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我想知道这是我的仪表还是存在一般ADC问题。
这就是发生的事情:如果我通过0.5V直流电源的1k电阻测量电流,我会在1mA电压下获得~418mA。 (实际上我已经测量了985.50欧姆的电阻和.497V的直流电源,但这已经足够接近我认为如果有人希望复制我的错误)。 如果我测量10mA电流下的电流,我得到的正确读数为.504安培。 所以你会认为我的1mA刻度存在问题,但是如果我测量刻度上的低电流,比如一个带有.5VDC电源的10k电阻,我会在1mA刻度上得到正确的读数。 当我向上移动时,错误会增加。 感谢您的时间和帮助,罗伯特 以上来自于谷歌翻译 以下为原文 I'm wondering if this is my meter or there is a general ADC problem. Here's what's happening: If I measure the current through a 1k resistor from a .5V DC source I get ~.418 mA's on the 1mA scale. (Actually I've measured the resistance at 985.50 ohms and the DC source at .497V but this is close enough I think if anyone wishes to duplicate my error). If I measure the current on the 10mA scale I get a correct reading of .504 amps. So you would think there's a problem with my 1mA scale but if I measure low current on the scale, say a 10k resistor with a .5VDC source I get the correct reading on the 1mA scale. As I move up the scale the error increases. Thanks for your time and help, Robert |
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2个回答
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您好,这不是异常,而只是由34410A内部分流电阻上的负载电压引起,必须考虑到这一点,就像任何其他DMM一样。
请参阅“用户指南”第131页的规格:直流电,分流电阻:1A为0.1欧姆,10mA为3A 2欧,100u为200欧,100mA,1mA在您的电路中,1mA范围内的总电阻为985 + 200欧姆= 1185欧姆。 计算电流:I = 0.497V / 1185欧姆= 0.419mA在10mA范围内,总电阻为987欧姆,因此电流计算为0.497 V / 987欧姆= 0.504 mA。 因此,您的乐器一切都很好。 坦率 以上来自于谷歌翻译 以下为原文 Hello, this is no anomaly, but simply caused by the burden voltage over the internal shunt resistor of the 34410A, which has to be taken into account, like on any other DMM also. See specification, page 131 of the 'User's Guide': DC Current, Shunt Resistor: 0.1 Ohm for 1A,3A 2 Ohm for 10mA, 100mA 200 Ohm for 100uA, 1mA In your circuitry, the total resistance on the 1mA range is 985 + 200 Ohm = 1185 Ohm. That calculates the current to: I = 0.497V / 1185 Ohm = 0.419mA On the 10mA range, the total resistance is 987 Ohm, so the current is calculated to 0.497 V / 987 Ohm = 0.504 mA. Therefore, everything is fine with your instrument. Frank |
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qwe043 发表于 2018-9-28 09:07 谢谢! 我知道阻抗在不同的尺度上发生了变化但从未计算出下降。 很高兴知道,我对读数会更加小心! 最好,罗伯特 以上来自于谷歌翻译 以下为原文 Thanks! I knew the impedance changed over different scales but never calculated the the drop. Good to know, I'll be more careful with my readings! Best, Robert |
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