I want to buy an inverter-generator to run my Trane heating system in case of a power outage. The Trane is on a 20 amp circuit. The sticker says it needs a minimum of a 15.2 amp circuit.

My question is, how do I determine (not estimate) the surge amperage of the 1 hp blower motor in my Trane heating system?

The inverter generators I have seen have max outputs of 2,000, 3,000, or 4,000 watts. I am trying to figure out if any of these will run my heating system. It seems the critical issue is the starting current of the 1 hp motor.

How might I go about measuring the starting current of my heating system?

Update: My HVAC 1 hp fan motor is constantly running- on low if not providing heat and on high if it is providing heat. Based on that, it seems to me that I need to use the in-rush amperage, since the theoretical calculations are all not relevant to my situation. Thanks.

  • The invertors peak kilowatt rating may be a good indication of it's motor starting capabilities. I have a 3kW invertor w/ a 5kW peak rating. It has no problem starting motor loads.
    – Kris
    Nov 22 '15 at 1:47
  • What voltage is your circuit? 120 or 240? Even at 240 volts your total is 4800 watts. If it is a gas furnace (which it sounds like) then it is 120 volts so that is only 2400 watts. And that is from the breaker size not the actual current draw. The code says a 1hp motor at 120 draws 16 amps that is only 1920 watts. I agree with Kris even a small 3000 watt generator should run your furnace without even grunting. You can measure the surge current with a peak-hold function on a clamp on ammeter.
    – ArchonOSX
    Nov 22 '15 at 2:17
  • @Kris 5k is a large peak for a 3k inverter. How long will it provide that peak? What brand/model?
    – Yehuda_NYC
    Nov 22 '15 at 2:17
  • Here's a link(wagan.com/wagan-tech/power-inverters/proline/…) to the one I use.
    – Kris
    Nov 22 '15 at 2:30
  • I'm not sure how long, but I would imagine it's only intended for a short period e.g. less than 5sec
    – Kris
    Nov 22 '15 at 2:48

Measuring the startup load is as easy as using a clamp on meter and turning the air handler on and off.

The startup load will be high for a split second, then drop down to a steady load.

Use Ohms Law to calculate the wattage.

Watts/Voltage = Current

As mentioned in my comment, the invertors peak load is a good indication of the invertors motor starting current.


To get the most accurate motor startup load readings from your meter, you may need to use one that supports "in-rush" readings.

Meters like the Fluke 374, 375 and 376 support "in-rush" readings.

Edit 2

I did some digging around and found the formula to calculate inrush current.

Take the NEMA assigned letter to your motor, in which case yours is letter B, and plug that into this equation:

Iinrush=(code letter value X horse power x 1000) /( √3 X Voltage)

You can ignore the square root of 3 if your voltage source is single phase.

in-rush = 3.54 x 1HP x 1000 / 120V = 29.5 Amps

Bottom line is your invertor needs a surge rating of at least 4000 Watts

4000 / 120 = 33 Amps

NEMA Motor Letter Table

enter image description here

  • Agree with the above. If you can borrow your intended generator, you'll see if it's big enough. (For instance, my honda eu2000i handles a brief 22A startup, but not a 30A.) Nov 22 '15 at 1:56
  • 1
    What clamp meter do you use that shows accurate readings for an event that lasts for a split second? Seems like you'd need either a peak reading, or a steady output of readings with dozens of samples a second to show such a short peak.
    – Johnny
    Nov 22 '15 at 5:47
  • A Fluke video says it averages the first 5 cycles rather than take the first, highest cycle. Is that good when trying to determined the specs for a generator-inverter? youtube.com/watch?v=aRJMNqffdlc
    – Yehuda_NYC
    Nov 23 '15 at 2:35
  • 2
    Fluke meters in 330 series replaced by 370 series: fluke.com/fluke/m3en/electrical-testers/clamp-meters/…
    – Yehuda_NYC
    Nov 23 '15 at 2:59
  • I've updated the answer with a in-rush formula.
    – Kris
    Nov 23 '15 at 17:51

The startup current for most correctly operating blower motors is low (usually less then 200% full load amperage), so the OP's concern should not be much of an issue. If it is a premium efficiency system it might use an ECM motor, which is electronically managed and also has a low startup current.

If one cares to measure it oneself, no special inrush meter is needed. Almost any clamp-on amp meter will do because blower motor startup is rather slow, typically taking several seconds.

  • That's what I thought but I wasn't 100% sure, as I measured mine that way with no in-rush setting.
    – Kris
    Nov 22 '15 at 18:50
  • I incorrectly remembered the startup current range for blower PSC motors in my original post, so I corrected that information. The rest of the original post was accurate.
    – user39367
    Nov 23 '15 at 0:52
  • @chris Good point it does take a second or two for a blower motor to get rolling, giving someone time to take an ammeter reading even with an old amprobe.
    – ArchonOSX
    Nov 23 '15 at 2:08
  • It does seem the start up of the blower is slow. Thanks.
    – Yehuda_NYC
    Dec 15 '15 at 15:56

A good rule of thumb for small motor loads is to multiply the max over current protection device by 1.25%. This will give you a good idea of what size of generated power or amperes needed to over come in inrush of current needed. 1.25 x 15.2= 19amperes. A source of power generated should follow the formulas stated in the calculations shown in other answers shown. "foot note: a 4500 watt 120/240 volt generator will handle the surge of the unit current. If you want to power up any other devices or lights , a larger wattage source is needed.

  • The 125% rule applies to sizing overcurrent protection for continuous loads, not for estimating inrush current. Also note that it's 125%, not 1.25%. If it was 1.25%, the calculation would look like this 0.0125 * 15.2 = 0.19 amperes.
    – Tester101
    Dec 10 '15 at 14:27
  • The original question has nothing to do with generators, but invertors.
    – Kris
    Dec 10 '15 at 23:19

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