What's TL431 Output voltage

Question

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:

Why is Vo 2V? How do I calculate this voltage?
When the TL431 conducts, why doesn't Vo connect to the ground like the third figure?
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.

@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. — Jitter456, Apr 15 '21 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. — Russell McMahon, Apr 15 '21 at 12:56
@bimpelrekkie see my above comment. Vout_low is clamped via an internal diode to Vref. — Russell McMahon, Apr 15 '21 at 12:59
@RussellMcMahon Can you explain in more detail, I don't understand how to calculate the Vo in this circuit. I am very confused . — Jitter456, Apr 15 '21 at 13:03
In my simulation result the Vo is 2V. That is then probably caused by the diode between REF and CATHODE. This causes the voltage at CATHODE (= 2.0 V) to be lower than the voltage at REF (2.5) because that the only way that the control loop can work. It is not how it SHOULD work. When working properly that diode (between REF and CATHODE) should be reversed biased. You forward bias it. — Bimpelrekkie, Apr 15 '21 at 14:05
You're not using the TL431 correctly, as you're moving around your schematics I'll write an answer to be more clear. — Bimpelrekkie, Apr 15 '21 at 14:06

JRE · Accepted Answer · 2021-04-15T15:08:04.613

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. — Jitter456, Apr 15 '21 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, Apr 15 '21 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, Apr 15 '21 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. — Jitter456, Apr 15 '21 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. — Jitter456, Apr 15 '21 at 14:46
@Jitter456: I've update my answer now that you've given the entire origial circuit and the goal you were trying to reach. Explain things in more detail the next time around. Explain what the goal instead of what you think is the way to reach the goal. — JRE, Apr 15 '21 at 15:10

score 1 · Answer 2 · answered Apr 15 '21 at 14:15

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 – Jitter456 Apr 15 '21 at 14:46 — Jitter456, Apr 15 '21 at 14:46

What's TL431 Output voltage

2 Answers2