How can I check TL431 to know it is damaged or not?

[Willen]

And can I use it as a Low Drop out Regulator? Now watching datasheet

 

[MrAl]

Hi,

Connect it as the simplest possible circuit shown on the data sheet and see if the voltages measure correctly. Check currents too maybe.

 

[alec_t]

Yes, it can be used as an LDO reg.

 

[Willen]

Yes, it can be used as an LDO reg.

But current consumption (battery power waste) may be high than LDO reg, isn't it?

I think it works as a zenere diode, so.

 

[alec_t]

But current consumption (battery power waste) may be high than LDO reg, isn't it?

Sorry, I don't understand that. All linear regulators dissipate power.

I think it works as a zenere diode, so.

Think of it as an adjustable zener diode.

 

[MrAl]

Hello again,


The TL431 is a precision shunt regulator. So yes that means it works similar to a zener diode just with much higher accuracy and temperature stability.

To get it to work as an LDO though would mean having a smallish cathode resistor so that the cathode resistor does not drop too much voltage. The problem is that with a small value cathode resistance the input voltage range will be severely limited. On the flip side is when we use an LDO we often have almost matched input/output voltages anyway, so it might work in some applications.

For example, say we have 5.1v minimum on the input and we want 5v output at 10ma. With a 10 ohm cathode resistor, since 0.1v/10 Ohms=10ma, that wont leave enough current for the proper operation of the chip itself, so lets say we use 5 ohms. Then 0.1/5=20ma which means the chip gets 10ma and the output load gets 10ma. However, if the input goes up to 5.2v, then we have 0.2/5=40ma input current and with 10ma output current that means the chip has to conduct 30ma. So we waste more power. At 5.3v input, we then have 0.3/5=60ma input current, so the chip gets 50ma and the output gets 10ma.
So you see where this is going. The chip is going to be using more power than the output load, so that kind of limits the usefulness. It does work yes, but as a zener diode it may eat up a lot of power in the process.

In contrast, a series regulator would limit the output to 5v at 10ma in this application, and that would mean the input would also see about 10ma even if the input voltage went up higher.

With some external transistors we should be able to get the TL431 to act as a series regulator with low dropout and much lower power dissipation, but that's extra parts then.

 

[Willen]

Looks complicated!

 

[MrAl]

Hi there Willen,


Sorry if i made it sound complicated. In a nut shell if you have low enough current and dont want to use external transistors then it is just a matter of calculating the correct resistor value for the shunt regulator.

The min shunt current is spec'd at 1ma, so if we call the input voltage Vin and the output voltage Vout and the load current Iout then we have:
0.001=(Vin-Vout)/R-Iout

or solving for R we get:
R=(Vin-Vout)/(Iout+0.001)

But also there is the constraint that the device can handle 100ma max. So you have to make sure the load always picks up enough of the load current.

If you have a specific application we could take a look at that in detail and see what the R would come out to if you like. We have to know the following:
1. The input voltage min and max.
2. The output voltage.
3. The load current min and max, or nominal if it does not change much.
4. Might be good to know the max ambient temperature just in case we have to derate the IC chip.

 

[Willen]

Wow! I want to use regulated supply for oscillator of tiny FM Transmitter for stability. So here is parameter as you asked:

1) Input ca be 6V to 9V. (ideal is 9V because will use 9V battery and it goes down after few hours of using to 6V and lower)
2) Output voltage needed 5V.
3) Current needed for oscillator is around 10mA.
4) My room temperature is normally....lets say 30 oC.

OK than can you show LDO schematics now using TL431?

 

[MrAl]

Hello again,


Here is a schematic for your purpose.
Let me know if you build it.

 

[alec_t]

Here's one possibility, with a ~0.1V dropout. But wouldn't it be easier just to buy a 5V LDO reg?

 

[MrAl]

Here's one possibility, with a ~0.1V dropout. But wouldn't it be easier just to buy a 5V LDO reg?
View attachment 74474

Hi alec,

Did you miss the schematic in post #10? That one doesnt need any transistors.

 

[alec_t]

Did you miss the schematic in post #10?

Yes. My page hadn't refreshed when I posted, so our posts crossed.
The shunt reg you posted is certainly simpler, albeit at the expense of higher current wastage (~34mA as opposed to ~2mA for the series reg) and a greater dropout voltage (~1V).

 

[Willen]

Here's one possibility, with a ~0.1V dropout. But wouldn't it be easier just to buy a 5V LDO reg?
View attachment 74474

Yes it would be easier to buy a LDO but I never found this LDO, so.

Wow! Droupout is ~0.1V? Better than real LDO regulator wich has LDO ~0.3V! Let me guide something-

1) What might be the max-min input voltage for this circuit?
2) I think I can use any general purpose PNP transistor like BC557 (?)
3) there is 470R as a "Load", should I have to remove it if I connected the circuit to a real load? (to oscillator)
4) what is the load output current
5) To get desired voltage output with LDO, should I have to use variable resistor as a R5?

Thank you.

 

[MrAl]

Hello Willen,


So you dont mind using external transistors then?

 

[Willen]

Hello Willen,


So you dont mind using external transistors then?

Hi,
I just need Low Drop Out Regulator. Is your simple circuit LDO? alec_t is saying that his little complicated circuit has ~0.1V drop out.

 

[MrAl]

Hi,
I just need Low Drop Out Regulator. Is your simple circuit LDO? alec_t is saying that his little complicated circuit has ~0.1V drop out.

Hello Willen and Alex,

I designed that simple circuit based on your specifications, therefore it works the way you said it should work. It works from at least 9.5v down to 6.0v and always provides an output of 5.0 volts.

Your stated specs do not require a 0.1v drop out voltage max, it requires a 1.0v drop out voltage max.

You stated:
1. 9v down to 6v
2. 5v output
3. 10ma output

The circuit i provided does that. If you have other specs to add then you'll have to consider what might change.

I might add that Alex's circuit is not a bad circuit so if you wish to provide two transistors you can get longer battery life.
So the choice is yours, to go with a very simple circuit which would draw more current from the battery, or go with a more complex circuit that will allow a 9v battery to run the transmitter for a longer time period. If it was mine, i would add the two transistors, or else use a switching regulator. I hate wasting battery energy
If you do add the two transistors check for oscillations at the output with various conditions of load just in case that circuit needs compensation.
Also, change R1 from 1.5k to 700 Ohms or close to that value.
Also, if you see any oscillation you can try placing a small value capacitor across R4, such as 0.001uf. That should help to compensate, but be aware these kinds of regulators are sometimes hard to compensate so lets hope we get lucky

Oh yeah, one small problem with your circuit Alex. The resistor R1 is not of sufficient size to allow the TL431 to work with it's specified minimum cathode current of 1ma. Decreasing this to around 700 ohms should solve this problem without upsetting anything else.

 

[alec_t]

Oh yeah, one small problem with your circuit Alex. The resistor R1 is not of sufficient size to allow the TL431 to work with it's specified minimum cathode current of 1ma. Decreasing this to around 700 ohms should solve this problem without upsetting anything else.

Good point, MrAL. I tried various R1 values in the sim and found values above ~2k didn't do the job. I forgot about the 1mA minimum current.

To answer your questions in post #14, Willen :-
1) If you want 5V out the minimum Vin is ~5.1V. The max Vin is limited by the voltage and power ratings of Q1.
2) Yes, but one with a low saturation voltage is best.
3) Yes, remove it (it just simulates your load). Replace it with the actual load, plus decoupling capacitors (say 10uF electrolytic and 0.1uF ceramic).
4) Regulation is maintained for load currents up to ~60mA. For higher currents you would need to (at least) reduce the value of R3. Higher currents raise the dropout voltage slightly.
5) R4 and/or R5 could be adjustable to change the output voltage; e.g. a 100k pot could replace both.

 

[MrAl]

Hi again,

i also now question whether it is worth building one or not since they make quite a few different low dropout regulators.
Also, i noticed that Willen does not really need a low dropout regulator, but just a pseudo low dropout regulator because according to his spec's his lowest input will be 6.0 volts DC. With 5.0 volts output, that leaves a full volt of overhead to play with.

 

[alec_t]

Agreed. Reading between the lines I think the OP wants to get the max life from his battery (don't we all ), and also to use the TL431 he's got. But like you, MrAl, I think it would be simplest to buy an LDO reg off the shelf.