732A output impedance - Page 1
Forum section: electronics / projects

author: kkayser
author_rank: Contributor
timestamp: April 20, 2016, 06:33:13 pm
content: I was getting impossible results when connecting my 732A and 732B to my 720 Kelvin Varley. Upon further investigation, I measured the output impedance of the 732A. The result was 200 milliohms. The spec is 5 milliohms. Has anyone ever heard of this?I periodically compare the 732A with the 732B using the 3458A as a transfer standard. All of these comparisons are well with the specs for the meter and the standards.

author: barry14
author_rank: Regular Contributor
timestamp: April 21, 2016, 12:01:19 am
content: Note that the output impedance is only 5 milliohms for the 10 volt output. How did you measure the impedance?

author: kkayser
author_rank: Contributor
timestamp: April 21, 2016, 12:46:11 am
content: I know that the lower voltages have much higher output resistance. I measured the 10 v output R by loading with 10k and 5k. With each load I measured the output voltage and compared it to the open circuit output voltage. I measure the voltages with a 3458A set to max input R. I measure each voltage with a 5 minute mean at nplc = 50. All 3 measurements are done within 20 minutes. Typical Standard deviations with this method are .250 microvolts or .025PPM.

author: zlymex
author_rank: Frequent Contributor
timestamp: April 21, 2016, 02:07:50 am
content: Where did you connect the voltage sense point on 732A?And where did you connect the resistor pins?

author: kkayser
author_rank: Contributor
timestamp: April 21, 2016, 02:10:53 pm
content: I'm not sure I understand your questions. The 10 volt output of the 732A and the load resistor and the 3458A meter are connected in parallel. The load resistor is an RSI RS925A decade box..

author: zlymex
author_rank: Frequent Contributor
timestamp: April 21, 2016, 02:55:10 pm
content: Quote from: kkayser on April 21, 2016, 02:10:53 pmAt what point those three things meet together?Do you use a pair of test leads for RS925A? Where did you plug the other end of the test leads?

author: TimFox
author_rank: Super Contributor
timestamp: April 21, 2016, 04:34:01 pm
content: If the output resistance is a real problem, you could connect an external semi-precision power supply through an appropriate resistor to the K-V divider input and adjust the external supply for a null reading at the terminals against your 732A output. That would greatly decrease the loading on the 732A.

author: kkayser
author_rank: Contributor
timestamp: April 21, 2016, 08:00:47 pm
content: Quote from: zlymex on April 21, 2016, 02:55:10 pmThe 732a is connected to the 3458A with a Pomona two wire shielded lead. The shield is grounded at the 732A. The load resistance is connected with the same type of lead piggy-backed into the lead on the 3458A. This shield is grounded at the resistance decade.Correction: I made an error when I calculated the output R. Output R is very close to 80 milliohms. And does not vary much with changes in load.

author: kkayser
author_rank: Contributor
timestamp: April 21, 2016, 08:56:23 pm
content: Quote from: TimFox on April 21, 2016, 04:34:01 pmThis type of measurement has problems, to say the least. You must have a power supply with noise and drift under 1 microvolt at 10 volts nominal. Furthermore it must have a resolution of at least 1PPM. I find these circuits pick up a lot of noise. I have not found a way to use this type of circuit in my lab.

author: zlymex
author_rank: Frequent Contributor
timestamp: April 22, 2016, 05:58:44 am
content: Quote from: kkayser on April 21, 2016, 08:00:47 pmThat could be the problem. I still don't have a clear picture of how you connect the load. If the load is piggy-backed into the meter side socket, them you actually measure the leads resistance from 3458A to 732A. If however your load is piggy-backed into the 732A side socket, then it is most likely that 732A is faulty, or the piggy-back has a larger than normal resistance.

author: TimFox
author_rank: Super Contributor
timestamp: April 30, 2016, 09:30:16 pm
content: Quote from: kkayser on April 21, 2016, 08:56:23 pmThis suggestion is not so critical as you say, but does require a decent power supply and maybe a G-R resistor box. The idea is to first adjust either the supply or the resistor for null against the 732A output, then remove the null meter and leave the power supply (maybe 20 V or so) and resistor (100 k for a 100 k K-V) connected, along with the 732A that will determine the 10 V into the K-V divider. Any error or drift in the current from the supply through the resistor will react with the small-ish output resistance of the 732, not the high-ish resistance of the K-V.With round numbers, if the 732A has an output resistance of 0.2 ohms, and if the external power supply is set to 20 V through a 100 kohm resistor to the 100 kohm input of the K-V divider, then to get an error of 1 microvolt (out of 10 V at the K-V input) would require an error (or drift) of 5 microamps through the external resistor, which would be 5% of the 100 microamp current of 10 V across 100 kohms, where 5 microamps X 0.2 ohms = 1 microvolt.

author: kkayser
author_rank: Contributor
timestamp: May 09, 2016, 04:50:00 pm
content: Quote from: zlymex on April 22, 2016, 05:58:44 amYou are correct; the problem is lead resistance. When we get to .1 PPM, little things mean a lot. The Pomona leads I was using are 90 milliohms, round trip, including the shorting bar. 90 mohm lead resistance adds to the 5 mohm 732B output resistance. This amount of output resistance is of minor concern with 10 megs or more of input resistance. But is, indeed, significant with 100K input resistance of the Kelvin Varley.The best solution is to connect the 3458A directly to the Kelvin Varley, not the 732B. an alternative solution is to use low resistance leads. I have some that are 4 feet long and have a 4 wire resistance of 8 mohms each, 8 mohms round trip. I expect a pair of 2 footers will have a total resistance of about 5 mohm.