Model AC Servomotor with Dynamic Load Torque

[Abdul Wali]

Hi,
i want to model an AC Servomotor where i assume that a dynamic load is attached to the shaft of AC Servomotor. the paper that i have attached to this post has ac servomotor model running without load that's why in equation 5 TL(s) (the load torque)=0 . Now i want to include TL(s) in the model which means that a load is connected to the shaft of the motor. in order to include TL(s) into the system i need to know the formulla of the TL(s) becouse i can not simply include TL(s) in the equation else i will not be able to get the final transfer function ( Theta(s)/E(s) ). So i am thinking to put in the components of the TL(s) into the equation such as T=KI or T=J(inertia) * a(accelaration).

So May someone please help me that what can be the correct equation of TL(s) in this case so i can subtitute in? or/and how can i include the load torque (TL(s)) into the motor model.

 

[Tom.G]

The sentence before eq.(5) has you ignoring the load to get the motor transfer function. Just re-insert the load terms to get the total transfer function.

Equation (2) in your attachment defines motor torque in terms of inertia (J), friction (B), and load torque (T_L).
Replace the (T_L) term with its component terms (load inertia (J_L) and load friction (B_L)), and carry them thru the balance of the calculations. Take the gearbox ratio into account so the inertia and friction are as seen by the motor shaft.

 

[Abdul Wali]

The sentence before eq.(5) has you ignoring the load to get the motor transfer function. Just re-insert the load terms to get the total transfer function.

Equation (2) in your attachment defines motor torque in terms of inertia (J), friction (B), and load torque (T_L).
Replace the (T_L) term with its component terms (load inertia (J_L) and load friction (B_L)), and carry them thru the balance of the calculations. Take the gearbox ratio into account so the inertia and friction are as seen by the motor shaft.

@Tom.G Thank you very much for your reply. so it will be something like TL=Ja(t) + Bv(t) where a=theta double dash and v = theta dash Right?

 

[Tom.G]

Yup, I think you've got it!

 

[Tom.G]

Take the gearbox ratio into account so the inertia and friction are as seen by the motor shaft.

I just realized, there is the possibility the original article already included the built-in gearbox when specifying the motor characteristics. I only scanned the article so you may want to dig out that possibility; especially if you are actually building the hardware!

 

[Abdul Wali]

I just realized, there is the possibility the original article already included the built-in gearbox when specifying the motor characteristics. I only scanned the article so you may want to dig out that possibility; especially if you are actually building the hardware!

i am only focusing on the software, so will it be precise if I neglect the gearbox? because I don't know how to take the gearbox ratio into account while modelling ? and how should it be indicated? @Tom.G

 

[Tom.G]

i am only focusing on the software, so will it be precise if I neglect the gearbox? because I don't know how to take the gearbox ratio into account while modelling ? and how should it be indicated?

Then just assume the numbers from the article apply at the output shaft. That is the most likely scenario.

As for taking account of a gear ratio between a motor and a load. Let's use a 2 to 1 reduction ratio as an example. That means for 2 revolutions of the motor shaft, the load make 1 revolution. There are two reasons you might want to do that. First, you might want the load to move slower. Second, maybe you need more torque to move the load. Since the power into and out of a gearbox is essentially identical, by halving the speed you double the torque available at the load.

Now let's look at it from the load back towards the motor, again keeping in mind that the power remains constant thru the gearbox. At any given load speed, the motor will be turning twice as fast. That indicates the torque at the motor side will be half the torque at the load. This increased speed and reduced torque are called the Reflected Load at the motor. This results in the inertia seen by the motor to be 1/4 the load inertia. Or in the general case, 1/(GearRatio²). In the Real World there are of course details to account for; for instance the inertia and friction of the individual gears, particularly important in a multistage or high ratio gearbox.

 

[Abdul Wali]

Then just assume the numbers from the article apply at the output shaft. That is the most likely scenario.

As for taking account of a gear ratio between a motor and a load. Let's use a 2 to 1 reduction ratio as an example. That means for 2 revolutions of the motor shaft, the load make 1 revolution. There are two reasons you might want to do that. First, you might want the load to move slower. Second, maybe you need more torque to move the load. Since the power into and out of a gearbox is essentially identical, by halving the speed you double the torque available at the load.

Now let's look at it from the load back towards the motor, again keeping in mind that the power remains constant thru the gearbox. At any given load speed, the motor will be turning twice as fast. That indicates the torque at the motor side will be half the torque at the load. This increased speed and reduced torque are called the Reflected Load at the motor. This results in the inertia seen by the motor to be 1/4 the load inertia. Or in the general case, 1/(GearRatio²). In the Real World there are of course details to account for; for instance the inertia and friction of the individual gears, particularly important in a multistage or high ratio gearbox.

@Tom.G Actually I don't understand that how to apply that gear ratio value to the output shaft in the mathematical model?

 

[Tom.G]

Actually I don't understand that how to apply that gear ratio value to the output shaft in the mathematical model?

Hmmm. I thought that was covered in posts #2,3,4.

Then just assume the numbers from the article apply at the output shaft. That is the most likely scenario.

That statement was supposed to say that the numbers in the article likely apply to the output shaft after the built-in gearbox, and that the load is directly attached to that shaft. In that case, just follow posts #2,3,4. The rest of post #7 was a procedure and explanation for adding a gearbox before the load.

Now if you are asking about how to do the math in the article, I'll have to leave that to others on this forum. It's has been too many decades since I've had to do that!

Any help? Or did I misunderstand what you are misunderstanding?

 

[Abdul Wali]

Hmmm. I thought that was covered in posts #2,3,4.

That statement was supposed to say that the numbers in the article likely apply to the output shaft after the built-in gearbox, and that the load is directly attached to that shaft. In that case, just follow posts #2,3,4. The rest of post #7 was a procedure and explanation for adding a gearbox before the load.

Now if you are asking about how to do the math in the article, I'll have to leave that to others on this forum. It's has been too many decades since I've had to do that!

Any help? Or did I misunderstand what you are misunderstanding?

i got it, thanks once again.

 

[Abdul Wali]

The sentence before eq.(5) has you ignoring the load to get the motor transfer function. Just re-insert the load terms to get the total transfer function.

Equation (2) in your attachment defines motor torque in terms of inertia (J), friction (B), and load torque (T_L).
Replace the (T_L) term with its component terms (load inertia (J_L) and load friction (B_L)), and carry them thru the balance of the calculations. Take the gearbox ratio into account so the inertia and friction are as seen by the motor shaft.

@Tom.G i appreciate your help and i have one more question. Is the following assumption correct? Motor inertia Jm and Load inertia JL are different from each other and they can not be added together but when the load inertia JL changes it will affect the Motor Inertia Jm and the same theory goes for Motor Friction Bm and Load friction coefficient BL. (so when i am modelling the motor mathematically i can not do the following Jm+jL = 2J or Bm+BL=2B) Right?

 

[Tom.G]

Manufacturers generally supply the characteristics referenced to the output shaft that they supply on the motor assembly. If the a manufacturer ships you a motor with a gearbox already attached (as in your simulation), then they have included the gearbox in the supplied Jm and Bm characteristics.

IF the load is connected directly to the supplied shaft, then Jl and Bl can be added directly Jm and Bm.

Only if you add a gearbox between the manufacturer supplied drive and the load do you have to account for it, per the procedure in post #7.

 

[Abdul Wali]

Manufacturers generally supply the characteristics referenced to the output shaft that they supply on the motor assembly. If the a manufacturer ships you a motor with a gearbox already attached (as in your simulation), then they have included the gearbox in the supplied Jm and Bm characteristics.

IF the load is connected directly to the supplied shaft, then Jl and Bl can be added directly Jm and Bm.

Only if you add a gearbox between the manufacturer supplied drive and the load do you have to account for it, per the procedure in post #7.

@Tom.G thanks once again, I read somewhere something like below, what is your opinion about this?

TL should have the component terms (load inertia and friction) and one more component that represents the torque demanded by the load during its operation. The load can be an elevator, for instance, and in this case this additional term is a function of the time that is constant to the weight lifted. Or it can be another kind of load, and in this case it can be whatever function of time. So this additional term that is a direct function of the time (and not indirect through velocity or acceleration) describe the very nature of the load attached to the motor drive. If it is not used the whole motor drive would be useless, it would drive something useless. Usually this term of the torque is called disturbance, because it tends to perturbate the regime that can be attained by any value of the main input, that is E. so the equation will be:
TL(t)=JLθ¨(t)+BLθ˙(t)+Tdist(t)
Looking forward for your reply.

 

[Tom.G]

Well, that's rather muddled!

After reading it four times it seems to be addressing (sort of) two different things. I don't see where there is a problem with an elevator. Design for maximum load and control the acceleration/deceleration and limit maximum speed.

I don't exactly understand what is meant by the "Tdist(t)" term. If it is an "Unknown, Random" upset, make sure the controller is stable in the sense of not oscillating or grossly over overshooting from a step upset.
Or perhaps they mean a very nonlinear, varying, load. In that case, gain switching or similar is used in the controller to better control the load in its different operating regions. This one is not for the faint of heart!