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I love a mystery

By Chuck Newcombe

Several years ago, I received an email from one of Fluke's offices in another country asking me to answer a customer's question about the use of a Fluke DMM, if I could. Always one to take on a challenge, I immediately dove into the information given.

It seems that a man had inherited the safety management position at a large industrial plant. As he was reviewing the procedures in place, he came upon one that required two dc voltage measurements to be taken at a specific location using a Fluke digital multimeter, the first with the DMM alone, and the second with a one megohm resistor placed across the meter's input. When the new manager did as instructed, he recorded 290 V with the DMM alone and 230 V with the resistor added. And, while the readings were within the tolerances specified in his document, it apparently wasn't clear to him just what safety parameter was being tested. The new manager asked if someone at Fluke could explain the possible reasons for the two readings, since the meter specified was a Fluke.

I immediately recognized the technique as one used to determine the source resistance of a measured voltage. It can be important, because if the resistance is too high it will affect the accuracy of the measurement.

In theory, you measure the open circuit voltage followed by the short circuit current. Then, using Ohm's Law, you can easily determine the series resistance present. But, while the measurement of the open circuit voltage is ok, you might be unpleasantly surprised at the circuit's reaction to a short circuit, particularly if the voltage is high and the series resistance is low, in which case this method is not recommended. A safer and more reasonable approach is to take two voltage readings using two different, but fairly high, meter input resistances as the circuit load.

How does it work? Well, let's start with what we know.

  1. The DMM input resistance is10 Megohms.
  2. The measured voltage using the meter alone is 290 V.
  3. Using Ohm's Law, the measurement circuit current is therefore 29 microamps.
  4. Using the formula for parallel resistors, with the 1 and 10 Megohm resistors in parallel, the equivalent meter input resistance is about 0.9091 Megohm.
  5. The measured voltage using the parallel combination is 230 V.
  6. Again using Ohm's Law, the circuit current is about 253 microamps.

To summarize, we now have two different load resistances, two different measured voltages and two calculated currents, and we're trying to determine some unknowns, the source resistance, and the original unloaded voltage present before we made our measurements (we'll assume the two unknowns didn't change over the time period involved.)

Using the readings given, I quickly (ok, it took awhile, I was a little rusty with my substitution algebra) determined that the source resistance was about 268 kohms, and the open circuit voltage was nearly 298 V for the case given.

So where was the measurement being taken? Apparently, it was on an insulated catwalk next to a series of "pots" in an aluminum smelter line. In the smelter in question there were about 200 pots, electrically connected in series. A dc current in excess of 100 kA flows through the pots, with a drop of about 5 V in each pot, for a total voltage from end to end of up to 1000 V! The measurements were being made to test the insulation of the catwalk from the pots, where the voltage present might range from a few volts to the maximum, depending on the technician's location.

The neat thing about the method used is that the resistance measurement depends only on the ratio between the two voltages, in our case 230/290, or 0.79. So readings of 10 V and 7.9 V, or 200 V and 178 V, or even 1000 V and 790 V elsewhere on the line would represent about the same resistance. The graph in figure 1 will give you an idea of the range of resistances that can be dealt with using this method.

While this method is effective, I feel obligated to offer a few cautions regarding its use.

First, adding external parallel resistances to your meter must be done with great care, especially when voltages greater than 30V are present.

  • The resistor must be rated for the voltage, and the power that it might have to dissipate. (1 megohm at 1000 volts will dissipate 1 watt.)
  • Shrouded test lead integrity should be maintained through the use of the appropriate banana plugs and jacks in the assembly of the paralleling accessory.

Second, the resolution of the meter limits the accuracy when the differences are small.

Oh yeah – about the open circuit voltage. It's a simple matter to compute that once you know the series resistance. Just use the current calculated for the highest voltage measurement, along with the resistance, to determine how much voltage drop occurred across the resistor. Add that onto the voltage reading to determine the voltage that was present before you connected the meter.

Got an idea or question for Chuck? Write in and let us know.

Chuck Newcombe has recently retired after a long career with Fluke during which he conducted extensive market research to aid in product definition and new product design. He particularly enjoyed "job shadowing" technicians in many different disciplines, to learn about new processes and the problems maintaining them. His experience ranges from automated test system deployment and applications software development, to production test system troubleshooting.

He has presented countless seminars on measurement methods and applications and has over ten years experience on the subject of power quality. He currently consults on measurement methods, including power quality, and continues to present seminars focusing on the effects of power harmonics. He's also guilty of TV and radio repair, automotive electrical system rebuilds, and some really off the wall stuff.