# Servomotor Power Supply Sizing

Correctly sizing a servomotor and drive power supply is critical to optimizing system cost. Power supply capacity is important in battery-powered mobile applications. To size servo drive power supplies, it is essential to understand power conversion and subsequent losses. This paper covers the power conversion process, losses and system efficiency; techniques to quantify electrical and mechanical power; an empirical method to size a power supply well as efficiency and efficiency design tradeoffs.

## Electrical Model

Fig. 1 shows the electrical model of a servomotor system. The power supply comprises a bridge rectifier and bus capacitor. For mobile applications, these would be replaced by a battery. Either way a DC voltage, VBUS, is supplied to the servo drive inverter power stage. The inverter is modeled as a direct connection from power supply to motor. The motor winding is modeled as a resistance (R) and inductance (L). The back-EMF voltage source represents the generator action of the motor as it rotates. Back-EMF increases with speed and opposes VBUS.

The magnitude of VBUS must be enough to drive current to produce the desired peak torque (T) while being opposed by the maximum back-EMF. Neglecting motor inductance, VBUS can be calculated to a close approximation as show below (ω is the maximum angular velocity of the application):

$$I_{\max }=\frac{T_{pk}}{K_T}$$

$$V_{BUS}\approx I_{\max }\ast R+EMF\approx I_{pk}\ast R+K_E\ast \omega$$

If KT and KE are specified as RMS values they should be converted to peak (crest) values.
To determine the required power capacity of the supply it is necessary to look at system losses and subsequent efficiency of both electrical and mechanical components.

## Power Conversion and Losses

Fig. 2 shows the chain of power conversion in the motion control system. The servo drive converts DC power to AC power. The servomotor converts electrical power to mechanical power. A gearbox converts low torque, high speed to high torque, low speed. Losses are inevitable during these conversions. The magnitude of the losses vary by technology and application, but typical values are shown in Fig. 2.

Primary losses in the conversion process are outlined below. Secondary effects include cable conduction losses and servo drive quiescent power consumption. These losses will be considered negligible in this discussion.

Drive Conduction Losses
Power transistors in a servo drive inverter are not ideal switches and have finite on-resistance. During the on-time of the PWM duty cycle, there are conduction losses in accordance with the power equation I2R.

Drive Switching Losses
PWM transitions are not instantaneous and there is conduction during the transistor on/off state changes. Drive switching losses are proportional to PWM frequency.

Motor Copper Losses Current
(IPH) in a motor phase winding of resistance (RPH) generates heat defined by the power equation, I2R. Copper losses increase with temperature since RPH increases with temperature.

Motor Core Losses
As the rotor turns the changing magnetic field induces eddy currents in the motor iron resulting in conduction losses. As the stator magnetic field rotates, reorientation of magnetic polarity in the motor iron requires energy and some energy is lost due to intermolecular friction. Motor core losses are proportional to speed and the number of rotor poles.

Current Ripple Losses
Current ripple resulting from pulse width modulation has an RMS component which causes heating in the motor winding. Current ripple is reduced as winding inductance and PWM switching frequency are increased. See Motor Inductance and PWM Frequency for further details.

Gearbox Losses
Gearbox losses are caused by gear tooth and shaft seal friction. There are many parameters to be considered: load torque, speed, temperature, lubricant, gear ratio. The specifications for gearboxes sometimes assume ideal conditions so care should be taken in determining losses.

## System Efficiency

The overall system efficiency is calculated by multiplying the individual component efficiencies. An example is shown in Fig. 3.

The servo drive datasheet typically specifies efficiency. The specification assumes a particular PWM frequency and commutation method. If the drive is configured to a higher PWM frequency, switching losses will be higher so the efficiency must be derated. If efficiency is specified for trapezoidal commutation and FOC is used, the efficiency must also be derated.

Motor efficiency is typically not specified and must be estimated from copper and core losses. Assuming FOC and SVPWM (Space Vector Pulse Width Modulation) copper losses can be calculated as follows:

$$Copper\ Losses=3\ast I_{PH\ RMS}^2\ast \ R_{PH}$$

For trapezoidal commutation:

$$Copper\ Losses=I_{DC}^2\ast R_{L-L}$$

Core losses are difficult to calculate as they vary with speed. An approximation of the peak core loss can be derived from the motor torque/speed curve. In Fig. 4, the torque rolloff as speed increases is due to core losses. At the maximum application speed (ω rad/sec), the peak core loss is defined as:  Fig. 4

Current ripple losses can generally be neglected unless the motor has very low inductance (< 1mH) or the PWM switching frequency is low (< 8 kHz). See Motor Inductance and PWM Frequency Tech Note for more details.

Gearbox efficiency is specified in the gearbox datasheet. If derating curves for load, speed and temperature are not provided it will be necessary to consult the gearbox supplier.

## Drive Electrical Power Output

Fig. 5 shows a servo drive inverter and three-phase brushless motor. The inverter generates sinewaves from VBUS (DC) by varying the PWM duty cycle in a sinusoidal pattern. In addition to sinusoidal modulation, the duty cycle is adjusted for the desired phase current magnitude. Three sinewaves, phase-shifted by 120°, are applied to the motor windings to create a rotating current vector. Modern servo drives use a variation of this technique, space vector modulation, which enables the application of 100% of VBUS to the motor windings.

RMS line-to-line voltage applied to the motor is shown below where “D” is the PWM duty cycle. The duty cycle varies between 0 and 1 depending on the desired VL-L (“1” applies maximum voltage)

$$V_{L-L}=\frac{V_{BUS}}{\sqrt{2}}\ast D$$

Phase voltage VPH can be derived from VL-L by vector addition:

$$V_{PH}=\frac{V_{L-L}}{\sqrt{3}}$$

Electrical power is the sum of the power in each motor phase. The magnitude of the voltage applied to each phase is the same. As the windings are balanced, the current in each phase is the same:

$$Drive\ P_{OUT}=3\ast \ V_{PH}\ast I_{PH}$$

$$Drive\ P_{OUT}=\sqrt{\frac{3}{2}}\ast V_{BUS}\ast I_{PH}\ast D$$

The PWM duty cycle is sometimes set to unity in the power equation, but this can be misleading. For example, if a higher KT winding (more turns per stator coil) is used, the current can be reduced for the same torque (T = KT * I). It is a common misconception that the required power is reduced proportionally. A higher KT winding also has a higher KE resulting in an increased back-EMF. To drive the same current, the duty cycle must be increased so that the average voltage applied can overcome the higher back-EMF. The electrical power is the same for both windings.

Resistance is increased in the higher KT winding with more turns, but copper losses are reduced due to the I2 term in the power equation (Copper Losses = I2R). Drive conduction losses are also reduced because of the lower current demand.

## Mechanical Power

Mechanical output power is the product of torque (T Nm) and angular velocity (ω rad/sec):

$$Mechanical\ P_{OUT}=T\ast \omega$$

This calculation is straightforward for constant or even variable speed applications, but it is more complicated when point-to-point moves are involved. Power is work done (energy) per unit time. In the acceleration period in Fig. 6 power is positive as the load is increasing in kinetic energy. During the run period, power is positive as energy is provided to overcome friction. During the deceleration period the kinetic energy of the load is returned to the drive so power is negative.

The kinetic energy of the motor is absorbed as electrical charge in the bus capacitor. For high kinetic energy (large inertia rotating at high speed), the bus capacitor may charge to a voltage which exceeds the rating of the power transistors. In this case the energy must be dissipated as heat in a resistor via a shunt regulator. The shunt regulator connects a power resistor across VBUS when the voltage gets too high.

The determination of peak mechanical power can be calculated in one of two ways as shown below:

$$Mechanical\ P_{OUT_PK}=T_{accel}\ast \omega$$

$$Mechanical\ P_{OUT_PK}=\ \frac{Kinetic\ Energy}{t}=\frac{J\ast \omega ^2}{2\ast t}$$

J = total inertia [motor, gearbox, load] (Kg.m2), ω = angular velocity (rad/sec), t = acceleration time (sec)
Average mechanical power calculation is more complex. The following equation assumes a repetitive move with a consistent dwell time between moves. The integral is the area under the curve over the period T. The equation assumes that regenerative energy is reabsorbed by the bus capacitor. This scenario is not typical which makes average power difficult to calculate. The problem is compounded in multi-axis applications with different axes in regeneration at different times. ## Electrical Input Power by Empirical Method

Ideally the required electrical input power, both peak and average, could be calculated as follows:

$$Electrical\ P_{IN}=\ \frac{Mechanical\ P_{OUT}}{System\ Efficiency}$$

Considering the complexities of calculating mechanical average power as well as the difficulty of quantifying some system losses, an empirical method is often used to determine power supply capacity. Fig. 7 shows the basic technique. An oversized test power supply is required. It should have the same voltage as the intended power supply. A digital storage oscilloscope (DSO) with current and voltage probes measures the input power (VBUS * I) to the drive throughout a machine cycle. The DSO may incorporate analysis algorithms to facilitate average power calculations.

*$$K_M=\frac{K_T}{\sqrt{R}}$$