Motor Copper Loss Calculator
Calculate winding I²R copper loss for DC motors, steppers, three-phase motors and BLDC phase references. MOT-006 focuses on the resistive heat produced inside energized windings, including hot resistance, allowed-current solves, explicit phase sums and copper-loss share.
This is not a full motor thermal, efficiency, inverter or drive simulation. Use measured RMS current and the winding resistance that matches the current path you are modeling.
Engineering tool
Motor Copper Loss Calculator
Calculate motor winding I²R copper loss for single windings, multiple windings, steppers, three-phase references, temperature rise, allowed-current limits, and loss share.
Calculation mode
Parameter panel
Result console
- Copper Loss
- 4W
- Voltage Drop
- 2V
- Current
- 2A
- Resistance
- 1Ω
Reference table
| Current Convention | RMS |
|---|
Motor Copper Loss Formula Audit
| Copper Loss Definition | Motor copper loss is winding resistive loss: Pcu = I²R. |
|---|---|
| Current Definition | Use DC current for steady DC windings or true RMS current for time-varying phase waveforms. |
| RMS Current Convention | Average current is not generally interchangeable with RMS current for I²R loss. |
| DC Current Boundary | For constant DC current, DC magnitude and RMS value are the same. |
| Single-Winding Formula | Pcu = I²R and Vdrop = IR. |
| Multiple-Winding Formula | Ptotal = Nenergized × I²R for equal energized windings. |
| Per-Phase Formula | Each phase contributes Iphase,rms²Rphase. |
| Explicit Phase Sum | Pcu = Ia²Ra + Ib²Rb + Ic²Rc. |
| Balanced Three-Phase Formula | Pcu,total = 3Iphase,rms²Rphase when current is true per-phase RMS. |
| BLDC Current Convention Boundary | BLDC six-step, sinusoidal and FOC definitions vary; the calculator does not infer phase RMS from bus current. |
| Line / Phase Current Boundary | No automatic √3 conversion is applied. |
| Per-Phase / Line-to-Line Resistance Boundary | The calculator expects the resistance represented by the selected current path. |
| Stepper One-Phase Model | One energized phase uses P = I²R. |
| Stepper Two-Phase Model | Two equal energized phases use P = 2I²R. |
| Microstepping Boundary | Ideal sine/cosine mode assumes equal phase resistance and Ia² + Ib² = Ipk². |
| Temperature Resistance Formula | R2 = R1[1 + α(T2 - T1)]. |
| Copper Temperature Coefficient | Default copper α is 0.00393 per °C. |
| Fixed-Current Hot-Loss Assumption | Hot loss is calculated at the same current unless the current is changed explicitly. |
| Solve-Current Formula | I = sqrt(Ptotal / (N R)). |
| Copper-Loss Share Definition | Pcu% = Pcu / Pin × 100%. |
| Remaining-Power Naming Boundary | Pin - Pcu is labeled power remaining after accounted copper loss, not shaft power. |
| PWM Ripple Boundary | PWM ripple changes RMS current; use true RMS when available. |
| AC Resistance Boundary | Skin effect and proximity-effect AC resistance are outside V1. |
| MOT-003 Scope Boundary | MOT-003 covers stall current and locked-rotor current stress. |
| MOT-005 Scope Boundary | MOT-005 covers overall efficiency; copper loss is only one loss component. |
Formula
Formula reference
Motor copper-loss formulas
The formulas are magnitude-based. Signed current direction is not used for copper-loss heating.
Pcu = I²RVdrop = I RPtotal = Nenergized I²RPcu = Ia²Ra + Ib²Rb + Ic²RcP3φ = 3 Iphase,rms² RphaseR2 = R1[1 + α(T2 - T1)]I = sqrt(Ptotal / (N R))Pcu% = Pcu / Pin × 100%Variable definitions
- Pcu
- motor winding copper loss
- I
- DC magnitude or true RMS current
- R
- winding or phase resistance for the selected current path
- Nenergized
- energized winding count
- α
- resistance temperature coefficient
- Pin
- electrical input power at the selected boundary
Motor Copper Loss Formula Audit
| Copper Loss Definition | Pcu is winding resistive loss. |
|---|---|
| Current Definition | Use DC magnitude for constant DC current and true RMS for time-varying current. |
| RMS Current Convention | I²R loss requires RMS current for waveforms. |
| DC Current Boundary | For constant current, DC magnitude equals RMS. |
| Single-Winding Formula | Pcu = I²R and Vdrop = IR. |
| Multiple-Winding Formula | Ptotal = Nenergized × I²R. |
| Per-Phase Formula | Each phase is calculated from its own RMS current and resistance. |
| Explicit Phase Sum | Pcu = Ia²Ra + Ib²Rb + Ic²Rc. |
| Balanced Three-Phase Formula | Pcu,total = 3Iphase,rms²Rphase. |
| BLDC Current Convention Boundary | No bus-to-phase current conversion is assumed. |
| Line / Phase Current Boundary | No automatic √3 factor is applied. |
| Per-Phase / Line-to-Line Resistance Boundary | Use resistance matching the current path. |
| Stepper One-Phase Model | P = I²R. |
| Stepper Two-Phase Model | P = 2I²R. |
| Microstepping Boundary | Ideal sine/cosine model uses Ia² + Ib² = Ipk². |
| Temperature Resistance Formula | R2 = R1[1 + α(T2 - T1)]. |
| Copper Temperature Coefficient | Default α = 0.00393 / °C. |
| Fixed-Current Hot-Loss Assumption | Hot loss holds current constant. |
| Solve-Current Formula | I = sqrt(Ptotal / (N R)). |
| Copper-Loss Share Definition | Pcu% = Pcu / Pin × 100%. |
| Remaining-Power Naming Boundary | Pin - Pcu is not labeled shaft power. |
| PWM Ripple Boundary | Ripple should be reflected in true RMS current. |
| AC Resistance Boundary | Skin and proximity effects are not modeled. |
| MOT-003 Scope Boundary | Stall-current analysis remains in MOT-003. |
| MOT-005 Scope Boundary | Overall efficiency remains in MOT-005. |
Worked Examples
| Example | Calculation | Result |
|---|---|---|
| I = 2 A, R = 1 Ω | Pcu = 2² × 1 | 4 W |
| I = 5 A, R = 0.2 Ω | Pcu = 5² × 0.2 | 5 W |
| Two windings, 2 A each, 1 Ω each | Ptotal = 2 × 2² × 1 | 8 W |
| Balanced 3-phase: Iphase = 10 A, Rphase = 0.1 Ω | 3 × 10² × 0.1 | 30 W |
| Explicit Ia = Ib = Ic = 10 A, R = 0.1 Ω | 10 + 10 + 10 | 30 W |
| Explicit Ia = 10 A, Ib = 8 A, Ic = 6 A, R = 0.1 Ω | 10²·0.1 + 8²·0.1 + 6²·0.1 | 20 W |
| Stepper one phase: I = 1.5 A, R = 2 Ω | 1.5² × 2 | 4.5 W |
| Stepper two phases: I = 1.5 A, R = 2 Ω | 2 × 1.5² × 2 | 9 W |
| Ideal sine/cos microstep: Ipk = 1.5 A, R = 2 Ω | R × Ipk² | 4.5 W |
| Current doubled | (2I)²R / I²R | Loss becomes 4× |
| Resistance doubled | I²(2R) / I²R | Loss becomes 2× |
| I = 2 A, R = 1 Ω voltage drop | Vdrop = IR | 2 V |
| 1 Ω from 20°C to 100°C, α = 0.00393 | Rhot = 1[1 + 0.00393(80)] | 1.3144 Ω |
| 2 A with hot resistance 1.3144 Ω | 2² × 1.3144 | 5.2576 W |
| Hot loss compared with cold loss | 5.2576 W > 4 W | Hot loss is higher at fixed current |
| Allowed loss 10 W, R = 1 Ω, N = 2 | I = sqrt(10 / 2) | 2.23607 A |
| Allowed loss 0 W | I = sqrt(0 / NR) | 0 A |
| Allowed current with R = 0 Ω | Denominator invalid | Rejected |
| Pin = 100 W, Pcu = 20 W | 20 / 100 × 100% | 20% |
| Pin = 100 W, Pcu = 20 W | 100 - 20 | 80 W remaining after accounted copper loss |
| 1000 mΩ | Unit conversion | 1 Ω |
| 1000 mA | Unit conversion | 1 A |
| Single winding vs N = 1 | 1 × I²R | Same result |
| Balanced 3-phase vs explicit symmetric sum | 3I²R equals Ia²Ra + Ib²Rb + Ic²Rc | Same result |
| Independent reference | 3 A and 500 mΩ | Pcu = 4.5 W; Vdrop = 1.5 V |
| Point comparison | 4 A vs 2 A at same resistance | Point B loss is 4× |
Engineering Notes
Motor Copper Loss
Copper loss is the heat generated by current flowing through winding resistance.
Winding Resistance
Low winding resistance can still create high loss because motor current may be large.
RMS Current
RMS current is the correct current for resistive heating when waveforms vary with time.
DC Current
A constant DC winding current can be used directly in the I²R formula.
Phase Current
For three-phase motors, use actual phase RMS current and per-phase resistance.
Line Current
Line current is not automatically the same as phase current in every connection and drive mode.
Line-to-Line Resistance
Measured terminal resistance may need interpretation before it becomes per-phase resistance.
Stepper Motor
Holding loss depends on how many phases are energized and how current is regulated.
Microstepping
Ideal sine/cosine microstepping keeps the current-vector magnitude constant, but real drivers and motors have ripple and tolerance.
Hot Resistance
Copper resistance rises with temperature, increasing fixed-current copper loss.
PWM Ripple
Ripple increases RMS current and therefore can increase winding heat.
BLDC Motor
Bus current, phase current, RMS current and peak current are different boundaries.
Thermal Limit
Copper loss must be reviewed with motor thermal resistance, cooling, duty cycle, and insulation class.
AC Resistance
High-frequency skin and proximity effects can make effective resistance higher than DC resistance.
Efficiency Boundary
Copper loss is only one part of motor efficiency and should not be treated as total loss.
Common Mistakes
- Using average current instead of RMS current for PWM or sinusoidal waveforms.
- Confusing DC bus current with BLDC phase RMS current.
- Using line-to-line resistance as per-phase resistance without checking the winding connection.
- Assuming every BLDC drive can use the same 3I²R interpretation.
- Ignoring hot winding resistance.
- Calling Pin - Pcu shaft power.
- Treating copper loss as total motor loss.
- Forgetting that copper loss increases with the square of current.
- Using negative current or signed current in a magnitude-only I²R calculation.
- Ignoring PWM ripple current.
- Ignoring skin effect or proximity effect in high-frequency windings.
- Using stepper holding-current rules without knowing one-phase, two-phase, or microstep operation.
- Assuming calculated tank or winding heat guarantees safe motor temperature without thermal validation.
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Open CalculatorFAQ
Support reference
FAQ
How do I calculate motor copper loss?
Use Pcu = I²R for each winding or phase, then sum the energized windings or phases that carry current.
Should I use RMS current or average current?
Use true RMS current for time-varying waveforms because copper loss depends on current squared. Average current is not generally interchangeable.
Can I use DC current for a brushed DC motor?
For steady DC current in a winding, the DC magnitude equals the RMS value, so Pcu = I²R applies directly.
How do I calculate copper loss for multiple windings?
For equal windings with equal current, multiply the per-winding I²R loss by the number of energized windings.
How do I calculate BLDC copper loss?
Use per-phase RMS current and per-phase resistance when known. Do not infer phase RMS current from DC bus current without a defined drive waveform and measurement boundary.
How do I calculate three-phase motor copper loss?
For balanced phases with true per-phase RMS current, use Pcu,total = 3Iphase²Rphase.
Can I use line current directly?
Only when line current is the same current that flows through the resistance represented in the model. This calculator does not apply automatic square-root-three conversions.
Can I use line-to-line resistance directly?
Use caution. Wye line-to-line resistance is often about twice per-phase resistance, while delta connections are more complex. This calculator expects the resistance for the selected current path.
How do I calculate stepper motor copper loss?
One energized phase uses I²R, two equal energized phases use 2I²R, and the ideal sine/cosine microstep model with equal phase resistance uses R Ipk².
Why does hot winding resistance matter?
Copper resistance rises with temperature, so the same current produces more copper loss as the winding heats.
What copper temperature coefficient should I use?
For copper, a common first-pass value is 0.00393 per °C near room temperature. Use winding material and temperature data when available.
How does PWM ripple affect copper loss?
Ripple increases RMS current relative to a smooth average current. If true RMS current is known, use it directly.
Is copper loss the same as total motor loss?
No. Total loss can also include iron loss, friction, windage, brush loss, driver loss, stray loss, and high-frequency AC effects.
Does copper loss determine motor efficiency?
Copper loss is one contributor to efficiency. Use the Motor Efficiency Calculator when you need the overall input/output efficiency boundary.
Can copper loss be negative?
No. If a calculation seems to imply negative loss or remaining power below zero, the input boundary or units are inconsistent.
Why does doubling current quadruple copper loss?
Because I²R loss depends on current squared, so 2× current produces 4× copper loss for the same resistance.
Engineering Disclaimer
This calculator provides first-pass copper-loss estimates. Production motor design should verify winding resistance, RMS current, thermal limits, drive waveform, cooling, insulation class, duty cycle, manufacturer data and measured temperature rise.