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LM317AEMP - setting of charge-current-limit

Dice-K
Mastermind 7500 points
Other Parts Discussed in Thread: LM317A

Hello,
I have two questions about LM317AEMP.
I use LM317AEMP for charging a lead-battery.
I read the Figure-53 of page-25 in the data-sheet(SNVS774O).

1.
Please teach me the value of charge-current-limit in case of the Figure-53(RS=0.2ohm).

2.
Please teach me the recommended value of RS.
 charge-current-limit : 0.35A
 battery voltage : 10.4V
 charging voltage : 13.7V
I want to know the equation of the setting of charge-current-limit

Regards,
Daisuke

  • 0
    TI__Expert 3920 points

    Hi,

    Please note that the application you refer to will lower the current as the battery load becomes charged. This means that the current limit will change dynamically until the battery voltage reaches the charge limit. Once it reaches that point, the charge current will be 0A. In your case, it seems you want to stop charging once battery reaches 13.7V, and charge at 0.35A when the battery voltage is 10.4V

    First, set your output to 13.7V. For example, this will give R1 = 240, and R2 = 2390. 

    In the case when connecting a 13.7V load (ie a fully charged battery), the current into the battery should be 0A. This means the drop across R1 = 1.25V. The drop across Rs should essentially be 0V, and therefore, no current is flowing to the battery load.

    In the case when you connect a 10.4V battery, The drop across R1 will be 0.95V.

    VR1 = 10.4 * (240/(240 + 2390)) = 0.95V

    This means that the drop across Rs will be 0.3V so that there is a voltage drop across VOUT and ADJ of 1.25V. (0.3V + 0.95V = 1.25V).

    RS = 0.3V / 350mA = 0.86 Ohms. With this value of Rs, the current into the battery should be around 350mA when the battery voltage is 10.4V.

    Note that Rs will need to be adjusted based on the type of battery load, so I would start with 1 Ohm and test on bench. You may need to slightly raise, or lower, to get closer to your target current limit, but 1 ohm should get you close as a starting point.

    Tim

  • 0 in reply to Tim Reyes
    Mastermind 7500 points

    Hello Tim,
    Thank you for your reply.

    I attach my circuit.
    Please teach me the value of current-limit in this case.
    I use a variable-resistance in substitution for a battery in this circuit.

    Actual value is 440mA.

    I want to know the method of a theoretical calculation.

    Best regards,
    Daisuke

    Visio-LM317A.pdf
  • 0 in reply to Dice-K
    Mastermind 7500 points

    Hello Tim,
    I supplement it about my question as follows.

    This is for a battery charging application.

    But, for an examination of factory shipment of the customer product, it is necessary to define the output electric current level by the this circuit.

    It is the reason why he uses resistance load for virtually.

    Best regards,
    Daisuke

  • 0 in reply to Dice-K
    Mastermind 7500 points

    Hello Tim,
    I ask you again.

    I attach my circuit.
    This is for a battery charging application.

    But, for an examination of factory shipment of the customer product, it is necessary to define the output electric current level by the this circuit.

    Please teach me the value of current-limit in this case.
    I temporarily use a variable-resistance in substitution for a battery in this circuit.
    I want to know the method of a theoretical calculation.

    Actual value is 440mA.

    I want to compare an actual value with a calculated value.
    And I want to confirm the validity of the actual value.

    Best regards,
    Daisuke

    Visio-LM317A.pdf
  • 0 in reply to Dice-K
    TI__Expert 3920 points

    Hi Daisuke,

    In the case that 13.84V is measured at V2 (which is shown in the diagram you attached), then there will be about 1.244V across R1.

    13.84V * (R1 / (R2 + R1)) = 1.244V.

    This means that RS (RS = 2.7 ohms || 2.7 ohns = 1.35 ohms) will have a 0.006V drop. This gives a current of 4.44 mA.

    0.006V / 1.35 ohms = 4.44 mA.

    So in the case you attached, you should expect current limit to be around 4.44 mA.

    Tim

  • 0 in reply to Tim Reyes
    TI__Expert 3920 points

    If you do this calculation for all of your values of V2 in your table, you should expect a close approximation to the actual current value.

  • 0 in reply to Tim Reyes
    Mastermind 7500 points

    Hello Tim,
    Thank you for your reply.

    I calculated current-limit when I changed V2 in an attached file.
    When V2 is less than 3.47V, it is different from actual values.
    Is 0.4A the maximum of this IC?
    It is 2.2A on page-6 of the data-sheet. (DCY, Vin-Vout<15)

    The maximum-rating of RS is 1W.

    Best regards,
    Daisuke

    LM317AEMP_current-limit.xlsx
  • 0 in reply to Dice-K
    TI__Expert 3920 points

    Hi Daisuke,

    I believe you are hitting thermal shutdown when Vout gets down that low. 300mA and a 10V drop from Vin to Vout will have the LM317A dissipate about 3W.

    Maximum current is 1.5A for the IC.

    Tim