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Need help with this voltage regulator circuit simulation

Hi, I'm trying to build a 12V, 3A voltage regulator. Attached here is the simulation schematic and LTSpice IV libraries for TL431 reference voltage IC. Can you please tell me what is going wrong here? The simulation takes ages. I have a feeling the transient analysis inputs are off but I'm not sure how and what values I should input here for a 5 second plot.

Any help would be appreciated. Thanks.
 

Attachments

  • LM1085-sim.zip
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Divide and rule.

I managed to simulate the rectifier. It will take some time to do 5 sec.
Add the next part of the circuit, that will slow it down.
 

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  • ep2.png
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Yes I did. Also, I found out that the current load at the end has to be marked "This is an active load" in its settings. The overvoltage protector by itself doesnt seem to work the way its intended (shutdown when anything above 13V is supplied at the input). Wonder why thats the way it is....
 
There is an IC missing in your diagram - LTspice does this.

13V as the input to what? The regulator should stop high voltages comming through from the rectifier.

I assume that that the FET gate is not dropped low enough to turn it fully off.
Why do you need all this protection? What does the Schottky diode do?
 
Which IC is this? I was hoping to power a sata hard-drive with this power supply and hence all the protection. There is a another similar one with 5V output. I tried a simulation with just the over-voltage regulator (i've attached the LTSpice file here) powered by a DC voltage source. I then varied the source in steps of 1V from 5V to 15V. At 15V the circuit does not trip as intended.

I found the circuit off the web where it had a nice simulation and a video about how it works. I'm not entirely sure what the schottky does but i think its an inline protection diode. I put in a schottky since the one in the LTSpice library with this current rating turned out to be a schottky... I sound terribly like a noob!
 

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  • vreg.zip
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Harald,

I looked at the diagram in TL431TI.asy and the first part of the circuit was there which I understood was the trouble. There was a regulator and then a blank area which pesumably originally had a chip of some sort.

Some time ago I tried to simulate a circuit with analog and digital components. The analog components would be saved but not the digital ones so these had to be reinstalled when the circuit was reloaded.

Perhaps I am doing things wrong, it happens !
 

Harald Kapp

Moderator
Moderator
Duke: The vreg.zip contains the symbol (TL431.asy) for the IC. If you put the ASY-file in the same directory as the ASC-file, LTSPICE finds the symbol and displays it:
attachment.php

You also put TL431.lib in the same directory. LTSPICE finds the files and simulates.

The simulation runs in no time at all on my PC. But there's no regulation at all. Vout follows Vin. I think that's no wonder because the TL431 is imho operated completely wrong. The ref-input receives a voltage proportional too V1 (divided by R3, R4). Therefore the voltage at the cathode of U2 follows the input voltage instead of being kept steady. Since this voltage controls the gate of M1, the output can't but follow V1.

To the OP: Have a loook at the TL431 datasheet . Figure 27 on page 30 shows how to use the TL431 plus a bipolar Transistor as a linear regulator. I think a bipolar transistor is the better choice over a Mosfet here.
 

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My apologies Duke, I should've been more clearer on how to use the TL431.lib. I struggled with finding and adding them to the library myself.

Thanks Harald!! I actually have a regulator based on the LM1085. The output of the regulator feeds into this circuit which is supposed to be an over-voltage protector. Figure 27 in the datasheet is a linear regulator by itself or would this also act as an over-voltage protection?
 

Harald Kapp

Moderator
Moderator
So now your circuit is no longer a voltage regulator but an overvoltage protection?

Why do you need an overvoltage protector when you already have a regulated output? Anyway. Have a look at this site. Circuit "c" is an overvoltage protection. It operates by triggering the triac in case of an overvoltage. The triac will short circuit the power supply, blowing the fuse and protecting the rest of the circuit.

If you want a non-destructive protection, you can try figure 12 from this datsheet. Set the shunt regulator's output voltage a bit above the nominal regulated input voltage. The shunt regulator will be inactive because Vin=Vout < Vset.
If Vin rises above Vset, the shunt transistor will become active, bypassing current from Vout to ground. This in turn should cause the current limiter in the first voltage regulator to reduce the output voltage in an azzempt to reduce the current. Without a fuse, however, this fault condition can be active for a long time, causing both the bypass transistor and the voltage regulator to become hot and eventually to become defect.
 
Thanks Harald! It seems the over-voltage protection part was making everything go wonky. Here is another schematic again using LM1085 but without the over-voltage protection bit. I've attached the LTSpice schematic here. Everything works fine and I can see a nice clean 12V too. How do I simulate a current load with this? I tried the load from the LTSpice library (and selected this is an active load) but when I do that, everything goes haywire.
 

Attachments

  • LM1085VoltReg.zip
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Harald Kapp

Moderator
Moderator
I added a load (500mA) and the circuit works fine with an output between 9.8 and 11 V.
Show the complete schematic including load. Do not check "step the load current source" in the simulation commands menu.
 
Here is the working schematic with a 2A load. Even though this works, is this correct? Can I build it in real-life and expect the same results?
 

Attachments

  • LM1085VoltReg-working.zip
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Harald Kapp

Moderator
Moderator
The circuit will work, but it is less than ideal.
The average input voltage of U1 is 17.5V. The output voltage being 12V meand that 5.5V are lost across U1. Together with the current fo 2A this makes 2A*5.5V=11W power lost across the regulator - lost and converted to heat. You will need a good heatsink.
It is better to reduce the input voltage. The 1085 operates down to 1V ionput-output differential voltage. Therefore 13V at the input is enough. Add some margin and design the transformer+rectifier for 14V input voltage to the 1085. The power lost in the regulator then is only ~4W.

You also don't need capacitors C1---C4.
 
Thanks Harald. I just realized that I need two similar 5V supplies (since I intend to power a hard-drive and a raspberry-pi with this. So the 12V+5V(1) would be for the HDD and the 5V(2) for the raspberry-pi) and I wont be able to use the 14-odd volts coming from the transformer+rectifier to power the 5V regulator too since the power dissipation on the 5V regulator would be massive. The idea was to use a single transformer and bridge rectifier to power all three. The schematic is attached here (the amplitude of the sine source is the peak value and not the rms value). Any way around this other than building separate supplies?
 

Attachments

  • LM1085VoltageRegTriple.zip
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Harald Kapp

Moderator
Moderator
You can use the circuit in this way, but you'll have to live with the heavy losses of the 5V regulators (~12.5V*Iload per regulator). And you still haev C1...C4 in place. Remove them.

Consider using a PC power supply. You can find rather inexpensice ones or maybe you can get hold of a used one? These supplies are switched mode supplies which means they have considerably smaller losses than a linear regulator.
 
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