Posted Thu, 23 Feb 2023 07:27:31 GMT by Li, Jeremy

Hello all,
according to Volume and Surface Resistivity Measurements of Insulating Materials Using the 6517B Electrometer/High Resistance Meter | Tektronix
 Icalc is a weighted average of the latest four current measurements, each at the end of a separate alternation.

Could you help to advise how to define the Weight of the four current measurement.  is it 0.25 for each current measurement or other?

And  I don't quite understand about below parameter for Standard method resistivity tests

PRE-DISCH: The predischarge time in seconds. 

BIAS-TIME: The bias time in seconds. 

MEAS-V: The measurement voltage. This is typically set to the same level as BIAS V. 
MEAS-TIME: The measurement time in seconds. 

what is BIAS-TIME, and what is the difference between predischarge time and measurement time

Posted Wed, 17 May 2023 05:20:31 GMT by Olsson, Thomas
Did you ever get a response?<br> <br> The manual says this about the standard method:<br> When this test is run, the V-source is initially set to source 0 V for a specified time (PREDISCH time) to allow any charge to dissipate. The V-source then applies a specified voltage (BIAS V) to the electrodes of the test fixture for a specified time (BIAS-TIME). This “bias” period allows currents in the test circuit to stabilize. The V-source then applies the test voltage (MEAS-V) and, after a specified delay (MEAS-TIME), the Model 6517B measures the resistivity of the sample and stores the reading in the buffer. Note that the test voltage (MEAS-V) is typically at the same level as the bias voltage (BIAS V).<br> <br> (I should maybe make my own post but thought I might have more of a chance to get an answer from you than from Tektronix since you have not had a reply in the forum for three months)<br> <br> What is hard to understand here is the difference between BIAS-TIME and MEAS-TIME, especially if the BIAS-V and MEAS-V are the same. The &quot;General measurement procedure&quot; for resistivity says that the bias time IS the&#160;electrification period, but then the measurement time must be zero, right? So if I need to electrify the sample for 60 seconds, do I set BIAS-TIME or MEAS-TIME to 60? Or do I set BIAS-TIME to 0 and MEAS-V to 60 if I have the same BIAS-V and MEAS-V? The standard method conforms to ASTM D257, but I cannot see anything there about the distinction between a bias voltage and measure voltage, so the use-case for this is unclear to me.<br> <br> &#160;
Posted Mon, 05 Jun 2023 19:33:00 GMT by McKinney, Ty
Hello,&#160;<br> <span style="font-size:11pt;"><span style="font-family:Calibri,sans-serif;">The calculation for Icalc is detailed in this white paper: <a href="" style="color:#0563c1;text-decoration:underline;"></a>. It uses second order binomial coefficients for 4 measurements so 1(I1) + 3(I2) + 3(I3) + 1(I4) all divided by 8. And the polarities of the current measurements are made to be all positive. </span></span><br> <span style="font-size:11pt;"><span style="font-family:Calibri,sans-serif;">The predischarge time is the time that 0V is applied to the sample before measuring to allow charges to dissipate.<br> The bias time is the time that the source voltage is applied before the measurement is taken, which would be for settling purposes.<br> The measurement time is the speed of the measurement aka NPLC. So say you have 10s predischarge, 60s bias and 1 NPLC, the sample would have 0 V for 10 s, then the source voltage for 60s then the source voltage would continue to be applied while the box takes 1 measurement at 1 NPLC- so 16.67 ms. </span></span>
Posted Thu, 29 Jun 2023 07:39:57 GMT by Li, Jeremy

Hi Ty,
Thanks for the reply.
As describe in the document:

Note that not all of these currents are of the polarity that might be expected, given the polarity of the voltages applied (+, –, +, –).

Is it possible that  the current to be a negagive value according to I calc=(I1+3(–I2) +3(I3) +(–I4))/8,  and then how should we do with it?

Posted Fri, 30 Jun 2023 07:27:08 GMT by Li, Jeremy
I tried to measured a 1M Ohm resistor, the voltage is +-50V, 15S, and the current is 50uA, -50uA,50uA, -50uA,50uA, -50uA,50uA, -50uA,50uA, -50uA,50uA, -50uA...... so according to &amp;#160;I calc=(I1+3(–I2) +3(I3) +(–I4))/8, the current&amp;#160; Icalc is 50uA or -50uA.&lt;br&gt; &lt;br&gt; sorry, overlook the information in your reply above. this seems to be ok for me to do the calculation.&lt;br&gt; &lt;code&gt;&lt;em&gt;1(I1) + 3(I2) + 3(I3) + 1(I4) all divided by 8. And the polarities of the current measurements are made to be all positive.&lt;/em&gt;&lt;/code&gt;<br> but it seems a bit different from the white pater

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