PROBLEM TO BE SOLVED: To measure electrostatic potential as well as to easily eliminate static electricity by providing a means for arithmetically operating a difference in an absolute value and/or the ratio of electric currents obtained by an ammeter for a (+) current and an ammeter for a (-) current. ;SOLUTION: Discharge is performed from a (+) discharge needle 22 and a (-) discharge needle 24 of a static eliminator 10 electrically connected to the human body 50 to eliminate static electricity of the human body 50. At the same time, this positive/negative discharge current is measured by ammeters 18, 20 to calculate a difference (I-J) between a (+) high tension current I and a (-) high tension current J and/or the ratio (I/J) by an operation circuit 32. Electrified electric potential of the human body 50 is determined from the polarity and the size, and a level of static electricity gathered in the human body 50 is displayed on a display device 34. For example, a red lamp 36 in the display device 34 is lighted in electrification of (+) static electricity of IJ, a green lamp 38 is lighted in nonelectrification of static electricity of I=J, and a yellow lamp 40 is lighted in electrification of (-) static electricity of IJ. Thus, electrostatic potential is constantly returned to 0 V.;COPYRIGHT: (C)2001,JPO
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机译:解决的问题:测量静电势,并通过提供一种对算术电流表的(+)电流和(+)电流的绝对值和/或电流比率之差进行算术运算的方法,轻松消除静电。电流表。 ;解决方案:从电连接到人体50的静电消除器10的(+)放电针22和(-)放电针24进行放电,以消除人体50的静电。通过电流表18、20测量正/负放电电流,以通过操作计算(+)高压电流I和(-)高压电流J之间的差(IJ)和/或比率(I / J)电路32。根据极性和大小确定人体50的电势,并且在显示装置34上显示人体50中聚集的静电水平。例如,显示器中的红灯36 I(J)的(+)静电带电时,设备34点亮; I = J的静电不带电,绿色灯38点亮; I(-)静电时,黄色灯40点亮。 展开▼