AC Line Impedance Measreument (Extech CT70)

[bgelb0]

Hello -

I am using an Extech CT70 outlet / circuit load tester to investigate some missing/poor grounding in my house.

The Extech device provides an impedance measurement (in ohms) for each of the hot, neutral and ground contacts of an outlet. I am trying to figure out what it actually measures and how the measurement is achieved.

manual: www.extech.com/resources/CT70_UM.pdf (see pg 8)


I'm imagining the impedance measurement should correspond to the resistance of the branch circuit wiring, however I don't really understand how its possible to come up with an isolated measurement for each of the ground, neutral and hot conductors. For example, the device could determine an equivalent resistance based on Vdrop measurements with load applied to hot/neutral, but I don't know how it would determine how much drop occurred on one conductor vs. the other.

The documentation didn't seem to offer any clues.

Any further insight from this forum on what exactly the device is supposed to be measuring and how it is producing the measurement?

 

[gar]

190102-2009 EST

bgelb0:

I quickly read the .pdf, and as an instrument it looks impressive.

It seems to imply that measurements are made at 15 A or thereabouts and calculated for others. Just looking at the device it is obvious that any measurement at that actual current value will have a dissipation of 1800 W (15 A at 120 V). If this is done for only one full cycle the average power over 1 second is 1800/60 = 30 W. That is too high for an instrument of that size. Even at a 10 second rep rate it is 3 W.

As an electrical engineer you should run experiments on the device to see what it is actually doing. You should be able to guess at what is done to get some of the results. Consider what you would do to design such an instrument.

.

 

[mivey]

Gar is trying to make you think rather than spoon-feed but if you want to cheat a little just read some of gar's past posts on circuit tests.

This device is very similar to the Ideal Suretest circuit analyzer.

Think about what info you would get with an unloaded receptacle followed by a real quick loading of the receptacle to 20 amps. Think about Ohm's Law.

If you can't get anywhere with that, ask.

 

[mivey]

To cut to the chase:

From the Ideal Sure Test manual (I added clarifications):

Steps, Load, Measurement, Current:
1, No-Load, V= No load Voltage H-N, N/A
2, Load H-N, V1= Voltage H-N, I1
3, Load H-N, V2= Voltage H-G, I2
4, Load H-G, V3= Voltage H-N, I3
5, Load H-G, V4= Voltage H-G, I3

Note steps 4 & 5 are the same but has two voltage readings and one current.


Calculated Impedance:
rZ* = (V-V1)/I1 = Total branch circuit Impedance
rN = (V2-V1)/I2 = Impedance of neutral conductor
rG = (V3-V4)/I3 = Impedance of the Ground wire
rH* = (rZ-rN)* = Impedance of the Hot or High conductor


*If the method used to measure current employs a shunt resistor the internal shunt resistance of the meter should be taken into account in your calculations. On most DMM’s this shunt resistance is around 0.1 ohm.

 

[mivey]

To cut to the chase:

From the Ideal Sure Test manual (I added clarifications):

Steps, Load, Measurement, Current:
1, No-Load, V= No load Voltage H-N, N/A
2, Load H-N, V1= Voltage H-N, I1
3, Load H-N, V2= Voltage H-G, I2
4, Load H-G, V3= Voltage H-N, I3
5, Load H-G, V4= Voltage H-G, I3

Note steps 4 & 5 are the same but has two voltage readings and one current.


Calculated Impedance:
rZ* = (V-V1)/I1 = Total branch circuit Impedance
rN = (V2-V1)/I2 = Impedance of neutral conductor
rG = (V3-V4)/I3 = Impedance of the Ground wire
rH* = (rZ-rN)* = Impedance of the Hot or High conductor


*If the method used to measure current employs a shunt resistor the internal shunt resistance of the meter should be taken into account in your calculations. On most DMM’s this shunt resistance is around 0.1 ohm.

Looking again, I1 and I2 would also be the same as we have just one load application. Wasn't looking too close since I basically cut and pasted with a little clean up.

 

[bgelb0]

Mivey, thanks for digging up this information and answering my question.

The main thing I was wondering about is whether ground was required (as a reference during H-N load test) in order to establish the H/N impedance values. I guess this means that the measurement would not be possible with an open ground.

I was thinking that with the quality of the ground connection unknown (as in my old house) I was unsure about relying on it to establish the other values. But actually I guess as long as there's no load being applied to it, even an improper/poor ground can still serve as a voltage reference to establish H-G and N-G voltage differences during the H-N load application (and therefore be able to separate the resistances of the H and N conductors), which is made clear by your set of equations.

Thanks,
Ben

To cut to the chase:

From the Ideal Sure Test manual (I added clarifications):

Steps, Load, Measurement, Current:
1, No-Load, V= No load Voltage H-N, N/A
2, Load H-N, V1= Voltage H-N, I1
3, Load H-N, V2= Voltage H-G, I2
4, Load H-G, V3= Voltage H-N, I3
5, Load H-G, V4= Voltage H-G, I3

Note steps 4 & 5 are the same but has two voltage readings and one current.


Calculated Impedance:
rZ* = (V-V1)/I1 = Total branch circuit Impedance
rN = (V2-V1)/I2 = Impedance of neutral conductor
rG = (V3-V4)/I3 = Impedance of the Ground wire
rH* = (rZ-rN)* = Impedance of the Hot or High conductor


*If the method used to measure current employs a shunt resistor the internal shunt resistance of the meter should be taken into account in your calculations. On most DMM’s this shunt resistance is around 0.1 ohm.

 

[mivey]

Mivey, thanks for digging up this information and answering my question.

The main thing I was wondering about is whether ground was required (as a reference during H-N load test) in order to establish the H/N impedance values. I guess this means that the measurement would not be possible with an open ground.

I was thinking that with the quality of the ground connection unknown (as in my old house) I was unsure about relying on it to establish the other values. But actually I guess as long as there's no load being applied to it, even an improper/poor ground can still serve as a voltage reference to establish H-G and N-G voltage differences during the H-N load application (and therefore be able to separate the resistances of the H and N conductors), which is made clear by your set of equations.

Thanks,
Ben

You could cheat and run a temporary ground wire back to the panel and just run steps 1, 2, and 3 since you would not care about the ground impedance. I'm thinking you want the reference ground bonded at the panel neutral since for this purpose it is just a very long meter lead.

Some amorphous/poor ground might not be stable enough. It could even (we would hope not) be part of the load path.

You will find some of this long reference lead stuff in some of gar's old posts. He is a great knowledge base.