TL431 in feedback loop

Hello Mark,

If you query cross regulation in our forum you get a few resources you can use, like here:
http://www.powerint.com/forum/power-supply-design/cross-regulation-smps-flyback-5-outputs
http://www.powerint.com/sites/default/files/product-docs/an22.pdf
You can implement some “unconventional” methods for “perfect” cross regulation. For example, you can use feedback to TL431 from V+ to V- to regulate the output voltage for double value. Now you can connect a simple class B amplifier (2 complementary transistors and 2 equal resistors) to V+ and V-, with the output connected to GND (virtual ground). This way the amplifier will source or sink current to balance the outputs as necessary. If 0.6V error voltage is too big, you can add a super-diode to polarize your amplifier in class AB.
The regulator is connected usually as a shunt regulator with a resistor to high voltage, the high voltage will drop on that resistor and the cathode is going to regulate to the programmed value; unless you have an exotic topology you’re safe. Just check the resistor to limit the current under the maximum allowed for your TL431.
The whole control loop is keeping TL431 in nominal range; hence the minimum 1mA is automatically delivered.

Cheers,
PI_Crusher

I saw AN-22 but it only shows a weighted regulation with several positive outputs. This is not the same as regulating positive and negative rails.
But this quotation from the other post:
"Generally one of the outputs is tightly regulated and the balance outputs are cross regulated that is, they are regulated using good coupling between the winding turns of the main regulated output and windings of these additional outputs."
answers to my first question; only one rail should be regulated (the other will be regulated using good coupling between windings). I even did a test. It works but I've got 1V difference between the rails. This may be due to "not perfect" coupling between windings, right :-)?
The rest of your post answers to my question. But just to be sure: do you mean that TL431 can even be connected to much higher voltages (e.g. 100V) as long as there is a resistor that limits the current? And how the minimum current is achieved? Automatically by TL431?

Mark

Hello Mark,

If you need further improvement, use an amplifier.
In shunt configuration the cathode to anode voltage is the one you programmed trough reference pin, let’s say 5V. All the remaining 95V voltage is on the resistor used to connect cathode to V+, just select one resistor big enough to not push the maximum current/power over the limit.
The minimum current is supplied to TL431 by the whole input/output loop, from the output voltage trough the feedback components, opto-coupler, to the switching regulator, trough transformer and output rectifier, to output voltage.

Cheers,
PI_Crusher

Halo PI_Crusher,

I am facing the same problem for the +V and -V cross regulation using TL431. I can't find one suitable reference design is about that. Actually your description had been very clear, but still I don't understand the actual components placement should be, especially the virtual ground...is that the TL431 reference point? Would you mind to have a simple sketch for the connections such that I can calculate for my own. Since I have no idea about the 0.6V error voltage and the function of the super-diode.

Thanks in advance.

Jun.

Hello Jun,

For your reference, here is a simplified schematic you can adapt to your design (attached). Adjust Rfb1 and Rfb2 ratio for total output voltage (40V total output, 2.5V on the divider). Adjust R5 and R6 for virtual ground position (middle in this case). Note, your actual feedback loop and output configuration is different; it is up to you to identify the proper changes. You can create any number of outputs from a single secondary winding by adding extra virtual ground points. Also, for improved efficiency when the outputs are heavily unbalanced you have to use one transformer winding with rectifier and filter per each output and use the push-pull circuit for balancing only.
If you want to adjust the output polarization current you have to insert a super-diode between the PNP and NPN transistors. If you do not know how to do that, get in touch with one of our design consultants, see:
http://www.powerint.com/en/community/design-consultants

Cheers,
PI_Crusher

PI_Crusher,

Thanks for the schematic but as you said this is different from what we have. Here you have just one output winding and the ground is created artificially by the transistors on the output. So being frankly this is just single output power supply and the regulation is typical for such supplies (you can see it if you imagine that the output transistors have dissapeared for a moment).
BTW, I assume that this is a LTSpice simulation. Can you post it (I learn LTSpice)? Of course including LT1009 and PC817 if they are not is standard installation.
On the schematic you can see that the feedback loop is not referenced to the ground so this is not exactly what we are looking for.
I found a slightly better schematic (see attachment). Will it work? I'm not sure how the voltage on the cathode of PC817 should be calculated (as a result how R1 and D1 voltage should be calculated).

Mark

Hi Mark,

For design help go here:
http://www.powerint.com/en/community/design-consultants

Regards,
PI_Crusher

hello,
maybe it is good for you to see "11 ways to generate multiple outputs" from "articles"...."control..." of switchingpowermagazine.com

you need to register for free then log in, to get the article

Hello PI_Crusher,

Thanks a lot for your virtual ground concept.

The +V and -V of my PS are separated. It works well on balanced load but having about 2.6V difference on un-balanced load in my application.

With referenced from DI-138, DI-179 and DER-43, I tried to make a push-pull circuit for the TL431 fedback but was unsuccessful. The difficuty is,,, I have to maintain the compensation reisitor value for the feedback loop but at the same time providing enough current to the shift up zener. Since I used a voltage mirror to drive the base of the emitter follower, this altered the overall feedback loop.

Since the feedback loop seems to be un-documented and no similar reference design is about push pull topology for TL431 feedback in V+ and V- o/p. Could you show me how should be the proper push pull config. for that?

Once again,, thanks for your help.

Goodday.

Jun.

Hello Jun,

Sure, I can help you but you need first to post your schematic. Only based on your specific application we can look into changes.

Cheers,
PI_Crusher

Thanks PI_Crusher,

The testing schematic and 2 modified circuits were attached,. For the testing cct, I think there is no sinking/sourcing action taking place. Besides, I can't make Q1 fully saturated, Vce always having voltage,.

I had proposed two other ccts, hopefully you could have comments on it.

Thanks a lot.,

Jun.

Hello Jun,

With or without separated V+ and V- the same circuit can be used, without changes. I updated the generic example for you. The added circuit has nothing to do with the original power supply or with the feedback loop. It is 100% independent, connected on outputs only. You can keep your TL431 reference from +15V to GND but I recommend using the reference from +15V to -15V (you have to adjust the resistive divider accordingly). It is an active balancing load and works only when the outputs are not balanced. When the outputs are balanced, both transistors are not polarized and are not in conduction. Actually, you can say it is not working at all on balanced loads. You need at least 0.6V difference in order to activate NPN or PNP transistor.
For improved performance you can add a polarizing super-diode, Darlington transistors, operational amplifier, etc. Pay attention to the voltage/current/power rating for these two transistors. You must make sure you can cover worst case scenario: zero load on one output, maximum on the other.

Cheers,
PI_Crusher

Halo PI_Crusher,

Thanks again for your schematic again but I need your further help to size the two transistors. Hopefully, your could help me again.

During the worst case, +Ve was dropped to 11.2V and -Ve was raised to 17.79V when taking 3.6 amp for both 14.80V output. The load is about 4 ohm and the difference is about 12W on each side. Therefore, I used the complementary darlington power transistors TIP3055/TIP2955 with two 1N4003 diodes in series to pre-conduct the transistors. The circuit did not work when 10K ohm biasing resistors were used or even 5k ohm. I have to use 2k to make the circuit works but the transistors getting very hot even on no-load.

Should I use a pre-amp for the power transistors? If the maximum current is 3.6A, for 15V o/p, is that meaned that the power transistors needed to withstand at lease 54W? OR,,, I should only use the 12W to calculate the components power handling and bias etc? If so,,, there will only be 0.8A required for the 12W compensation. Pls advise.

THanks.
Chun.

Hello Chun,

You should decide for yourself when too much is too much. If you need to keep high efficiency then only a second switching stage could help you. I hope you paid attention to what was discussed about cross regulation. Have you checked AN-22? Are you using stacked windings?
.
If using a balancing load is acceptable, then your transistor should take the difference: 17.8V-14.8V=3V for a current of 3.5A, hence 10.5W power dissipation under worst case scenario. Total output power in your case is 100W, hence you will burn 10% as heat. If this regime is only present for very short periods of time, this power loss can be acceptable providing you use adequate heat sinking.
.
Double-check your transistors choices, you are not using darlington. TIP3055/TIP2955 are ordinary transistors with hfe as low as 20. For ON-Semi TIP100TIP105 transistors you get hfe=2500 and this is normal for darlington. You can use 10k resistors with that hfe.
.
The transistors are getting very hot without load because your complementary transistors are not matched (so much so with low hfe). Measure the voltage on the base and make sure it is equal with output voltage at no load conditions. If necessary, adjust one of these resistors to force the base voltage to be equal with 0V, the middle voltage at no load (to account for small PNP to NPN differences). Only then you have zero current trough balancing transistors at no load. When base voltage is zero, you need at least (Vbe) for unbalanced difference before conduction on any output transistor. With darlington I think you do not need this fine-tune resistor adjustment but you have to verify.

Cheers,
PI_Crusher

Hello PI_Crusher,

Thanks for your helps, advice, comment, infos, recommandations and knowledge etc,,, but i have to bordering you again.

I am using separated windings for the +Ve and -Ve, one diode on positive and another on negative. I used a link to tied their returns together to be a common return. I fully understand that to simply the solution, making two individual PS will be more easy. But I have no choice since I didn't know my appliciation is in this unbalanced condition. I did try to measure the resistance of the load side before hand but in vain... it looks like opened circuit.

Actually, the PS works very well even on 50% unbalanced resistive load. Just,, once I connect to the load side,,, there is about 1.0V difference in between..ie. +14.40 and -15.40. I can't tune up the voltage further as the OVP of my application will trip off the input power when the output voltage exceeds +/- 1V. The actual load current is not 3.6A online ....this is the bench test max. figure I tested the PS. As I have to make sure the PS can work under the max. limit.

Thanks again for your recommandation. I will look for more info/knowledge about power electronics. Hopefully, I can fix it.

Thank you very much.

Brgds,
Jun.

Hello Jun,
Please pay attention to AN-22. It is trivially simple to stack the two windings (still using 2 diodes) changing connections only, no other design tricks (see attached).
On the converter side you have 0V, +20V, and +40V. On the load side you have -20V@L, 0V@L, and +20V@L. All you have to do is to connect 0V to -20V@L, +20V to 0V@L, and +40V at 20V@L. Your converter side ground it is different from your load side ground but is absolutely normal. The final general ground is your load side GND.
Cheers,
PI_Crusher

Hi PI_Crusher,

Now I know the winding technique, order and especially the ground are all important. But I need some time to see how to change the connections since the PCB layout had been fixed. And, do u think the 1V different at the output can be eliminated without using the active balancing cct?

Thanks and best regards.

Jun.

Hi Jun,
The cross regullation it is going to be better. Even if you still have to keep the output balancing transistors, at least the power dissipation under worst case conditions will be reduced significantly. Be carfull what you call worst case scenario, function of the real output load and how long that situation can be present. If it is a very short time, you can use the transistors without heat-sink. Just run the tests you need.
Cheers,
PI_Crusher

Hi PI_Crusher,

Thanks and appreciated for your indeed.

Cheers,

Jun.

Hello PI_Crusher,

I have made the circuit to reduce the voltage different to about 0.5V. The effect is good, just the transistor getting very hot. The power dissipation is about 8W but the adhoc heatsink temperature is about 105C under worst case. As my observation, my load side is very closed to this condition. Actually, it is working is ok, I just want to reduce the temperautre.

Can I use MOSFET to replace the transistors? But the VI product seems the same. Or,,can I use another swith for this purpose? What should be if there is similar circuit? Pls advise.

Thank you very much,

Brgds,
Jun.