2013 leaf 6.6kw

[smkettner]

NEC requires #8 with a 40a breaker. More important to change the wire than the breaker.

A Leviton 160 or Legrand will work on #10 30a as it will limit current to 16a. Only opportunity charging would be a full speed assuming 30+ amp evse.

ClipperCreek has a 20 amp EVSE that will supply 4.8kW and is fine on 30 amp circuit. At least it will be a little faster at home.
http://www.clippercreek.com/products.html

 

[EdmondLeaf]

Specification Ev se-Rs
Connector SAE J1772 compliant
Voltage 208VAC to 240VAC

Output Current 30A max
Input Current 40A max

I guess your evse (and mine) is able to deliver 30A or bit more to the car but to be safe 8 wire is needed

 

[gbarry42]

Where do you suppose the missing 10A goes :?:

 

[EdmondLeaf]

this is from AV spec I believe all will be 40A rated, or input is 40A rated for safety

 

[smkettner]

Where do you suppose the missing 10A goes :?:

NEC requires 80% max load for extended (3hr+) service connections.

 

[mitch672]

Where do you suppose the missing 10A goes :?:

It's not "missing", its called a "safety margin", using the %125 rule. If you draw a continuos 30AMPS for over 3 hours, it's classified as a "continuous load", and the breakers are supposed to be rated at %125 of the ampacity. 30A * 1.25 = 37.5A, since they don't make 38 AMP breakers, 40AMP is the closest value that will work.

BTW, this is the same for other circuits as well, for example, the standard 12A draw of the Leaf on the 120V unmodifed EVSE, the breaker should be rated at 12 * 1.25, or 15AMPs if used more than 3 hours, this is for breaker heating, and to elliminate false tripping.

 

[kmp647]

If I wanted I could sell my aero unit and purchase a SPX or some other lower amp or adjustable unit
one that just needs a 30amp breaker
I would be able to charge at 27 amps or whatever
otherwise the aero unit just assumes its on a 40a circuit and pulls up to the max

public charging then L2 would deliver the 32a

 

[planet4ever]

Too many percentages and numbers being thrown around here. Many of them right, but quite a few wrong. First, it may seem illogical if you don't have a mathematical bent, but the following is true:

• Start with any number and add 25% to it (i.e. multiply by 1.25).
• Now subtract 20% from the result (i.e. multiply the result by 0.8).
• The answer will be the same number you started with.

This is what is going on with the limits on continuous loads. Looked at from the breaker end, you have to derate by 20%. Thus a 30A breaker can support only a 24A (not 25.5A or 27A) continuous load. Looked at it from the EVSE end, a 30A EVSE must be connected to wire and breakers that can support 37.5A. a 32A EVSE requires 40A wire and breakers.

Back to the original question, our present 3.3kW charger is rated based on its output, not its input, and presumably that would also be true for a 6.6kW charger. The 3.3kW charger pulls up to 3.84kW, but you can't just double that, because there appears to be a constant 300w cooling load for our present charger, regardless of voltage or amperage. If that cooling system would be adequate for the 6.6kW charger, and if the efficiency of the charger itself was the the same, it would be pulling 7.38kW (3.84 x 2 - 0.3). But we don't know about those two "if" conditions.

Ray

 

[TEG]

Note, I recall heaing that the 2013 6.6kW is to be "optional".
It may be that you get a choice of sticking with 3.3kW or pay extra to get an upgraded unit.
(And again, cannot be retrofitted to 2011/2012s...)

 

[mitch672]

See this article, Nissan predicting as many as 60,000 Leaf sales for 2013

http://www.myelectriccarforums.com/nissan-predicting-leaf-sales-to-increase-sixfold-by-2013/" onclick="window.open(this.href);return false;

 

[smkettner]

Back to the original question, our present 3.3kW charger is rated based on its output, not its input, and presumably that would also be true for a 6.6kW charger. The 3.3kW charger pulls up to 3.84kW, but you can't just double that, because there appears to be a constant 300w cooling load for our present charger, regardless of voltage or amperage. If that cooling system would be adequate for the 6.6kW charger, and if the efficiency of the charger itself was the the same, it would be pulling 7.38kW (3.84 x 2 - 0.3). But we don't know about those two "if" conditions.

I assume the cooling load is to run the vehicle electronics and pump the coolent. These would be 12v loads supplied from the main battery through the DC/DC converter.

Charger output 3.3kW to the battery would then be 3.0kW net to the battery. 6.6kW would be 6.3kW net to the battery. Charger input might be exactly double but I think the larger unit may have small gain in efficiency.

 

[KeiJidosha]

Back to the original question, our present 3.3kW charger is rated based on its output, not its input, and presumably that would also be true for a 6.6kW charger. The 3.3kW charger pulls up to 3.84kW, but you can't just double that, because there appears to be a constant 300w cooling load for our present charger, regardless of voltage or amperage. If that cooling system would be adequate for the 6.6kW charger, and if the efficiency of the charger itself was the the same, it would be pulling 7.38kW (3.84 x 2 - 0.3). But we don't know about those two "if" conditions.

Ray

What would be preferable is the charger regulates current, the limiting factor from the EVSE, not power. If the charger pulls 32amps, it will charge at 7.7kW from 240V. It appears the ActiveE does this, and the MINI E did as well. That's 15% faster.

 

[planet4ever]

I assume the cooling load is to run the vehicle electronics and pump the coolent. These would be 12v loads supplied from the main battery through the DC/DC converter. Charger output 3.3kW to the battery would then be 3.0kW net to the battery. 6.6kW would be 6.3kW net to the battery. Charger input might be exactly double but I think the larger unit may have small gain in efficiency.

Actually, the pump runs at the high voltage, powered by the traction battery. (The same pump also cools the DC/DC converter, so you had that part backwards.) But I have to admit your basic point sounds like it is likely to be correct.

Ray