U

makes TIMER2 trigger every 2083 CPU cycles. Calculating

52.083 us / 2083 = 25 ns
1 / 25 ns = 40 MHz

we can conclude that the processor is probably running at FCY = 40 MHz in the example.

The letter u in PR2 = 2038u; does not mean microseconds; it is a C language syntax that makes the integer literal unsigned. See Signedness (Wikipedia).

Answer 2

Setting PR2 to 2083 means that the timer triggers every 2084 (not 2083) clock cycles. When you calculate a timer period, you always have to subtract 1 because the timer value is zero-based.