HC12 - Two Choices of Clock for PWM

1. PCLKx = 0 (Clock A or B):
   • ${\rm Clk}_{\rm {x}} = 0.125 \mu{\rm {s}} \times 2^N$
   • ${\rm Per}_{\rm {x}} = 0.125 \mu{\rm {s}} \times 2^N \times ({\tt {PWPERx+1}})$
2. PCLKx = 1 (Clock S0 or S1):
   • ${\rm Clk}_{\rm {x}} = 0.125 \mu{\rm {s}} \times 2^{(N+1)} \times (M+1)$
   • ${\rm Per}_{\rm {x}} = 0.125 \mu{\rm {s}} \times 2^{(N+1)} \times (M+1) \times ({\tt {PWPERx+1}})$

$N$ = PCKA 2:0 for PWM 1 and 0, $N = 0, 1, \cdots, 7$
$N$ = PCKB 2:0 for PWM 3 and 2.

$M$ = PWSCAL0 7:0 for PWM 1 and 0, $M = 0, 1, \cdots, 255$
$M$ = PWSCAL1 7:0 for PWM 3 and 2.

Normally want PWPERx = 200 $->$ 255.

Example: Want 5 kHz PWM - Per$_{\rm {x}}$ = 200 $\mu$s = 1,600 P clock cycles..

1. Try PCLKx = 0 and PWPERx = 255
   
   $$1600 = 2^N \times ({\tt {PWPERx+1}})$$
   
   
   
   
   $$2^N = \frac{1600}{256} = 6.25$$
   
   
   Not possible. Try $2^N = 8$, $N = 3$
   
   $$1600 = 2^N \times ({\tt {PWPERx+1}}) = 2^3 \times ({\tt {PWPERx+1}})$$
   
   
   
   
   $${\tt {PWPERx+1}} = \frac{1600}{2^3} = 200$$
   
   
   
   
   $${\tt {PWPERx}} = 199$$
   
   
   Exactly 5 kHz, get 1 part in 200 accuracy

2. Try PCLKx = 1 and PWPERx = 255
   
   $$1600 = 2^{(N+1)} \times (M+1) \times ({\tt {PWPERx+1}})$$
   
   
   
   
   $$2^{(N+1)} \times (M+1) = \frac{1600}{256} = 6.25$$
   
   
   Not possible. Use a slightly different frequency - let
   
   $$2^{(N+1)} \times (M+1) = 6$$
   
   
   
   
   $$N = 0 {\rm\ and \ } M = 2$$
   
   
   This gives
   
   $${\rm Per}_{\rm {x}} = 2^{(0+1)} \times (2+1) \times (255+1) = 1536 {\rm ~P~clocks} = 192 \mu{\rm {s}}$$
   
   
   for a 5.2 kHz frequency