What don't you mean besides the typo?
Some rocket science going on here Besides the fact my question was directed to PW,what i was wanting to know,is why would the 50ohm resistor within the FG,show a more accurate P/in value than an external CVR attached to the DUT?. The 50 ohm resistor within the FG ,would be just as susceptible to inductance,phase shift,and the like's,just as much as an external CVR. What you would be doing in this case,is also adding the voltage drop from the FG lead's into the equation. So in the end,it would actually show a more incorrect measurement than that of the way we have been doing it. Brad
« Last Edit: 2017-05-09, 10:24:16 by TinMan »
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