# How to determine the correct capacity for a sub-panel for a shop?

I plan to install a sub-panel for my woodworking shop, and would like to determine the total capacity that should supply that sub-panel. These are the main items that will likely be running in the shop, possibly simultaneously:

• table saw, 240v, 14 amp
• dust collector, 120v, 9 amp
• freezer, 14 cu ft (I don't have the amperage)

(The lights are run on their own circuit.)

What is an appropriate capacity for the sub-panel?

NEC 2008 gives us an easy way to do things in residential.

220.82 Dwelling Unit.

(B) General Loads. The general calculated load shall be not less than 100 percent of the first 10 kVA plus 40 percent of the remainder of the following loads:

(1) 33 volt-amperes/m2 or 3 volt-amperes/ft2 for general lighting and general-use receptacles. The floor area for each floor shall be calculated from the outside dimensions of the dwelling unit. The calculated floor area shall not include open porches, garages, or unused or unfinished spaces not adaptable for future use.

(2) 1500 volt-amperes for each 2-wire, 20-ampere small appliance branch circuit and each laundry branch circuit covered in 210.11(C)(1) and (C)(2).

(3) The nameplate rating of the following:

a. All appliances that are fastened in place, permanently connected, or located to be on a specific circuit

b. Ranges, wall-mounted ovens, counter-mounted cooking units

c. Clothes dryers that are not connected to the laundry branch circuit specified in item (2) d. Water heaters

(4) The nameplate ampere or kVA rating of all permanently connected motors not included in item (3).

So we can use 220.82 (B)(2) to figure for the dust collection, freezer, and an additional circuit for receptacles.

`1500VA * 3 = 4500VA / 120V = 37.5 Amperes`

You'll then have to use the values from the nameplate on the table saw to figure for that (A Volt-Ampere value should be listed on the nameplate, use that number for more accurate calculations). You could also use this method for the dust collection system and freezer since they are both "permanently connected, or located to be on a specific circuit".

`3360VA / 240V = 14 Amperes`

`37.5A + 14A = 51.5A`

So This is what our subpanel will look like.

• 60A double pole breaker in the main panel.
• 6 AWG feeder cable for a run up to 75 ft., 4 AWG feeder cable for a run up to 150 ft.
• 60A main breaker in the subpanel.
• 20A double pole breaker for table saw.
• 20A single pole breaker for dust collector.
• 20A single pole breaker for convenience receptacles.

## Notes:

Don't forget to balance your loads between the two legs in the subpanel.

• If we assume the loads are balanced perfectly, wouldn't the total draw be 14A + (37.5 / 2)A per leg? (33A per leg?) If so, this setup could be done with 40A main breakers, correct? I know I should design in extra capacity; I just want to make sure that my understanding is correct. Thanks :) – bitsmack Nov 30 '15 at 20:02

This is not an answer related to code but simply doing the layman's math.

First off the Freezer should be on a dedicated circuit. Not because of any excessive amperage draw, but because if you fire up the table saw and the dust collector at the same time and it pops the breaker, you don't want the freezer being shut off for potentially an undetermined period of time, ruining whatever's inside.

If you figure 20A breaker for the table saw (another dedicated circuit would be wise), 15 for the dust collector and 15 for the freezer, plus another 15 for assorted outlets, you're looking at a 100A sub panel to allow for the 65A you're needing, plus to allow room for expansion (required by code in CA, not sure about US but you generally don't want to max the panel out from the beginning).

Which raises the question - what kind of amperage does your MAIN panel carry - you're adding a 100A sub, which means even if your house panel is 200A it may not be enough.