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

[Willen]

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.

Hi, are there any simple equation of R4 and R5 (I think it is divider) to make different output voltage regulator myself? If it is complicated, leave it. because main problem has been solved.
Do not focus on my spec. I have a 9V small battery and want to got 5V out as long as possible. So I have to say that- min and max input voltage is 5V to 9V.

 

[MrAl]

Hi Willen,


Ok you changed your specifications a bit here now. So it appears you want the best low dropout regulator you can get.

So now you have to decide what you really want to do. Still use the TL431 or use that plus transistors or go to a ready made solution or use a solution that gets so close to 0 volts dropout that you would hardly know the difference.

But my recommendation is not to worry so much about the 'lowness' of the dropout. That's because there isnt that much energy left in a 9v battery anyway at 5v, and at 6v there is only a little more left too. So the difference in total energy output from the battery at 5v wont be hugely different if you chose the cutoff to be 6v for example. Yes it is a little bit more, but is it really worth going after? And bring this down to 5.5v and you really dont get much more out of the battery after that anyway.

You can test this by taking a 9v battery and running it down using a 10ma constant current load. You'll see it takes a long time for the voltage to reach 6v, but it takes a relatively much shorter time to reach 5v once the 6v point has been passed. You'll have to sacrifice one battery to do this test though
Alternately use an AAA battery and we can extrapolate the results to a 9v battery.

BTW i assume you are using alkaline.

 

[alec_t]

are there any simple equation of R4 and R5 (I think it is divider) to make different output voltage regulator myself?

Yes. See the TL431 datasheet (but for R1 and R2 read R4 and R5)!

 

[Willen]

OK, I will use 12V 9A battery of my motor bike which is about to die because It cannot give enough current to bike while starting, So I have to kick to start . But it can light the headlight (of bike's) very well. I think the battery gives good performance for 50 or 100mA load.

Another thing- I learnt 'voltage forces to device so we have to limit it using voltage regulator, but current won't forces to device. Device will take suitable current itself.' But which is the situation, where we have to use current limiter circuit?

 

[alec_t]

I learnt 'voltage forces to device so we have to limit it using voltage regulator, but current won't forces to device. Device will take suitable current itself.'

I think you'd better un-learn that . Some devices are designed to have a certain range of supply voltage, other devices are designed to draw a certain range of current, all are designed to handle a certain maximum power.

 

[Willen]

Confusing.... Sorry!
Lets say- My small spy transmitter draws 15mA current at 9V. I have 9V 45Amp large battery. But it don't need 15mA current limitter. Device picks up 15mA itself among 45A battery.

Can you show me any examples?

 

[alec_t]

Lets say- My small spy transmitter draws 15mA current at 9V. I have 9V 45Amp large battery. But it don't need 15mA current limitter. Device picks up 15mA itself among 45A battery.

Correct. That device is designed to have a 9V supply and will draw whatever current it needs.
Now consider a constant-current source for, say, charging a NiCd battery. The source is designed to provide a certain current regardless (within reason) of the input supply voltage.

 

[Willen]

Correct. That device is designed to have a 9V supply and will draw whatever current it needs.
Now consider a constant-current source for, say, charging a NiCd battery. The source is designed to provide a certain current regardless (within reason) of the input supply voltage.

Here NiCd battery won't pick up needed current itself? (So current limiter needed?)

Did you mean- I can't charge 3.6V NiCd battery with 5V 45Amp source without a current limiter. Will it damage the NiCd battery?

 

[alec_t]

Here NiCd battery won't pick up needed current itself? (So current limiter needed?)

Correct.

Did you mean- I can't charge 3.6V NiCd battery with 5V 45Amp source without a current limiter.

Yes, I do mean that.

Will it damage the NiCd battery?

Definitely!

 

[Willen]

Ops! Mistakly pressed 'Vote No' instead 'Vote Yes'.
Thank you for your nice help!

 

[Willen]

Hello again,


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

Hi MrAl,
Need your great help!
Few weeks later today I calculated all resistors, voltage and currents of this simple circuit designed by you. But I am being puzzle for Series Resistor (you wrote R=90 ohms). I tried a formula given by you but got different value.

Please can you show me this few steps of calculation please?

(I tried with simulation but puzzled more)

 

[tvtech]

I feel I have to say this.

I don't always look at all the help that certain Members give here to those that want to learn. I stumbled across this thread and and I am gobsmacked

To alec and Mr Al..you guys are both worth Gold for having the patience and time to basically teach here. And teach properly with infinite patience.

I admire both of you. I wish I had the patience you both possess in helping interested people.

All the best,
tvtech

 

[MrAl]

Hi MrAl,
Need your great help!
Few weeks later today I calculated all resistors, voltage and currents of this simple circuit designed by you. But I am being puzzle for Series Resistor (you wrote R=90 ohms). I tried a formula given by you but got different value.

Please can you show me this few steps of calculation please?

(I tried with simulation but puzzled more)

Hello Willen,

The forumla on the schematic (post #10) shows:
Vout=(1+R2/R1)*Vref

so when R1=R2 this comes out to:
Vout=2*Vref

The chip has a built in reference that is set to 2.5v, so the output will be two times 2.5v which comes out to 5 volts. So that's how you get the output voltage.

Now to set the series resistor R (shown as 90 ohms in the schematic) you have to know the load current Iout and the minimum input voltage Vmin, and of course the output voltage Vout. Assuming R1 and R2 are large (like 10k each) we use the load current to calculate the voltage drop across R (the 90 ohm resistor) and make sure that with that amount of voltage drop we still have enough voltage to reach at least 5 volts for the output.
The voltage drop is:
Vdrop=R*Iout=90*0.010 (90 ohms times 10ma)=0.9 volts

and since the minimum input voltage is 6 volts this gives us a non regulated output voltage of:
Vmin-Vdrop=6-0.9=5.1 volts.

Now since 5.1 volts is greater than what we want which is 5.0 volts, this value of resistor is ok and leaves a little room for error.

To calculate this value or R exactly, we would use this:
R=(Vmin-Vout)/Iout
so we would get:
R=(6-5)/0.010=100 ohms

so 100 ohms would be the more exact value, but 90 ohms allows a little room for error.

You also want to calculate the chip current when the input is at Vmax to make sure it does not exceed the requirements of that chip. The max for that chip is 100ma i think (double check this on the data sheet). And also, check the power dissipation in the series resistor R and the chip too.

So first lets calculate the current through the chip at max input of 9.5 volts...
Ichip=(Vmax-Vout)/R-Iout
so:
Ichip=(9.5-5.0)/90-0.010=0.040 amps

and this is only 40ma so the design works so far because this is less than 100ma.

Now the power dissipation in the chip is:
Pchip=Ichip*Vout
so:
Pchip=0.040*5=0.2 watts, which is 200milliwatts.

That sounds ok but check the data sheet to make sure the package can handle 200mw without a heatsink or else add a heatsink.

Now the power in the resistor R is:
Pr=(Vmax-Vout)^2/R
so:
Pr=(9,5-5.0)^2/90=0.225 watts which is 225 milliwatts.

Now a 1/4 watt resistor is 250mw and we have a possible 225mw in the resistor, so maybe we should use a 1/2 watt resistor instead which is 500mw which would be much better.

You may also want to think about the efficiency. We have here 5*0.010 or 50mw output and at times we might have 225mw dissipated in the resistor and 200mw in the package, so there will be times when the efficiency is quite low meaning a large part of the energy stored in the battery goes to waste. This is the basic nature of the shunt regulator.

Is this design process more clear now?

 

[alec_t]

That sounds ok but check the data sheet to make sure the package can handle 200mw without a heatsink or else add a heatsink.

Depending on the package type the maximum allowed power dissipation for the IC is 770mW or 1000mW. A heatsink may not be necessary, but allow space for one and fit it if the IC runs hot to the touch.

 

[Willen]

...Is this design process more clear now?

WOW! How great and clear calculation! That is why I love electronics

Oh, then... I can say 225mW power througn series resistor and 200mW power (40mA) throught chip is 'battery power waste'. Oh it's very high, I think it is good to use with infinite power source (I mean at 220V distribution line using 12V transformer.) instead of small battery.

Another thing- Can I charge 1Ah small battery without current limiter using this shunt regulator? (regulator should give approx 100mA ideal current to charge 1Ah i think). But I guessed not because battery try to take high current (more than capacity of chip) then chip will burn, isn't it?

 

[MrAl]

Hi Willen,


Yes that is true, that many people would see that much wasted power as too much and go with something more reasonable if the input voltage is going to remain high. When the input voltage is lower the waste can be much less, but for higher voltages we see lots of waste. If the load is 10ma at 5v that's only 50mw so wasting even 250mw is already five times more than the load itself. In these cases a series regulator is usually employed or if need be a switching regulator. A series regulator would supply 50mw and waste maybe 40mw, not perfect but better than with the shunt regulator. Shunt regulators are usually used when there is a good reason to do so. So you may want to look into a series regulator next.

Current limiting a battery charger isnt that hard to do when the current isnt too high. I am not sure what you mean by 100ma ideal current though. But a series regulator is usually better for that purpose too (charging a battery) than a shunt regulator.

So the more we look at this the more it looks like you should abandon the original idea of using a shunt regulator and go with a series regulator. There are many options here too.

 

[alec_t]

There is also the matter of what type of battery do you want to charge? Lead-acid, NiMH, Li-ion etc all have very different charging profiles for efficiency and safety reasons.

 

[Willen]

-Simply I mean- Can I normally charge 1Ah 3.6V battery using this 5V shunt regulator? (accepting few missmatch on V and I). Battery is NiCd.

-Um.... I found charging current ratio of Lead Acid battery-
Charging current= Ah/10

Then please tell me others' charging current ratio- Li-ion, LiMh, NiCd etc.

 

[tvtech]

Sigh

 

[tvtech]

I believe Basics are the problem with this kind of stuff.

If the person you are trying to teach does not know the fundamental rules.....you will never ever get anywhere.
A hard road for any teacher to travel.

Just knowing Ohms Law for example would be a solid start for anyone REALLY interested in Electronics.

I am not getting involved here.

But yes, I had to chirp as usual. Solid Basics are the secret. Ohms Law is ingrained in my Brain. V over R X I. Many people sit and look amazed when my little brain can calculate stuff....and the answer is given to them without the use of a Calculator.

Only the Basics. I can make that Formula talk( if it could) because I know it so well.

Regards,
tvtech