MOTOR SPEED CONTROLLER. Ben Hwang Pavan Bhandiwad April 30, 2003. Introduction. Many motors lose speed due to counter torque in the opposite direction Will cause drastic losses in efficiency of the motor Might damage components attached to the motor if not running at expected speed
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April 30, 2003
Kp + Ki / s = (Kp*s + Ki) / s.
Where Kp is the proportional constant and Ki is the integration constant.
% overshoot = exp(-π / (1-²)1/2) = 0.001.
Solving for we get 0.83.
tr = 1 = (1+0.7+1.6²) / n.
Solving for n we get 2.68.
For the DC test, e is zero. Thus all impedances caused by inductances are equal to zero.
From the circuit above, we can calculate Rs from the following formula:
Rs = Vdc / Idc.DC Test
Since Rr+j*e*Llr << j* e*Lm, the equivalent impedance of the circuit is:
Z = Rs + Rr + j* e*(Lls +Llr) = Vas / Is.Blocked Rotor Test
Im(Z)/(2*377) = Lls = Llr
Re(Z) – Rs = Rr
At a slip of zero, Rr/s is infinite and thus an open circuit occurs.
The current in the right branch is zero. The equivalent impedance of the circuit is then:
Z = Rs + j*e*(Lls + Lm).No Load Test
Im(Z)/377 – Lls = Lm
Rs = 1.4 , Rr = 0.56 , Lm = 0.089 H, Llr = Lls = 0.0073 H.
We will know define some constants to help us with our motor function calculation.
(1) ’dr = -Rr/Lr*dr – Np*r* qr+ Rr*Lm/Lr*Ids
(2) ’qr = -Rr/Lr*qr + Np*r* dr+ Rr*Lm/Lr*Iqs
(3) I’ds = Lm*Rr// Lr2* dr + Np*r*M/ / Lr*qr- *Ids+ Vds/
(4) I’qs = Lm*Rr// Lr2* qr - Np*r*M/ / Lr*dr- *Iqs+ Vqs/
(5) ’r = 1.5*Np*Lm/J (dr Iqr-qr Ids) – Tl/J
If we let the matrix X’=
X’1 = ’dr
X’2 = ’qr
X’3 = I’ds
X’4 = I’qs
Let the matrix X =
X1 = dr
X2 = qr
X3 = Ids
X4 = Iqs
And let the matrix U =
U1 = Vds
VqsLinearizing the Last Equation
r = 1.5*Np*Lm/J [X4 -X3 -X2 X1] *X
Evaluated at the equilibrium points this equals
r = 1.5*Np*Lm/J [2.586-2.586-.0012 -.0012] *X
X’=A*X + B*U
Y = C*X + D*U
Where X is the flux linkages and currents, U is the input voltages, and Y is the output speed/voltage.
Using the Matlab ss2tf function we can convert state space form to a transfer function.
(.75*s + 952) / (.0255*s + 1.271)
As stated earlier, our PI controller function can be represented as (Kp*s+Ki)/s.
Since we have a unity feedback system, finding an expression for our overall transfer function is relatively simple.
TF = (MF)(PI) / [1 + (MF)(PI)]
Where MF is the motor function and PI is the PI control function.
Our overall transfer function will be in terms of Kp and Ki and will look the following:
(As2 + Bs + C) / (s2 + Ds + C) = out / in