我想我会在这里给出更全面的答案。对于 84MHz 时钟，有多种预分频器和周期组合都可以使用。这里仅仅是少数：
PSC ARR F ERROR EXACT
1 41999 1000.000000 0.0000000000 YES
2 27999 1000.000000 0.0000000000 YES
3 20999 1000.000000 0.0000000000 YES
4 16799 1000.000000 0.0000000000 YES
5 13999 1000.000000 0.0000000000 YES
6 11999 1000.000000 0.0000000000 YES
7 10499 1000.000000 0.0000000000 YES
9 8399 1000.000000 0.0000000000 YES
11 6999 1000.000000 0.0000000000 YES
13 5999 1000.000000 0.0000000000 YES
14 5599 1000.000000 0.0000000000 YES
15 5249 1000.000000 0.0000000000 YES
19 4199 1000.000000 0.0000000000 YES
我怎么想出这些？即使是像 MikroElektronica 这样的商业工具也只能提供一种精确（或不精确）的组合。如何找到它们？我只是编写了一个 python 程序来计算它们。它将每个分类为精确，或记录那些不精确的相对误差。通过更改程序顶部的容差，您可以根据需要“收紧”或“放松”计算。
这是整个程序：
import numpy as np
import pandas as pd
TARGET_F = 1000 # In Hz so 50.0 is 0.020 seconds period and 0.25 is 4 seconds period
CLOCK_MCU = 84000000
TOLERANCE = 0.0001
# -----------------------------------------------------
def abs_error(num1, num2):
return abs((num1 - num2) / num1)
def hertz(clock, prescaler, period):
f = clock / (prescaler * period)
return f
def perfect_divisors():
exacts = []
for psc in range(1, 65536):
arr = CLOCK_MCU / (TARGET_F * psc)
if CLOCK_MCU % psc == 0:
if arr <= 65536:
exacts.append(psc)
return exacts
def add_exact_period(prescaler):
entries = []
arr = CLOCK_MCU / (TARGET_F * prescaler)
if arr == int(arr):
entry = [prescaler, arr, TARGET_F, 0.0]
entries.append(entry)
return entries
def possible_prescaler_value():
possibles = []
for psc in range(1, 65536):
if psc in exact_prescalers:
continue
h1 = hertz(CLOCK_MCU, psc, 1)
h2 = hertz(CLOCK_MCU, psc, 65536)
if h1 >= TARGET_F >= h2:
possibles.append(psc)
return possibles
def close_divisor(psc, tolerance):
arr = CLOCK_MCU / (TARGET_F * psc)
error = abs_error(int(arr), arr)
if error < tolerance and arr < 65536.0:
h = hertz(CLOCK_MCU, psc, int(arr))
return psc, int(arr), h, error
else:
return None
# ------------------------------------------------------------------------
# Make a dataframe to hold results as we compute them
df = pd.DataFrame(columns=['PSC', 'ARR', 'F', 'ERROR'], dtype=np.double)
# Get exact prescalars first.
exact_prescalers = perfect_divisors()
exact_values = []
for index in range(len(exact_prescalers)):
rows = add_exact_period(exact_prescalers[index])
for rowindex in range(len(rows)):
df = df.append(pd.DataFrame(np.array(rows[rowindex]).reshape(1, 4), columns=df.columns))
# Get possible prescalers.
poss_prescalers = possible_prescaler_value()
close_prescalers = []
for index in range(len(poss_prescalers)):
value = close_divisor(poss_prescalers[index], TOLERANCE)
if value is not None:
close_prescalers.append((value[0], value[1], value[2], value[3]))
df = df.append(pd.DataFrame(np.array(close_prescalers).reshape(len(close_prescalers), 4), columns=df.columns))
# Adjust PSC and ARR values by -1 to reflect the way you'd code them.
df['PSC'] = df['PSC'] - 1
df['ARR'] = df['ARR'] - 1
# Sort first by errors (zeroes and lowest errors at top of list, and
# then by prescaler value (ascending).
df = df.sort_values(['ERROR', 'PSC'])
# Make and populate column indicating if combination is exact.
df['EXACT'] = pd.Series("?", index=df.index)
df['EXACT'] = np.where(df['ERROR'] == 0.0, "YES", "NO")
# Format for output.
df['PSC'] = df['PSC'].map('{:.0f}'.format)
df['ARR'] = df['ARR'].map('{:.0f}'.format)
df['F'] = df['F'].map('{:.6f}'.format)
df['ERROR'] = df['ERROR'].map('{:.10f}'.format)
output = df.to_string()
print(output)
print()
print('these are the ', df.shape[0], ' total combination meeting your tolerance requirement')
exit(0)
使用这个程序，每个人都可以自信地计算这些值。我希望它被证明是有用的。