About the Example Model: S12ZVM FOC Sensorless Motor Control

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About the Example Model: S12ZVM FOC Sensorless Motor Control

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edenli
Contributor V

Hi Guys,

There is an example about the sensorless speed control,the link is https://community.nxp.com/docs/DOC-332663 , and Can provide a basic theory document about the example?

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dumitru-daniel_
NXP Employee
NXP Employee

Hi Eden Li, 

I have received the material for #1

Stage 1: represent the closed loop control implementation. The QUESTION ??? is how to compute the controller parameters if you know the plant model?

1.JPG

Stage 2: Based on control system theory both the plant==motor electrical model==Grl(s) and the current controller==Gpi(s) are 1st order transfer functions

2.JPG

Stage 3: if you perform the computations shown in https://community.nxp.com/thread/450679#comment-904019 

then you'll get the closed loop transfer function for each current loop.

3.JPG

Best regards,
Daniel

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dumitru-daniel_
NXP Employee
NXP Employee

Please have a look here: http://www.nxp.com/files-static/training/doc/ftf/2014/FTF-AUT-F0232.pdf 

Hope this helps!

Best regards,
Daniel

306 Views
edenli
Contributor V

Hi Daniel,

My friend ,First,thank you for your reply!

Next,i have some doubts,and hope you can solve it for me.

#1:I download  the file that the link you sent me,and the content in the page 40 (shown as below),is not so clear  and there is some overlap.

and if you can send a clear file to me?

pastedImage_1.png

#2: I don't know how to deduce the transfer function in the Part  A (shown as below).

pastedImage_2.png

#3:About the model of S12ZVM_FOC_Sensorless,shown as below, i wanna make sure the IdEObsrv(Part A shown) and IqEObsrv(Part B shown) is EMF or not ?  i  thought is EMF, my reason ,EMF contains the position and speed of rotor, to get the Postion of the rotor using the arctan() Function,namely Theat=arctan(EAlpha/EBeta),

and there is another doubt that the -1*2^15 in Part C,why we multiplied the -1*2^15 with IdEObsrv?

pastedImage_3.png

#4:Shown as below,how to deduce the id_frac_est = (ud_frac*K1_n + w_n*iq_frac*K2_n + ed_frac*K3_n + id_frac_est*K4_n)? 

I don't know how to deduce it,and the id_frac_est  is deduced form  EMF?

pastedImage_10.png

and the attachment is the model of S12ZVM FOC Sensorless Motor Control.

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dumitru-daniel_
NXP Employee
NXP Employee

Hi Eden Li, 

For #1, i'm still waiting for the originator of the presentation to sent me the ppt version since that is an animation.

For #2, that transfer function represents the closed loop computation for a first order RL system controlled via a PI controller (1st order)

Since it's quite a pain to write all the math in html i'm attaching a picture with computations. Please excuse my hand-writing.

IMG_20170509_223419.jpg

For #3 and #4 i will reply separatly

306 Views
edenli
Contributor V

Hi Daniel,

Thank you for your reply, i very appreciate with you.

Best Regrads!

Eden Li

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dumitru-daniel_
NXP Employee
NXP Employee

Hi Eden Li, 

I have received the material for #1

Stage 1: represent the closed loop control implementation. The QUESTION ??? is how to compute the controller parameters if you know the plant model?

1.JPG

Stage 2: Based on control system theory both the plant==motor electrical model==Grl(s) and the current controller==Gpi(s) are 1st order transfer functions

2.JPG

Stage 3: if you perform the computations shown in https://community.nxp.com/thread/450679#comment-904019 

then you'll get the closed loop transfer function for each current loop.

3.JPG

Best regards,
Daniel

View solution in original post

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dumitru-daniel_
NXP Employee
NXP Employee

Eden Li, 

Can you please open separate questions for #3 and #4 ? 

Thank you!

Daniel

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edenli
Contributor V

Hi Daniel,

OK,i will separate questions for #3 and #4.

Best Regards!

Eden Li.

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dumitru-daniel_
NXP Employee
NXP Employee

Hi Eden Li, 

Please allow me some time to find the appropriate files and to prepare the answers to your questions. 

Stay tune!

Best regards,
Daniel

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306 Views
edenli
Contributor V

Hi Daniel,

Thank you very much!

Best Regards!

Eden Li

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