Minimum value R1, LM317 voltage regulator

Hello. This is a first post from a complete novice about a first circuit. I have made a small board to reduce a DC voltage from a nominal 12v to a nominal 9v using a LM317 voltage regulator. Having made the circuit, with 13.5v going in, I am getting 7.3v coming out. This is without any load other than my meter.

I wonder if there is something about the resistor choice, other than their relative values, that is important. Not knowing that there was anything other than a ratio between the two

(from the formulaVOUT = 1.25 * ( 1 + R2/R1 )
)

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I went for 10ohm and 62ohm as this gave me the 6.2 (R2) to 1 (R1) ratio I needed.

In addition to the 2 resistors, there are 2 Tant Capacitors (0.1uf and 1uf) across input and output.

I have since found sites indicating that R1 should not be less than 100Ohms -eg
LM317 Voltage Calculator - Electric Circuit
The product information gave no minimum value for R1.

Can anyone help me with this please?

thanks

Coltsfoot

There is almost no limitation on the value of that resistor ..
Just keep in mind that the voltage across it is constant – very close to 1.25V – so the current through it will be determined by 1.25/R formula, which in fact is used in the “constant current” configuration ..
But if you use smaller resistor than is required for normal operation you will allow more current to flow through the wasteful path ..

IanP
:wink:

Thanks for that. In the meantime, I popped out to my local supplier and picked up some 240ohm and 1.5kohm resistors, and substituted them into my circuit. I now have 9.1v output, so I am happy (although I don't understand why the voltage has picked up).

Coltsfoot

There's an explanation in this video: EEVblog #158 – AVR ISP MK2 + LM317 Regulator Tutorial @ EEVblog – The Electronics Engineering Video Blog.

hi,
in the datasheet he has suggested to always use 240ohms resistor as R1(upper resistor of the divider) and vary the other value. If you are using other value then reference voltage 1.25 there is there in the formula 1.25(1+R2/R1) will not be there so the formula will not hold good.

sunil_n said:

in the datasheet he has suggested to always use 240ohms resistor as R1(upper resistor of the divider) and vary the other value.

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Look closely again at the datasheet. 240 ohms is used with the more expensive LM117. 120 ohms is used with the cheaper LM317 to provide a 10mA load so that the output voltage does not rise (explained in the datasheet) when there is no other load.

OK, my circuit has now died, and certainly the resistors have had it, as they are open circuit. How do I calculate the power specification I need for the resistors in a circuit like this LM317 Voltage Calculator - Electric Circuit? Sorry but be easy with me as I am a total novice and gave up physics too early at school (wish I hadn't). I need to deliver 1.5Amps over the cicuit and maybe the TS317 is a bit under spec.d for this, as it will not exceed 1.5Amps. My input voltage is 12v (up to 14.5v in reality) and output 9v. thanks in advance

coltsfoot said:

OK, my circuit has now died, and certainly the resistors have had it, as they are open circuit. How do I calculate the power specification I need for the resistors?

Click to expand...

R1 has 1.25V across it. If it is a 120 ohm resistor then it heats with (1.25V squared)/120= 0.013W which is almost nothing.
For 9V output then R2 is 744 ohms (use 750 ohms), has 7.75V across it and it heats with only 0.08W.

Thanks very much. So if my resistors were up to it (I think they were both 0.6W) what else would cause them to open circuit? I had bolted on a big heat sink onto the regulator and this was only just warm to the touch.

---------- Post added at 20:59 ---------- Previous post was at 20:51 ----------

Well I guess that is probably too broad a question to ask, and I should do some research first. So from Wikipedia....

Damage to resistors most often occurs due to overheating when the average power delivered to it (as computed above) greatly exceeds its ability to dissipate heat (specified by the resistor's power rating). This may be due to a fault external to the circuit, but is frequently caused by the failure of another component (such as a transistor that shorts out) in the circuit connected to the resistor. Operating a resistor too close to its power rating can limit the resistor's lifespan or cause a change in its resistance over time which may or may not be noticeable. A safe design generally uses overrated resistors in power applications to avoid this danger.

Click to expand...

If my LM317 voltage regulator has failed (and that has then cooked the resistors) how would I test it with a multimeter?

thanks

,

An LM317 automatically shuts down if it overheats. when it cools then it starts working again. Too many thermal cycles like this will destroy it.

An LM317 is destroyed if its pins are connected backwards. Then the 120 ohm resistor might have 14.5V across it and will burn out with (14.5V squared)/120= 1.75W.

The 750 ohm resistor will not burn out with 14.5V across it because its heating will be only 0.28W.

You cannot test an LM317 with a multimeter.