Motor Starting Voltage Dip Calculator

The lights dim when the compressor starts. Everybody has seen it and almost nobody has put a number on it, and the number is worth having, because the same dip that makes a fluorescent flicker for half a second is also what decides whether the motor accelerates or sits there drawing locked-rotor current until something opens. Torque falls as the square of the voltage, so a dip that looks trivial costs roughly twice as much torque as it costs volts.

Line to line, measured at the panel the motor is fed from, with the shop running normally.
From the nameplate. Often a little below the supply — a 460 V motor on a 480 V system is the usual pairing.
The nameplate LRA, or the locked-rotor current the manufacturer publishes for this motor. Not a figure from a table of typical values.
Used only to show the ratio between running and starting current, and the running load on the supply.
Across the line is 1.00. For a reduced-voltage starter or a soft starter, this is the fraction of locked-rotor current it actually pulls from the line, and it comes from the starter data, not from a rule of thumb.
From the transformer nameplate. Transformer mode only.
The percent impedance stamped on the transformer nameplate. Transformer mode only. This ignores the impedance of everything upstream of the transformer, which makes the dip look slightly smaller than it is.
From the utility, from an engineering study, or from the label on the equipment. Short-circuit current mode only.
Short-circuit kVA mode only.
Your own figure, or one your equipment supplier or utility has given you. This page has no view on what is acceptable.
Motor Starting Voltage Dip Calculator — Inrush vs SourceBuildFigure

Two numbers decide the whole thing

How hard the motor pulls, and how hard the supply pushes back. The first is the locked-rotor kVA: 1.732 times the voltage times the locked-rotor amps. At the defaults, a nameplate 363 A at 460 V sitting on a 480 V supply pulls 378.8 A, and 1.732 times 480 times 378.8 is 315 kVA.

The second is the short-circuit strength of the supply. A 500 kVA transformer with 5.75 percent impedance can deliver 500 divided by 0.0575, which is 8,696 kVA into a bolted fault. That is the number that says how stiff the bus is.

The dip is the first divided by the sum of both: 315 over 9,011, which is 3.49 percent. The bus falls from 480 V to 463.2 V for about a second.

Torque is the number that hurts

3.49 percent sounds like nothing. But induction motor torque follows the square of the applied voltage, so 96.5 percent of the voltage leaves 93.1 percent of the torque. The torque penalty is 6.9 percent against a 3.5 percent voltage penalty — very nearly double.

The multiple runs the opposite way to intuition, and it is worth being precise about because the page prints it. Torque loss divided by voltage loss is exactly two minus the dip. So it is 1.97 at a 3.5 percent dip, 1.80 at 20 percent, 1.50 at 50 percent and 1.10 at 90 percent. The proportional penalty is at its harshest where the dip is small enough to be dismissed, and it eases off as things get genuinely bad. In absolute points the gap does grow for a while — 3.4 points at a 3.5 percent dip, 16 points at 20 percent — but it peaks at a 50 percent dip and shrinks after that. At a 20 percent dip you have 80 percent of the voltage and only 64 percent of the torque, which is the case that actually strands motors.

That is the mechanism behind the motor that starts fine on a quiet morning and refuses to accelerate when the rest of the shop is running. Nothing failed. The bus was two percent softer and the torque margin was already thin.

What a reduced-voltage starter changes, and what it does not

Setting the starter fraction to 0.33 drops the current the line sees to 125 A and the dip to 1.18 percent, which is the entire reason those starters exist. What the page deliberately does not do is tell you the torque the motor makes behind that starter, because that depends on the type. An autotransformer starter, a wye-delta starter and a soft starter all reduce line current and motor torque by different relationships, and the numbers come from the starter data sheet rather than from this arithmetic.

The general shape holds though: everything that reduces starting current also reduces starting torque, and the trade is the whole design problem. A starter that solves a flicker complaint by halving the inrush has also taken most of the torque, and if the load was already marginal it will now not start at all.

Where the model is optimistic and where it is pessimistic

Optimistic in two places. It treats the starting kVA and the source strength as plain magnitudes and ignores the angle between them, and it stops at whatever point you measured the supply — the feeder, the disconnect and the starter between there and the motor all add their own drop at locked-rotor current, which is five or six times running current, so a feeder that drops two percent running drops ten or twelve while the motor is starting. Run that separately on the voltage drop page at the starting current, not the running current.

Pessimistic in one: the angle it ignores works in your favour for a motor starting at a low power factor into a mostly reactive source, so the real dip is usually somewhat smaller than the figure here. Engineers use this approximation anyway because it errs the safe way and because the inputs to a better model are rarely available.

Generators are a different problem

A utility transformer has a fixed impedance. A generator does not behave that way at all under a step load: the dip depends on its subtransient reactance, on the excitation system and on how fast the governor recovers, and a generator that can carry a motor all day may still dip fifteen or twenty percent when it starts. If the supply is a genset, the manufacturer has motor starting figures for it, and they are the ones to use.

Questions people ask

How much voltage dip does a motor start cause?

Divide the motor starting kVA by the sum of the starting kVA and the short-circuit kVA available at the connection point. At the defaults, 314.9 kVA of inrush against 8,696 kVA of supply gives 3.49 percent. The two inputs are the nameplate locked-rotor amps and either the transformer kVA and impedance or an available fault current figure from the utility.

Why does a small voltage dip cost so much torque?

Because induction motor torque goes as the square of the applied voltage. A 3.5 percent dip leaves 96.5 percent of the voltage and 96.5 squared is 93.1, so 6.9 percent of the torque is gone. The torque penalty is roughly double the voltage penalty for small dips and grows faster than that as the dip deepens.

How do I get short-circuit kVA from a transformer nameplate?

Divide the transformer kVA by its percent impedance expressed as a decimal. A 500 kVA unit at 5.75 percent gives 500 over 0.0575, which is 8,696 kVA. That ignores the impedance of everything upstream of the transformer, which makes the answer slightly optimistic — the utility can give you a figure that includes it.

Will a soft starter fix my flicker problem?

It will reduce the dip, because it reduces the current the line sees, and the page will show you by how much once you enter the fraction of across-the-line inrush the starter data says it draws. What it will also do is reduce the starting torque, and if the load was already hard to start that trade may not be available. The starter manufacturer has both halves of it.

Does this work for a generator supply?

Not well. A generator does not present a fixed impedance the way a transformer does — the dip under a step load depends on its subtransient reactance, its excitation system and its governor response, and it is usually much deeper than a utility supply of the same kVA would give. Use the genset manufacturer motor starting data instead.

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