# tl431 calculations

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

1. ### 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

2,342
665
Jun 20, 2015
What is the load current ?

3. ### 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.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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:

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.

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5. ### 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

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.

MaxVsup (non regulated voltage)
Min Vsup (non regulated voltage)

once you know the above
IakMin =1ma (from data sheet) ;

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

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7. ### 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..