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Category Archives: 软硬兼施

使用树莓派驱动超声测距模块

2015 年 07 月 18 日 – 上午 3:31

这里用的模块，是SRF04超声测距模块，淘宝上买的，10块钱，加八块运费

测距的时序图如下：

想给触发信号trig一个约10us的高电平，模块会发送测距信号，回响信号echo会输出一个高电平，高电平的次序时间，与检测距离成正比，这就是该模块超声定位的基本原理
先面是相关的pytho代码

#srf04.py coding:utf-8 srf04超声测距模块驱动
import RPi.GPIO as GPIO
import time
from time import ctime, sleep
 
__gproTrig = 16    trig
__gproEcho = 12    echo
#初始化输入、输出gpio口
def init():
    GPIO.setmode(GPIO.BCM)
    GPIO.setup(__gproTrig, GPIO.OUT)
    GPIO.setup(__gproEcho, GPIO.IN)
 
def measure():
    #确保启动前为低电平
    GPIO.output(__gproTrig, False)
    time.sleep(0.0001)
    #启动测距
    GPIO.output(__gproTrig, True)
    time.sleep(0.00001) #10us
    GPIO.output(__gproTrig, False)
    #开始记时
    start   = 0
    stop    = 0
   #         print("measure--1")
    while start == 0 or stop == 0:
        val     = GPIO.input(__gproEcho)
        if stop == 0 and start == 0 and val == 1:
   #         print("measure--2")
            start = time.time()
        if start > 0 and stop == 0 and val == 0:
      #      print("measure--3")
            stop = time.time()
    #根据驱动公式计算测距结果
    elapsed = stop - start
    distance = (elapsed * 34300)/2
    return distance
 
if __name__ == '__main__':
    init()
    for i in range(300)://五分钟测试
     #   print("srf05-start:", i)
        dis = measure()
        print('--------------dis:', dis)
        sleep(1)

over~
~~~~~~后记
其实这个超声模块已经到手很久了，但是看到驱动要使用10us信号后，就一直纠结怎么来做10us的时钟，测试用python的sleep实现时钟，又特别不精确，那么，就需要100MHz的晶振，可晶振怎么接到树莓派上。。。于是就卡到这里了。
后来偶然搜到了类似模块的驱动方式，才发现其实是我想多了，要什么晶振，sleep足够了~~~嘿嘿，照着这种思路做下去，居然真的成功了
很多事情就是这样：先有一个大概的思路，没有更好地办法的情况下，就先照着这个思路走，大部分时候都能走的过去

——本文为原创，转载请注明出处昆仑的山头

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树莓派GPIO使用练习–点亮LED点阵

2015 年 04 月 15 日 – 上午 6:29

这里只是为了练习GPIO的使用，没有加任何的译码设备，也没加任何的放大设备，直接用树莓派的16个gpio引脚，连接一个8*8LED点阵的16个引脚，考虑到树莓派所能提供的电流不能太大，所以统一在点阵的8个阳极各加了一个1kΩ的电阻

8*8电阻的16个引脚，8个阳极，8个阴极，我们可以第一为R1~R8、C1~C8（考虑到点阵是方形的，所以不去区分行列，只区分阴阳，上面的R/C和“阴/阳”的对应关系，请根据实际共阴还是共阳区分清楚）

当我们给R1/C1两个管脚上电时（即在树莓派中，将R1对应的GPIO口拉到高电平，C1拉到低电平），第一行、第一列的这个点就会亮起来，同理，给R3/C5上电，[3, 5]的这个点会连起来

于是，我们可以通过这种方式依次点亮所有的点
。。。

于是，问题来了

如果我想同时点亮[3, 5]/[3, 6]/[4, 6]三个点，就需要同时拉高C3/C4，拉低R5/R6，可是，如果我做了这个操作，[4, 5]这个点也会亮起。。。这怎么办哪？

答案时，依次先点亮[3, 5]/[3, 6]，熄灭这两个点，再点亮[4, 6]，。。。熄灭[4, 6]，点亮[3, 5]/[3, 6]，如此往复，当这种切换的频率足够高时，由于认得视觉暂留，看到这三个点就是同时亮起的，也就是说，同一时刻，自会有单独一行、或者一列有点在发光

呵呵，很熟悉吧，电影、大屁股显示器什么的，和这个原理类似

好的，明白了原理，就可以动手写程序吧

这里使用树莓派最主流的编程语言：python，显示的东西，是“Hello word”，左右移动显示

import RPi.GPIO as GPIO  #导入必要的python库
import time
 
#Hello World 字模
vhws = [
  [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
  [0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0],
  [0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,1,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0],
  [0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,1,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,1,0,1,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0],
  [0,0,0,0,0,0,0,0,1,1,1,1,1,0,1,1,1,1,1,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,1,0,0,0,1,0,0,0,0,0,1,1,1,0,0,0,0,0,0],
  [0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,1,0,0,0,0,0,0],
  [0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,1,0,0,1,0,1,0,0,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,0,1,0,1,0,0,1,0,0,1,0,0,0,0,0,0],
  [0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,1,1,1,0,0,0,0,0],
]
 
#左移还是右移
isRig = 1
vhwi = 0
#此函数返回当前时间点当前帧需要显示的8*8数组
def hwv():
  global isRig
  global vhwi
  l = len(vhws[0])
  if isRig &gt; 0:
    vhwi += 1
  else:
    vhwi -= 1
  if vhwi &gt;= (l - 8):
    isRig = 0
  if vhwi  0):
    if timeLine % 10 == 0:
      nowV = hwv()
    #双层循环实现的扫描显示  
    for i in range(0, len(nowV)):
      allOver()
      #拉低所有阴极
      GPIO.output(rs[i], GPIO.LOW)
      for j in range(0, len(nowV[i])):
        if nowV[i][j] &gt; 0:
          GPIO.output(cs[j], GPIO.HIGH) #点亮这个点
        else:
          GPIO.output(cs[j], GPIO.LOW)  #不点亮
        time.sleep(sleepTime)
    timeLine = timeLine - 1
  return
#ledtest 17 27 22 5 6 13 19 26 18 23 24 25 12 16 20 21
#这里是管脚对应关系
rs = [21, 20, 16, 12, 25, 24, 23, 18] #R1:13 R2:3 R3:4 R4:10 R5:6 R6:11 R7:15 R8:16
cs = [26, 19, 13, 6, 5, 22, 27, 17] #C1:5 C2:2 C3:7 C4:1 C5:12 C6:8 C7:14 C8:9
#开始
def startVs():
  GPIO.setmode(GPIO.BCM)
  for i in xrange(8):
    GPIO.setup(cs[i], GPIO.OUT)
    GPIO.setup(rs[i], GPIO.OUT)
  allOver()
  return
#关闭
def stopVs():
  GPIO.cleanup()
  return
#将所有管脚置为反响，即熄灭所有
def allOver():
  for i in xrange(8):
    GPIO.output(cs[i], GPIO.LOW)
    GPIO.output(rs[i], GPIO.HIGH)
  return
 
#timeLine = 1000
frames = 60
#显示一个8*8数组
def viewArr():
  timeLine = 1000
  sleepTime = 0.0001 #1 / (frames * 8)
  while(timeLine > 0):
    if timeLine % 10 == 0:
      nowV = hwv()
    for i in range(0, len(nowV)):
      allOver()#每次点亮前，先熄灭所有
      GPIO.output(rs[i], GPIO.LOW)
      for j in range(0, len(nowV[i])):
        if nowV[i][j] &gt; 0:
          GPIO.output(cs[j], GPIO.HIGH)  #点亮
        else:
          GPIO.output(cs[j], GPIO.LOW)   #不点亮
        time.sleep(sleepTime)  #扫屏的时延
  timeLine = timeLine - 1   #显示时常控制
return
 
if __name__ == '__main__':
  startVs()
  viewArr()
  stopVs()

下面是实际效果的几个截图：

下面两个是线路连接：

转载请注明出处：昆仑的山头

Tags python, 树莓派 | Permalink | Comment (0)

Category Archives: 软硬兼施

使用树莓派驱动超声测距模块

树莓派GPIO使用练习–点亮LED点阵

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时间线

链接表

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Category Archives: 软硬兼施

使用树莓派驱动超声测距模块

树莓派GPIO使用练习–点亮LED点阵

五花八门

分门别类

时间线

链接表

七嘴八舌

近期发布

发布日历