# What's TL431 Output voltage

I am using PSIM to simulate the TL431, and I have some questions.

This is my PSIM circuit:

• When Vref is less than 2.5V, the TL431 doesn't conduct, it looks like an open circuit so Vo is almost 5.1V.
• When Vref is higher than 2.5V, the TL431 conducts and Vo is 2V.

I re-drew the TL431 circuit.

This is the TL431 circuit:

This is the TL431 equivalent circuit:

This is current flow in the TL431 circuit when V+ is more than 2.5V:

This is an info sheet about the TL431:

There are my questions:

1. Why is Vo 2V? How do I calculate this voltage?
2. When the TL431 conducts, why doesn't Vo connect to the ground like the third figure?
3. I see someone says when the TL431 conducts then Vo is 2.5V. I don't know why.

This figure is from Ti Design a Flybuck Solution With Optocoupler to Improve Regulation Performance.

• 1) In the 2nd circuit, why is there a diode between Vo and the + input of the opamp? I do not see this diode being present in the TL431's datasheet. 2) in the 3rd schematic there's now a short between the output of the TL431 (collector of the NPN) and the REF input, WHY?. Now that circuit does not correspond with the 1st circuit anymore. If you connect the TL431 like that (CATHODE and REF shorted) is behaves as a 2.5 V zener diode. 3) explain what the circuit is designed to do, does it regulate a voltage or compare a voltage? Commented Apr 15, 2021 at 12:22
• @Bimpelrekkie 1.) I see this in the Christophe Basoo APEC 2011 document, I put the figure in there. 2.) Because the diode in conduct so I short them3.)I want to design the flyback compensator, but I don't understand how to decide the Vo when the TL431 conduct. Commented Apr 15, 2021 at 12:49
• YOU must DESIGN what Vout_on is. The spec sheet says Vout should be by design ,>= Vref. In practice Vout will fall to a minimum of 1 diode drop below VRef.this SEEMS to do no harm but violates specs. Commented Apr 15, 2021 at 12:56
• @bimpelrekkie see my above comment. Vout_low is clamped via an internal diode to Vref. Commented Apr 15, 2021 at 12:59
• @RussellMcMahon In my simulation result the Vo is 2V. I Commented Apr 15, 2021 at 13:01

Now that you've actually named what you were trying to do, the situation becomes very different.

In comments, you've explained that all you really want is to calculate the feedback resistors for a particular circuit switching regulator using the TL431.

The circuit you referred to is from Design a Flybuck Solution With Optocoupler to Improve Regulation Performance.

Turn to page 7 of that document and use the equations there to calculate the feedback resistors.

$$\V_{REF} = V_{OUT2} \times \frac {R_{FB3}}{R_{FB3} + R_{FB4}} \$$

• $$\V_{REF}\$$ is the fixed 2.495 V reference of the TL431.
• $$\V_{OUT2}\$$ is the output voltage you are designing for.

Solved for $$\R_{FB4}\$$, that's $$\R_{FB4} = \frac {V_OR_{FB3}}{V_{REF}} -R_{FB3} \$$

Pick $$\R_{FB3}\$$, calculate $$\R_{FB4}\$$.

Original answer, outdated by further information from the OP:

Start with the datasheet instead of some random pages on the internet.

Page 17 shows the typical TL431 shunt regulator circuit, and includes the equation for determining R1 and R2:

First, note the difference between that circuit and yours. You have R1 and R2 connected to the power source directly, but they should be connected between the cathode of the TL431 and ground. Your circuit is incorrect, so none of the normal equations from the datasheet will work.

If you correct the error in your circuit, then you can calculate the output voltage as described in the datasheet. The equation in the picture is a simplfied calculation. The datasheet goes into a more detailed version that is more precise.

For the simple version:

• $$\V_O = (1 + \frac{R1}{R2}) \times V_{ref}\$$
• $$\V_O\$$ is the output voltage you want to have.
• $$\R1\$$ and $$\R2\$$ are resistors that you must set to get the desired $$\V_O\$$

$$\V_{ref}\$$ is a fixed value of 2.495V. This is given by the design and construction of the TL431.

You re-arrange the equation to solve for R1, then pick a value for R2 and calculate R1.

$$\R1 = R2 \times (\frac {V_O}{V_{ref}} -1) \$$

In your PSIM example, you have $$\R1 = R2 = 10k\$$ and $$\V_{SUP}\$$ = 5.1V. From the above equation, you come up with $$\V_O = 4.99V\$$.

You have to correct your circuit for that to hold.

I'd expect the output to be somewhat lower than the calculated value. The TL431 requires some current to operate. It gets its current depending on the difference between $$\V_{SUP}\$$ and $$\V_O\$$. Since that is very small in your circuit, I'd expect the output to drop to a point that the TL431 can get its needed current.

You'll probably have to increase the input voltage if you really want 5V out of the TL431.

• But I want to use it in the flyback, the figure you post is not the same as the flyback compensator structure. the difference between the Ti document and my circuit is I don't put R1, C1 In my circuit. Commented Apr 15, 2021 at 14:34
• Right. But you can't simulate just the TL431 the way you did. A flyback circuit has R1 and R2 connected to the flyback output and V_O driving the regulator feedback. Your simplification doesn't do what you think it does.
– JRE
Commented Apr 15, 2021 at 14:37
• The difference between the TI schematic and yours is that V_O drives the regulator through the op-amp. That is a far cry from your simplification which has R1 and R2 "nailed" to the input voltage.
– JRE
Commented Apr 15, 2021 at 14:40
• I got it. Thanks Because I want to know what's the magnitude of the R3 I should put in my circuit，if I want to drive the TL431, so I need to know what's the voltage of Vo. Commented Apr 15, 2021 at 14:43
• right now I know why the Vo is 2V in this circuit, so if the TL431 conduct the Vo is above 2V. Commented Apr 15, 2021 at 14:46

Using the TL431 like this:

Cannot work.

Why?

To regulate a voltage, the TL431 needs a feedback path from CATHODE to REF.

Your schematic doesn't have this path. OK, not really true, there's an internal diode between REF and CATHODE that's now conducting. But this isn't how the TL431 should be used. At the output you will get about Vref - Vdiode = 2.5 V - 0.6 V = 1.9 V. You measure/simulate 2.0 V, that's close enough.

I wrote "this isn't how the TL431 should be used" as that diode between REF and CATHODE should be reverse biased. That means that the voltage at CATHODE must be equal to or larger than 2.5 V.

Look in the schematic in the datasheet (also in JRE's answer) how to use the TL431. Your upper-right 10 k ohm resitor now connects to the 5.1 V supply voltage. If you connect that resistor instead to the CATHODE of the TL431 you will get the schematic from the datasheet and you will get 5.0 V at Vo.

• Thanks, after you and JRE, I understandnd Commented Apr 15, 2021 at 14:46