[问答]

OHOMF测量3458A的测量顺序是什么

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问答对人有帮助，内容完整，我也想知道答案 0 OHOMF测量3458A的测量顺序是什么，OCOMP和AZERO都是ON？我在OHMF模式下测量一个3458A的电阻，我对集成过程中电压的路径感兴趣，因为这个电阻实际上是一个热敏电阻，这意味着在积分期间电阻变化很大。这就是我用另一个3458A对OHMF-DMM的电压（感应）进行采样的原因。我正在使用OCOMP和AZERO来进行更高质量的电阻测量。为了同步，第二（采样）DMM由第一个的孔径波形触发，因此电压测量与电阻测量相关。我注意到每个电阻读数，我从采样DMM（MEM FIFO）得到四个读数，这可能是实际值（用于最终计算电阻），OCOMP读数和一些AZERO。问题是，我不知道哪个是哪个。该手册自相矛盾地说，没有源电流的OCOMP在测量之前完成（第209页），而第62页则说明真正的读数（电流开启）先完成，然后是OCOMP（关闭电流）。我的测量数据显示前两个采样电流的样本，然后是两个没有电流的样本。有谁知道测量序列的实际顺序？到目前为止，我最好的猜测是首先是真正的读数，然后是AZERO（当前两者都有），然后是另一个AZERO（当前关闭）的OCOMP。编辑：vcshamoen于2013年11月27日上午8:58 以上来自于谷歌翻译以下为原文 What is the measurement sequence of an OHMF measuring 3458A with OCOMP and AZERO both ON? I'm measuring a resistor with with an 3458A in OHMF mode and I'm interested in the path the voltage takes during the integration, as this resistor is actually a thermistor, which means the resistance varies strongly during integration. That's why I'm sampling the voltage(sense) of the OHMF-DMM with another 3458A. I'm using both OCOMP and AZERO for better quality resistance measurements. For the sake of synchronization, the second (sampling) DMM is triggered by the aperture waveform of the first, so that the voltage measurements coïncide with the resistance measurement. I noticed per resistance reading, I get four readings from the sampling DMM (MEM FIFO) which are probably the actual value (used to eventually calculate the resistance), the OCOMP reading and some AZERO's. The problem is, I don't know which is which. The manual contradicts itself saying the OCOMP without source current is done prior to the measurement (page 209), while it says on page 62 that the genuine reading (current on) is done first, and followed by an OCOMP (cuurent off). My measurement data reveals first two samples with current on, followed by two samples without current. Does anyone know the actual order of the measurement sequence? My best guess so far is the genuine reading first, followed by an AZERO (current on with both), then the OCOMP with yet another AZERO (current off). Edited by: vcshamoen on Nov 27, 2013 8:58 AM 0
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答案对人有帮助，有参考价值 0 您好，要回答您的问题，了解您的测量目标非常重要。目标是？：1。减少热敏电阻的自热2.提高测量速度为了减少热敏电阻的自热，建议使用自动量程进行标称测量，然后选择并使用下一个更高的量程（即上行）。通过向上范围，自加热减少了100倍（I ^ 2 ^ R）。较高范围（读数百分比+范围百分比）存在较大误差，但必须考虑相对于自热减少的误差。如果目标是提高测量速度，那么减小孔径时间将提高读取速度。使用第二个3458A进行采样不会提高测量性能，因为在同步期间无法考虑内部延迟。建议使用单个3458A进行测量，同时根据需要设置范围或孔径时间。关于每个电阻测量确定的四个测量，测量及其发生的顺序是：离子： 1.HI_sense到LO 2. LO_sense到LO Vmeas _Ion = HI_sense到LO - LO_sense到LO * Ioff：* 3。 HI_sense到LO 4. LO_sense到LO Vmeas _Ioff = HI_sense到LO - LO_sense到LO读取= Vmeas_Ion - Vmeas_Ioff 以上来自于谷歌翻译以下为原文 Hello, To answer your question it is important to understand your measurement objective. Is the objective to?: 1. Reduce self-heating of the thermistor 2. Increase measurement speed To reduce self-heating of the thermistor, the recommendation is to make a nominal measurement using auto-range and then select and use the next higher range (i.e. up-range). By up-ranging, self-heating is reduced by a factor of 100 (I^2^R). There is greater error on the higher range (% of reading + % of range), but that must be considered relative to the reduction in self-heating. If the objective is to increase measurement speed, then reducing the aperture time will increase reading speed. Using a second 3458A for sampling will not improve measurement performance as internal delays cannot be accounted for during synchronization. The recommendation is to using a single 3458A to make and process the measurements, while setting the range or aperture time as appropriate. Regarding the four measurements identified per resistance measurement, the measurements and the sequence in which they occur are: Ion: 1. HI_sense to LO 2. LO_sense to LO Vmeas _Ion = HI_sense to LO - LO_sense to LO Ioff: 3. HI_sense to LO 4. LO_sense to LO Vmeas _Ioff = HI_sense to LO - LO_sense to LO Reading = Vmeas_Ion – Vmeas_Ioff

2019-3-8 06:27:02 评论举报薄坤坤