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 <title><![CDATA[模拟量问题]]></title>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446.html]]></link>
 <description><![CDATA[Latest 50 replies of 模拟量问题]]></description>
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 <description><![CDATA[难道不用看不同的PLC的使用说明吗？俺用过不同的国产PLC，有直接是读数最高值对应20MA，但是0对应4MA，也有4MA是对应最高值/20*4的。分辨率不同。]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2034492.html]]></link>
 <author><![CDATA[静夜思飘]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Fri, 22 Mar 2019 00:02:55 +0000]]></pubdate>
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 <description><![CDATA[分辨率除以最大测量值就是倍率]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2034410.html]]></link>
 <author><![CDATA[心碎了无痕]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Thu, 21 Mar 2019 09:17:24 +0000]]></pubdate>
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 <description><![CDATA[其实就是数学公式里，2点坐标确定一条直线的那个公式]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2034147.html]]></link>
 <author><![CDATA[lzmtxp99]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Wed, 20 Mar 2019 03:55:53 +0000]]></pubdate>
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 <description><![CDATA[有的时候可以忽略掉电压这个中间环节，把两个终端的值直接换算]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2034038.html]]></link>
 <author><![CDATA[ymmfa]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Tue, 19 Mar 2019 12:15:01 +0000]]></pubdate>
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 <description><![CDATA[模拟量转换，实际上就是初中数学的内容啊 [s:17]]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2034034.html]]></link>
 <author><![CDATA[宿迁电工]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Tue, 19 Mar 2019 12:08:49 +0000]]></pubdate>
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 <description><![CDATA[约0.78MPa]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2034021.html]]></link>
 <author><![CDATA[hgz284631157]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Tue, 19 Mar 2019 10:45:07 +0000]]></pubdate>
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 <description><![CDATA[[quote]引用第4楼hong0618于2019-03-16 13:56发表的&nbsp;&nbsp;:
如果设定峰值是1MP，计算应是25000/32000×1=0.78125Mp
[/quote]
那就是说倍率和0--10V是一样的算法？]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2033477.html]]></link>
 <author><![CDATA[心碎了无痕]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Sun, 17 Mar 2019 05:07:26 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[[quote]引用楼主心碎了无痕于2019-03-15 09:04发表的 模拟量问题 :
模块分辨率为32000，使用4--20ma,测量1MP的气压，则测量值25000为多少MP？

32000/20=1600
1600*4=6400
32000-6400=25600
.......
[/quote]

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S7-1200编程手册截图
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 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2033388.html]]></link>
 <author><![CDATA[wanzhen2006]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Sat, 16 Mar 2019 08:40:42 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[如果设定峰值是1MP，计算应是25000/32000×1=0.78125Mp]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2033360.html]]></link>
 <author><![CDATA[hong0618]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Sat, 16 Mar 2019 05:56:11 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[请问为什么要这么算呢&nbsp;&nbsp;
32000/20=1600
1600*4=6400
32000-6400=25600
25600/1=25600（倍率）
25000-6400=18600
18600/25600=0.73MP&nbsp;&nbsp;&nbsp;&nbsp;]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2033137.html]]></link>
 <author><![CDATA[zy123518763]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Fri, 15 Mar 2019 05:35:34 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[谢谢！！！！！]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2033110.html]]></link>
 <author><![CDATA[心碎了无痕]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Fri, 15 Mar 2019 03:50:59 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[[quote]引用第1楼于1970-01-01 08:00发表的&#160;&#160;:

[/quote]
对的]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1664446#2033104.html]]></link>
 <author><![CDATA[sss6491]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Fri, 15 Mar 2019 03:40:39 +0000]]></pubdate>
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