ptonsparky
Tom
- Occupation
- EC - retired
Well, that didn't work....
Which one is correct?
- Thread starter ptonsparky
- Start date
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ptonsparky
Tom
- Occupation
- EC - retired
oldsparky52
Senior Member
- Location
- Wilmington, NC USA
Why can't they all be correct?
ptonsparky
Tom
- Occupation
- EC - retired
ptonsparky
Tom
- Occupation
- EC - retired
ptonsparky
Tom
- Occupation
- EC - retired
Just to check, I’d like to see the readings with fresh batteries.
we see odd readings on ours with weak batteries. I know the dummy icon isn’t on to tell you the battery is low.
neither are ours....
when we get odd readings, we change batteries no matter what icon is showing.
we see odd readings on ours with weak batteries. I know the dummy icon isn’t on to tell you the battery is low.
neither are ours....
when we get odd readings, we change batteries no matter what icon is showing.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
191107-2436 EST
ptonsparky:
I really don't understand your experiments and results.
I need one experiment at a time, and a precise definition of the conditions.
If AC measurements are made along a low resistance conductor that has a high current flow, then the measuring loop may see a substantial induced voltage from that current that may produce a large error in the intended measurement. A foot or so of copper water pipe with 10 A flowing thru it won't show much voltage drop. To get a 5 mV drop at 10 A requires the resistance to be 5/10,000 ohm. #10 copper wire is 1 ohm per 1000 ft, or 1/1000 ohm per foot. I estimate that 1/2" copper water pipe is about 1/10 that of #10 copper wire. Or the water pipe would be about 1/10,000 ohms for 1 foot.
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ptonsparky:
I really don't understand your experiments and results.
I need one experiment at a time, and a precise definition of the conditions.
If AC measurements are made along a low resistance conductor that has a high current flow, then the measuring loop may see a substantial induced voltage from that current that may produce a large error in the intended measurement. A foot or so of copper water pipe with 10 A flowing thru it won't show much voltage drop. To get a 5 mV drop at 10 A requires the resistance to be 5/10,000 ohm. #10 copper wire is 1 ohm per 1000 ft, or 1/1000 ohm per foot. I estimate that 1/2" copper water pipe is about 1/10 that of #10 copper wire. Or the water pipe would be about 1/10,000 ohms for 1 foot.
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ptonsparky
Tom
- Occupation
- EC - retired
Been there. It is the reason I have two of the 87s. I thought the first had gone toes up before I noticed the battery symbol.
The 87s I checked to see if the batteries were up and the other two I knew were within the last couple months.
ptonsparky
Tom
- Occupation
- EC - retired
The first picture shows M2 to M4 all connected to a grounded/bonded tank of the water heater. Each shows relatively the same voltage. M1 shows the voltage between the floating storage tank and the WH tank.
Each subsequent single meter test shows the voltage to the floating ST, with M1 showing what the voltage between tanks is raised to as each meter is tested.
later...
Each subsequent single meter test shows the voltage to the floating ST, with M1 showing what the voltage between tanks is raised to as each meter is tested.
later...
ptonsparky
Tom
- Occupation
- EC - retired
Using a wiggy type tester between the power and the floating ST would probably have shown nothing but the ST to HWT voltage would have gone up to the 124. I don't have one that I use.
Placing two of the T+ in parallel dropped the voltage reading to about 12 volts. Not shown.
The meters all showed the correct voltage in the context that they were used, but voltage to a floating piece of cu pipe, or anything else is meaningless.
Placing two of the T+ in parallel dropped the voltage reading to about 12 volts. Not shown.
The meters all showed the correct voltage in the context that they were used, but voltage to a floating piece of cu pipe, or anything else is meaningless.
PaulMmn
Senior Member
- Location
- Union, KY, USA
- Occupation
- EIT - Engineer in Training, Lafayette College
Do a search for >voltmeter ohms per volt<. The impedance of the meter makes a difference! I remember one of my professors mentioning that 1970s era voltmeters were usually 20,000 ohms/volt. He added that earlier voltmeters had a lot less resistance (impedance?), and that some circuit diagrams for electronic things had notes that 'volt readings are with 1000 ohms/volt meters' or some such, as the quote, below, explains.
Quoting from AntiqueRadios.com (https://www.antiqueradios.com/forums/viewtopic.php?t=129660):
Quoting from AntiqueRadios.com (https://www.antiqueradios.com/forums/viewtopic.php?t=129660):
1000 ohms per volt, if you do the math (Ohm's Law, E/R, 1/1000) is equivalent to 1mA. 1mA was the standard range for a panel meter in the days before alnico magnets, so all the radio VOMs were 1000 ohms per volt, and the schematics and voltage tables were written with that assumption.
In the late 30s, when alnico became available, 50µA meters were possible, which translates to 20,000 ohms per volt.
If you have a meter that draws 50 microamps, but want it to draw 1 milliamp, you just shunt it with a suitable resistor.
In the late 30s, when alnico became available, 50µA meters were possible, which translates to 20,000 ohms per volt.
If you have a meter that draws 50 microamps, but want it to draw 1 milliamp, you just shunt it with a suitable resistor.
ptonsparky
Tom
- Occupation
- EC - retired
The analog Sympson meters were nice. If the needle stayed in relatively the same portion of the meter as you changed scale you knew the voltage reading was bogus. Ghost Voltage is what the term is now. It changed the ohms/volt.
hbiss
EC, Westchester, New York NEC: 2014
- Location
- Hawthorne, New York NEC: 2014
- Occupation
- EC
ptonsparky
Tom
- Occupation
- EC - retired
I agree, it’s a mess...:ashamed1:
That and taking a voltage measurement to some floating object means nothing, other than the impedance differences stand out.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
191108-2433 EST
ptonsparky:
An experiment with a pipe (tube). Stainless steel about 7 ft long 3/4" OD. Wall thickness not as thin as copper tubing. Stainless has about 10 times more resistance than steel, and steel about 3 times that of copper.
This was just a quick experiment. One voltmeter lead was run thru the tube. The other voltmeter lead was clipped to the opposite end of the tube from where the internal lead was connected. This configuration produces close to the least magnetic coupling from the test current thru the stainless tube to the one turn loop of the meter test leads. In any measurement of voltage it is not possible to eliminate the measuring one turn loop. Thus, you try to make that loop have the least possible cross-sectional area.
With a 10 A 60 Hz sine wave test current thru the tube the voltage drop over the tube length was 4.9 mV.
With some external dangling meter leads between the same two points on the tube the voltage drop measured about 9.3 mV. This additional voltage was a result of the added error signal from a magnetically induced voltage in the more open 1 turn loop added to the resistive voltage drop.
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ptonsparky:
An experiment with a pipe (tube). Stainless steel about 7 ft long 3/4" OD. Wall thickness not as thin as copper tubing. Stainless has about 10 times more resistance than steel, and steel about 3 times that of copper.
This was just a quick experiment. One voltmeter lead was run thru the tube. The other voltmeter lead was clipped to the opposite end of the tube from where the internal lead was connected. This configuration produces close to the least magnetic coupling from the test current thru the stainless tube to the one turn loop of the meter test leads. In any measurement of voltage it is not possible to eliminate the measuring one turn loop. Thus, you try to make that loop have the least possible cross-sectional area.
With a 10 A 60 Hz sine wave test current thru the tube the voltage drop over the tube length was 4.9 mV.
With some external dangling meter leads between the same two points on the tube the voltage drop measured about 9.3 mV. This additional voltage was a result of the added error signal from a magnetically induced voltage in the more open 1 turn loop added to the resistive voltage drop.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
191109-1535 EST
A continuation from my above experiment.
A resistive voltage drop will have the measured voltage in perfect phase and waveform relationship with the current thru the resistance.
An induced voltage in a coil will have a 90 degree phase shift relative to the inducing current. This comes from e = K*N*dPhi/dt . Phi is the magnetic flux which is proportional to the inducing current, and the derivative of a sine wave is a negative cosine wave. Thus, the 90 degree shift in the induced voltage.
In this experiment we have two sine wave shapes of exactly the same frequency and some phase releationship. The sum of two sine waves is another sine wave of some phase angle relative to one of the originating sine waves as a reference.
When the measuring loop area is very small, then the induced voltage in the measuring loop is very small from varying magnetic fields coupling with the measuring loop.
In my previous experiment I first used a very small measuring loop by running one meter lead thru the tube. The second measurement was with a much larger loop by randomly running the measuring lead somewhere outside the tube. Also note that the current thru the tube was from an external randomly located wire, thus, no significant cancelation of magnetic fields from the 10 A current in the region of interest.
Today I reran the experiments with a scope as the measuring instrument and synchronized from the current thru the tube. As expected in the first experiment the measured voltage was in phase with the tube current, and in the second there was a phase shift in the measured voltage, and its amplitude was greater. This provides verification of why the second experiment had a larger voltage.
None of these experiments had any problem from loading of the source by the instrumentation because of the very low impedance of the source being measured.
When making measurements you have to understand your instruments, and the characteristics of what is being measured.
.
A continuation from my above experiment.
A resistive voltage drop will have the measured voltage in perfect phase and waveform relationship with the current thru the resistance.
An induced voltage in a coil will have a 90 degree phase shift relative to the inducing current. This comes from e = K*N*dPhi/dt . Phi is the magnetic flux which is proportional to the inducing current, and the derivative of a sine wave is a negative cosine wave. Thus, the 90 degree shift in the induced voltage.
In this experiment we have two sine wave shapes of exactly the same frequency and some phase releationship. The sum of two sine waves is another sine wave of some phase angle relative to one of the originating sine waves as a reference.
When the measuring loop area is very small, then the induced voltage in the measuring loop is very small from varying magnetic fields coupling with the measuring loop.
In my previous experiment I first used a very small measuring loop by running one meter lead thru the tube. The second measurement was with a much larger loop by randomly running the measuring lead somewhere outside the tube. Also note that the current thru the tube was from an external randomly located wire, thus, no significant cancelation of magnetic fields from the 10 A current in the region of interest.
Today I reran the experiments with a scope as the measuring instrument and synchronized from the current thru the tube. As expected in the first experiment the measured voltage was in phase with the tube current, and in the second there was a phase shift in the measured voltage, and its amplitude was greater. This provides verification of why the second experiment had a larger voltage.
None of these experiments had any problem from loading of the source by the instrumentation because of the very low impedance of the source being measured.
When making measurements you have to understand your instruments, and the characteristics of what is being measured.
.
synchro
Senior Member
- Location
- Chicago, IL
- Occupation
- EE
Another advantage provided by the approach gar just described for measuring resistances or impedances is that he's using a 4-terminal or Kelvin method of measurement. This has one pair of wires for applying the test current, and a different pair of wires for measuring the voltage that is developed across the conductor being tested by this applied current. This essentially eliminates the effect of the test lead impedance from the measurement, making it much more accurate when the test lead impedance is not negligible when compared to that of the conductor being measured. Or in other words, when very low resistances or impedances are to be accurately measured.
https://en.wikipedia.org/wiki/Four-terminal_sensing
https://en.wikipedia.org/wiki/Four-terminal_sensing
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
:thumbsup:
Another way of saying "they changed the ohms/volt" is to say that for a high impedance source like capacitive coupling the current through the meter would be almost independent of the meter resistance. And what you were measuring on the analog meter was really the current through the meter with a series resistor So same meter needle position corresponds to same current through the meter movement.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
191109-2025 EST
Original Simpson 260s were 20,000 ohms/volt on DC and 1000 ohms/volt on AC. I believe the rectifier was copper-oxide. These would be 1940s into 1950s. I don't remember when the change to 5000 ohms/volt for AC occurred, but was probably mid to late 1950s. I have an about 1946 or 1947 260. Then late 50s and early 60s I bought several 270s.
In 1952 I bought a Simpson VTVM that was packaged in the 260 case. 10 megohm input impedance. Still works. So does the 1940s 260 still work.
Ohms/volt has to be understood as to what it means. Such a voltmeter is based upon an analog micro or milliampere meter movement. In the old days 1 mA full scale, and starting in the 1940s 50 microamperes full scale. The ohms/volt rating is a measure of the required series resistance to obtain the meter full scale. 250 V full scale at 20,000 ohms/volt requires a 5 megohm series resistance. On this range almost as high an internal impedance as a run of the mill DVM.
.
Original Simpson 260s were 20,000 ohms/volt on DC and 1000 ohms/volt on AC. I believe the rectifier was copper-oxide. These would be 1940s into 1950s. I don't remember when the change to 5000 ohms/volt for AC occurred, but was probably mid to late 1950s. I have an about 1946 or 1947 260. Then late 50s and early 60s I bought several 270s.
In 1952 I bought a Simpson VTVM that was packaged in the 260 case. 10 megohm input impedance. Still works. So does the 1940s 260 still work.
Ohms/volt has to be understood as to what it means. Such a voltmeter is based upon an analog micro or milliampere meter movement. In the old days 1 mA full scale, and starting in the 1940s 50 microamperes full scale. The ohms/volt rating is a measure of the required series resistance to obtain the meter full scale. 250 V full scale at 20,000 ohms/volt requires a 5 megohm series resistance. On this range almost as high an internal impedance as a run of the mill DVM.
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