{"id":3507,"date":"2022-03-09T19:10:54","date_gmt":"2022-03-09T19:10:54","guid":{"rendered":"https:\/\/novanta.com\/robotics-automation\/technical-paper\/power-budgeting-and-motor-sizing-for-battery-applications\/"},"modified":"2026-04-28T05:59:25","modified_gmt":"2026-04-28T05:59:25","slug":"power-budgeting-and-motor-sizing-for-battery-applications","status":"publish","type":"novanta_tech_paper","link":"https:\/\/novanta.com\/robotics-automation\/technical-paper\/power-budgeting-and-motor-sizing-for-battery-applications\/","title":{"rendered":"Power Budgeting and Motor Sizing for Battery Applications"},"content":{"rendered":"\n
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Introduction<\/h2>\n\n\n\n

While motor output power is an important consideration for all applications, motor efficiency is equally important for power-limited systems such as battery-operated applications. This paper will provide the equations needed to calculate input power, output power, motor power losses, and efficiency for properly sizing a motor, servo drive, and power supply for your application. It will also provide helpful examples of calculating motor power while clearing up common misconceptions about power supply and motor current.<\/p>\n\n\n\n

Conservation of Power<\/h2>\n\n\n\n

It is important to understand that the power supplied to a motor is conserved, as electrical energy is converted to mechanical and thermal energy. That means that the difference in the power supplied to the motor and the power output from the motor is power lost through inefficiencies like heating and frequency-dependent hysteretic losses. A 100% efficient motor would be able to convert 100% of the input power to output power. However, all motors experience some degree of power loss. Let\u2019s break down the dominant power components in motor operation.<\/p>\n\n\n\n

Conservation of Power<\/h3>\n\n\n\n

All power to the motor is ultimately sourced from the bus power supply unit. If the goal is to choose a motor given an existing power budget, meaning the power supply voltage and current are predetermined, the available input power to the motor can be calculated as follows.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.1<\/p><\/p>\n\n\n\n

where:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.2<\/p><\/p>\n\n\n\n

Alternatively, if a specific application torque-speed point must be achieved and drive efficiency is known, we can work backwards to size the correct power supply unit using the equations below.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.3<\/p><\/p>\n\n\n\n

where:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.4<\/p><\/p>\n\n\n\n

If we assume a 100% efficient drive and therefore no power losses are present, supply power is equal to motor input power.<\/p>\n\n\n\n

A common misconception is that motor current and power supply current are the same value. This can be true, but only when the motor is in the stalled condition and no output power is being produced. In this case, it also requires that the motor\u2019s thermally limited stall current is equal to the bus voltage divided by the motor\u2019s phase-to-phase resistance. In the stalled condition, equation 4 becomes:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.5<\/p><\/p>\n\n\n\n

Using equation 5 and substituting power supply current for motor current gives:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.6<\/p><\/p>\n\n\n\n

Isolate motor current in equation 6:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.7<\/p><\/p>\n\n\n\n

The power supply current and motor current are only equivalent when the motor\u2019s thermally limited stall current is equal to the power supply voltage divided by the phase-to-phase resistance of the motor winding. As stated above, although we show that this situation is in fact possible, it is often not the case, as motors in most applications are not operating at the stall condition, nor is it common that equation 7 is true.<\/p>\n\n\n\n

Motor Output Power<\/h3>\n\n\n\n

Output power of a motor is the resulting power that is available to move the load after all system losses have been accounted for.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.8<\/p><\/p>\n\n\n\n

Output power can be calculated by multiplying the application speed \"Application and torque (\ud835\udc41\ud835\udc5a).<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.9<\/p><\/p>\n\n\n\n

Motor Power Losses<\/h3>\n\n\n\n

There are two types of power losses in a motor \u2013 core losses and copper losses. While power losses in a torque motor are dominated by copper losses, it is important to understand what causes each source and how they affect the output power of a motor.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.10<\/p><\/p>\n\n\n\n

Core losses are caused by eddy currents in the stator\u2019s iron core. While providing a rigid support for the copper windings and completing the magnetic circuit of the motor, the iron core also allows for eddy currents which create torque ripple in the motor. This is minimized by constructing the stator stack out of iron laminations that are bonded together. Eddy current losses are frequency-dependent and increase in high-speed applications. For certain motor designs where high speeds are targeted (>5 kRPM), frequency-dependent losses should be considered when calculating motor efficiency however, they are negligible to copper losses in our torque motors. Therefore, we will assume that motor power losses are equal to copper losses for the rest of this paper.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.11<\/p><\/p>\n\n\n\n

Copper losses are caused by heating, as current flows through the windings of a motor. Three-phase brushless motor theory tells us that power loss can be calculated using the equation below where the single-phase resistance is known.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.12<\/p><\/p>\n\n\n\n

Equation 12 assumes that the single-phase resistance can be measured. However, generally the neutral tap of wye-wound motors is not accessible \u2013 making it impossible to measure the single-phase resistance. Therefore, we must introduce phase-to-phase resistance into the equation above.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.13<\/p><\/p>\n\n\n\n

If current is not specified in units of \ud835\udc34\ud835\udc45\ud835\udc40\ud835\udc46<\/sub>, use the equations below to convert current to units \ud835\udc34\ud835\udc5d\ud835\udc52\ud835\udc4e\ud835\udc58 \ud835\udc5c\ud835\udc53 \ud835\udc60\ud835\udc56\ud835\udc5b\ud835\udc52<\/sub> or \ud835\udc34\ud835\udc37\ud835\udc36<\/sub>.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.14<\/p><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.15<\/p><\/p>\n\n\n\n

Now plug \ud835\udc3c\ud835\udc37\ud835\udc36<\/sub> or \ud835\udc3c\ud835\udc5d\ud835\udc52\ud835\udc4e\ud835\udc58 \ud835\udc5c\ud835\udc53 \ud835\udc60\ud835\udc56\ud835\udc5b\ud835\udc52<\/sub> into equation 13: <\/p>\n\n\n\n

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\"\"<\/figure>\n\n\n\n

Eq.16<\/p><\/p>\n<\/div>\n\n\n\n

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\"\"<\/figure>\n\n\n\n

Eq.17<\/p><\/p>\n<\/div>\n<\/div>\n\n\n\n

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For more information on motor unit conversion, please read the technical paper linked here: Motor Unit Conversions – What They Mean and How They are Used<\/a> – Applimotion Direct Drive Motors<\/a> (celeramotion.com)<\/p>\n\n\n\n

<\/strong>Overall power losses can also be calculated if the motor input and output powers are known.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.18<\/p><\/p>\n<\/div>\n<\/div>\n\n\n\n

Motor Efficiency<\/h3>\n\n\n\n

Motor efficiency is the ratio of the motor\u2019s output power to input power. While 100% efficiency is ideal, in application, all motors will experience power losses that decrease the overall efficiency of the motor.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.19<\/p><\/p>\n\n\n\n

While motor efficiency is not commonly specified on motor datasheets, it is synonymous with the motor constant (\ud835\udc3e\ud835\udc5a<\/sub>). The motor constant reflects the motor\u2019s ability to output torque given power losses.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Eq.20<\/p><\/p>\n\n\n\n

Therefore, if you compare two motors of the same frame size and one has a higher \ud835\udc3e\ud835\udc5a<\/sub>, that means for the same power loss, the motor with a higher \ud835\udc3e\ud835\udc5a<\/sub> has a higher torque output. Alternatively, a higher Km motor can output the same torque with lower power losses. For more information on the motor constant, please read our technical paper “Why Motor Constant Matters in Thermally Limited Applications”<\/a>.<\/a><\/p>\n\n\n\n

To determine what motor constant is needed for a specific torque requirement, first, calculate the dissipated power, then use equation 20 to calculate \ud835\udc3e\ud835\udc5a<\/sub>.<\/p>\n\n\n\n

Test Motors:<\/p>\n\n\n\n