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authorheyuanjie87 <[email protected]>2015-09-09 10:26:42 +0800
committerheyuanjie87 <[email protected]>2015-09-09 10:26:42 +0800
commit984a7ba05128992d8b89fc689f5f5e4f135e2f1b (patch)
tree5e8474d1aa8c44af226ad5a2fd29cee21a3ba2cf /components/drivers
parentefc1b058ac553db24232aa0f49c7b573f3a498c2 (diff)
修改说明文档
Diffstat (limited to 'components/drivers')
-rw-r--r--components/drivers/hwtimer/README_CN.md7
1 files changed, 2 insertions, 5 deletions
diff --git a/components/drivers/hwtimer/README_CN.md b/components/drivers/hwtimer/README_CN.md
index 8ee6f3df06..fb3b4fa60a 100644
--- a/components/drivers/hwtimer/README_CN.md
+++ b/components/drivers/hwtimer/README_CN.md
@@ -94,8 +94,5 @@ void timer_irq_handler(void)
假设计数器最大值0xFFFF,计数频率1Mhz,定时时间1秒又1微秒。
-由于定时器一次最多只能计时到65535us,对于1000001us的定时要求
-就需分15次完成。误差时间: 1000001 - (65535*15) = 16976us。
-这个误差太大,为了减小误差最理想的情况就是整除。实际中不可能
-所有数都被整除(1不合适),所以就只有最大限度逼近理想情况。例如
-上式中以 50000us定时20次即可将误差减小到1us(理论值,忽略指令时间)
+由于定时器一次最多只能计时到65535us,对于1000001us的定时要求。
+可以50000us定时20次完成,此时将会出现计算误差1us。