BLDC Efficiency Analyzer
Analyze BLDC motor efficiency at any operating point. Breaks down copper, iron, and mechanical losses. Finds the optimal current and RPM for peak efficiency.
Formula
Reference: Hanselman, D. — Brushless Permanent Magnet Motor Design
How It Works
This calculator breaks down BLDC motor losses into copper, iron, and mechanical components to determine efficiency across the operating range. Drone designers, EV engineers, and robotics developers use it to find the optimal current for maximum flight time or minimum thermal stress.
Total motor loss has three components. Copper loss dominates at high current and scales quadratically. Iron loss follows the Steinmetz equation: , where hysteresis loss ( term) dominates below 500 Hz and eddy current loss ( term) dominates above. For typical silicon steel laminations, and iron loss is roughly proportional to RPM. Mechanical loss from bearing friction and windage is approximately constant for a given speed.
The efficiency curve peaks at a specific current. Per Krishnan (2010), the optimal current for maximum efficiency is , where is the speed-dependent no-load loss and is phase resistance. This occurs when copper loss equals the sum of iron and mechanical losses -- the equal-loss principle.
No-load current measured at operating voltage directly gives (since copper loss at no-load is negligible). This single measurement anchors the entire efficiency model. Per IEC 60034-2-1, the preferred method for small motors is loss segregation from no-load and locked-rotor tests.
Worked Example
Analyzing a 2806.5 drone motor (Kv=1300) on 4S LiPo at hover. Specs: = 0.065 ohm (wye), = 1.8 A at 14.8V, hover throttle draws 8.5 A.
Step 1 -- Determine no-load losses: = = 14.8 x 1.8 = 26.6 W This includes iron loss + bearing friction + windage at operating speed
Step 2 -- Calculate copper loss at hover: Phase current (wye, trapezoidal drive): = 8.5 A = = 3 x 8.5 x 0.065 = 14.1 W Note: using 3 phases conducting simultaneously (simplified 6-step model)
Step 3 -- Total loss and efficiency: = = 26.6 + 14.1 = 40.7 W = 14.8 x 8.5 = 125.8 W = 125.8 - 40.7 = 85.1 W = 85.1 / 125.8 = 67.6%
Step 4 -- Find peak efficiency current: = where = 3 x 0.065 = 0.195 ohm = = 11.7 A At : = 11.7 x 0.195 = 26.7 W (equal-loss point) = 14.8 x 11.7 = 173.2 W, = 173.2 - 53.3 = 119.9 W = 119.9 / 173.2 = 69.2%
Result: Peak efficiency is 69.2% at 11.7 A. At 8.5 A hover the motor runs at 67.6% -- close to optimal. No-load losses (26.6 W) dominate at light loads, making this motor oversized for sub-5A applications.
Practical Tips
- ✓Measure no-load current at the actual operating voltage and RPM -- I0 varies significantly with speed because iron loss scales with frequency; a measurement at 50% throttle does not predict losses at 100% throttle
- ✓Measure phase resistance at operating temperature, not cold: copper resistance increases 0.393% per degree C, so a motor at 100C has 30% higher resistance than at 25C -- use
- ✓Operate the motor between 20-80% of the peak efficiency current -- below 20% no-load losses dominate (efficiency drops rapidly) and above 80% copper losses grow quadratically, both wasting battery energy
Common Mistakes
- ✗Measuring winding resistance with the motor hot after a flight and using it as the baseline: Phase resistance at 80C is 22% higher than at 25C, leading to overestimated copper losses in efficiency calculations -- always record temperature alongside resistance
- ✗Ignoring iron losses by assuming all electrical loss is I-squared-R: In high-Kv motors above 20,000 RPM, iron loss can exceed copper loss at moderate currents -- the Steinmetz eddy current term scales with frequency squared, making it the dominant loss mechanism at high speed
- ✗Running the motor continuously near stall current expecting it to survive: At stall, 100% of input power becomes heat in the windings with zero mechanical output -- even 5 seconds at stall can exceed the winding insulation temperature rating and cause permanent demagnetization of the rotor magnets
Frequently Asked Questions
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