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TL431 bias resistor

This question is answered
Hithesh
Posted by Hithesh
on Jul 04 2011 15:26 PM
Expert2000 points

How do you decide the bias resistor for TL431?

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  • Ron Michallick
    Posted by Ron Michallick
    on Jul 05 2011 16:22 PM
    Verified Answer
    Verified by Ron Michallick
    Mastermind30410 points

    Hello Hithesh,

    Figure 1 in the TL431 data sheet (page 20) has a resistor. The Vka output goes to a load not shown.

    The minimum value for this resistor is  (Input - Vka) / (100mA + "minimum load current')
    The maximum value for this resistor is  (Input - VKa) / (1mA + "maximum load current')
    Vka = 2.495V, Input = input voltage.

    This will keep the TL431 cathode current with in the allowable range of 1mA to 100mA.

    Regards,
    Ron Michallick

     

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

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  • Hithesh
    Posted by Hithesh
    on Jul 06 2011 11:41 AM
    Expert2000 points

    Ron,

    I read the datasheet again after posting here.

    I thought its pretty simple. My supply is 5V and Vref required is 2.5V.

    Ik min is 1mA. I thought I'll make this 5mA.

    So a 1K resistor should do it, right?

     

    I don't understand what you mean by Vka output goes to a load not shown.

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  • Ron Michallick
    Posted by Ron Michallick
    on Jul 06 2011 14:21 PM
    Mastermind30410 points

    HIthesh,

    The voltage on the 1k resistor is 2.5V (5V-2.5V); so the resistor current is 2.5mA. This is a good value.

    The TL431 has 2.5V across it. Something in your application is connected to this 2.5V node (other than the 1k resistor). That 'something' is the 'load'.

    Regards,
    Ron Michallick

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

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  • Saurabh Bansal
    Posted by Saurabh Bansal
    on Nov 15 2011 07:56 AM
    Prodigy30 points

    Hello Ron Michallick,

    this calculation is for circuit of Fig. 1 in the TL431 datasheet, what if the circuit of Fig. 2 is to be used, how to calculate the R1 & R2 values.....

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  • Ron Michallick
    Posted by Ron Michallick
    on Nov 15 2011 12:05 PM
    Mastermind30410 points

    Saurabh,

    Figure 2 has a formula for calculating the resistors. The constants (nominal) are Vref=2.495V and Iref=2uA.
    For more accurate result use the equations in this app note. Setting the shunt voltage on an adjustable shunt regulator

    Regards,
    Ron Michallick

     

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

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