# Motor constants

The motor size constant (${\displaystyle K_{\text{M}}}$) and motor velocity constant (${\displaystyle K_{\text{v}}}$, alternatively called the back EMF constant) are values used to describe characteristics of electrical motors.

## Motor constant

${\displaystyle K_{\text{M}}}$ is the motor constant[1] (sometimes, motor size constant). In SI units, the motor constant is expressed in newton metres per square root watt (${\displaystyle {\text{N}}{}\cdot {}{\text{m}}/{\sqrt {\text{W}}}}$):

${\displaystyle K_{\text{M}}={\frac {\tau }{\sqrt {P}}}}$

where

• ${\displaystyle \scriptstyle \tau }$ is the motor torque (SI unit: newton–metre)
• ${\displaystyle \scriptstyle P}$ is the resistive power loss (SI unit: watt)

The motor constant is winding independent (as long as the same conductive material is used for wires); e.g., winding a motor with 6 turns with 2 parallel wires instead of 12 turns single wire will double the velocity constant, ${\displaystyle K_{\text{v}}}$, but ${\displaystyle K_{\text{M}}}$ remains unchanged. ${\displaystyle K_{\text{M}}}$ can be used for selecting the size of a motor to use in an application. ${\displaystyle K_{\text{v}}}$ can be used for selecting the winding to use in the motor.

Since the torque ${\displaystyle \tau }$ is current ${\displaystyle I}$ multiplied by ${\displaystyle K_{\text{T}}}$ then ${\displaystyle K_{\text{M}}}$ becomes

${\displaystyle K_{\text{M}}={\frac {K_{\text{T}}I}{\sqrt {P}}}={\frac {K_{\text{T}}I}{\sqrt {I^{2}R}}}={\frac {K_{\text{T}}}{\sqrt {R}}}}$

where

• ${\displaystyle I}$ is the current (SI unit, ampere)
• ${\displaystyle R}$ is the resistance (SI unit, ohm)
• ${\displaystyle K_{\text{T}}}$ is the motor torque constant (SI unit, newton–metre per ampere, N·m/A), see below

If two motors with the same ${\displaystyle K_{\text{v}}}$ and torque work in tandem, with rigidly connected shafts, the ${\displaystyle K_{\text{v}}}$ of the system is still the same assuming a parallel electrical connection. The ${\displaystyle K_{\text{M}}}$ of the combined system increased by ${\displaystyle {\sqrt {2}}}$, because both the torque and the losses double. Alternatively, the system could run at the same torque as before, with torque and current split equally across the two motors, which halves the resistive losses.

## Units

The motor constant may be provided in one of several units. The table below provides conversions between common SI units

${\displaystyle 1}$ ${\displaystyle {\sqrt {2}}}$ ${\displaystyle 1}$ ${\displaystyle {\frac {1}{\sqrt {2}}}}$ ${\displaystyle {\frac {\pi }{30}}}$ ${\displaystyle {\frac {\pi }{30{\sqrt {2}}}}}$ ${\displaystyle k_{t},{\frac {Nm}{A_{pk}}}}$ ${\displaystyle k_{t},{\frac {Nm}{A_{RMS}}}}$ ${\displaystyle k_{v},{\frac {V_{LL,pk}}{\frac {rad}{s}}}}$ ${\displaystyle k_{v},{\frac {V_{LL,RMS}}{\frac {rad}{s}}}}$ ${\displaystyle k_{v},{\frac {V_{LL,pk}}{rpm}}}$ ${\displaystyle k_{v},{\frac {V_{LL,RMS}}{rpm}}}$

## Motor velocity constant, back EMF constant

${\displaystyle K_{\text{v}}}$ is the motor velocity, or motor speed,[2] constant (not to be confused with kV, the symbol for kilovolt), measured in revolutions per minute (RPM) per volt or radians per volt second, rad/V·s:[3]

${\displaystyle K_{\text{v}}={\frac {\omega _{\text{no-load}}}{V_{\text{peak}}}}}$

The ${\displaystyle K_{\text{v}}}$ rating of a brushless motor is the ratio of the motor's unloaded rotational speed (measured in RPM) to the peak (not RMS) voltage on the wires connected to the coils (the back EMF). For example, an unloaded motor of ${\displaystyle K_{\text{v}}}$ = 5,700 rpm/V supplied with 11.1 V will run at a nominal speed of 63,270 rpm (= 5,700 rpm/V × 11.1 V).

The motor may not reach this theoretical speed because there are non-linear mechanical losses. On the other hand, if the motor is driven as a generator, the no-load voltage between terminals is perfectly proportional to the RPM and true to the ${\displaystyle K_{\text{v}}}$ of the motor/generator.

The terms ${\displaystyle K_{\text{e}}}$,[2] ${\displaystyle K_{\text{b}}}$ are also used,[4] as are the terms back EMF constant,[5][6] or the generic electrical constant.[2] In contrast to ${\displaystyle K_{\text{v}}}$ the value ${\displaystyle K_{\text{e}}}$ is often expressed in SI units volt–seconds per radian (V⋅s/rad), thus it is an inverse measure of ${\displaystyle K_{v}}$.[7] Sometimes it is expressed in non SI units volts per kilorevolution per minute (V/krpm).[8]

${\displaystyle K_{\text{e}}=K_{\text{b}}={\frac {V_{\text{peak}}}{\omega _{\text{no-load}}}}={\frac {1}{K_{\text{v}}}}}$

The field flux may also be integrated into the formula:[9]

${\displaystyle K_{\omega }={\frac {E_{\text{b}}}{\phi \omega }}}$

where ${\displaystyle E_{\text{b}}}$ is back EMF, ${\displaystyle K_{\omega }}$ is the constant, ${\displaystyle \phi }$ is the flux, and ${\displaystyle \omega }$ is the angular velocity.

By Lenz's law, a running motor generates a back-EMF proportional to the speed. Once the motor's rotational velocity is such that the back-EMF is equal to the battery voltage (also called DC line voltage), the motor reaches its limit speed.

## Motor torque constant

${\displaystyle K_{\text{T}}}$ is the torque produced divided by armature current.[10] It can be calculated from the motor velocity constant ${\displaystyle K_{\text{v}}}$.

${\displaystyle K_{\text{T}}={\frac {\tau }{I_{\text{a}}}}={\frac {60}{2\pi K_{\text{v(RPM)}}}}={\frac {1}{K_{\text{v(SI)}}}}}$

where ${\displaystyle I_{\text{a}}}$ is the armature current of the machine (SI unit: ampere). ${\displaystyle K_{\text{T}}}$ is primarily used to calculate the armature current for a given torque demand:

${\displaystyle I_{\text{a}}={\frac {\tau }{K_{\text{T}}}}}$

The SI units for the torque constant are newton meters per ampere (N·m/A). Since 1 N·m = 1 J, and 1 A = 1 C/s, then 1 N·m/A = 1 J·s/C = 1 V·s (same units as back EMF constant).

The relationship between ${\displaystyle K_{\text{T}}}$ and ${\displaystyle K_{\text{v}}}$ is not intuitive, to the point that many people simply assert that torque and ${\displaystyle K_{\text{v}}}$ are not related at all. An analogy with a hypothetical linear motor can help to convince that it is true. Suppose that a linear motor has a ${\displaystyle K_{\text{v}}}$ of 2 (m/s)/V, that is, the linear actuator generates one volt of back-EMF when moved (or driven) at a rate of 2 m/s. Conversely, ${\displaystyle s=VK_{\text{v}}}$ (${\displaystyle s}$ is speed of the linear motor, ${\displaystyle V}$ is voltage).

The useful power of this linear motor is ${\displaystyle P=VI}$, ${\displaystyle P}$ being the power, ${\displaystyle V}$ the useful voltage (applied voltage minus back-EMF voltage), and ${\displaystyle I}$ the current. But, since power is also equal to force multiplied by speed, the force ${\displaystyle F}$ of the linear motor is ${\displaystyle F=P/(VK_{\text{v}})}$ or ${\displaystyle F=I/K_{\text{v}}}$. The inverse relationship between force per unit current and ${\displaystyle K_{\text{v}}}$ of a linear motor has been demonstrated.

To translate this model to a rotating motor, one can simply attribute an arbitrary diameter to the motor armature e.g. 2 m and assume for simplicity that all force is applied at the outer perimeter of the rotor, giving 1 m of leverage.

Now, supposing that ${\displaystyle K_{\text{v}}}$ (angular speed per unit voltage) of the motor is 3600 rpm/V, it can be translated to "linear" by multiplying by 2π m (the perimeter of the rotor) and dividing by 60, since angular speed is per minute. This is linear ${\displaystyle K_{\text{v}}\approx 377\ ({\text{m}}/{\text{s}})/{\text{V}}}$.

Now, if this motor is fed with current of 2 A and assuming that back-EMF is exactly 2 V, it is rotating at 7200 rpm and the mechanical power is 4 W, and the force on rotor is ${\displaystyle {\frac {P}{V*K_{\text{v(SI)}}}}={\frac {4}{2*377}}}$ N or 0.0053 N. The torque on shaft is 0.0053 N⋅m at 2 A because of the assumed radius of the rotor (exactly 1 m). Assuming a different radius would change the linear ${\displaystyle K_{\text{v}}}$ but would not change the final torque result. To check the result, remember that ${\displaystyle P=\tau \,2\pi \,\omega /60}$.

So, a motor with ${\displaystyle K_{\text{v}}=3600{\text{ rpm}}/{\text{V}}=377{\text{ rad}}/{\text{V·s}}}$ will generate 0.00265 N⋅m of torque per ampere of current, regardless of its size or other characteristics. This is exactly the value estimated by the ${\displaystyle K_{\text{T}}}$ formula stated earlier.

EXAMPLE: Torque applied at different diameters, ${\displaystyle K_{\text{v (rpm/V)}}}$= 3600 rpm/V ≈ 377 rad/s/V , ${\displaystyle K_{\text{T}}}$ ≈ 0.00265 N.m/A (each calculatable if one is known), V = 2 v, ${\displaystyle I_{\text{a}}}$= 2 A, P = 4 W , (any 2 makes the 3rd, ${\displaystyle P=VI}$)
diameter = 2r r = 0.5 m r = 1 m r = 2 m Formula (${\displaystyle K_{\text{v(rpm/V)}}}$) Formula (${\displaystyle K_{\text{v(rad/s/V)}}}$) Formula (${\displaystyle K_{\text{T}}}$) shorthand
${\displaystyle \tau }$ = motor torque (N.m/s) 0.005305 N·m 0.005305 N·m 0.005305 N·m ${\displaystyle {\frac {30I}{\pi K_{\text{v(rpm/V)}}}}}$ ${\displaystyle {\frac {I}{K_{\text{v(rad/s/V)}}}}}$ ${\displaystyle K_{\text{T(N.m/A)}}*I}$ ${\displaystyle K_{\text{T}}*I}$
linear ${\displaystyle K_{\text{v}}}$ (m/s/V) @ diameter 188.5 (m/s)/V 377.0 (m/s)/V 754.0 (m/s)/V ${\displaystyle {\frac {\pi rK_{\text{v(rpm/V)}}}{30}}}$ ${\displaystyle rK_{\text{v(rad/s/V)}}}$ ${\displaystyle {\frac {r}{K_{\text{T(N.m/A)}}}}}$ ${\displaystyle K_{\text{v}}*r}$
linear ${\displaystyle K_{\text{T}}}$ (N.m/A) @ diameter 0.005305 N·m/A 0.002653 N·m/A 0.001326 N·m/A ${\displaystyle {\frac {30}{\pi rK_{\text{v(rpm/V)}}}}}$ ${\displaystyle {\frac {1}{rK_{\text{v(rad/s/V)}}}}}$ ${\displaystyle {\frac {K_{\text{T(N.m/A)}}}{r}}}$ ${\displaystyle K_{\text{t}}/r}$
speed m/s @ diameter

(linear speed)

377.0 m/s 754.0 m/s 1508.0 m/s ${\displaystyle {\frac {\pi rVK_{\text{v(rpm/V)}}}{30}}}$ ${\displaystyle VrK_{\text{v(rad/s/V)}}}$ ${\displaystyle {\frac {rV}{K_{\text{T(N.m/A)}}}}}$ linear ${\displaystyle K_{\text{v}}*V=K_{\text{v}}*Vr}$
speed km/h @ diameter

(linear speed)

1357 km/h 2714 km/h 5429 km/h ${\displaystyle {\frac {3\pi rVK_{\text{v(rpm/V)}}}{25}}}$ ${\displaystyle 3.6VrK_{\text{v(rad/s/V)}}}$ ${\displaystyle {\frac {3.6rV}{K_{\text{T(N.m/A)}}}}}$ linear ${\displaystyle K_{\text{v}}*V*{\frac {3600}{1000}}}$
torque (N.m) @ diameter

(linear torque)

0.01061 N·m 0.005305 N·m 0.002653 N·m ${\displaystyle {\frac {30I}{\pi rK_{\text{v(rpm/V)}}}}}$ ${\displaystyle {\frac {I}{rK_{\text{v(rad/s/V)}}}}}$ ${\displaystyle {\frac {K_{\text{T}}*I}{r}}}$ ${\displaystyle {\frac {\tau }{r}}}$
shorthand half diameter = half speed

* double torque

full diameter = full speed

* full torque

double diameter = double speed

* half torque

${\displaystyle K_{\text{v(rad/s/V)}}={\frac {2\pi K_{\text{v(rpm/V)}}}{60}}}$${\displaystyle K_{\text{T(N.m/A)}}={\frac {60}{2\pi K_{\text{v(rpm/V)}}}}}$ ${\displaystyle K_{\text{v(rpm/V)}}={\frac {60K_{\text{v(rad/s/V)}}}{2\pi }}}$${\displaystyle K_{\text{T(N.m/A)}}={\frac {1}{K_{\text{v(rad/s/V)}}}}}$ ${\displaystyle K_{\text{v(rpm/V)}}={\frac {60}{2\pi K_{\text{T(N.m/A)}}}}}$${\displaystyle K_{\text{v(rad/s/V)}}={\frac {1}{K_{\text{T(N.m/A)}}}}}$ ${\displaystyle 1={\displaystyle K_{\text{v (rad/s/V)}}}*{\displaystyle K_{\text{T (N.m/A)}}}}$${\displaystyle 1=linear{\displaystyle K_{\text{v(m/s/V)}}}*linear{\displaystyle K_{\text{T (N.m/A)}}}}$

## References

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3. ^ "Brushless Motor Kv Constant Explained • LearningRC". 29 July 2015.
4. ^ "GENERAL MOTOR TERMINOLOGY" (PDF), www.smma.org
5. ^ "DC motor model with electrical and torque characteristics - Simulink", www.mathworks.co.uk
6. ^ "Technical Library > DC Motors Tutorials > Motor Calculations", www.micro-drives.com, archived from the original on 2012-04-04
7. ^ "Home". www.precisionmicrodrives.com. Archived from the original on 2014-10-28.
8. ^
9. ^ "DC motor starting and braking", iitd.vlab.co.in, archived from the original on 2012-11-13
10. ^