• State TI Thinks Resolved
  • Locked Locked
  • Replies 1 reply
  • Answers 1 answer
  • Subscribers 30 subscribers
  • Views 48 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

DRV8462: The application issues of chips

Tonyx.wu
Genius 3915 points
Part Number: DRV8462

Tool/software:

I've been reading the manual of RV8426 again these two days. Regarding the automatic torque section, the red box is my personal understanding of the value of ATQ_LRN_MIN_CURRENT. Could you please check if it's like this for me?

Because the register bits of ATQ_LRN_MIN_CURRENT are only 5 bits, dividing by 8 is exactly the 8-bit data shifted to the right by 3 bits. Conversely, ATQ_LRN_MIN_CURRENT*8 means padding a 5-bit value to 8 bits, that is, 32*8=256





Calculation formula

The actual initial current value =ATQ LRN_MIN_CURRENT ×8

The actual final current value = the initial current value +ATQLRN STEP(the increment configured by the register). - Example:

If ATQ LRN MIN CURRENT=24(binary 11000b), then the initial current =24×8=192mA;

If ATQ_LRN_STEP=32(configuration value 10b), then the final current =192+32=224mA.

2. Core setting basis: (1) The current range matches the working load

- Initial current: It should be selected between 30% and 50% of the maximum operating current of the motor.

Reason: This range can ensure that the torque of the motor is moderate when starting under no-load conditions, avoiding learning failure due to too small current or overheating caused by too large current.

- Final current: It should be between 80% and 100% of the maximum operating current and must not exceed 255(register upper limit). - Typical scenario: If the maximum current of the motor is 2A, the initial current should be selected as 0.6A to 1.0A(corresponding.

The CURRENT value of ATQ LRNMIN is 75-125.

If the maximum current is 2A, take 40% of the maximum current as the starting current, which is 0.8A. In binary, that is 255*40%=102,102÷8=12.75, the value is 13, which is the decimal value corresponding to ATQLRN MIN CURRENT, that is, ATQ_LRN_MIN_CURRENT=01101

Please check the parts marked in red to see if they are correct. If there are any incorrect parts, please provide your complete, detailed explanations and correct information. Thank you!

  • 0
    TI__Mastermind 44425 points

    Hi Tony,

    Thanks for your question. Did you also get a chance to check out this application note, https://www.ti.com/lit/an/slvaff1/slvaff1.pdf? 

    The learning is done to compute constant losses in the system mostly due to I2R in the motor coils. This should be performed with the currents of interest such that the algorithm can interpolate the losses at current between lower limit and upper limit. Prior to this it will be helpful to determine the ATQ_TRQ_MIN and ATQ_TRQ_MAX values which are 8-bit scalar for the range in which you'd want the auto torque to maintain the current hence torque of the stepper. Let's assume you have a 2 A maximum current required to support the highest load torque in your system. This would define the ATQ_TRQ_MAX value. To get maximum dynamic range of the auto torque you can scale the VREF itself to output 2 A coil current while ATQ_TRQ_MAX = 255 which also would be the setting for this parameter. Next, you'll have to determine the lowest could current that can be used in your system that can support the lowest torque requirement in your system. Let's assume this turns out to be 1 A. Then ATQ_TRQ_MIN would have to support 1 A which would be 128. 

    Using this you can set the learn min and learn step. Choose these two current coordinates as wide as possible but must be within the bounds of the ATQ_TRQ_MIN and MAX. To calculate the LRN_MIN_CURRENT, divide ATQ_TRQ_MIN/8; 128/8 = 16. Now, we need to set the LRN_STEP. It should be set to a value such that the final value is close to 255, not higher than that value. If you use 128 you'd get 256 which is outside the range. So you can pick 64, the final value would be 128 + 64 = 192. The losses for the higher values up to 255 would be internally calculated by the algorithm. 

    Your thought process followed the above description. You were on the right track. Thank you.

    Regards, Murugavel