Enphase Combiner box neutral size

[JoeNorm]

I am using the enphase branded combiner box for an enphase system. It houses the monitoring device which requires a neutral.

My question is about the size of this neutral. As far as I know it is only being used for one thing, the electronics inside the box. So am i allowed to downsize it to a 14awg wire since the controls circuit is is 15amp?

That would mean if I am backfeeding, say 50 amps to the main panel I would run my two larger L1+L2 with a smaller neutral. Is this allowed?

I am sure there is a code section on this, thanks

 

[ggunn]

I am using the enphase branded combiner box for an enphase system. It houses the monitoring device which requires a neutral.

My question is about the size of this neutral. As far as I know it is only being used for one thing, the electronics inside the box. So am i allowed to downsize it to a 14awg wire since the controls circuit is is 15amp?

That would mean if I am backfeeding, say 50 amps to the main panel I would run my two larger L1+L2 with a smaller neutral. Is this allowed?

I am sure there is a code section on this, thanks

You can only make it as small as the EGC.

 

[jaggedben]

I think it's allowed, with ggunn's clarification, even though I've seen so many full size neutrals run.

See 220.61 and 215.2(B).

 

[JoeNorm]

Thanks, that's good to know. For my last job I had to drill a 10" concrete wall and then do a feeder tap so I was interested in the smallest volume setup I could get away with. Most of the time it is less critical.

 

[ING23]

By the way , and talking about neutral and Enphase System , Any one know how The microinverters will feed a house load of 120 Volt without going to the neighborhood transformer, How current flow knowing from the Roof there is no neutral just 240 V?

 

[ggunn]

By the way , and talking about neutral and Enphase System , Any one know how The microinverters will feed a house load of 120 Volt without going to the neighborhood transformer, How current flow knowing from the Roof there is no neutral just 240 V?

If the inverters are grid tied they are connected to the neighborhood transformer. In the absence of any export limitations, which direction the current flows through the transformer depends on the loading on the secondary.

 

[wwhitney]

By the way , and talking about neutral and Enphase System , Any one know how The microinverters will feed a house load of 120 Volt without going to the neighborhood transformer

They can't, the house loads depend on the utility transformer for the unbalanced portion of the 120/240V loading that ends up on the neutral.

Cheers, Wayne

 

[Zee]

I have also only used a neutral size=EGC.
If the broader question is how small of a conduit can you get away with, remember the Consumption CT coil wiring! I usually connect the 2 twisted pairs with WAGOs to a communication cable (like for ethernet).

 

[jaggedben]

By the way , and talking about neutral and Enphase System , Any one know how The microinverters will feed a house load of 120 Volt without going to the neighborhood transformer, How current flow knowing from the Roof there is no neutral just 240 V?

For a grid tied system, the neighborhood transformer is necessary as Wayne said.

For systems that go off-grid with the Enphase System Controller, that is why they have an autotransformer in the System Controller to derive the 120V.

 

[ING23]

Thanks for yours replay

 

[ING23]

For a grid tied system, the neighborhood transformer is necessary as Wayne said.

For systems that go off-grid with the Enphase System Controller, that is why they have an autotransformer in the System Controller to derive the 120V.

We could said the current is going outside into the Utility transformer just for a load of 120V ( let said a light inside the house) . This current is going to the transformer passing through the METER , How come is not measured by the Utility?

 

[wwhitney]

We could said the current is going outside into the Utility transformer just for a load of 120V ( let said a light inside the house) . This current is going to the transformer passing through the METER , How come is not measured by the Utility?

Let's say your 240V 2-wire microinverters are producing 10A on each of L1 and L2. Let's say your loads are resistive and consist of 5A connected L1-N and 20A connected L2-N.

So what are the currents passing through the meter? On L1, you have 5A from the load, less 10A from the PV, so you get -5A going into the house, or 5A going out of the house. On L2, you have 20A from the load, less 10A from the PV, so you get 10A going into the house. Then on N, you can compute it two different ways: it's either the difference of L1 and L2 at the meter, i.e. 10 - (-5) = 15A, or it's the difference in the two loads, i.e. 20 - 5 = 15A.

So how much are you going to be charged? If the L1-N voltage = L2-N voltage = 120V, then you'll be using 120V * 10A = 1200W on the L2-N, and you'll be exporting 120V * 5A = 600W on L1-N. So you're using 600W net, and if you have full net metering and maintain this situation for 100 minutes, you'll be charged for 1 kWh.

Cheers, Wayne

 

[jaggedben]

We could said the current is going outside into the Utility transformer just for a load of 120V ( let said a light inside the house) . This current is going to the transformer passing through the METER , How come is not measured by the Utility?

First, single phase meters do not measure the current on the neutral. They don't need to. They just need to measure current on the hots in sync with the voltage in order to measure all the power.

It is actually possible for the 240V inverter to feed current from say L1 to the load, then the neutral, then out to the transformer, from where the current returns to the inverter on L2. This actually reverses the direction of current on L2 from what the meter expects, so the meter measures a negative power reading which is interpreted as power export. (For a load connected to L2, just switch L1 and L2.) It's complicated and I'm way oversimplifying the description, but it works.

 

[ING23]

First, single phase meters do not measure the current on the neutral. They don't need to. They just need to measure current on the hots in sync with the voltage in order to measure all the power.

It is actually possible for the 240V inverter to feed current from say L1 to the load, then the neutral, then out to the transformer, from where the current returns to the inverter on L2. This actually reverses the direction of current on L2 from what the meter expects, so the meter measures a negative power reading which is interpreted as power export. (For a load connected to L2, just switch L1 and L2.) It's complicated and I'm way oversimplifying the description, but it works.

I am trying to understand how the meter will skip that reading , the energy is used on the premises , is not exported. I agreed with you it has to interact with the Utility transformer to get the neutral . could you draw on this improvised line diagram how the current flows?

 

[ING23]

Let's say your 240V 2-wire microinverters are producing 10A on each of L1 and L2. Let's say your loads are resistive and consist of 5A connected L1-N and 20A connected L2-N.

So what are the currents passing through the meter? On L1, you have 5A from the load, less 10A from the PV, so you get -5A going into the house, or 5A going out of the house. On L2, you have 20A from the load, less 10A from the PV, so you get 10A going into the house. Then on N, you can compute it two different ways: it's either the difference of L1 and L2 at the meter, i.e. 10 - (-5) = 15A, or it's the difference in the two loads, i.e. 20 - 5 = 15A.

So how much are you going to be charged? If the L1-N voltage = L2-N voltage = 120V, then you'll be using 120V * 10A = 1200W on the L2-N, and you'll be exporting 120V * 5A = 600W on L1-N. So you're using 600W net, and if you have full net metering and maintain this situation for 100 minutes, you'll be charged for 1 kWh.

Cheers, Wayne

Still melting my brain trying to understand the direction of the current. (I understand is AC ) I think your explanation was helpful.

 

[jaggedben]

Still melting my brain trying to understand the direction of the current. (I understand is AC ) I think your explanation was helpful.

Yes you're diagram is correct.

If the inverter were absent, your import would be (5A+20A)×120V=3000W
With the inverter supplying 10A, your import is 10A×120V=1200W and your export is 5A×120V=600W. Import minus export is 600W total. And the difference between load and import is 2400W which is your inverter at 10A×240V, so it checks out.

It's an important point that the meter reading may actually be the sum of imports on one leg and exports on the other. With more inverter output, you could have exports on both legs.

(I think you understand this is an extremely simplified way of looking at it, but if we properly did all the trigonometry for the AC sine waves it would also work out.)