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Suppose I have the following output stage of a flyback:

schematic

simulate this circuit – Schematic created using CircuitLab

The schematic is a bit simplified but the idea is here. Do not pay attetion to the values of the different components. They are not right. What I wanted to show is that the feedback is function of Vout1 and Vout2. The reference pin of the TL431 is about 2V5. When the reference pin is exceeded, the LED of the optocoupler begins to conduct as a rough approximation. In any case, there are many solutions on Vout1 and Vout2 for having 2V5 to the reference pin. So how could be define Vout2? or Vout1? At a certain point an equilibrium between the two outputs will be reach, but this equilibrium could be unstable? How does it work?

As the transformer is stacked, the ouput voltage Vout1 is function of the output voltage Vout2. Actually suppose there is the same number of windings between the two outputs, Vout1 is equal to 2 times Vout2. Then an equilibrium could be reached as there is only one unknown parameter into the previous equation for determining Vout1 (2Vout2) and Vout2 with the reference voltage of the TL431. Nevertheless if Vout1 = 2Vout2. Why do we need to add Vout1 to the feedback? As it seems that Vout1 is regulated through Vout 2? And this what we supposed by remplacing into the previous equation for determining Vout1 and Vout2 with the reference pin voltage.

The flyback is working in DCM.

Thank you very much and have a nice day!

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  • \$\begingroup\$ Classic issue! How much leakage inductance do you have in each output winding and what’s your worst case unbalanced load? \$\endgroup\$ – winny Jun 16 at 18:16
  • \$\begingroup\$ The transformer has no leakage inductance and you can say that the load of Vout1 is equal to R1 and the load of Vout2 is equal to R2 \$\endgroup\$ – Jess Jun 16 at 19:25
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    \$\begingroup\$ Check the components' connections around TL431 in your schematic. R5-R6 junction should be connected to REF pin. \$\endgroup\$ – Rohat Kılıç Jun 16 at 19:36
  • \$\begingroup\$ In that ideal case, you have an easy job and can regulate on one rail only. \$\endgroup\$ – winny Jun 16 at 19:42
  • \$\begingroup\$ Thank you for your comment Winny. I try to understand even it seems an easy job (at least to you) . \$\endgroup\$ – Jess Jun 17 at 9:37
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Actually suppose there is the same number of windings between the two outputs, Vout1 is equal to 2 times Vout2.

Without cross-regulation, nope.

Imagine you regulate only Vout1. Vout1's load current flows through Vout2's winding. So Vout2 will vary with Vout1's load. One common solution is using discrete windings and connecting the bottom end of Vout2's winding next to Vout2's rectifier diode (i.e. cathode).

Cross regulation with combined feedback kinda solves this problem. Please note that both outputs cannot be regulated tightly compared to regulating a single rail.

When there's only one output voltage to be regulated tightly using a shunt regulator (e.g. TL431), we all know that the output voltage is set by the divider resistors:

schematic

simulate this circuit – Schematic created using CircuitLab

Fig.1: Secondary-side regulation with a shunt regulator for single output voltage

Let's assume that the bias current through REF pin is zero. $$ \mathrm{ I_{R6} = \frac{V_{REF}}{R6} = \frac{V_{o1} - V_{REF}}{R5} \\ \\ \therefore V_{o1}=V_{REF}(1 + \frac{R5}{R6}) } $$


If there are multiple outputs to be regulated with a shunt regulator, things change a bit:

schematic

simulate this circuit

Fig.2: Secondary-side regulation with a shunt regulator for multiple output voltages

To determine the output voltages, a "regulation factor" should be defined for each output. These factors should be between 0 and 1, and the sum of the factors should be 1.0. If an output voltage should be regulated more tightly then its regulation factor should be greater than all the others.

Let's make the calculations for two output voltages as shown in OP's question.

$$ \mathrm{ I_{R6} = \frac{V_{REF}}{R6} = I_{R5} + I_{R7} \\ I_{R5} = \frac{V_{o1} - V_{REF}}{R5} \\ I_{R7} = \frac{V_{o2} - V_{REF}}{R7} } $$

Assuming Vo1 should be regulated more tightly than Vo2. Let the regulation factor for Vo1 be \$K_{Vo1}=0.7\$. So, the regulation factor for Vo2 will be \$K_{Vo2}=1 - 0.7 = 0.3\$. This means that the current flowing through R5 should be higher than that flowing through R7:

$$ \mathrm{\frac{I_{R5}}{I_{R7}} = \frac{K_{Vo1}}{K_{Vo2}} = \frac{0.7}{0.3} } $$

So, $$ \mathrm{\\ I_{R6} = I_{R5} + I_{R7} \\ I_{R5} = 0.7 \ I_{R6} \\ I_{R7} = 0.3 \ I_{R6} } $$

The rest is simple: Pick a reasonable value for \$\mathrm{I_{R6}}\$ then calculate the rest.

EXAMPLE

We want to regulate Vo1 = 5V and Vo2 = 12V with single TL431. And the 5V-output should be regulated more tightly.

Let's pick \$\mathrm{K_{5V}}=0.6\$. So, \$\mathrm{K_{12V}}=0.4\$.

Let's pick \$\mathrm{I_{R6}= 0.5mA}\$, so we can calculate R6: \$\mathrm{R6=2.5V/0.5mA = 5k\Omega}\$.

We obtain \$\mathrm{I_{R5}=0.6\ I_{R6} = 0.3mA}\$ (1) and \$\mathrm{I_{R7}=0.4\ I_{R6} = 0.2mA}\$ (2).

Finally;

using (1), we obtain \$\mathrm{R5 = \frac{5V-2.5V}{0.3mA}=8.3k\Omega}\$.

and using (2), we obtain \$\mathrm{R7 = \frac{12V-2.5V}{0.2mA}=47.5k\Omega}\$.


FINAL NOTES

  • The greater the regulation factor, the better the regulation.

  • As I stated earlier, both outputs cannot be regulated tightly compared to regulating a single rail. In practice, there'll be slight fluctuations.

  • For single-output converters, the regulation factor for that output is 1.0.

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  • \$\begingroup\$ You re right ! The voltage drop across the diode depends on the load. In any case what would be Vout1 and Vout2 in the case of this kind of feedback. If you can show some maths it would be nice to you. \$\endgroup\$ – Jess Jun 16 at 20:12
  • \$\begingroup\$ @Jess I'm on mobile, so I cannot show any mathematical details. But you can do it by yourself: The REF pin voltage is 2.5VDC. Assumming the pin bias current is zero, the current that flows through R6 is 2.5/R6=(Vo1-2.5)/R5+(Vo2-2.5)/R7. To solve this, you should define an "importance factor" for each output and the sum of the factors should be 1. For example, if you want Vo1 to be regulated more tightly then the current that flows through R5 should be greater than the other, so its imp factor should be greater (e.g. 0.7 for Vo1, 0.3 for Vo2). I'm sure that you got the idea. The rest is simple. \$\endgroup\$ – Rohat Kılıç Jun 17 at 1:13
  • \$\begingroup\$ Thank you I got the idea. Nevertheless you have one equation and to unkown outputs Vout1 and Vout2. I do not know how you can find Vout1 and Vout2 so. The two outputs are linked together. Obviously it miss something ... Otherwise I could say Vout1 = 0V and Vout2 = X in order to have the reference voltage equal to 2V5. \$\endgroup\$ – Jess Jun 17 at 9:30
  • \$\begingroup\$ @Jess see my edit then. \$\endgroup\$ – Rohat Kılıç Jun 19 at 10:46
  • \$\begingroup\$ I appreciate all your effort for answering the question but I do not understand why the outputs will reach 5V and 12V because you decide to apply a coefficient equal to 0.6 and 0.4. Why the physics should apply your coefficients ? The physics could stabilize one output at 0V, no current will flow through your resistor and the other output will go up in order to make flowing enough current trough your resistor and set the reference to 2V5. And the controller will have no reason to change its duty cycle as the reference is at 2V5 \$\endgroup\$ – Jess Jun 19 at 19:51
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It's all about load regulation and the different voltage dropped in the diode rectifiers when different loads are applied on V1 and V2.

The transformer has no leakage inductance

This simplifies things. However, full loading on one output will cause a volt drop due to the dynamic resistance of rectifier diode. If that output (V2 for instance) was exclusively used as the feedback then it would stay perfectly regulated but the other output would rise slightly.

  • V2 remains stable but V1 would rise by X % when fully loading V2's output

If both outputs were summed (using the correct proportionate resistors) to produce the feedback then there is a compromise: -

  • V2 would fall X/2 % and V1 would rise by X/2%

Why do we need to add Vout1 to the feedback?

You don't have to - you could decide that V2 is the most important rail to stabilize and therefore you live with an X % rise on V1 when V2 is stable but fully loaded.

There are ways to compensate for this using series magnetic cross-coupling - if the main regulation is for V2 and it draws more load current, an anti-phase voltage is applied in series with V1's winding to counteract the X % rise in V1. The anti-phase voltage is produced by the AC current taken from V2's secondary winding: -

enter image description here

I've seen it done once and it worked but it does complicate the parts list.

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  • \$\begingroup\$ Thank you Andy for your answer. It is interesting ! I gave the answers to Rohat Kilic even if it is not what I am looking for. I will try to find my own way :) \$\endgroup\$ – Jess Jun 19 at 19:57
  • \$\begingroup\$ Maybe explain what you are looking for. \$\endgroup\$ – Andy aka Jun 19 at 19:58
  • \$\begingroup\$ What I want to say but it is probably not clear is why the relation Vref = Vout1/K1 + Vout2/K2 shoud give you the desired Vout1 and Vout2. There is one equation and 2 unknown parameters (Vout1 and Vout2). So there is a lot of possibility on Vout1 and Vout2 to obtain a constant equal to Vref as K1 and K2 are fixed. It miss an equation.And the other equation is probably the relation between Vout1 and Vout2 which are linked with the turn ratio of the transformer. \$\endgroup\$ – Jess Jun 20 at 8:56
  • \$\begingroup\$ OK so you are after a relationship formula. Why don't you make a simple simulation to demonstrate this to yourself? \$\endgroup\$ – Andy aka Jun 20 at 9:08
  • \$\begingroup\$ I will try ! (I already tried, it was not a success, but I will try again :D) \$\endgroup\$ – Jess Jun 20 at 9:35

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