I am designing a 12lb hobby robot with a 1000Kv BLDC motor at 14.8V nominal voltage and stall torque of 30A. How do I solve for the torque constant. I have seen equations where Kt(torque constant) = 1/Kv. And also equations where Kt = 30/pi(Kv). Which is the most accurate equation? Thank you,
How to calculate the torque constant for a BLDC motor
brushless-dc-motorrobottorque
Related Solutions
Brushless AC
Lets start with a brushless AC machine first & Field Orientated Control.
So BLAC machine's have their stator windings sinusoidally distributed (higher concentration close to the tooth, lower for the outer turns)
Now to control such machine is relatively complex and you can use a field orientated control ( F.O.C. ).Using Park's transform (Clark + rotation) the 3phase sinus stator current's can be transformed 1st into a rotating 2phase representation
Park Transforms - General
$$ I_{\alpha \beta 0 } = \frac{2}{3}\begin{bmatrix} 1 & \frac{-1}{2} & \frac{-1}{2} \\ 0 & \frac{\sqrt{3}}{2} &\frac{-\sqrt{3}}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix} $$
or simply
\$I_\alpha = I_a \$
\$I_\beta = \frac{2I_b + Ia}{\sqrt{3}} \$
And these phasors can further be reduced to two DC quantities via a rotating frame of reference transform
$$ \begin{bmatrix} I_d\\ I_q \end{bmatrix} = \begin{bmatrix} Cos(\Theta ) & Sin(\Theta )\\ -Sin(\Theta ) & Cos(\Theta ) \end{bmatrix} \cdot \begin{bmatrix} I_\alpha\\ I_\beta \end{bmatrix} $$
These two DC terms (with Id usually controlled to 0, unless field weakening is desired) are simple inputs to two classic PI loops to control Id and Iq (the output of which is Vd & Vq).
This Vd & Vq, via inverse Clark&PArk produce the 3phase voltage that will be applied to the stator (insert SVM or SPWM)
Great, it works
Brushless DC & general Permanent Magnet machine equations
Thing is BLDC machines produce higher torque than a sinusoidally wound stator (downto the higher concentration of winding around the teeth to produce the flattened line-line profile). The downside is the torque ripple. This is mostly due to simplifying the control downto which equally limits the effective bandwidth of the controller.
A park-like transform can be applied to overcome some of these shortcomings as the aim is to produce a stator stimulus closer to the airgap profile - tpyically a Quazi squarewave (30-120-30) is superimposed onto a trapezoidal backEMF (60-60-60).
First of all you need the machine equation for a BLDC machine (which is the same for a BLAC machine). $$ \begin{bmatrix} V_a\\ V_b\\ V_c \end{bmatrix} = R_s \begin{bmatrix} i_a\\ i_b\\ i_c \end{bmatrix} + L\frac{\mathrm{d} }{\mathrm{d} t}\begin{bmatrix} i_a\\ i_b\\ i_c \end{bmatrix} + \begin{bmatrix} e_a\\ e_b\\ e_c \end{bmatrix}$$
\$e_a, e_b, e_c\$ are the 3 backEMF's that are generated for a rotating rotor. Via Faraday's law $$ \varepsilon = - \frac{\mathrm{d} \Phi _B}{\mathrm{d} t} $$
BackEMF is the rate of change of flux
$$ \frac{\mathrm{d} \Phi }{\mathrm{d} t} = \frac{\mathrm{d} \Phi }{\mathrm{d} \Theta}\frac{\mathrm{d} \Theta }{\mathrm{d} t} = \frac{\mathrm{d} \Phi }{\mathrm{d} \Theta}\omega $$
$$ e = \omega {\Phi}' $$
Thus: $$ \begin{bmatrix} e_a\\ e_b\\ e_c \end{bmatrix} = \omega_e \frac{\mathrm{d} }{\mathrm{d} \vartheta_e}\begin{bmatrix} \Phi_a\\ \Phi_b\\ \Phi_c \end{bmatrix} = \omega_e \begin{bmatrix} {\Phi}'_a\\ {\Phi}'_b\\ {\Phi}'_c \end{bmatrix}$$
As the total torque produced is the summation of the 3 phases producing torque (factoring in the pole-pair count):
Torque equation
\$T_e = P({\Phi}'_a i_a + {\Phi}'_b i_b + {\Phi}'_b i_b)\$
Now, the Park transform can be applied to voltage, currents and flux, thus:
$$ T_e = \frac{3}{2}P({\Phi}'_\alpha i_\alpha + {\Phi}'_\beta i_\beta + {\Phi}'_0 i_0) = \frac{3}{2}P({\Phi}'_d i_d + {\Phi}'_q i_q + {\Phi}'_0 i_0) = \frac{3}{2}P{\Phi}'_q i_q $$
For a BLAC machine \${\Phi}'_q\$ is a simple sinus profile (and thus a lookup table, CORDIC... can be used) but for a BLDC an appropriate lookup table of flux vs angle is required specific for the machine being used.
You can find the answer to the question at Understanding D.C. Motor Characteristics.
What you are missing in this problem is the curve between Torque and Speed that you can find at section 3.1.
The maximum torque is \$T=8 Ncm\$ (with the gear ratio - \$T_r=160 Ncm\$) and the no load velocity is \$w = 80rad/s\$ (with the gear ratio \$w_r=2rad/s\$). You need to find \$T_f=60 Ncm\$ and the formula is:
$$T_f = T_r - w_f \cdot \frac{T_r}{w_r}$$
where \$w_f\$ is your angular velocity where the torque is \$60Ncm\$
Best Answer
The equation for the torque constant of a BLDC motor: \$K_{\tau} = \frac{60}{2{\pi}K_{v(RPM)}} = \frac{1}{K_{v(SI)}} \$, where \$K_{\tau}\$ is the torque constant in \$\frac{N \cdot m}{A}\$, and \$K_v\$ is the speed constant, in either rpm or rad/s.
The speed constant given to you is almost certainly in rpm, not rad/s, so you would use the first equation with the unit conversion factor. You can verify the speed constant with an encoder/resolver, another motor, and an oscilloscope. Spin the test motor up to a constant velocity using the other motor (applying a constant voltage/duty cycle to the drive motor likely sufficient). Measure the induced backemf voltage from test motor with your oscilloscope, and the steady-state velocity of the test motor with the encoder/sensor (or by using frequency detection on your oscilloscope and scaling it based on the number of pole pairs). The steady state speed divided by the induced backemf will be the the Kv value of the motor (and will likely be more accurate than the one provided by the motor manufacturer/seller).
EDIT: The answer above is guilty of a common misconception regarding the measurement of the motor torque constant. The Kv rating commonly provided by motor sellers is obtained with this method, but it differs from the true value by a factor of \$ \sqrt{2}\$ for Wye wound motors, and \$ \sqrt{\frac{2}{3}} \$ for delta wound motors. See this video for further details.