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tl431 calculations

Discussion in 'General Electronics Discussion' started by docb, Nov 18, 2015.

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  1. docb

    docb

    131
    2
    Feb 11, 2010
    Using a TL431 to limit a 29v source to about 5v.

    I see I calculate the two divider resistors to set the vref.

    There's one other resistor that limits the current.
    Any suggestions how to determine that value?
     
  2. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    Are you talking about Rsup?
    What is the load current ?
     
  3. docb

    docb

    131
    2
    Feb 11, 2010
    There's a current limiting resistor before the voltage divider in the sample circuits. I don't know if that what you mean by Rsup.

    Load current about 20-30mA.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,500
    2,840
    Jan 21, 2010
    The minimum current through a TL431 is around 0.5mA from memory, but let's make it 2mA.

    Thus, there are three places for the current to go:

    1) through the load (30ma)
    2) through the tl431 (min 2mA)
    3) through the voltage divider across the TL431.

    The TL431 has a reference voltage of about 2.5V, so the voltage divider for 5V is simply 2 resistors of the same value. Let's use 10k.

    So 5V across 20k is 0.25mA

    The total current is 32.25mA.

    The series resistor needs to drop no more than Vin - Vout at 32.25mA.

    If we want the circuit to operate down to 10V input voltage, then the required series resistance is (10 - 5)/0.032225 = 155 ohms (let's say 150 ohms).

    If the maximum input voltage can go up to 14V and the load is disconnected, then the current through the 431 will be ((14 - 5)/150) - 0.00025 = about 60mA and thus the dissipation of the TL431 will be 300mW and the series resistor will dissipate 0.54W

    under these circumstances the 431 will probably need a heatsink, and the series resistor should be rated at 1W to be safe.

    Change the figures to suit your application. OOPS I used 12V as your source voltage, so you'll have to recalculate anyway. do that and we'll check you got it right.
     
    Tha fios agaibh likes this.
  5. docb

    docb

    131
    2
    Feb 11, 2010
    "The TL431 has a reference voltage of about 2.5V, so the voltage divider for 5V is simply 2 resistors of the same value. Let's use 10k."

    If it's 10k each, then 1 x r1/r2=2.5v. Shouldn't it be a 20k and 10k, so 1 x r1/r2=5v?
     
  6. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    Vout=Vref(1+R1/R2)
    for R1=R2 you get
    Vout=Vref*2=5V
    So R1=R2=10k ohm is correct like steve suggested


    Rsup is the series resistor .
    In order to calculate it you need to know the following info.

    Maxiload=Max load current (Not "about")
    Miniload=Min load current (Not "about")
    MaxVsup (non regulated voltage)
    Min Vsup (non regulated voltage)

    once you know the above
    IakMin =1ma (from data sheet) ;
    RsupMin=MinVsup/(IakMin+Maxiload+5/20K)

    IakMax= 100ma (from data sheet);
    The maximum AK current in the circuit is
    Iakmax=MaxVsup/RsupMin should be smaller than 100ma



    431-shunt.jpg
     
    Tha fios agaibh and (*steve*) like this.
  7. docb

    docb

    131
    2
    Feb 11, 2010
    I see. The divider is calculated so there's always 2.5v at REF?
    That sets the Vo, and the Rsup is for current..
     
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