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TMS320F280049C: TIDM-2008/TIDM-1007

Yale Li 李耀先
Expert 4660 points
Part Number: TMS320F280049C
Other Parts Discussed in Thread: TIDM-1007

Hi Team,

In TIDM-2008/TIDM-1007, Lab2.

1. Why the input of Current Loop is ac_cur_meas - ac_cur_ref_inst? This is the opposite of actual Compensator error calculation method.

2. The output of duty1PU is Current Loop plus voltage feed forward. Voltage feed forward = ac_vol_sensed / vBus_sensed. Why the output of Current Loop need to divide by  vBus_sensed?

Thanks & Regards

Yale Li

  • +1
    TI__Intellectual 1945 points

    Hi Yale

    My apology for the late reply. 

    1. The calculated dutyPU goes to synchronous FET. Lets say the calculated duty is 0.3, the effctive duty ration for the active switch will be 0.7. Basically above formula gives d'.

    2. The feed forward is based on the formula: d'= input voltage/output voltage for a boost converter. That is why it is divided by the vBus_sensed.

    Please let me know if you have further questions.

    Thank you

    Amir Hussain

  • 0 in reply to Amir Hussain
    TI__Expert 4660 points

    Hi Amir,

    Thanks for your reply!

    For question 1, we understand clearly.

    For question 2, the output for current loop as output's adjusted part has no relationship with feed forward's calculation formula. It's just added to feed forward to adjust and fix the actual output. actual output d' = Gi + Dforwardfeedback, but Dforwardfeedback = input voltage/output voltage. Could you please explain it further?

    Thanks & Regards

    Yale Li

  • 0 in reply to Yale Li 李耀先
    TI__Guru 50430 points

    Hi Yale,

    First of all, my sincere apology for late response. Just checking with you if the customer has figured Q#2, or we still need support from SME?

    Regards, Santosh

  • 0 in reply to Santosh Jha
    TI__Expert 4660 points

    Hi Santosh,

    My customer still need support.

    Thanks & Regards,

    Yale Li

  • 0 in reply to Yale Li 李耀先
    TI__Intellectual 1945 points

    Hi Yale

    My apology for the late response. I have been on leave for last 20 days. I will respond on the above question by Nov 23.

    Thanks

    Amir 

  • +1 in reply to Yale Li 李耀先
    TI__Intellectual 1945 points

    Hi Yale

    I did not understand what do you mean by "the output for current loop as output's adjusted part has no relationship with feed forward's calculation formula".

    Just to add to my previous answer, the feedforward is also necessary to accommodate for the negative half cycle. The controller is designed in such a way that the d' should be positive in the positive half of the input and d' should be negative in the negative half of the input voltage. The feedforward part helps achieve that.  

    Thank you

    Amir Hussain