Electronic – TL431 output voltage

integrated-circuittl431

Recently, I use the Tl431 to be a regulator, and I want to change the output of TL431.

I read the TI datasheet if I want to change the output I need to change the divider resistor, and I also use PSIM to simulate it, it can works.
but I don't know how to analyze it, I want to know why the output voltage can be written as Vref*(1+R1/R2)+Iref*R1

can someone give me some advice?

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Best Answer

The TL431 is a high gain device with an internal voltage reference. When the external input is slightly above the reference voltage, cathode and anode are effectively connected together with a low resistance. When the external input is slightly below the reference input, cathode and anode are, in effect open circuit.

So, if you use two resistors to form a potential divider from cathode to anode and connect the middle node to the reference input, due to negative feedback and internal high-gain, the reference input acquires the same voltage as the internal reference voltage (2.495 volts typically).

So, we have a potential divider where the TL431 adjusts/shunts \$V_{KA}\$ to make \$V_{REF}\$ = 2.495 volts hence, the resistive potential divider equation is: -

$$\dfrac{V_{REF}}{V_{KA}} = \dfrac{R2}{R1+R2}$$ $$$$

$$V_{KA}= V_{REF}\cdot \dfrac{R1 + R2}{R2} = V_{REF}\cdot\left(1 + \dfrac{R1}{R2}\right)$$

$$$$

That's the basic formula but if you want to factor in the slight error caused by the reference input bias current (\$I_{REF}\$) then that formula becomes as stated in your question and the data sheet.

Proof that imagines R2 to be in parallel with \$R_{REF}\$ where \$R_{REF} = V_{REF}/I_{REF}\$: -

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