BLDC Motor Torque-Speed Envelope Calculator
BLDC Motor Torque-Speed Envelope Calculator
The motor sizing plot: the continuous and peak torque a brushless machine and its drive can hold at every speed, your load curve drawn on the same axes, and the speed where the two cross — which is where the machine will actually settle. Below base speed the drive holds rated torque and is limited by current; above it the back-EMF has reached the available bus voltage and torque falls.
BLDC torque-speed envelope
a 48 V bus, a 100 rpm/V machine with 0.08 Ω terminal to terminal, 30 A continuous and 90 A peak, space-vector PWM with 95% headroom, field weakening to 2.0× base speed, driving a fan that needs 1 N·m at 3,000 rpm
Two limits, and the corner between them
Vav = kmod × headroom × Vdc
ωbase(I) = (Vav − I·R) ÷ Ke T = Kt·I below it
above it, either T(ω) = Kt·(Vav − Keω) ÷ R (voltage-limited)
or T(ω) = Tbase·ωbase ÷ ω (constant power, field-weakened)
Trms = √( Σ Ti²·ti ÷ Σ ti )
- Kv
- no-load speed per volt applied across two of the three terminals, in rpm/V
- R
- winding resistance measured terminal to terminal, warm
- kmod
- peak line-to-line voltage the inverter can make, as a fraction of the bus: 1 for six-step and for space-vector PWM, 0.866 for sine-triangle PWM
- Trms
- the root-mean-square torque over a repeating cycle. Winding heating follows I²R, so RMS current — and therefore RMS torque — is what the continuous limit has to cover, not the average
Worked example
a 48 V bus, a 100 rpm/V machine with 0.08 Ω terminal to terminal, 30 A continuous and 90 A peak, space-vector PWM with 95% headroom, field weakening to 2.0× base speed, driving a fan that needs 1 N·m at 3,000 rpm
Ke = 60 ÷ (2π × 100) = 95.49 mV·s/rad, and Kt is the same number in N·m/A
Available voltage = 48 × 1.00 × 0.95 = 45.6 V; base speed at 30 A is (45.6 V − 30 × 0.08) ÷ 95.49 mV·s/rad = 452.4 rad/s = 4,320 rpm
The corner point is 2.865 N·m at that speed, which is 1.296 kW of shaft power — and that is the constant-power level the field-weakening branch holds above it
The fan needs 1 N·m at 3,000 rpm, so its coefficient is 10.13 µN·m·s² and its torque rises as the square of speed. It would reach the continuous torque at 5,078 rpm, which is above base speed — so the crossing is in the field-weakening region
There, T·ω = 1.296 kW and T = c·ω², so ω = (P ÷ c)1/3 = 503.9 rad/s = 4,811 rpm, at 2.572 N·m and 1.296 kW
At the 4,000 rpm operating speed the machine still has 2.865 N·m continuous against the fan's 1.778 N·m — a margin of 1.61×. Over the stated duty cycle (4.5 N·m for 6 s, then 1.0 N·m for 24 s) the RMS torque is 2.202 N·m, which is 76.9% of the continuous limit
What the envelope above base speed is really doing
| Region | What limits the machine | What the page draws | What to watch |
|---|---|---|---|
| Below base speed | the current limit — of the drive, or of the winding’s temperature | a horizontal line at Kt × I | torque is flat, so power rises linearly with speed. The thermal limit is an RMS limit, not a peak one |
| At base speed | back-EMF plus IR drop has reached the available voltage | the corner | base speed is different for every current limit: at peak current the IR drop is larger, so the peak envelope turns down sooner than the continuous one |
| Above base speed, no field weakening | voltage. The machine sits on its own line, exactly as a brushed DC motor does | T = Kt·(Vav − Ke·ω) ÷ R, falling to zero at the no-load speed | that line is steep. The whole speed range above base speed is Vav ÷ (Vav − I·R), which for a low-resistance machine is a few per cent |
| Above base speed, field weakening | voltage again, but d-axis current is used to cancel part of the magnet flux | the constant-power hyperbola T = P ÷ ω, to the speed multiple you set | it needs current control, spare current and voltage margin. Lose the current regulator at speed and the back-EMF rectifies through the body diodes into the bus, which is the failure that destroys drives |
Reading the envelope, and where the load meets it
A brushless motor and its drive have two separate limits, and the plot on this page is the boundary between them. Below base speed the limit is current. Torque is Kt times current, the drive can make as much voltage as the machine needs, and the envelope is a flat line at Kt × I. Above base speed the limit is voltage. The back-EMF has grown to the point where it plus the IR drop uses up everything the bus can supply, and the available torque falls. Where exactly the corner sits depends on which current you ask about: at the peak current the IR drop is larger, so the peak envelope turns down at a lower speed than the continuous one. That is why the two lines on the chart do not have their corners above one another.
What happens above the corner is the part worth being careful about. The textbook picture is a clean constant-power hyperbola running off to the right, and it is an approximation with conditions attached. It assumes the drive has current control, that it will push d-axis current into the machine to cancel part of the magnet flux, and that it has the current and the voltage margin to keep doing so. A surface-magnet machine driven trapezoidally does almost none of this: its honest envelope above base speed is its own voltage-limited line, T = Kt(Vav − Keω) ÷ R, which falls to zero at the no-load speed and is steep. This page draws whichever of the two you choose and reports the no-load-to-base-speed ratio so you can see how little room the voltage-limited line really gives. There is also a reason to care beyond sizing: a field-weakened machine that loses its current regulator at speed rectifies its own back-EMF through the inverter’s body diodes straight into the DC link, and that is the fault that destroys drives.
The load is the other half of the picture. A hoist is constant torque, a mixer is roughly proportional to speed, and a fan, pump or propeller goes as the square of speed — which is what makes the crossing interesting, because a square-law load rises to meet a falling envelope at a well-defined point. That crossing is where the machine settles: below it the motor has torque to spare and accelerates, above it the load wins and it slows down. The page solves for that point in closed form in every combination of load law and envelope model.
And then the duty cycle. An intermittent load is not judged by its peak torque or by its average; it is judged by its RMS, because winding heating follows I²R and the continuous limit is a thermal limit. A cycle that pulls 4.5 N·m for six seconds and 1 N·m for the next twenty-four has an average of 1.7 N·m and an RMS of 2.2 N·m, and it is the second number that has to fit inside the continuous envelope. That comparison is only fair if the cycle is short compared with the motor’s thermal time constant — minutes for a small machine, much longer for a large one. For the inverter’s own losses at the operating point use the BLDC inverter loss calculator, for the torque-power-speed arithmetic on its own the motor power, torque and speed calculator, and for the brushed equivalent of this whole picture the DC motor calculator, which draws the same straight line from stall to no-load.
The load line itself is on the chart if you ask for it. T = Kt(Vav − Keω) ÷ R is what the machine and its bus can do before any current limit is applied, and it is what a load trajectory has to be compared against. Switch the third line to it and you will usually find it towering over the envelope, because a low-resistance servo machine’s stall torque is many times the current its drive will allow — which is worth seeing once, because it says that current, not voltage, is what bounds the machine over almost its whole speed range. For an actuator following a sinusoidal command, the trajectory that has to fit under that line is a closed curve rather than a point: the electromechanical actuator load locus calculator builds it from the command, the screw and the load.
Frequently asked questions
What is base speed?
The highest speed at which the drive can still force the current you asked for. At that speed the back-EMF plus the IR drop equals the voltage the inverter can make from the bus, so there is nothing left to push current with. It is different for every current: base speed at peak current is lower than base speed at continuous current, because the IR drop is bigger.
Why does my motor not reach its Kv times the bus voltage?
Three reasons, all on this page. The inverter cannot use the whole bus — sine-triangle PWM gives only 0.866 of it as peak line-to-line voltage — device drops and dead time take more, and the current regulator needs margin to work at all. Kv × V is the no-load speed of an ideal drive with no load and no losses. The figure this page calls the no-load speed uses the voltage actually available.
Is the constant-power region real?
It is real for a machine with field-oriented control, current to spare and usually some rotor saliency; it is close to fiction for a small surface-magnet motor driven with six-step commutation and no current regulator. The page offers both models because the difference is not small: without field weakening the entire speed range above base speed is the ratio of the available voltage to the available voltage minus IR, which for a low-resistance machine is a few per cent.
My load is a propeller. Where will it actually run?
At the speed where its torque curve crosses the continuous envelope, which is the headline figure. A propeller’s torque goes as speed squared, so it rises steeply into a falling envelope and the crossing is sharp — which is also why a small change in bus voltage or in propeller pitch moves the operating point much less than people expect.
Should I size on peak torque or RMS torque?
Both, against different limits. The peak of your cycle has to fit inside the peak envelope or the drive will current-limit and the move will take longer than planned. The RMS of your cycle has to fit inside the continuous envelope or the winding will overheat. The average torque is not a limit against anything and is the number most often quoted.
Does this page handle a gearbox?
Not directly — enter the load reflected to the motor shaft. A gear ratio divides the load torque and multiplies the speed, and it divides a translating load’s inertia by the square of the ratio; work that out first with the gear ratio calculator and then bring the result here.
Related calculators
References
- Krishnan R. Permanent Magnet Synchronous and Brushless DC Motor Drives. CRC Press, 2010. Chapters on the PM synchronous machine’s flux-weakening operation and on the constant-torque and constant-power regions of the torque-speed envelope.
- Mohan N, Undeland T M, Robbins W P. Power Electronics: Converters, Applications and Design, 3rd ed. Wiley, 2003. The chapters on motor drives, for the constant-torque and constant-power regions of a drive’s envelope, and the inverter chapters for the voltage a six-step or PWM bridge can synthesise from a given DC bus. Edition and publisher verified by search; chapter numbers are not quoted because they were not.
- Infineon Technologies. Linear Mode Operation and Safe Operating Diagram of Power-MOSFETs, application note AP99007, V1.1, May 2017 (J. Schoiswohl) — the inverter-side limit behind the peak current figure, and why a drive’s peak rating is a short-pulse rating.