E=1/2LI^{2}.

Energy has units of Joules and a Joule equals a Watt*second. A Volt*Ampere equals a Watt.

The switched inductance is half the motor inductance so the energy stored in the inductor is:

E=1/4LI^{2}.

where L is the motor inductance.

Energy is required at every commutation, which occurs at a frequency of:

F_{com} = N[kRev/min]*1000/k*min/(60sec)*Poles/2[elec rev/mech rev]* 6/elec rev

F_{com} = N[kRev/min]* Poles*50

The power used to provide energy to the top switched inductance is:

P = E* F_{com }= 1/4LI^{2}*F_{com } = 1/4LI^{2}* N[kRev/min]* Poles*50

The average voltage used drive the switch inductance is.

V = P/I = 1/4LI* N[kRev/min]* Poles*50

This average voltage must be subtracted from the supply voltage in the equation for speed.

Substituting we get an equation with N on both sides:

N = (V-12.5*L*I*Poles*N-I*R)/K_{e}.

Solving for N:

N = 2*(V-I*R)/(2*K_{e}+25*L*I*Poles).

The average current supplied by the energy stored in the inductance released from the power supply rail is:

I_{ave} = P/V = 1/4LI^{2}* F_{com}/V

The average measured current is then:

I_{m} = (I-L*I^{2}*F_{com})/(4*V)

Next: Final comments on the motor model

]]>The red curves are the motor performance curves with the effects of inductance included. If your design requires your motor to operate above the speed curve that includes inductance you will be disappointed. The curve that includes inductance has a steeper slope caused by the inductance. Since the equation for N (no inductance effects):

N=(V-IR)/Ke

shows that the slope is due to motor resistance, people have calculated the resistance that fits the actual performance curves and termed this Reff, the effective resistance. However this is misleading on multiple counts. The calculated effective resistance is supply voltage (speed) dependent and the motor I^2R losses will be overestimated.

As a three-phase brushless DC motor rotates, voltage is applied to the leads in a particular pattern, termed a 6 Step Drive. The frequency that the motor is commutated as is termed the commutation frequency. At each commutation, one motor lead that is attached to a supply rail is open-circuited and another lead that was previously open circuited is attached to a supply rail. The lead that was previously attached to a supply rail has current flowing in it and this current continues to produce useful torque. The lead that previously was open circuited has no current flowing in it and voltage must be applied across it to overcome the winding inductive impedance.

When the lead with current flowing is open circuited, the phase voltage flies from the supply rail it was attached to by the drive, to the other supply rail, where it is clamped by the drive diodes. Since the back EMF curve is almost at the same level as when the lead was energized, and the torque constant is proportional to the back emf constant, the torque produced by this freewheeling current is almost the same magnitude as produced prior to switching.

In effect the motor inductance acts as a switch inductance resistor, consuming voltage and reducing peak motor speed on one hand, while producing torque and reducing the measured motor current on the other. Since the freewheeling current is not measured, a torque constant calculated from an actual motor performance curves will overestimate the torque constant.

Actual measurements of brushless DC motor stators has shown that there is very little inductance coupled between motor phases, therefore the inductance switched at each commutation point can be approximated by the motor inductance divided by two. Since the effects of inductance is already a secondary effect, the effects of coupled inductance and of losses in laminations due motor lead commutation can be safely ignored. In other words, we will assume no coupled inductance and that the inductor is lossless.

Many drive related effects reduce the motor speed, such as EMI filter resistance, attachment wire resistance and switching transistor resistance. To predict the actual expected motor performance all must included in your motor model. The effects of motor inductance has a **significant** effect and also must be included in your motor model. Unfortunately, the speed torque curves supplied by manufacturers are regularly incorrect, predicting results that are unattainable in practise. Care is required in any design where minimum speeds are a critical specification.

Next: Inductance Effects: Quantitative Discussion

]]>That is, when all losses are ignored, the motor speed is independent of torque and purely a function of the applied voltage. While this equation is too simple to be useful, it highlights an important concept, that the motor speed is controlled by the supply voltage. Performance curves define the motor’s overall operating envelope. The controller varies the voltage to obtain the desired motor speed.

When winding losses, the largest source of motor losses, are added to the torque to the simplest equation we get:

We see that as the motor current increases with increasing load, the motor speed decreases.

Next lets add some additional sources of losses, losses due to motor friction tf and losses due to windage, lamination hysteresis, lamination eddy current losses, and fields circulating in the motor housing, which are all lumped together as Kd, which has units on in-oz/kRPM.

Doing some equation manipulation we get for N:

And for I:

All these equations are plotted below for a motor (no drive losses are considered) operating from a 140 VDC supply with a Kt of 17.5 in-oz/A and therefore a Ke of 12.94V/kRPM.

The black traces are the simplest equations, the red traces are the most complex form. We see the effects of winding losses have significant impact while that of tf and Kd are minor. This is important because adding Kd to the equation complicates things, and we do not want things to be overly complicated. We will lump Kd effects into tf by adding a torque of Kd*V/Ke, removing the N term in the equation for current. By doing so, we simplify the equations, allowing for reasonable closed form solutions for N and I when we add the effects of motor inductance, which have a major effect on the curves but is often neglected.

Next: The effects of motor inductance

]]>Most systems operate at one more speed torque points. Typically a system has three main operating points:

1. A low load-high-speed operating point.

2. The primary operating point, which is system operates at the highest duty cycle, and

3. A high load operating point, which the system operates at low duty cycle.

These operating points are static speed operating points. All systems have some dynamic requirements, at a minimum, some acceleration is required; however dynamic requirements will be discussed at a later time.

Systems that decelerate [i.e. all systems] and systems that lower loads need to have controllers that are capable of handling the regenerated energy. Deceleration produces limited energy but load lowering can regenerate substantial energy. This energy must be dissipated or be returned to the power supply. Typically, if regenerator energy is not properly handed, the regenerated energy increases the controller internal supply voltage, commonly known as DC link voltage, until part failures occur.

Next: Performance curves

]]>The most important motor size constant is K_{m}, the motor constant. It has units of torque per square root of Watts. Since torque is proportional to the current, and winding losses are proportional to **current squared**, K_{m} cannot have units of torque per Watts since then can K_{m} would change with the specific torque point. **K**_{m}** is useful since you can determine the minimum K**_{m}** required based upon two speed torque requirement points. K**_{m}** is key to motor selection**

Another size constant which is much less important is K_{d}, which has units of torque per speed. It represents the motor’s internal speed related losses such as air friction, lamination hysteresis losses, and lamination eddy current losses.

One last motor size constant is T_{f}, the motor friction constant with units of torque. Keep in mind that friction torque is not necessarily the same as the torque measured to rotate the motor. The torque required to rotate the motor may include substantial cogging torque. Cogging torque is caused by the interaction between the motors magnets and the motors laminations. It is not a loss; any energy used over coming a cogging is returned in passing the cotton point.

Next: Defining the requirements

]]>The relationship between the new motor winding constants and the previous constants is really quite simple: the new back EMF constant, Ke prime, is equal to the old back EMF constant, Ke, times the ratio of the new turns divided by the old turns. The same equation holds true for the torque constant. Both the resistance and the inductance vary as the squared of the ratio of the new turns divided by the old turns.

To most people the fact that the inductance varies with the square of the turns ratio makes a great deal of sense while resistance varying with the turns ratio squared seems a bit puzzling. However think of it this way, by doubling the number of turns the length of the wire has doubled, doublilng the length and therefore the resistance. Also, to fit twice the turns into the same cross-sectional area you now need to half the cross-section of each wire, again doubling the reistance. The result of this is that the resistance varies by the square of the turns ratio. This holds true for resistance in any device where the cross-sectional of area available for winding is fixed such as in inductors, solenoids or solenoid brakes.

One of the things that becomes obvious is that the electrical time constant of the motor is a size constant since both the inductance and resistance vary by the same ratio.

]]>Size constants are constants that are fully defined by the motor size. These constants are the peak torque, thermal impedance and the motor constant, K_{m}, which has units of torque per square root of watts. K_{m} is an important constant in sizing motors and will be discussed later.

Winding constants depend on how the motor is wound, specifically, the gauge of the wire and the number of turns. Winding constants are K_{t}, the torque constant, K_{e}, the back EMF constant, L the inductance, and R the resistance.

Size constants are fixed by the motor since each motor has a fixed amount of lamination steel, magnet and volume for the copper winding. No matter how you fill the available winding area, motor size constants do not change. On the other hand, winding constants change depending upon how the motor is wound.

A motor manufacturer could have in stock a single design of lamination stack and rotor for brushless DC motors waiting to be wound for customers. When properly wound these motor components could be assembled into motors that have a wind range of back emf constants. Because laminations require special tooling it’s typical for motor manufacturers to have a number of **different lengths** of lamination stacks using the same lamination and thereby providing **different size motors**. Each different lamination stack size provides a different size motor and

The reason they are proportional is because the same flux and turns that produces back emf voltage also produce torque. Showing in detail why this is so would be too complicated for a short article. Instead an example will serve as proof, and also show you how you can calculate the coversion factor.

Let us consider a motor that happens to have an efficiency of 100%, just to simplify our discussion. With this motor:

**Pin = Pout**

Arbitrarily we can assume a power of 1 watt, and equally arbitrarily we can assume the supply is a one volt one ampere supply. So Pin is 1 V x 1 A = 1 W. Then:

Pin = 1 W = Pout = Speed * Torque = 1V/Ke * 1A*Kt * C

Where C is a conversion factor that depends on the units of Kt and Ke. For example if Kt has units of N-m/A and Ke has units of V/s (radians is unitless) then the torque will have units of N-m and the speed will have units of s^-1, and C is 1. Since 1 N-m/s = 1 W, a little algebra shows the **with these units the value of Kt must equal Ke**.

The metric system is so easy….

If Kt has units of ozf-in/A and Ke has units of V/kRPM then the torque will have units of ozf-in and the speed will have units of kRPM, and C is:

2*pi radians/rev*1000/k*60s/minute / 141.612 ozf-in/N-m

The first part converts kRPM to s^-1 and equals 104.72. The second part converts ozf-in to N-m.

Keeping in mind that 1 N-m/s = 1 W, C is then 1/1.352,or **the value of Kt, when Kt has units ozf-in/A, is 1.352 times the value of Ke, when Ke has units of V/kRPM.** Any motor data you see where this is not true is due to errors in actual measurements.

And the laws of physics are safe for another day.

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