From motor requirements
to the right power supply.

1. Define the wheel requirements

I use two operating points for the worked example.

The first point requires 10 Nm at 200 rpm. The second requires at least 2 Nm at 300 rpm. I assume this minimum torque is sufficient to rotate the wheel with its rim and attached button box.

I do not require the full 10 Nm at 300 rpm. Maximum torque and maximum speed do not need to occur simultaneously.

This does not mean the controller automatically limits torque to 2 Nm above 200 rpm. The two points define minimum design requirements, not the complete torque characteristic.

2. Enter the motor parameters

The example uses measurements from my hoverboard motor together with several design assumptions.

I assume three symmetrical star-connected windings and torque-producing current defined using the amplitude convention for Iq, with Id approximately zero.

The values 85%, 95%, 5 W and 50°C are assumptions of my model. They are not universal parameters for all motors and controllers.

3. Calculate the current needed for torque

For a BLDC/PMSM motor controlled using FOC, torque is approximately proportional to torque-producing current Iq.

T is torque in Nm, Kt is the torque constant in Nm/A, and Iq is torque-producing current in A. Rearranging gives the required current:

3.1. Current required for 10 Nm

3.2. Current required for 2 Nm

This is the current used by the FOC model to calculate torque, not the current drawn from the DC power supply.

For a FFBeast configuration where 100% Power Limit corresponds to 37 A, I can calculate the firmware setting:

For my motor, achieving 10 Nm therefore requires a setting of approximately 48% Power Limit.

4. Account for winding resistance and its temperature rise

Winding resistance is one of the most important supply-sizing parameters. It directly affects both heat produced in the motor and the voltage needed to maintain the required current.

4.1. Why does resistance matter so much?

Motor windings are made from copper wire. Copper is a good conductor, but it is not perfect: it has electrical resistance. When current flows through a winding, some electrical energy becomes heat.

The Joule–Lenz relationship describes this:

P is heat dissipation in W, I is conductor current in A, and R is resistance in Ω. The key point is that losses increase with the square of current. Doubling current quadruples winding losses if resistance stays constant.

For example, for a conductor resistance of 0.175 Ω:

These are example losses in one winding at the stated RMS current. I calculate total three-phase losses later, accounting for all three windings. This already shows why increasing motor torque can increase heat generation so quickly.

4.2. Resistance also affects the required voltage

There is a second, equally important effect. Maintaining current through a winding requires voltage. The steady-state resistive voltage drop is:

Higher current or resistance means a higher voltage drop. In a BLDC motor, the controller therefore needs enough voltage to produce the current required for torque. If voltage is insufficient, it cannot maintain the commanded current and available torque falls.

This is especially important at high speeds, when the rotating motor generates additional voltage. Section 6 explains that effect.

4.3. Why does resistance increase with temperature?

In metals such as copper, electrical resistance increases with temperature. In simplified terms, stronger lattice vibrations in heated copper make electron flow more difficult. A hot winding therefore has more resistance than the same winding at room temperature.

Over the temperature range relevant to my project, I use a linear approximation:

• Rcold: resistance at the measurement temperature [Ω].
• Rhot: resistance after heating [Ω].
• α: copper temperature coefficient [1/°C].
• ΔT: temperature rise relative to the measurement temperature [°C].

Near the reference temperature, this corresponds to approximately 0.393% more resistance per additional degree Celsius.

4.4. Calculation for my motor

A 50°C rise increases line-to-line resistance from 0.35 Ω to approximately 0.419 Ω. The percentage increase is:

My motor resistance therefore increases by almost 20%.

4.5. How does the resistance increase affect performance?

It causes two unfavorable effects at once.

First, copper losses increase. At the same current, the motor produces more heat. A 19.65% resistance increase also raises copper losses by 19.65% at unchanged current, adding to the thermal load on the windings.

Second, the required voltage increases. To maintain the same current, the controller must overcome a larger winding voltage drop. At low speeds, ample voltage headroom may make this unimportant. At high speeds, a motor already near the voltage limit may lose the ability to maintain current as resistance increases. Available torque, and therefore achievable mechanical power, then falls.

This is why the motor can behave differently when cold and after prolonged operation, even with the same supply voltage and Power Limit setting. My calculator accounts for this when calculating minimum supply voltage and power.

4.6. Line-to-line resistance vs. individual winding resistance

My motor has three symmetrical star-connected windings. I measured resistance between two motor wires. In this connection, the measurement current passes through two windings in series.

I continue using line-to-line resistance in subsequent calculations because the formulas already include the conversion factors. This lets me use the measurement directly, provided its definition remains consistent.

5. Calculate the equivalent speed constant — the Kt/Kv relationship

Two useful constants describe a BLDC/PMSM motor: Kt, the torque constant, and Kv, the speed constant. They describe the same electrical machine from different perspectives.

5.1. What does Kt mean?

Kt specifies torque per unit of torque-producing current. For my motor:

In simplified terms, each ampere of Iq produces approximately 0.56 Nm. A larger Kt gives more torque at the same current.

5.2. What does Kv mean?

Kv describes the relationship between motor speed and an appropriately defined voltage. With sufficient voltage, a larger Kv allows higher speed at the same reference voltage.

A higher Kv is not always better. A motor designed to achieve high speed at low voltage generally requires more current for the same torque.

5.3. Why are Kt and Kv related?

In a permanent-magnet motor, the same electromagnetic construction produces torque from current and generates voltage when the rotor turns. The torque constant and voltage constant are therefore closely related.

With consistent SI units and appropriately matched voltage and current definitions:

Ke is the voltage constant in the corresponding SI convention. Expressing speed in revolutions per minute gives:

This is the relationship for the equivalent constant used in my model. Other conventions also use:

The difference between 9.55 and 8.27 comes from different voltage definitions, not different motor physics. Before using a formula, I need to know whether Kv refers to phase voltage, line-to-line voltage, voltage amplitude or the DC bus voltage. I cannot mix these conventions arbitrarily.

5.4. What happens if I change Kt?

The constants are inversely proportional: if Kt increases, the corresponding Kv decreases, and vice versa.

A larger Kt produces more torque at the same current but requires more voltage at the same speed. A smaller Kt requires more current for a given torque, but an appropriately designed motor can achieve higher speed at the same voltage. This is a fundamental motor-design tradeoff.

Changing the physical winding can also change resistance and other parameters. Merely changing Kt mathematically in the calculator does not imply that all other real motor parameters would stay identical.

5.5. Calculation for my motor

I use this value next to calculate the voltage needed to maintain torque at the required speeds.

6. Calculate the voltage needed for the required torque and speed

This is a key calculation. A motor may produce high torque at low speed but fail to maintain it at higher speed because of back-EMF, or back electromotive force.

6.1. What is back-EMF?

A BLDC/PMSM motor has a permanent-magnet rotor and a wound stator. Controlled winding current produces a magnetic field that interacts with the rotor magnets to generate torque.

Rotor rotation also has another effect. As magnets move relative to the stator windings, their linked magnetic flux changes. Faraday’s law of induction states that changing magnetic flux induces voltage. The rotating motor therefore generates voltage in its own windings.

This happens whether the rotor is turned by the controller or by an external force. During electrically driven operation, this induced voltage opposes the voltage used by the controller to establish current, hence the name back electromotive force.

6.2. Why does back-EMF rise with speed?

Faster rotor rotation changes the linked flux more quickly, increasing the generated voltage. The simplified relationship is linear:

E is generated voltage, Ke is the motor voltage constant, and ω is angular speed. At zero speed, ω = 0 and E = 0. At higher speed, back-EMF rises in proportion to rotational speed.

6.3. Why does back-EMF limit torque?

Torque primarily depends on current. Maintaining torque means maintaining winding current, which requires voltage to overcome both winding resistance and back-EMF.

UEMF is the back-EMF-related voltage term; UR is the resistive voltage term. At low speed, back-EMF is small and the controller has ample voltage headroom for current regulation. At high speed, rising back-EMF consumes more of the available voltage budget.

Once the controller reaches its voltage limit, it cannot maintain the commanded current and available torque falls. This is why supply voltage matters so much for maintaining torque at high speed. A large supply current rating alone is insufficient: the controller also needs enough voltage to establish the required current.

6.4. Calculate the back-EMF voltage term

In this model, the coefficient converts the assumed voltage convention to line-to-line voltage amplitude.

The motor-generated voltage at 300 rpm is approximately 50% higher than at 200 rpm.

6.5. Calculate the resistive winding voltage drop

I also account for the voltage drop caused by current through the copper resistance. In the assumed line-to-line voltage convention:

This coefficient accounts for the individual winding resistance and the phase-to-line voltage relationship.

6.6. Calculate total required voltage

Although the second point requires five times less torque, it requires more voltage because back-EMF is much higher at 300 rpm. The highest required torque does not necessarily mean the highest required voltage.

7. Account for PWM limits and select the supply voltage

The 20.02 V and 21.61 V calculated above are required voltages in the motor model convention. They are not yet supply nameplate voltages. To size the supply, I need to consider how the controller generates winding voltages.

7.1. How does a FOC controller power the motor?

The MKS XDrive Mini uses a three-phase MOSFET power bridge. Each phase is driven by a pair of transistors: the high-side transistor connects the phase node to the positive supply rail, and the low-side transistor connects it to the negative rail.

The controller does not directly generate an ideal analog sine-wave voltage. Instead, it rapidly switches the transistors using pulse-width modulation (PWM). By changing their on-times, it adjusts average winding voltage. Coordinated switching of all three phases creates the rotating voltage vector needed to regulate current and torque.

7.2. What is SVPWM?

Space Vector Pulse Width Modulation combines switching states of the three bridge legs. Within each short PWM period, it selects the state durations required to produce the desired average voltage vector. This uses the DC bus voltage efficiently to control the three-phase motor.

Modulation capability is finite. Eventually, the requested vector becomes too large to reproduce with the chosen modulation method and measurement requirements, and the controller reaches its voltage limit.

7.3. Why not use the full 100% of DC voltage?

Bridge hardware and timing impose practical limits. MOSFETs require time to turn on and off. A short dead time separates switching of the high-side and low-side transistors in the same leg, preventing simultaneous conduction that could short the supply rails.

Gate-signal propagation delays and current-measurement constraints also matter. My controller measures phase currents using two shunt resistors.

7.4. How do current shunts work?

A shunt is a very low-resistance resistor that develops a small voltage drop when current flows through it:

The measurement circuit amplifies that signal, and the microcontroller ADC converts it to a digital value. The controller can then calculate actual current and compare it with the FOC current command.

With two shunts, measuring two branch currents is sufficient because the third can be reconstructed from:

7.5. Why can current sensing limit modulation?

For shunts in the low-side bridge legs, reliable current measurements are available only in suitable switching states. In particular, current through a conducting low-side transistor can pass through that leg’s shunt. The controller must provide an adequate measurement window.

Immediately after switching, noise and ringing can disturb the signal. It needs time to settle, followed by the ADC sampling time. At very high PWM duty cycles, low-side conduction may become too short for reliable measurement, making phase-current reconstruction difficult.

This matters especially for one- or two-shunt low-side sensing. Practical modulation range can therefore also be limited by current-sampling requirements. However, not every two-shunt design has the same limit: measurement electronics, PWM frequency, sampling strategy and control algorithm all affect it.

7.6. Why use 85% in the calculator?

I treat this as a conservative voltage utilization factor, leaving headroom for SVPWM operation, switching times, dead time, current-sampling windows and the chosen control method.

85% is not a universal physical maximum for SVPWM. It is my design assumption, used to avoid operating at the idealized bridge limit.

7.7. Calculate the minimum supply voltage

Both conditions must be met, so I choose the larger voltage. This is the requirement on the available motor voltage budget, not yet the minimum nominal DC supply voltage. Divide it by voltage utilization:

The required minimum supply voltage is approximately 25.4 V, not 21.61 V. The two values describe different calculation stages and must not be confused.

The displayed 25.43 V uses unrounded intermediate values, matching the calculator. Dividing the rounded 21.61 V by 0.85 gives approximately 25.42 V.

7.8. Select a 36 V supply

I do not want to select voltage exactly at the calculated limit, so I choose a standard 36 V DC supply.

This provides a 30.6 V budget against a requirement of approximately 21.61 V, leaving headroom to maintain current and torque at the specified speeds.

Higher voltage does not automatically increase torque at standstill, because torque still primarily depends on current. It does allow that current to be maintained at higher speed as back-EMF increases.

8. Calculate mechanical power

With the minimum voltage established and a supply voltage selected, I can calculate power. First, I need to explain the sources of the motor’s energy demand.

8.1. What makes up the power draw?

An electric motor converts electrical energy into mechanical energy, but the process is not ideal. Some energy becomes mechanical work and the rest is lost, mainly as heat. My model separates four terms:

• Mechanical power, Pmech: power delivered to the shaft, determined by torque and speed.
• Copper losses, PCu: heat generated by current flowing through winding resistance.
• Controller losses, Pdriver: assumed power-stage losses, including MOSFET conduction and switching.
• Electronics draw, Pelectronics: power needed by the microcontroller, sensing circuits and other controller electronics.

This is the power the supply must deliver at the motoring operating point being analyzed.

8.2. What is mechanical power?

Power is the rate of doing work. In rotation, work depends on torque and the angle through which the shaft turns. For constant torque:

W is work in J, T is torque in Nm, θ is rotation angle in rad, and ω is angular speed in rad/s. Because angular speed is the time derivative of angle, mechanical power is torque times angular speed.

Here, 2π radians is one full turn, and dividing by 60 converts minutes to seconds.

8.3. Why does high torque not always mean high mechanical power?

A stationary motor can generate high torque without doing mechanical work through shaft rotation.

It can still draw electrical energy: winding current produces copper losses. A motor holding 10 Nm at zero speed can therefore heat up significantly despite zero mechanical output. If the same torque is maintained while speed rises, mechanical power increases in proportion to speed.

8.4. Mechanical power at 10 Nm and 200 rpm

8.5. Mechanical power at 2 Nm and 300 rpm

The first point requires more than three times the mechanical power. Although the second point needs more voltage, the first requires more power in the operating scenario I have defined.

9. Calculate copper losses

Not all electrical input becomes mechanical output. Winding resistance dissipates some as heat, as explained in Section 4:

For three star-connected windings with balanced phase currents and Id approximately zero:

When Iq denotes current amplitude in the assumed FOC convention, and resistance is measured between two motor wires, conversion gives:

Where does 0.75 come from?

1. The RMS value of a sinusoidal current equals its amplitude divided by √2.
2. Individual star-winding resistance is half the measured line-to-line resistance.

For operating losses I use the temperature-adjusted line-to-line resistance Rhot.

9.1. Copper losses at 10 Nm

9.2. Copper losses at 2 Nm

The current values shown are rounded; the loss results use the unrounded currents calculated from torque.

10. Account for controller losses

The controller also consumes power. Its power-stage losses include transistor conduction and switching. In my simplified model, power-stage efficiency is:

I do not simply multiply the output power by 5%. For 95% efficiency, the more accurate loss factor relative to that output is:

10.1. Controller losses at 10 Nm / 200 rpm

10.2. Controller losses at 2 Nm / 300 rpm

In both cases, I add a separate fixed 5 W for the controller electronics.

11. Calculate total required supply power

I can now combine all four terms:

Pmech is mechanical power, PCu is copper loss, Pdriver is controller power-stage loss, and Pelectronics is the remaining electronics draw, assumed to be 5 W.

11.1. Power balance at 10 Nm / 200 rpm

11.2. Power balance at 2 Nm / 300 rpm

The first point is more demanding in supply power. I therefore use it for the following calculation:

This estimates power for the defined motoring scenario, not the absolute maximum the controller can draw in every possible condition. Totals use unrounded values, so sums of displayed rounded terms may differ slightly.

12. Add a margin and determine the supply rating

I do not want to size the supply exactly at the calculated demand, so I add a 20% margin:

For these requirements, I would therefore look for a supply rated at at least approximately 400 W, provided its selected voltage is sufficient.

12.1. What supply current rating is needed?

For the selected 36 V supply, including the power margin:

This is the required DC supply current rating, not motor phase current. Earlier, I calculated approximately 17.86 A of Iq for 10 Nm. There is no contradiction: the FOC controller converts DC bus energy into the appropriate phase voltages and currents. DC supply current depends on total electrical power and supply voltage.

13. Interpret the final result

The example assumes 10 Nm maximum target torque, at least 2 Nm during self-rotation, operating speeds of 200 and 300 rpm respectively, Kt = 0.56 Nm/A, line-to-line resistance 0.35 Ω, winding temperature rise 50°C, controller efficiency 95%, voltage utilization 85% and a 20% power margin.

Important model limitation

I size power for a defined scenario: 10 Nm at 200 rpm and at least 2 Nm at 300 rpm. “At least 2 Nm” does not mean the controller cannot produce more torque at 300 rpm.

For example, at 36 V the model also allows 10 Nm at 300 rpm. If the controller commands the full 10 Nm in the direction of rotation there, demand rises to approximately 441 W before the margin, or approximately 529 W including 20%.

To guarantee operation at any available torque throughout the speed range, I would need to analyze the full torque–speed characteristic.

During braking, when motor torque opposes rotation, energy can flow back toward the DC bus. That requires a separate regeneration analysis and must not be confused with positive supply power draw.

The principle behind the calculator

First, determine the voltage needed to maintain the required torque at the required speed. Then calculate the power needed for mechanical work and electrical losses.

This avoids sizing the supply solely from maximum motor current or no-load speed. It accounts for the practical requirements of a Direct Drive wheel: high torque during fast hand movements, plus adequate speed and torque during self-rotation.

The result is a starting point for informed supply selection, with explicit assumptions, equations and limits of the simplified model.