Motor Control Tools

Inverter Voltage & Base Speed

Estimate available inverter voltage under SPWM/SVPWM and the resulting ideal vs. loaded motor base-speed limits.

Modulation method
V
Wb

Optional — for a loaded operating estimate

Ω
A
mH

Adds the reactive voltage drop.

mH

Accepted, but has no effect under the Id = 0 assumption used here.

Available fundamental phase voltage27.713V peak
Available line-to-line voltage33.941V RMS
Voltage utilization115.5%

Relative to the Vdc/2 (SPWM) baseline.

Ideal / no-load theoretical limit

5513.289RPM

Zero-current upper bound — never an achievable loaded operating point.

Practical loaded operating estimate

RPM

Enter phase resistance and current to compute a loaded estimate.

Formulas

Linear-region modulation limit

SPWM: Vphase,peak,max = Vdc / 2 SVPWM: Vphase,peak,max = Vdc / √3

Vdc
DC bus voltage [V]
Vphase,peak,max
Maximum fundamental phase voltage before overmodulation [V]

SVPWM extends the linear range by 2/√3 ≈ 1.1547× over SPWM by exploiting the zero-sequence freedom in a two-level three-phase inverter. Six-step overmodulation can exceed both limits at the cost of low-order current harmonics and is not modeled here.

Loaded speed limit (Id = 0)

Vd = −ωe·Lq·Iq Vq = Rs·Iq + ωe·λm Vd² + Vq² = Vphase,peak,max²

ωe
Electrical angular speed at the voltage limit [rad/s]
Rs
Stator phase resistance [Ω]
Lq
Q-axis inductance [H]
Iq
Phase current (assumed pure Iq, Id = 0) [A]

Solved as a quadratic in ωe. With Id = 0, Ld does not appear — a supplied Ld value has no effect on this estimate. Without Lq supplied, this reduces to ωe = (Vmax − Rs·Iq) / λm (resistive drop only).

Ideal vs. loaded — why both numbers matter

The ideal figure answers "what's the absolute physical ceiling this bus and modulation method can ever produce, with zero current?" — useful for sizing and sanity checks, but never something the motor will actually reach under load. The loaded figure answers a different, more practical question — "roughly how fast can this motor spin at this current before the inverter runs out of voltage headroom?" — and will always be lower than the ideal figure once current is nonzero. Neither number is "the" maximum speed of the motor: the ideal figure ignores load entirely, and the loaded figure is a simplified, Id = 0 estimate that a real drive using flux-weakening could exceed.

Worked example

Vdc = 48 V, SPWM, λm = 0.012 Wb, 4 pole pairs:

  • Available phase voltage: 48 / 2 = 24 V peak
  • Ideal / no-load limit: 24 / 0.012 = 2000 rad/s (electrical) ≈ 4775 RPM
  • With Rs = 0.2 Ω and Iq = 10 A: loaded limit = (24 − 0.2×10) / 0.012 ≈ 1833 rad/s ≈ 4377 RPM — about 8% lower than the ideal figure at this current.