Implementation
Calculation example of the PWM frequency
Example for 2 Hz PWM signal with 50% duty cycle:
›
Period time of 2 Hz signal is
0.5 s
›
High level time is
0.25
s (period * duty cycle)
›
Within
0.5 s
, the timer T12 is incremented
390625
times (0.5 s / 1.28 µs) and has
5
overflows
›
Within
0.25 s
, the timer T12 is incremented
195312
times (0.25 s / 1.28 µs) and has
2
overflows
›
Parameters available inside the ISR:
–
Counter value of timer T12 one period ago:
1000
–
Current counter value of timer T12:
63955
–
Counter value when the falling edge occured:
65244
–
Amount of overflows during one PWM period:
5
–
Amount of overflows during PWM high level time:
2
›
Calculation of total amount of increments:
–
Increments before the first overflow: 65534
– 1000 =
64534
–
Increments during the second and the last overflow: (5 - 1) * 65534 =
262136
–
Increments after the last overflow:
63955
–
Total amount of increments:
64534
+
262136
+
63955
=
390625
›
Calculation of increments during high level time:
–
Increments before the first overflow: 65534
– 1000 =
64534
–
Increments during the second and the last overflow: (2 - 1) * 65534 =
65534
–
Increments after the last overflow:
65244
–
Total amount of increments:
64534
+
65534
+
65244
=
195312
›
Calculation of the PWM frequency by dividing the frequency of timer T12 by the total amount of
increments during one PWM period:
–
781250 Hz / 390625 =
2 Hz
›
Duty cycle (increments during high level time / increments during total period) = 195312 / 390625 =
50%
11
2020-01-17
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