Problem with LM317 adjustable regulator

[godfreyl]

Re: problem with regulator

R2 can't create this problem, even if it is set to 0 the output voltage will be still be 1.25v , the voltage across R1 will be equal to Vref (1.25v) while the regulator is operating properly....

Yes, but the resistor burned, so there was obviously more than 1.25V across it. That means the regulator wasn't working properly. It also means the high current that burned R1 was also flowing through the pot, as there's no other path for it.

 

[jeffrey samuel]

Re: problem with regulator

Yes, but the resistor burned, so there was obviously more than 1.25V across it. That means the regulator wasn't working properly. It also means the high current that burned R1 was also flowing through the pot, as there's no other path for it.

does providing a voltage drop using a regulator suit the need before the voltage crosses the res i mean

 

[mailus]

TRY this.....
make as per circuit diagram...then tell what happened......

 

[alexan_e]

@thebadtall
can you please use a resistor in place of R2 (for example 1K) and measure the voltage at the ADJ pin and the output and report back?

 

[thebadtall]

TRY this.....
make as per circuit diagram...then tell what happened......

why should I also ground adj with 100k ?
won't it affect the Vout ?

@thebadtall
can you please use a resistor in place of R2 (for example 1K) and measure the voltage at the ADJ pin and the output and report back?

I am afraid to do with the current circuit, because it destroyed 2 ic's so far. (i dont want to waste another good one) I will make a new with new parts , and will report !

 

[mailus]

why should I also ground adj with 100k ?
won't it affect the Vout ?

no it doesn't affect the output...it just parallel two resistors((100k*2k)/(100k+2k)=1.9K)

 

[thebadtall]

and I would use it to allow more current ?

 

[jeffrey samuel]

you are just dividing the current between and so the effective current passing through the resistor as the effective resistance drops admittance increases resulting in more current through the other half of the ckt

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and I would use it to allow more current ?

to the ground yes

 

[Baas Rietrot]

Maybe the power rating of your resistor is too low.

 

[FvM]

Maybe the power rating of your resistor is too low.

I hear a some confusion about maximum expectable "R1" (resistor between Vout and Adj, assumed as 220 ohm) current.

In my view, there are three options to consider:

- healthy LM317, minimum operating current below 5.5 mA: R1 voltage will be always <=1.25 V, power dissipation about 7 mW

- healthy LM317, worst case minimum operating current of 10 mA: No correct regulation, R1 voltage drop <= 2.2 V, power dissipation 22 mW

- defective LM317, R1 voltage only limited by input voltage and R2.

P.S.: Special problem, how do we get a defective LM317 (or a circuit behaviour that looks like)?

At the end of the day, an "unexepectable" circuit short, input overvoltage or other issue will reveal. Nothing that can be seen from a distance.

 

[jeffrey samuel]

the power slip in the resistors can be controlled and so if you reduce it will suffice

 

[jiripolivka]

I saw a similar problem long ago with 7805 regulators. The Absolute Maximum input voltage is 40 V with LM317 and 35 V with 7805. By exceeding it, one breaks the internal Zener diode.
As you feed the regulator directly from the AC transformer and rectifier, use an oscilloscope to measure the input voltage peaks: without load it must not exceed 40 V peak, otherwise LM317 dies. Reduce the AC output from the transformer by loading it with a suitable lamp, or, use a 12 or 24 V lamp in series with the primary.

 

[alexan_e]

I saw a similar problem long ago with 7805 regulators. The Absolute Maximum input voltage is 40 V with LM317 and 35 V with 7805. By exceeding it, one breaks the internal Zener diode.

That is not correct, in LM317 there is no max input because there is no ground connected directly to the regulator.
There is only a max allowed in/out voltage difference

Alex

 

[jiripolivka]

That is not correct, in LM317 there is no max input because there is no ground connected directly to the regulator.
There is only a max allowed in/out voltage difference

Alex

I may be am not correct but nobody above mentioned the possibility of the overvoltage and this is usually the killer . Of course, overheating can be, too.

 

[jeffrey samuel]

in the case over voltage is just may be cause for blowing the ic but not the resistors in the ckt as the badtall says so there is over current producing higher heat in the ckt

he needs a way of controlling the current and thus removing the threat

there are many possibilities one is to provide a least resistive path to the gnd
and other is to increase the wattage of the resistor thereby it can withstand higher current

 

[FvM]

I may be am not correct but nobody above mentioned the possibility of the overvoltage and this is usually the killer . Of course, overheating can be, too.

I think all possible and impossible causes haven been mentioned, also overvoltage. See post #30.

 

[thebadtall]

I may be am not correct but nobody above mentioned the possibility of the overvoltage and this is usually the killer . Of course, overheating can be, too.

voltage is in the operating range of this IC (and i think that it has heat protection)

in the case over voltage is just may be cause for blowing the ic but not the resistors in the ckt as bad tall says so there is over current producing higher heat in the ckt

i am not bad,im just tall !

 

[jeffrey samuel]

i am not bad,im just tall !

sorry i never meant it really i better edit it now sorry

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it is done

 

[thebadtall]

i was definately joking ..... (i use this nick since forever)

 

[betwixt]

Nobody has mentioned the other possibility. Is the metal tab on the LM317 isolated ??

Brian.