Motor starting & voltage sag
The key idea
When you switch on an induction motor, it can demand six times its normal current just to get moving. That sudden demand pulls down the local voltage, so every other device connected to the same bus feels the dip too.
What happens when the motor starts?
At standstill, a motor is not behaving like a normal rotating machine yet. Electrically, it looks more like a transformer with its secondary short-circuited: only the winding impedance limits the current. A typical motor therefore draws around six times its rated current when started direct on line, and most of that starting current is reactive.
How the local network responds depends on its strength:
- A stiff grid has a high fault level and low internal impedance. It can supply the starting current with only a small drop in voltage.
- A weak grid has a low fault level and higher internal impedance. The same motor start pulls its voltage down much further.
This temporary drop is the voltage sag. Lighting, contactors, drives, computers, and nearby machinery all share the same bus voltage, so they experience the sag for as long as the motor takes to accelerate. That usually means seconds, not milliseconds.
The solution — and the catch
The obvious way to reduce the sag is to stop the motor taking so much current at once. Reduced-voltage starters do exactly that. A star–delta starter connects the windings in star during the start, while a soft starter raises the applied voltage electronically. Both reduce the current drawn from the grid.
But there is a catch: motor torque follows the square of the applied voltage. Star–delta reduces the direct-on-line starting current to one third, but it also reduces starting torque to one third.
That can work perfectly for a light load such as a ventilation fan, which needs very little torque at zero speed. A loaded conveyor or crusher is different. If the motor cannot develop more torque than the driven machine demands, it does not accelerate. It simply sits at or near zero speed, drawing starting current until the protection trips.
Try it in the sandbox
Start with the defaults: a 2 MW motor on a 100 MVA bus, direct on line. The bus dips by roughly 12% and recovers after a few seconds.
Next, reduce the grid fault level toward 20 MVA and watch the trace stop recovering. Finally, switch to star–delta on that weak bus. The voltage sag becomes shallower, but check whether the motor now has enough torque to start.
during the start: 0.876 pu
- Starting current
- 6× rated · 14.1 MVA
- Bus voltage during start
- 0.876 pu · dip 12.4% · below 0.90 pu
- Torque margin
- torque 1.23 pu vs load 0.30 pu
The model treats the motor as a fixed impedance while it accelerates. The bus voltage is therefore a pure-reactance divider: v = S_fault / (S_fault + k · S_motor). Here S_motor = MW / 0.85, and k is the starting multiple: 6 direct on line, 2 in star–delta and 3 on the soft starter. Torque follows v² times the method factor, against a 0.30 pu fan load. The widget draws the run-up as a fixed 6 s. A real study integrates the torque curves of the motor and the load to find the true duration.
Why this matters
Motor-start limits are a design condition
Network operators usually limit how far one motor start may pull down a shared bus—often to only a few percent. If a proposed start exceeds that limit, the design may need a reduced-voltage starter, a dedicated transformer, a smaller motor, or a stronger grid connection.
One start can stop an entire plant
If the bus voltage falls far enough, other equipment begins to fail. Contactor coils can drop out, variable-frequency drives can trip on undervoltage, and lighting can extinguish. One large motor start can therefore trigger a plant-wide stoppage.
Reduced voltage trades current for torque
There is no free lunch. A reduced-voltage starter protects the grid by giving the motor less starting torque. Before choosing a starting method, you must check the torque the driven machine needs from standstill through the full run-up.
A starter cannot create grid strength
No starting method can manufacture power that the supply cannot deliver. If the grid is fundamentally too weak, reducing the starting current may also leave too little torque to accelerate. The real fix may be a stiffer supply, a smaller motor, or a drive that can control torque throughout the start.
The math, if you want itOptional — the page reads completely without it
The widget models the motor during the start as a fixed impedance drawing k times the motor's rated apparent power, and the network as a pure reactance set by its fault level. The bus voltage is then a simple divider:
bus voltage during the start
v = SfaultSfault + k · Smotor
Here Smotor = Pmotor / 0.85, with the 0.85 folding power factor and efficiency into one number. The multiple k is 6 direct on line, 2 in star–delta, and 3 for a soft starter limiting the current to half the direct-on-line value.
The motor then develops this starting torque, in per unit of rated torque:
torque at reduced voltage
T = 1.6 · v² · m
where 1.6 pu is the locked-rotor torque at full voltage and m is the method factor: 1 direct on line, 1/3 in star–delta, and (3/6)² = 0.25 for the soft starter, whose torque falls with the square of its current limit. The load demands 0.30 pu at all speeds, a fan-type demand, and the motor stalls when T falls to that value.
The star–delta factors come from the same 3× relationship the wye and delta page derives: in star, each winding receives VLL/√3, so it draws 1/√3 of the current at 1/√3 of the voltage, which is one third of the power and one third of the line current. And since torque follows the voltage squared, one third of the torque too.
Two approximations are worth naming. The model treats the motor as a constant impedance for the whole start, though the impedance and power factor really change continuously with speed. And it draws the run-up as a fixed 6 seconds, where the real duration is the integral of motor torque minus load torque over the speed range. A real motor-starting study does that integration with both torque curves and the machine's inertia. It is what tells you whether a marginal start takes 8 seconds or never finishes.
See it in Phasor
In Phasor, a motor start is a power flow with the motor at its starting impedance. Add the machine and solve: the voltage results show the sag at every bus, not just the motor's, with the transformer between motor and source and the loads already drawing current all in the model. It shows you how far across the network the sag reaches.