Hello everyone,
if you look inside books for a motor control, you will find always control circuits with a PI-current control. It does not really make sense to me, because it means that you have always a steady-state error if you do not compensate the back-emf voltage somehow.
A simple example:
The motor (dc-motor or non-salient synchronous motor in q-reference system) can be modelled as:
Vin = (L*s+R)*I+K_v*K_m*I/(J*s)
L:=motor inductance
R:= winding resistance
K_m:=factor Torque/Current
K_v:=factor (Back-emf voltage)/speed
J:=inertia
s:=Laplacian variable
Vin:=Input voltage (Uq)
I:=current
It results in the following system:
G(s) = I/Vin=J*s/((L*s+R)*(J*s)+K_m*K_v)
The PI-controller has the function:
K(s)=K_p*(1+K_i/s)
The closed-loop transfer function is:
T(s)=K(s)*G(s)/(1+K(s)*G(s))
It results in a closed-loop bode plot and step function as attached.
Why is the steady-state error disregarded? How is it done in professional inverter systems?
Note: Friction is disregarded in this case. If you consider friction, you will shift the zero in the open-loop transfer function. But sometimes friction is very small and can be disregarded. Is it maybe a problem for this kind of modelling?
Thank you for your answer.
Sebastian
if you look inside books for a motor control, you will find always control circuits with a PI-current control. It does not really make sense to me, because it means that you have always a steady-state error if you do not compensate the back-emf voltage somehow.
A simple example:
The motor (dc-motor or non-salient synchronous motor in q-reference system) can be modelled as:
Vin = (L*s+R)*I+K_v*K_m*I/(J*s)
L:=motor inductance
R:= winding resistance
K_m:=factor Torque/Current
K_v:=factor (Back-emf voltage)/speed
J:=inertia
s:=Laplacian variable
Vin:=Input voltage (Uq)
I:=current
It results in the following system:
G(s) = I/Vin=J*s/((L*s+R)*(J*s)+K_m*K_v)
The PI-controller has the function:
K(s)=K_p*(1+K_i/s)
The closed-loop transfer function is:
T(s)=K(s)*G(s)/(1+K(s)*G(s))
It results in a closed-loop bode plot and step function as attached.
Why is the steady-state error disregarded? How is it done in professional inverter systems?
Note: Friction is disregarded in this case. If you consider friction, you will shift the zero in the open-loop transfer function. But sometimes friction is very small and can be disregarded. Is it maybe a problem for this kind of modelling?
Thank you for your answer.
Sebastian