EEVblog #409 - EDMI Smart Meter Teardown

[gxti]

V is constant, right?  It's 220 or 240.  The product converts the amps into volts with those coils so you're really only sampling the Amps (as volts) and the real volts going into the meter is probably programmed into the firmware or an option set by the tech when it's installed.

Hah, no. Not only is the RMS voltage not constant to anywhere near the precision required for billing purposes, but the meter needs to measure real power which you can't do just by measuring RMS volts let alone programming a value in at the factory. True RMS meters measure voltage and current simultaneously (or very closely, as is being discussed above) and multiply them to get instantaneous power. Then that is averaged over time to get average power or just summed to get total energy usage.

If you're just measuring voltage occasionally then you are only measuring apparent power.

[cengland0]

V is constant, right?  It's 220 or 240.  The product converts the amps into volts with those coils so you're really only sampling the Amps (as volts) and the real volts going into the meter is probably programmed into the firmware or an option set by the tech when it's installed.

Hah, no. Not only is the RMS voltage not constant to anywhere near the precision required for billing purposes, but the meter needs to measure real power which you can't do just by measuring RMS volts let alone programming a value in at the factory. True RMS meters measure voltage and current simultaneously (or very closely, as is being discussed above) and multiply them to get instantaneous power. Then that is averaged over time to get average power or just summed to get total energy usage.

If you're just measuring voltage occasionally then you are only measuring apparent power.

But there was nothing in the circuitry to measure voltage -- just amps.  So how do you believe they are doing it?  I still believe it has to be a setting either in the firmware or in the tech setup.

[lewis]

There is something in there to measure the voltage, the 3.3M resistors see the vid at 23.49

[adh]

But there was nothing in the circuitry to measure voltage -- just amps.  So how do you believe they are doing it?  I still believe it has to be a setting either in the firmware or in the tech setup.

Voltage measurement is the main reason why most of the meter circuitry is connected directly to mains. Probably there is voltage divider between phases and neutral, one would assume that the large resistors next to each current transformer are exactly for this.

You cannot measure real power without measuring voltage (or at least for crude approximation doing measurement synchronized to line frequency, just measuring voltage is mostly simpler to do). Nominal voltage times integrated current is either apparent power or complete nonsense (depending on how you integrate the current).  Due to various things connected to mains that either have complex impedance (motors) or that are completely nonlinear (switching power supplies, dimmers, motors that are spinning up...) current waveform does not exactly follow voltage waveform and thus no energy is transferred while there is (often significant) current flowing through the wires. Thus there is concept of real power (RMS instantaneous U*I, useful energy) and apparent power (nominal U * RMS I, what you dimension wiring for), with so called power factor being the ratio between these two things, with only linear loads it is cosine of phase shift between U and I (and thus it is commonly called $cos \phi$).

[cengland0]

There is something in there to measure the voltage, the 3.3M resistors see the vid at 23.49

As I understand it, the coils will convert the amps to volts and then you need a couple resistors so you can measure the voltage drop across them.  You're technically measuring volts but it's a representation of the Amps that have built up a magnetic field around the coil and the coil then converted those back to volts measured across the resistor.

[cengland0]

Voltage measurement is the main reason why most of the meter circuitry is connected directly to mains. Probably there is voltage divider between phases and neutral, one would assume that the large resistors next to each current transformer are exactly for this.

You cannot measure real power without measuring voltage (or at least for crude approximation doing measurement synchronized to line frequency, just measuring voltage is mostly simpler to do). Nominal voltage times integrated current is either apparent power or complete nonsense (depending on how you integrate the current).  Due to various things connected to mains that either have complex impedance (motors) or that are completely nonlinear (switching power supplies, dimmers, motors that are spinning up...) current waveform does not exactly follow voltage waveform and thus no energy is transferred while there is (often significant) current flowing through the wires. Thus there is concept of real power (RMS instantaneous U*I, useful energy) and apparent power (nominal U * RMS I, what you dimension wiring for), with so called power factor being the ratio between these two things, with only linear loads it is cosine of phase shift between U and I (and thus it is commonly called $cos \phi$).

First, wouldn't you agree that there were no physical connections to the mains?  All you could see was the mains completely shorted from the input to the output but the huge copper wire went through a coil to pick up the magnetic field produced when AC flows through a wire.  With such a simple circuit as this, how do you suppose they are measuring both current and voltage?

Second, the device that you're using inside the house will not affect the voltage coming in on the mains.  You can run a huge air conditioner or a small FM radio and the mains will still remain at 220 or 240 volts.  Peak to peak or RMS doesn't really matter here because the calculation is done in software -- not measured.

I understand what you're saying about switching power supplies.  This is why you buy battery backups with VA ratings instead of Watt ratings.  All the devices in my house that are plugged in the wall, use the voltage provided by the electric company as the source voltage and they will draw some amps based on that fixed voltage. If the voltage drops at my house, it's dropping for everyone in the neighborhood -- at least everyone on the same side of the huge transformer anyway.

[adh]

First, wouldn't you agree that there were no physical connections to the mains?  All you could see was the mains completely shorted from the input to the output but the huge copper wire went through a coil to pick up the magnetic field produced when AC flows through a wire.  With such a simple circuit as this, how do you suppose they are measuring both current and voltage?

Second, the device that you're using inside the house will not affect the voltage coming in on the mains.  You can run a huge air conditioner or a small FM radio and the mains will still remain at 220 or 240 volts.  Peak to peak or RMS doesn't really matter here because the calculation is done in software -- not measured.

I understand what you're saying about switching power supplies.  This is why you buy battery backups with VA ratings instead of Watt ratings.  All the devices in my house that are plugged in the wall, use the voltage provided by the electric company as the source voltage and they will draw some amps based on that fixed voltage. If the voltage drops at my house, it's dropping for everyone in the neighborhood -- at least everyone on the same side of the huge transformer anyway.

Mains is connected to the board for measurement through the three small wire nuts marked as no-connect that are internally shorted to input mains by the screws that Dave removed when he first tried to remove the main board (which is explained, and also why it is done this way, in follow up video linked by somebody in youtube comments).

Even when household loads will not significantly influence mains voltage (and they often influence and sometimes in surprising ways) they still will not draw current that exactly follow mains voltage and that is why you need to measure mains voltage. To be clear, when you do this digitally you will measure voltage and current significantly more often than 50/60Hz as to have multiple samples for each mains voltage period. Also the nominal mains voltage has pretty large tolerance, today effectively 220, 230, 240V are the same thing only with differently defined tolerance, long wire runs with significant resistances (rural areas...) are another thing.

It is not easy to visualize why capacitive or inductive loads behave as they do, but it is pretty easy to visualize that there will be current flowing through capacitor connected to AC voltage when it crosses zero (as the capacitor discharges back into the AC source) and this is the whole idea.

And as for how current transformer works: It is transformer as each other. On the coil there is precise amount of windings (often 1000 or 500) and the whole thing transforms current to current (divided by number of windings). Voltage plots in datasheet as (IIRC) shown in video are referenced to specific value of resistor across secondary that is recommended by CT manufacturer (often 100R). Large current transformers are prone to generating dangerous voltages on secondary winding when left open that potentially could also cause isolation breakdown inside the coil, generally there should be pretty small resistance across it, often current shunt in ampermeter. Small CT like these are often explicitly designed to survive open secondary (and sometimes even have characterized performance in that condition) but still using burden resistor of relatively small value is recommended, certainly not anything on the order of 3M3.

In all: In AC circuits power is not U*I.

[lewis]

First, wouldn't you agree that there were no physical connections to the mains? 

No - there are taps off the incoming phases and neutral to read the voltage. Please watch the video again. The three big bits of copper wire go through three current transformers, which is how the meter simultaneously measures the current.

Second, the device that you're using inside the house will not affect the voltage coming in on the mains. 

Yes, it can! Because of non-zero supply impedance. Also, the mains voltage varies throughout the day. Mine goes from 242V in the morning to 255V in the evening. But this is not the main reason why the meter needs to measure the phase voltage. To measure the average power, and hence energy, the meter needs to know how the current waveform corresponds with the voltage waveform. It is not simply enough for the meter to know that 'the mains is about 240V', it needs to know the relationship between the voltage waveform and the current waveform. Different loads cause different relationships. See below

...the calculation is done in software -- not measured.

Yes, it is measured.

If the voltage drops at my house, it's dropping for everyone in the neighborhood -- at least everyone on the same side of the huge transformer anyway.

You see, the voltage isn't fixed!


So let's assume the incoming voltage is 240V and you have a 10A load. What power is that?

2400W?   Not necessarily.

If you have a nice resistive load like a heating element, then the voltage waveform is perfectly in sync with the current waveform and the power is indeed 2400W. But if you have a capacitive or an inductive load like a motor in your vacuum cleaner, then you need to know the power factor (PF) before calculating the power. PF is cos(phi), phi being the phase angle between voltage and current. If cos(phi) is 0.5, then the power for your motor is now 240V x 10A x 0.5 = 1200W. So measuring the current and assuming a constant voltage is not enough, you need to know the relationship between the voltage and current waveforms (phi) in order to calculate the power (and hence energy).

But the meter doesn't work like that, I was using this as an illustrative example. It will use the following equation to ensure it accurately measures non-linear loads:



Basically that means 'average out the instantaneous voltage multiplied by the instantaneous current over a period of time'. The 'period of time' perhaps being one mains cycle. The effect is the same, the meter needs to know how the voltage and current waveforms correspond.

There' more here: http://en.wikipedia.org/wiki/AC_power

[cengland0]

Mains is connected to the board for measurement through the three small wire nuts marked as no-connect that are internally shorted to input mains by the screws that Dave removed when he first tried to remove the main board (which is explained, and also why it is done this way, in follow up video linked by somebody in youtube comments).

No - there are taps off the incoming phases and neutral to read the voltage. Please watch the video again. The three big bits of copper wire go through three current transformers, which is how the meter simultaneously measures the current.

Yes, it can! Because of non-zero supply impedance. Also, the mains voltage varies throughout the day. Mine goes from 242V in the morning to 255V in the evening.

Oops, you're both right.  I forgot about those taps he encountered when the screws were removed.  Now that I remember watching that part, I have to admit that could be used to measure the voltages.  I hate it when I'm wrong.  But seriously, without those, you're only measuring current though those coils.

Regarding your voltages changing from 242 to 255V throughout the day, what country are you in?  In the US, it seems our tolerances are much better as it fluctuates from 115 to 117V which is insignificant when you're billing for energy usage.  Just calculate based on 115 and don't worry about having to measure.  However, if your voltages vary as much as you're saying, you will definitely want to measure that especially since you pay per kilowatt hours instead of just plain amp hours.

[notsob]

Just a comment on the voltage fluctuations re 242V - 255V, with the advent of so much roof installed solar power in OZ, there is a lot of 'uncontrolled' power being fed into the grid. So daytime Voltage levels do tend to be higher than a few decades ago.

[TerraHertz]

Just a comment on the voltage fluctuations re 242V - 255V, with the advent of so much roof installed solar power in OZ, there is a lot of 'uncontrolled' power being fed into the grid. So daytime Voltage levels do tend to be higher than a few decades ago.

I've been wondering about this. What happens when a neighborhood with a lot of grid-connected solar systems gets isolated from the grid by some line fault? it's going to happen sooner or later.

I imagine the solar systems are designed to stop outputting and 'stay dead' if the line voltage goes below a certain value, so they won't continuously try to energize a shorted-out grid.
But what if there's enough solar capacity in an area to completely maintain the local load, and that section of the grid is disconnected from the main system?
Sounds like an interesting problem in chaotic systems control. Will the solar systems all try to 'track' each other? What happens to the line voltage and frequency?

Another aspect of this problem, is what happens when someone reconnects the breakers isolating a solar-driven section of grid to the main grid. The two are going to be out of phase - this can't be a good scenario.
I think the solar system inverters would be the likely losers of that particular conflict.

[lewis]

But seriously, without those, you're only measuring current though those coils.

That's right, but they are there! Without them the meter would be an ammeter, not a watt-hour meter.

Regarding your voltages changing from 242 to 255V throughout the day, what country are you in?

My lab is on an industrial estate in a rural backwater of Britain. We have a single phase supply feeding a large house and an industrial estate through a small substation transformer off an 11kV pole which in turn feeds a small village and several farms. Supply impedance is fairly high, about 0.3R, and the commercial fitters opposite can drop the mains voltage by 5V or so just by starting his compressor.

...which is insignificant when you're billing for energy usage.

If each house is drawing 10A that's a possible error of 20W. There's about 130 million households in the US, so that's an error of 2.6GW. Suddenly that's significant!

Just calculate based on 115 and don't worry about having to measure.

You can't do that because you need to know the relationship between the two waveforms, not just the numbers.

[lewis]

I've been wondering about this. What happens when a neighborhood with a lot of grid-connected solar systems gets isolated from the grid by some line fault? it's going to happen sooner or later.

I imagine the solar systems are designed to stop outputting and 'stay dead' if the line voltage goes below a certain value, so they won't continuously try to energize a shorted-out grid.
But what if there's enough solar capacity in an area to completely maintain the local load, and that section of the grid is disconnected from the main system?
Sounds like an interesting problem in chaotic systems control. Will the solar systems all try to 'track' each other? What happens to the line voltage and frequency?

Another aspect of this problem, is what happens when someone reconnects the breakers isolating a solar-driven section of grid to the main grid. The two are going to be out of phase - this can't be a good scenario.
I think the solar system inverters would be the likely losers of that particular conflict.

I've been wondering about this too. Next time I see s grid-tied inverter I'll investigate. Also, I've always wondered about linesman's safety when working on a main with grid-tied inverters connected to it, how do they reliably isolate the bit they're working on?

[cengland0]

If each house is drawing 10A that's a possible error of 20W. There's about 130 million households in the US, so that's an error of 2.6GW. Suddenly that's significant!

But at an average cost (in the US) of $100 per MWh, that's only $0.10 per Kilowatt hour.  At an error of 20 Watts, that is 20 x 24 hours * 30 days =  Watt hours 14.4 Kilowatt hours or a potential maximum billing error of $1.44 per month.  That's insignificant in my opinion and the "Customer Charge" that they impose even if you don't use any electricity can make up for that.

Anyway, others have stated their voltages vary significantly and I wasn't aware some countries had such poor tolerances in their electricity.  So with that, I agree it's important to measure the voltage too and it seems it's possible now that I noticed those little connectors from the mains to the PCB under the screws Dave removed.

[lewis]

If each house is drawing 10A that's a possible error of 20W. There's about 130 million households in the US, so that's an error of 2.6GW. Suddenly that's significant!

But at an average cost (in the US) of $100 per MWh, that's only $0.10 per Kilowatt hour.  At an error of 20 Watts, that is 20 x 24 hours * 30 days =  Watt hours 14.4 Kilowatt hours or a potential maximum billing error of $1.44 per month.  That's insignificant in my opinion and the "Customer Charge" that they impose even if you don't use any electricity can make up for that.

Anyway, others have stated their voltages vary significantly and I wasn't aware some countries had such poor tolerances in their electricity.  So with that, I agree it's important to measure the voltage too and it seems it's possible now that I noticed those little connectors from the mains to the PCB under the screws Dave removed.

  Look.... do the math properly. This is real back-of-the-envelope stuff. 20W error = 175.2kWh per year per household. That's 22.8 billion kWh error for the whole of the US, or $2.28billion in potential billing error, annually. If that's 'insignificant in your opinion' then I'll happily accept your cheque (or as you would call it: check).

These meters need to be accurate!

[cengland0]

Look.... do the math properly. This is real back-of-the-envelope stuff. 20W error = 175.2kWh per year per household. That's 22.8 billion kWh error for the whole of the US, or $2.28billion in potential billing error, annually. If that's 'insignificant in your opinion' then I'll happily accept your cheque (or as you would call it: check).

These meters need to be accurate!

Wait, this was a maximum error I was calculating on one house.  Since this voltage rises and falls, it will average out to not having much of an error.  In other words, sure you might be charged an extra 5 cents today but then you might get a 5 cents break tomorrow. 

You're assuming that the voltage will always be high and always cost the electric company extra but that's not realistic. 

Besides, my electric company charges me $14 every month regardless if I use electricity or not.  It's called a customer charge.  So if all companies did something similar (I don't know but they probably do), that's 130 million houses * 14 * 12 months = 21.84 billion in extra profit they are getting so they are still winning even if they take a 2.28 billion loss.  This doesn't count the profit they make on the electricity itself either and the 2.28 billion you quote is not realistic because it should average out close to zero.

This discussion is all in vain because I admit that they are probably testing the voltage as evidenced by the additional connectors to the mains that I did not remember until it was pointed out to me.

[NickS]

Besides, my electric company charges me $14 every month regardless if I use electricity or not.  It's called a customer charge.

That charge is for the maintenance of the poles and wires that supply your house. Even if you don't use any electricity, they still need maintenance.
That is why it is separate to the kWh billing which charges for the production of the electricity its self, not the infrastructure.

You also still haven't quite gotten that the voltage always needs to be read even if it never varied at all.
What you are saying is correct for DC voltage, but AC voltage has additional things like power factors which are very important when measuring power used.

[digsys]

I've been wondering about this. What happens when a neighborhood with a lot of grid-connected solar systems gets isolated from the grid by some line fault? it's going to happen sooner or later.
I imagine the solar systems are designed to stop outputting and 'stay dead' if the line voltage goes below a certain value, so they won't continuously try to energize a shorted-out grid.
But what if there's enough solar capacity in an area to completely maintain the local load, and that section of the grid is disconnected from the main system?
Sounds like an interesting problem in chaotic systems control. Will the solar systems all try to 'track' each other? What happens to the line voltage and frequency?
Another aspect of this problem, is what happens when someone reconnects the breakers isolating a solar-driven section of grid to the main grid. The two are going to be out of phase - this can't be a good scenario.
I think the solar system inverters would be the likely losers of that particular conflict.

I have friends in the government regulatory body who deal with this specifically - plus we sponsor scholarships in power engineering,
and have "looked after" PhD students also working on the problems. In a nutshell - it's an absolute MESS !! :-)
1/ Grid connect devices are supposed to be smart enough to drop-off when they detect an outage, and it mostly works.
They now "short" the cables they are working on.
2/ So person A connects his solar panels to the grid - he HAS to go in at say 0.5V GREATER than the supply to "source" energy.
neighbour B now connects - HIS system has to go 1.0V GREATER to become a "source". Person A is now screwed. ETC ETC
The Inverters now keep hopping up / down to try and "win". They are legally allowed to go to 155VAC in Aussie, but Inspections have
found tampered units that go to 275VAC (the max you can modify the Inverter).  The system is a mess - and idiots did not think it through.
3/ Most Inverters do manage to "lock" on to system phase OK. It's a design criteria. Apparently, they DON'T need to LOCK on to the
power factor though, but I can't understand why not? I have had it explained, but it got pretty deep :-)
4/ I haven't even started on ALL the problems the switch-mode NOISE is creating !! And how it starts all new skin-effect issues !!

The "ideal" system, as used in Europe etc, is for Solar farms with their own motor-generator / sub-station arrangement.
What we have now is an idiotic knee-jerk "solution" that's ONLY going to get worse ! AND it has bugger all benefit to the problem.
When I get time, I'll post the - base load / peak load / boost stats. Don't get me started :-)

[QSmits]

Just curious; I've got a 'smart meter' similar to that in my home. It is able to measure power consumed as well as power produced (it also has a plug to connect a gas meter to).

How would it differentiate between power consumed and produced? Something to do with looking at which way the voltage drop goes?

[gxti]

Yes, effectively. Although with a current transformer there is no "voltage drop". But either way the current waveform is inverted (or 180 degrees out of phase) with respect to the voltage. Since P=I*V, when they are opposite signs the power is negative.

[Pilot3514]

Several people have noted that they too had developed their own power meters.  I too am interested in monitoring my power usage.  However, I am hitting some problems in my design.  I would love to see Dave, or someone as interesting, do a step by step design and discus the design options and considerations.  There are always trade offs and it is interesting to hear different options.

My design effort began with a Current Transformer.  I was starting with rather low current so I was looking at something like a TA12L.  My understanding of a current transformer is that the current in the primary induces a current in the secondary that is proportional to the turns ratio.  So in a 1000:1 transformer, 1 mA will be induced for every amp passing in the primary.

Here is were the problems begin.  My little micro controller (insert your favorite) has a built in ADC but it reads volts, not amps.  So I need to convert volts to amps.  This is done all the time, I need a resistor.  So if a put a 1K resistor across the transformer, I get 1 volt per amp in the primary.  This looks promising.

The next problem is that the MC (Micro Controller) does not read AC it only reads DC.  So if a ground one side of the resistor and place a diode on the other side leading into the ADC of the MC I should be able to read the voltage that is proportional to the current.

Two more problems.  There is forward bias voltage to the diode so my reading will be off by half and amp or so.  Also, this is AC so my voltage will be zero half the time.  When I have a none zero reading it could be anything from zero to the  peak current value.

I need to get past these problems.  How do I get rid of the diode error and how do I calculate the RMS?

I was toying with the idea of putting a full wave bridge rectifier (4 diodes) between transformer and the resistor.  The diodes would pass the current and the voltage drop would not affect current through the resistor.  To get RMS I was thinking about taking as many samples as possible and calculate the RMS values from the samples.  I could also look for the peak value and calculate the RMS from there (assuming a sine wave).

Any comments?  Did I see a shining object and take of on an impractical tangent?  Can someone point me to a working design that I can plagiarize steal get inspiration from?

[HKJ]

The next problem is that the MC (Micro Controller) does not read AC it only reads DC.  So if a ground one side of the resistor and place a diode on the other side leading into the ADC of the MC I should be able to read the voltage that is proportional to the current.

You need to sample a couple of times (Say 100) for each period and multiply it with the voltage at that moment.
To get around the AC problem, you add a DC offset.

[digsys]

The AC issue is quite easy - make up a dual rail and use a dual rail op-amp to convert to precise 0V Ref. You can use something like
an LTC1144, which are very easy to use with few parts. You only need a few mA -ve
Edit: For the RMS - there are plenty of single chip True RMS converters out there. They do a MUCH better job than trying to write
your own code, and are fast. I've made power monitors and done exactly what I've posted.

[Pilot3514]

To get around the AC problem, you add a DC offset.

That is easy to say but it is not obvious to me what that means.

The AC issue is quite easy - make up a dual rail and use a dual rail op-amp to convert to precise 0V Ref. You can use something like
an LTC1144, which are very easy to use with few parts. You only need a few mA -ve

I don't see how using a switched capacitor charge pump to make a negative rail will get me any closer to a solution.  I still can not feed a negative value to the ADC of my microcontroller.  I was thinking that if I use split supplies of 2.5 volts each I could run the controller across them.  Then I got to thinking that if I just used a voltage divider that might get me to the same point.

Edit: For the RMS - there are plenty of single chip True RMS converters out there. They do a MUCH better job than trying to write
your own code, and are fast. I've made power monitors and done exactly what I've posted.

No matter how cheap, an RMS chip will still cost more that software.  And I can create one but not the other.  The one I create may be a piece of shit but it only costs my time.

[lewis]

I still can not feed a negative value to the ADC of my microcontroller.

Just think about what you mean by 'negative'. Negative relative to what? Relative to the VSS pin of the micro? Not necessarily...

If you make a potential divider between the micro's VDD and VSS of, say, 1K for the top resistor and 1K for the bottom resistor, and let's assume the micro is powered from 5V, then you have 2.5V at the junction of the two resistors with respect to VSS.

Connect a burden resistor in parallel with your CT. Then, one end of this connects to the potential divider, the other end connects to an ADC input (let's call it AN1) of the microcontroller. Another ADC input (let's call it AN2) connects to the potential divider junction.

So now, when the voltage across the burden resistor is +1V, you have 3.5V going into the AN1 input of the micro with respect to VSS. When the voltage across the burden resistor is -1V, you have 1.5V going into AN1. I mentioned connecting AN2 to the potential divider junction. This is for increased accuracy because the potential divider voltage may vary. In your code, the voltage across the burden resistor is determined by calculating: AN1 minus AN2. Use signed arithmetic, and perform the ADC conversions as close together as possible. Simple!

This is pretty much what HKJ means by 'adding DC offset'.