怎么读取8位控制器上的开放式ADC端口

亲爱的所有人，在我试图找到一个简单而可靠的方法来实现一个“触摸”控制通过一个8位PIC我绊倒以下：我删除了输入从我现有的ADC输入端口（电位器），只留下附加10厘米的电线。当测量输入引脚上的电压（不是通过ADC而是通过电压表外部）时，我得到如下结果。当我不接触导线时，在ADC输入引脚上得到0，5伏，当接触导线的末端时，引脚上的电压下降到0，1伏。我首先给连接的电线加上一个5M6电阻，后来我的灯的金属部分和效果保持完全相同。这是一个单一事件（例如，在一个错误的测试板上的泄漏电流）还是有一个合理的方式来解释为什么会发生这种情况。如果这是一个100%可重复的效果，它将是通过设备的导电部分实现触摸功能的最简单的方式，因为它应该是没有问题的。通过ADC读出电压DROPP。无论是关于PIC的架构还是在这个设置背后的电子规则，任何反馈都是受欢迎的（实际上我会说，通过触摸电线，我添加了一个接地连接，因此电压下降。但是为什么一开始就有电压呢？当我开始测试时，我的第一个错误是检查我是否能在触摸电线时确保引脚上的电压增加，并惊奇地发现了相反的效果……）



      以下为原文

    Dear all,

During my attempt to find a easy and reliable way to implement a "touch" controll via a 8 Bit PIC I stumbled over the following:

I removed the input from my existing ADC input port (a potentiometer), leaving only attached 10 cm of wire. When meassuring the voltage on the input pin (not via ADC but external via voltmeter) I got the following results. When I do not touch the wire, I get 0,5 Volt on the ADC input pin, when touching the end of the wire the voltage on the pin dropps to 0,1 Volt.

I first added a 5M6 resistor to the connected wire and later the metal parts of my lamp and the effect stayed exactly the same. Is this some single event (e.g. by leak currents on a faulty test board) or is there a reasonable way to explain why this would happen.

If this is a 100% reproducable effect it would be the most easy way to implement a touch function via a conducting part of the device as it should be no problem to readout the voltage dropp via the ADC.

Any feedback, both regarding the architecture of the pic or the electronical rules behind this setup are welcome (actually I would say by touching the wire I add a ground connection and therefore the voltage dropps ... but why is there any Voltage in the first place? My initial idia when starting the tests was to check if I can meassure a voltage increase on the pin when touching the wire and was surprised by finding this opposite effect...)

best regards
Michael