Starting Current Calculation for ATS Starter

The ATS will normally have tappings at 50%, 65% and 80% voltage levels, but there may be more on some units. Depending on which tapping is selected, starting current will be reduced by 75%, 58% and 36% respectively of that for DOL starting.

As a percentage of FL current, this would be around 150%, 253% and 384% respectively.

The formula for this is:- Ist = (Tr/100)²(Si)

Where Ist =Starting current, Tr = Tapping ratio %, Si = DOL starting current.

Eg, assuming a motor with a full load current of 50 amps and a DOL starting current of 6x or 300 amps and using a 65% tap,

The formula would be:- Ist = (65/100)²(300), = (0.65)² x 300, = 126.75 amps, which is (126.75/50) = 253% of FLC

Starting torque will be reduced by the square of the applied voltage - ie. 25%, 42% and 64% respectively, so the appropriate tapping will need to be selected based on the motor's ability to overcome the load's inertia and to reach around 80% full speed on that tapping. This selection will determine the ultimate starting current for your motor.

ATS = Automatic Transfer Switch.

Did you mean an RVAT (Reduced Voltage AutoTransformer) starter? If so, then spades' answer is correct, at least in THEORY. In reality, those would be the MAXIMUM starting current. If the load on the motor does not need all of that to accelerate, it would be lower and conversely if the load REQUIRES more than that, it will fail to accelerate and when the starter transitions, the current can still surge to 500-600%, maybe more if the transition is open. In other words it is load dependent.

You're right about the correct Acronym, but considering that the OP was talking about a motor starter and not a genset, I figured it was safe to assume that he was referring to an auto transformer starter which, although correctly referred to as RVAT starters, are commonly just called ATSs, when talking about starters as the reduced voltage part is a given.

You are also correct about the load dependency, but again the OP was referring to full load current, so I thought it reasonable to assume that full load would be applied.

Maybe I assume too much.

And sometimes I too pedantic.