Using TL431 for set-points

Discussion in 'General Electronics Chat' started by Dritech, May 28, 2014.

1. Dritech Thread Starter Well-Known Member

Sep 21, 2011
761
5
Hi all,

I am using the TL431AC to stable the supply voltage (approx. 5V) which will be used for reference set-points. I noticed that the input supply has to be 200mV higher than the input for it to output the calculated voltage. Is this normal? If not, what can cause this? is it the biasing current?

I now changed the resistor values. I used TI calculator for obtaining the values ( http://www.ti.com/tool/tl431calc ).

With reference to the schematic in the link below, R1 is 1.8ohms, R2 is a combination of 1K pot and 2.7Kohms, and R3 is 3.3Kohms. Are the values correct (especially R1) to obtain an output of 4.81V without having the "drop-down" voltage?

Note: At the output I will be connecting four potentiometers with different values in parallel, which will be used to set different reference-points. The total parallel resistance is equal to 511ohms.

Apr 5, 2008
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3. Dritech Thread Starter Well-Known Member

Sep 21, 2011
761
5
Hello Bertus,

In the thread you mentioned that R2 and R3 are low values. Also, R in my case is 1.8ohms which is quite low when compared to online circuits including the one on that thread.
Should I trust TI calculator, which is were I obtained these values? I used the following parameters in the calculator:

Vin MIN: 4.88V
Vin Max: 5.1V
Vo: 4.81V
Io: 9.4mA (4.81V from TL431 output divided by 511ohms)
Idiv/Ired: 200 (???)

4. ronv AAC Fanatic!

Nov 12, 2008
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R needs to be pretty low because it has to supply the 9.4 ma and at least 1 ma for the TL431 with only .07 volts difference between input and output.
I didn't see where it said the input had to be 200 mv higher than the output.

5. Dritech Thread Starter Well-Known Member

Sep 21, 2011
761
5
Hi,

Dividing 0.07V by 1.8ohms gives 40mA. Is that ok or shell I reduce the current?

6. ronv AAC Fanatic!

Nov 12, 2008
3,657
2,800
Ahh, I see. The calculator..

Here is how I would do it. You need to supply about 10 ma to the load and over 1 ma for the 431 to regulate. So lets use 12 ma. That would make the R 5.8 ohms at low input. So lets use 5.6 ohms - a common value that would give 12.5 ma. Now at high input if the load is disconnected the 431 has to take all the current - about 52 ma.
With the very small resistor the 431 would fry if the load were removed, but it is ok here.

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