# Metering a Single Phase 208 VAC Circuit

aemla
TL;DR Summary
How would one measure (with a meter) active power of a single phase 208 VAC circuit where L1 and L2 are the only wires?
Good day,

I hope to understand how to use a 3 phase energy meter to measure single phase 208 VAC circuit.
The basic idea seems simple, yet, complex meters don't seem to have this ability.

Would anyone be able to describe what kind of information one would need from a digital 3 phase meter to calculate/determine active power of single phase 208 VAC circuit?

Upstream power transformer: Delta to Wye 120/208 VAC

Per my understanding, the most complex piece of data I would need is the phase angle between resultant voltage wave (between L1-L2) and current.

Most industrial meters provide phase angles for individual phases and phase angles between different voltage hot legs. Literature also seems confusing since two different angles are referred to as the phase angle. Some meter documents call the phase angle between voltage and current as the "displacement angle" and phase angle between different voltage hot legs as the "phase angle".

Any sample calculation or a link to a meter that has been proven to work would be greatly appreciated.
Thank you!

Mentor
If you only have two wires, you only have one phase and no angles are involved. What are you using for your voltage and current measurement instruments? Are you using a shunt or CT to measure the current? What range of currents do you need to measure, and what accuracy do you want for the power measurement?

aemla
If you only have two wires, you only have one phase and no angles are involved. What are you using for your voltage and current measurement instruments? Are you using a shunt or CT to measure the current? What range of currents do you need to measure, and what accuracy do you want for the power measurement?
Good morning,
The angles I'm referring to is the phase shift between voltage and current waves of a single phase circuit. This can be used to calculate the power factor, thus, real power. Current and voltage measurements will be used for power billing. I'm still researching for the required accuracy, so far, it seems to be +/- 2%.
Thank you.

Staff Emeritus
You have it right, the key is the phase difference between voltage and current. If phase zero, then you have unity power factor.

But in post #1, you said you were just trying to understand, but in #3 you mention doing this for billing purposes. The usual way of doing accurate watt-hour readings for billing is using a watthour meter rather than voltage and current measurements (counter intuitive, I know.) If someone presented me a bill based on voltage and current, my reaction would be "I don't believe it." The operating principle of the watthour meter is complicated, but the point is that they can be accurately calibrated.

https://en.wikipedia.org/wiki/Electricity_meter

The most common type of electricity meter is the electromechanical watt-hour meter.

On a single-phase AC supply, the electromechanical induction meter operates through electromagnetic induction by counting the revolutions of a non-magnetic, but electrically conductive, metal disc which is made to rotate at a speed proportional to the power passing through the meter. The number of revolutions is thus proportional to the energy usage. The voltage coil consumes a small and relatively constant amount of power, typically around 2 watts which is not registered on the meter. The current coil similarly consumes a small amount of power in proportion to the square of the current flowing through it, typically up to a couple of watts at full load, which is registered on the meter.

The disc is acted upon by two sets of induction coils, which form, in effect, a two phase linear induction motor. One coil is connected in such a way that it produces a magnetic flux in proportion to the voltage and the other produces a magnetic flux in proportion to the current. The field of the voltage coil is delayed by 90 degrees, due to the coil's inductive nature, and calibrated using a lag coil. This produces eddy currents in the disc and the effect is such that a force is exerted on the disc in proportion to the product of the instantaneous current and instantaneous voltage. A permanent magnet acts as an eddy current brake, exerting an opposing force proportional to the speed of rotation of the disc. The equilibrium between these two opposing forces results in the disc rotating at a speed proportional to the power or rate of energy usage. The disc drives a register mechanism which counts revolutions, much like the odometer in a car, in order to render a measurement of the total energy used.

• • jim mcnamara, sophiecentaur, DaveE and 2 others
Mentor
What is the make and model of the meter? Even an intended 3-phase meter might have diagrams in the instruction manual for how to wire up various single-phase applications.

aemla
Thank you for the replies.
The main priority is to have a good understanding of how such wiring configuration can be metered. Eventually, I will be applying this knowledge.

The final bill can be presented in terms of real power. I believe as long as calculated real power is within the accuracy standard (may not even be possible), it would be okay to use for billing.

I read up on the electromechanical meters you've mentioned, are they still being used for new installations? I would think having only a local display would be very inconvenient. Even with these meters, I still can't find something that measures L1-L2 without neutral.

Meter I'm Using:
AI Energy Meter 480VAC ST (6ES7134-6PA20-0BD0)

Just like with electromechanical meters, it's very difficult to find digital meters capable of measuring single phase 208 VAC. Digital meters provide much more data than electromechanical meters, however, I yet to see any meter capable of measuring phase angle between UL1-L2 voltage wave and current wave.

Staff Emeritus
If you're into digital meters, then rather than measure phases of V and I it would be better to measure real power P and volt*amps VA. You can then calculate the phase angle and Q from P and VA. The reason is that measuring or inferring phase angle directly provides a signal with a lot of noise.

I don't know about the available voltage ranges.

• DaveE
aemla
Thanks anorlunda,
Yes, I agree, measuring real power/energy would be simpler and is the end result that's needed anyway.
Unfortunately, you can't do that with traditional/standard meters since the "total active power" will be wrong. Standard meters measure active power of each phase with respect to neutral and then sum everything up. I just found a meter capable of actually measuring L-L... this was extremely hard to find, partly because I didn't understand what I needed.
Question, however, still remains (although slightly altered) if active (L-L) power can be calculated from individual phase measurements?

Staff Emeritus
Question, however, still remains (although slightly altered) if active (L-L) power can be calculated from individual phase measurements?

Of course. Three phase power is ##P=3*V*I*cos(\theta)## , where VLN= line to neutral voltage, and I is phase current, and ##\theta## is the phase angle between VLN and I. Also, ##V_{LN}=\frac{V_{LL}}{\sqrt3}##.

If you assume balanced three phase, then it doesn't matter which pair of phases you choose for VLL.

I don't know how you are measuring the phase between V and I. Be careful to not get confused by the 60 degree shift between VLL and VLN. Just remember that in most cases, power factor is close to unity so that ##\theta## must be close to zero.

I wish our late great favorite @jim hardy was still around to answer these questions. He was much more hands on than I am.

• DaveE
aemla
Of course. Three phase power is ##P=3*V*I*cos(\theta)## , where VLN= line to neutral voltage, and I is phase current, and ##\theta## is the phase angle between VLN and I. Also, ##V_{LN}=\frac{V_{LL}}{\sqrt3}##.

If you assume balanced three phase, then it doesn't matter which pair of phases you choose for VLL.

I don't know how you are measuring the phase between V and I. Be careful to not get confused by the 60 degree shift between VLL and VLN. Just remember that in most cases, power factor is close to unity so that ##\theta## must be close to zero.

I wish our late great favorite @jim hardy was still around to answer these questions. He was much more hands on than I am.
I currently have a standard meter and can only measure phase angles at individual phases. With equation above, I would need UL1-L2 to IL1 phase angle. Thus, giving another unknown.

For those who have a similar question and are simply trying to measure active power with single phase L-L 208 VAC circuit, I have confirmed that PowerLogic PM5300 will work.

Staff Emeritus
I currently have a standard meter and can only measure phase angles at individual phases.
Wait a minute, that makes no sense. To measure voltage, you need two points. What are the two points where your voltage leads go?

Edit: And how does your meter measure the phase? Phase relative to what?

aemla
I should of clarified. Individual phases relative to the neutral. That is, measurements between L1-N, L2-N, or L3-N in a 3 phase configuration. In such configuration, register (data tag in the meter) named "Phase Angle L1" would mean the phase shift between the current wave that's flowing between L1 and N AND the voltage wave that's between L1 and N. Most meters I've encountered only measure phase angles with respect to the neutral. Most of them can actually measure voltage between voltage phases, that is, voltage L1 to voltage L2. But they cannot measure current that's flowing from L1 to L2, thus, cannot measure power or energy or phase angles. For that you would need a meter that's capable of such measurement, such as one mentioned above. So that the phase angle of two waves can be measured: voltage wave between L1 and L2 versus current wave between L1 and L2.

I was wondering if there was a way to derive, mathematically, that phase angle (or current) between L1 and L2 with values from a meter that only measures with respect to the neutral.

I'm not actually sure exactly how digital meters measure phase angles but I imagine it's similar to an oscilloscope (aka measuring instantaneous voltages and currents [not RMS]).

Regards.

Staff Emeritus
I'm not actually sure exactly how digital meters measure phase angles but I imagine it's similar to an oscilloscope (aka measuring instantaneous voltages and currents [not RMS]).
No. Digital meters usually do measure RMS voltage, not phase angle. Even instantaneous samples, can't give you phase without a reference.

I hope you understand that the phase of a sinusoid is meaningful only relative to a reference sinusoid with the same frequency. Just like it takes two wires, not one, to measure a voltage difference, it takes two sin waves to define a phase difference.

• russ_watters
aemla
Right, they also measure RMS voltages and currents, I was saying in addition to that. Reference for instantaneous measurements would be the same as with RMS measurements. Yes, I understand that you need two items to compare. The two waves that we are discussing are the voltage wave and the current wave. You can define voltage wave as the reference to the current wave OR current wave as the reference to the voltage wave.

Gold Member
Standard meters measure active power of each phase with respect to neutral and then sum everything up.
I may be misunderstanding here, but I'm pretty sure standard watt-hour meters measure line to line voltage and current in each line. The old mechanical meters have NO connection to the neutral.

• Tom.G
aemla
By standard, I was referring to the most common digital meters I came across online. I can list some references if it'll help. If you don't mind, please list some meters that you are referring to. Not sure about electromechanical since they seem to have a local display only (aka outdated for more modern applications).

Gold Member
I can list some references if it'll help.

Argh - It is an Energy meter and will all of the work for you. So this is just a connection question.

The meter you referenced ( 6ES7134-6PA20-0BD0 ) is a siemens analog input module.

DS: https://ca-en.alliedelec.com/m/d/e079437aab96ab803936d6cfddcbd5d9.pdf

It needs to be accurate for any load connected in a 3 phase system, including single phase loads. Ideally you should have access to the Neutral, even if not connected to the load. IF you have the neutral - then you should have 3 Wires L1-L2-N - and 2 phase CTs)

Connect exactly as Fig 3.1 - just ignore the UL3 Voltage and the IL3 CT connections.

IF you do not have the Neutral THEN confirm the Transformer Secondary WYE IS Grounded.

Then connect UL1 to you L1, UL2 to your L2, N to a local GND in the panel - And IL1 to CT 1 and IL2 to CT2.

IF - the trans is not grounded ( doubtful) OR you do not have a solid panel ground ( sorry but for lask of a better term "what the..." ) - then you need a different meter - or contact Siemens for their blessing.

I would not connect a leg of 208 to the N terminal becasue I do not know if the Neutral terminal can support the voltage relative to GND. Actually this would be a very bad idea to do anyway- as terminals marked N should not be energized (it is an operational safety hazard to service personnel)

------

As for your PHASE ANGLE questions - is a generic term referencing ANY two inputs, so yes, it will apply between voltages, and voltages vs currents, etc. But in this case with this meter - you do not need to deal with that.

The meter gives you "Measured variable active energy (metering class) 0.5"

• • russ_watters and Tom.G
aemla
1. Siemens AI Energy Meter 480VAC ST
2. PAC4200
3. Class 1000 Single-Phase kWh Meter E-Mon D-Mon
4. Energy Management Energy Analyzer Type EM210

aemla
Argh - It is an Energy meter and will all of the work for you. So this is just a connection question.

The meter you referenced ( 6ES7134-6PA20-0BD0 ) is a siemens analog input module.

DS: https://ca-en.alliedelec.com/m/d/e079437aab96ab803936d6cfddcbd5d9.pdf

It needs to be accurate for any load connected in a 3 phase system, including single phase loads. Ideally you should have access to the Neutral, even if not connected to the load. IF you have the neutral - then you should have 3 Wires L1-L2-N - and 2 phase CTs)

Connect exactly as Fig 3.1 - just ignore the UL3 Voltage and the IL3 CT connections.

IF you do not have the Neutral THEN confirm the Transformer Secondary WYE IS Grounded.

Then connect UL1 to you L1, UL2 to your L2, N to a local GND in the panel - And IL1 to CT 1 and IL2 to CT2.

IF - the trans is not grounded ( doubtful) OR you do not have a solid panel ground ( sorry but for lask of a better term "what the..." ) - then you need a different meter - or contact Siemens for their blessing.

I would not connect a leg of 208 to the N terminal becasue I do not know if the Neutral terminal can support the voltage relative to GND. Actually this would be a very bad idea to do anyway- as terminals marked N should not be energized (it is an operational safety hazard to service personnel)

------

As for your PHASE ANGLE questions - is a generic term referencing ANY two inputs, so yes, it will apply between voltages, and voltages vs currents, etc. But in this case with this meter - you do not need to deal with that.

The meter gives you "Measured variable active energy (metering class) 0.5"
Good afternoon,
It is currently connected as you've descripted. Yes, secondary of wye is grounded as well. The problem is that in such wiring configuration with this particular meter, meter measures power and energy with respect to the neutral. For example, register that reads "Apparent Power L1" is a result of a multiplication of current on L1 and the voltage difference between L1 and N (120VAC).

The goal/question is to determine/calculate active power (I mentioned phase angle earlier because I think phase angle needs to be determined first) between L1 and L2.

Earlier, I was questioning if connecting L2 to N would work and came to the same realization, potential internal shorts may occur resulting a safety hazard. Thank you for confirming.
Thank you.

It seems all of your questions are in the ET 200SP Manual that I linked to.

The connection diagrams ARE in there. Section 3.2
ALL of the date you are looking for IS available. Appendix B

Yes power is reported per phase - but also total.

It is doing everything you need.

• russ_watters
aemla
I may need to start a new thread and clarifying the question in much more detail.
I have already learned a lot with this discussion, so thank you!
None of the connection diagrams fit the actual circuit. I have reviewed the manual several times now.

Total power is calculated by adding individual phases in three phase configuration (meter setting).
For example (I actually need active power, but using apparent power for the example since it's very easy to calculate yourself):
---- Meter Calculation (3 phase config)----
L1 power: 120VAC * 1Amp = 120 VA
L2 power: 120VAC * 1Amp = 120 VA
L3 power: 0VAC * 0 Amps = 0 VA
Total power: 120VA + 120VA + 0VA = 240 VA
*This could also be 0 VA depending on how CT is wired.
---- Hand Calculation (1 phase)----
L1 to L2 Voltage: 208 VAC
Current: 1Amp
Total power: 208 VAC*1Amp = 208 VA

You can see that two calculations above don't match, meter calculation is not correct. Since apparent power isn't correct, active power won't be either.

This meter (and others that I've referenced earlier) does not support single phase L-L configuration.
The only single phase configuration that it supports is L-N, and as mentioned before wiring second L to N is not a good idea.

So, besides getting a new meter, I was wondering if there is a way to calculate active power for single phase L-L configuration by using data provided by three phase configuration (such as inflow/outflow energies, phase angles, frequencies, PFs, etc.).

I hope this clarifies my question. Thank you.

Gold Member
If you don't mind a DIY project, see this for a DC version of a power meter.
https://www.physicsforums.com/posts/5890314

It would be modifiable to AC by biasing the op amp appropriately. You will probably want to use a Current Transformer instead of the Shunt that is shown, also the voltage input would need a step down transformer. Cheers,
Tom

Gold Member
@aemla, I had a quick look at the link provided by @Windadct. I'm not going to pretend to know exactly what that unit is doing internally. I see how the connection diagrams all connect to the neutral. I suspect this connection has less to do with measuring power (true or apparent) than you think.
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Not sure you grasp how phases and voltages relate. You likely won't believe this, but often times 3 phase power is measured and billed using only two old style mechanical spinning disk type watt-hour meters. There is one voltage coil and two current coils per meter. NO neutral connection.

aemla
If you don't mind a DIY project, see this for a DC version of a power meter.
https://www.physicsforums.com/posts/5890314

It would be modifiable to AC by biasing the op amp appropriately. You will probably want to use a Current Transformer instead of the Shunt that is shown, also the voltage input would need a step down transformer.

View attachment 285487
Cheers,
Tom
Perfect, thank you Tom.
Would you happen to know how active power is determined?

Gold Member
Would you happen to know how active power is determined?
For the circuit as drawn, it is an inherent characteristic assuming that the frequency response is flat & phase shift is negligible at all frequencies of interest.

The output is proportional to the brightness of the LED multiplied by the voltage applied to the LDR.

You will have to calibrate the circuit. There is a non-linearity of the LED, especially at lower currents (it is super-linear, i.e. more sensitive to current variation at low levels, below 20% of rated current); and don't forget they need a few volts to turn on.

The LDR resistance curve flattens at high illumination, which may or may not be a problem. You could just use the datasheet values, but is probably easier to calibrate with resistive loads.

You will also need to account/compensate for phase shifts in any current or voltage transformers you decide to use. Probably best to avoid transformers if you can, but that means the circuit is at line voltage, which could make things difficult. Perhaps connect an Optical Coupler to the output (look for techniques to linearize them, I seem to recall a single-OpAmp circuit).

Cheers,
Tom

aemla
@aemla, I had a quick look at the link provided by @Windadct. I'm not going to pretend to know exactly what that unit is doing internally. I see how the connection diagrams all connect to the neutral. I suspect this connection has less to do with measuring power (true or apparent) than you think.
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Not sure you grasp how phases and voltages relate. You likely won't believe this, but often times 3 phase power is measured and billed using only two old style mechanical spinning disk type watt-hour meters. There is one voltage coil and two current coils per meter. NO neutral connection.
Probably not very well.

Thanks for mentioning that, did some more research. In a three phase configuration, that makes sense. Summing up instantaneous currents of a three phase balanced load will result in 0, thus, no need for the neutral. However, things change when you have an unbalanced load and you would then need a neutral. This would also be true when only 2 phases are wired. So the meter would assume that the rest of the current is going through the neutral. Total power would be a result of calculated power summation of two phases with respect to the neutral. This, of course, would give a higher final result (as shown earlier) than what it actually is when single phase L-L is used. This is also what I'm seeing when reading the data. This is my attempt at understanding field results, is there a reason why it wouldn't be used on an unbalanced load?

I don't see why such completed meters wouldn't be able to measure single phase L-L. I might be missing a setting but I also don't see anything in the data tables that would suggest such measurement/calculation is made.

aemla
For the circuit as drawn, it is an inherent characteristic assuming that the frequency response is flat & phase shift is negligible at all frequencies of interest.

The output is proportional to the brightness of the LED multiplied by the voltage applied to the LDR.

You will have to calibrate the circuit. There is a non-linearity of the LED, especially at lower currents (it is super-linear, i.e. more sensitive to current variation at low levels, below 20% of rated current). The LDR resistance curve flattens at high illumination. You could just use the datasheet values, but is probably easier to calibrate with resistive loads.

You will also need to account/compensate for phase shifts in any current or voltage transformers you decide to use. Probably best to avoid transformers if you can, but that means the circuit is at line voltage, which could make things difficult. Perhaps connect an Optical Coupler to the output (look for techniques to linearize them, I seem to recall a single-OpAmp circuit).
Would calibration with the resistive loads account for the phase shift? I suppose not if transformers are part of the calibration circuit.

Gold Member
However, things change when you have an unbalanced load and you would then need a neutral.
This is not true. 3 phase delta connected secondaries can have a different load on each phase with no neutral required.

The issue here is yu need to see the vector relationship of the two phases - you can not just ADD apparent power like that, sorry that is just not how it works.

The 1 A you show as L-N is not the same phase angle as L-L.

The wiring of the meter - and the specific connection of the Voltages and the CTs - sets an expected phase shift between the measurements.

In a meter set up with Phase CTs ( current) and L-N Voltages the calculation per phase has no phase shift due to the connections - BUT when you look at Phase to Phase (L-L) voltages vs the Phase CT - there is a 30 Degree shift in the SENSING - that the meter accounts for. I am quite sure if you look at REAL power (active) using data from the meter - they WILL sum properly.

aemla
This is not true. 3 phase delta connected secondaries can have a different load on each phase with no neutral required.
You're right, I was referring to the 3 phase wye, should of read up on delta too:0

aemla
The issue here is yu need to see the vector relationship of the two phases - you can not just ADD apparent power like that, sorry that is just not how it works.

The 1 A you show as L-N is not the same phase angle as L-L.

The wiring of the meter - and the specific connection of the Voltages and the CTs - sets an expected phase shift between the measurements.

In a meter set up with Phase CTs ( current) and L-N Voltages the calculation per phase has no phase shift due to the connections - BUT when you look at Phase to Phase (L-L) voltages vs the Phase CT - there is a 30 Degree shift in the SENSING - that the meter accounts for.

View attachment 285556

I am quite sure if you look at REAL power (active) using data from the meter - they WILL sum properly.
I attached an example of real measurements (please ignore the highlights and incorrect L2 data [must of been a bad CT installation]). Point being, meter summed up individual phase results for the final totals.

Are you referring to phase shift of two voltage waves? I was referring to the phase shift between a voltage and a current wave. I suspect that's why there is an actual reading for "Phase Angle L1" and "...L2". Gold Member
Since you are using a WYE-connected source, is there a reason you can not connect that meter 'N' to the center (common) of the source?

According to the datasheet for the meter, the Neutral is used only for its internal power supply. Then you can use a connection as shown on the pages around page 22 on the datasheet.

For this to work the current measurement would have to sense only the desired load, that is measure at or in the specific load in question. Since Current Transformers are assumed, you would need only signal wires back to the meter.

Cheers,
Tom

Gold Member
I would have expected the results your attachment shows. If you connect a load between L1 and N that is what it is. Between L2 and N and that is also whatever it is. The power dissipated in the two separate loads add up. There is no reason to worry about phase.
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I'll throw another curve at you. In the real world example in question, there is neutral current. However, it will not be the difference in current between leg one and leg two. Phase difference will come into play here. In a previous post you said that current imbalances in 3 phase wye guarantees neutral current. This is also not necessarily true. If the loads are connected between the legs and neutral then yes, current imbalances will cause current to flow in the neutral. But if the loads are connected only between the legs, it doesn't matter. There will be no neutral current.

aemla
Since you are using a WYE-connected source, is there a reason you can not connect that meter 'N' to the center (common) of the source?

According to the datasheet for the meter, the Neutral is used only for its internal power supply. Then you can use a connection as shown on the pages around page 22 on the datasheet.

For this to work the current measurement would have to sense only the desired load, that is measure at or in the specific load in question. Since Current Transformers are assumed, you would need only signal wires back to the meter.

Cheers,
Tom
I believe below is what you are describing. 