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DRV8243-Q1: Bulk Capacitor Sizing and Placement

Anjali Karmude
Prodigy 30 points
Part Number: DRV8243-Q1

Tool/software:

Hi everyone,

I am Anjali from Merai Newage Pvt. Ltd. Chennai. We are designing a board to drive two brushed DC motors in two linear actuators used to lift some weight. We are using the DRV8243 SPI (S) variant in the application.

In the first attempt, we used the same bulk capacitor as given in the evaluation board, 10µF, which we realized was not sufficient for the motors inrush current.

Below are the observed motor characteristics.

Peak continuous current consumed by motor (I) :     2.5 A

Inrush current at beginning (I'):       6 A

Duration of inrush peak (dt):           15 ms

Voltage (VM):                                  24 V

Based on above values, I calculated the required capacitance to be 1880µF.  [  C * V = ∆I * dt   ]

The datasheet recommends to reduce the path of high current to minimize inductance. However, due to the huge size of bulk capacitors, the length of high current path on our PCB increases.

 

Seen above, white motor connector and 4 bulk capacitors in parallel giving equivalent capacitance 1880µF.

Please let me know if my calculations of the bulk capacitors are correct. Please also suggest a way to reduce the high current path.

Thank-you and Warm Regards,

Anjali Karmude

  • +1
    TI__Mastermind 45125 points

    Hello Anjali,

    Thank you for your post.

    Anjali Karmude said:
    In the first attempt, we used the same bulk capacitor as given in the evaluation board, 10µF, which we realized was not sufficient for the motors inrush current.

    This will be an useful application note to help understand bulk capacitor sizing for DC motor drive applications, https://www.ti.com/lit/an/slvaft0/slvaft0.pdf. This also has some calculation formulae. While it may be useful to size the capacitor with the inrush current especially if your power supply cannot support this high level of continuous current like you did with your calculation and found it to be 1880 uF, if your power supply can support the peak then the capacitor value needed can be optimized based on the ripple that can be tolerated while driving with PWM - run current.  

    Anjali Karmude said:

    Based on above values, I calculated the required capacitance to be 1880µF.  [  C * V = ∆I * dt   ]

    The datasheet recommends to reduce the path of high current to minimize inductance. However, due to the huge size of bulk capacitors, the length of high current path on our PCB increases.

     

    Seen above, white motor connector and 4 bulk capacitors in parallel giving equivalent capacitance 1880µF.

    While this type of layout may be OK a better setup would be to have the four capacitors within a square which will reduce the inductance of the traces between them. Thank you.

    Regards, Murugavel