Mk 2 PV Router Circuit Board power source

Have not seen any reference to what source the MK 2 gets powered from. It would seem that it makes a great deal of difference since the MK 2 needs to get a specific reference of V and I for comparison.

Energy from the grid normally flows into the premises where consumption is taking place (when importing power). In this situation, the voltage and current waveforms are in-phase; they always have the same polarity.

The situation is reversed whenever surplus PV becomes available. To get energy to flow away from the premises (when exporting power), the instantaneous voltage at the PV’s inverter must always be slightly greater than that of the mains. While in this state, the voltage and current are 180° out-of-phase; their polarities are always opposite.

So, since the Export V and I are 180 degrees out of phase, this would NOT be a good reference for the MK 2 PV Router pcb then?..Am I thinking correctly here?

To detect whether energy is being imported or exported at any given moment, it is only necessary to ascertain whether the instantaneous value of Power at the grid connection point (i.e. Voltage × Current) is positive or negative.

And at what point between the mains in and the PV output can any change in polarity be recognized since they are both always different?

I don’t think it cares where it is powered from. It does care where the CT loop is installed and that should be somewhere on the connection from the grid where the import or export current can be measured reasonably accurately. Definitely gridwards of the point-of-connection of PV panels and all major loads.

That is only because of voltage drops within your installation. Exactly the same applies within your country where the power flow on the transmission lines connecting the various generating stations varies according to where in the country the maximum demand comes from.

I’m not sure what your thinking is. Why isn’t the system voltage a good reference - it is what your supplier’s meter uses and hence what they charge you for.

Can you explain that a little more clearly? What polarity do you think that quote refers to? It is power flow: Robin counts exported power/energy as positive and imported power/energy as negative (the opposite of OEM).

The Mk2 PV Router gets its voltage signal from an on-board mains transformer. The amplitude is further reduced by a pair of dropper resistors, and the instantaneous value of the resulting sinusoid is repeated measured by the processor. A similar signal is obtained from the CT (or CTs when used in the US).

When energy flows into the premises from the grid, the two signals (for V and I) are in phase. When there is surplus power and energy is being exported, the signal for current becomes inverted with respect to the signal for voltage.

When a Mk2 PV Router is actively diverting surplus power, the signal for current will alternate between being in-phase and inverted with respect to the signal for voltage. This is because the load is brought on in bursts so that the overall consumption exactly matches the overall amount of generation.

With my energy bucket description, the energy level rises when energy is exported to the grid, and falls when energy is consumed. This is indeed the opposite way around to the usual convention. But provided that the same convention/logic is used for both import and export, the ‘polarity’ of that convention will make no difference to the outcome. For datalogging purposes, average power data as recorded by the Mk2 PV Router is inverted before transmission to the receiving entity. Such data is then in the same form as that obtained from an emonTx.

So, in-phase Mains or out-of-phase PV export makes no difference for a PV diverter to get what it needs and do what it needs to do, or is there really just no difference either way.

I still don’t understand your concept of phase. Have you looked at the ‘Learn’ section and in particular the article Learn→Electricity Monitoring→AC Power Theory→Introduction→An Introduction to AC Power?

Phase is purely relative, and in one sense it is completely the wrong word to use when you’re talking about the power flow reversing. If you look at the last diagram on that page, you can just as easily - and I’d argue more correctly - say that the polarity of the current wave had changed, what was a positive half-cycle at the top of the page is a negative half-cycle at the bottom. Power is the product of voltage and current, so in the top diagram power is always positive, and in the bottom power is always negative.

In short, and using voltage as the reference, the current changes sign and so does the power.

I think I would reword that as:
“The sign of the current and the power is exactly what a PV diverter uses to do what it needs to do.”

So it is vitally important that it can detect the relative sign/phase of the current in relation to the voltage.

I’m just reading and asking for confirmation from what I quoted from the source reference from the very beginning.
My questioning was vague in my first post, sorry. Now you have answered that all clearly in the above reply.

Thanks to all whom relied!