Winding Resistance vs Temperature
Calculate motor winding resistance at operating temperature using the copper temperature coefficient of resistance.
Formula
How It Works
This calculator determines DC motor winding resistance and its effect on copper losses and speed regulation. Motor repair technicians, quality control engineers, and drive system designers use it to diagnose winding faults and predict performance variation with temperature. Winding resistance measurement is the primary diagnostic test for detecting shorted turns, open windings, and connection problems.
Per IEC 60034-4, armature resistance (R_a) includes conductor resistance plus brush contact resistance for brushed motors. Copper resistance follows the temperature coefficient equation: R(T) = R_25 × [1 + 0.00393 × (T - 25)], where 0.00393/°C is copper's resistance coefficient per IEC 60028. At typical operating temperature of 100°C, resistance increases 29.5% above the 25°C value.
Resistance directly impacts three performance metrics: (1) Copper losses P_Cu = I²×R_a, comprising 30-60% of total motor losses per IEEE 112; (2) Speed regulation—voltage drop I×R_a reduces back-EMF voltage and speed; (3) Maximum current at stall I_stall = V/R_a. A motor with 2Ω armature resistance on 24V supply draws 12A stall current—this determines fuse sizing and driver current capability. Per NEMA MG-1, winding resistance tolerance is ±10% from nameplate value at 25°C.
Worked Example
A 48V BLDC motor for an e-scooter has phase resistance 0.15Ω (line-to-line) at 25°C. Operating winding temperature reaches 110°C. Rated current is 30A continuous.
Step 1 — Calculate hot resistance: R_hot = R_25 × [1 + 0.00393 × (T - 25)] R_hot = 0.15 × [1 + 0.00393 × (110 - 25)] R_hot = 0.15 × [1 + 0.334] = 0.15 × 1.334 = 0.200Ω
Step 2 — Calculate copper losses at rated current: P_Cu_cold = I² × R = 30² × 0.15 = 135W P_Cu_hot = 30² × 0.200 = 180W Hot operation increases copper loss by 33%
Step 3 — Assess speed regulation impact: Voltage drop cold: I × R = 30 × 0.15 = 4.5V (9.4% of supply) Voltage drop hot: 30 × 0.200 = 6.0V (12.5% of supply) Speed reduction due to temperature: additional 3.1% at full load
Step 4 — Verify stall current capability: I_stall_hot = V / R = 48 / 0.200 = 240A Controller must handle 240A peak or implement current limiting
Result: At 110°C, winding resistance increases 33% from 0.15Ω to 0.20Ω. This raises copper losses from 135W to 180W and reduces loaded speed by an additional 3.1%. Design thermal management to limit temperature rise or derate continuous current.
Practical Tips
- ✓Per IEEE 1415 motor diagnostics, resistance 10% below datasheet indicates shorted turns (lower impedance path); 10% above indicates high-resistance joints, broken strands, or brush wear
- ✓Always normalize measurements to 25°C reference: R_25 = R_measured / [1 + 0.00393 × (T_measured - 25)] per IEC 60034-1 to compare with specifications
- ✓For BLDC motors, measure line-to-line (phase-to-phase): star-wound motors read 2× single-phase resistance; delta-wound read 2/3× per phase—verify winding configuration before calculating
Common Mistakes
- ✗Using standard multimeter for low resistance: Per IEC 60034-4, contact resistance and meter error introduce ±0.1-0.5Ω error; use 4-wire Kelvin measurement for resistances below 10Ω to achieve ±1% accuracy
- ✗Ignoring brush resistance in brushed DC motors: Carbon brush contact adds 0.1-0.5Ω total (0.05-0.25Ω per brush per Mersen brush specifications)—this is part of effective armature circuit resistance
- ✗Assuming cold and hot resistance are equal: At 100°C, copper resistance is 29.5% higher than at 25°C per IEC 60028; neglecting this causes 30% underestimate of hot copper losses and speed regulation
Frequently Asked Questions
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