# LM317 (not Buck) Linear Adjustable Regulator problems

• John1397
In summary, the LM317 will not work with a buck regulator because there is not a large enough voltage difference between what is going in and what is wanted out. You will need to use a higher voltage regulator with a higher input voltage to take into account any voltage fluctuations. If you want to use a 10K potentiometer, it will need to be able to handle the minimum current that the LM317 will pump out.

#### John1397

I purchased a buck regulator I apply 12 volts dc to in and get 11.5 volts out turn screw on pot makes no changes measured pot goes from 4k to zero ohms if one unhooks connection to terminal 1 adjustment to LM317 will this give you the minimum voltage of about 1.5-1.6 volts? I want 12 volts in and 1.5 to 1.6 volts out.

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A question well stated is half answered.
I can't really decipher your run-on sentence.

Learn to write one thought per sentence.
Place a period at the end of each sentence.
Place separate thoughts on separate lines.
That's showing consideration to the people from whom you're asking help.

John1397 said:
if one unhooks connection to terminal 1 adjustment to LM317 will this give you the minimum voltage of about 1.5-1.6 volts?
No it won't.
Have you read the LM317 datasheet ?
It's at https://www.onsemi.com/pub/Collateral/LM317-D.PDF
Here's the root of your problem.

old jim

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dlgoff, davenn and berkeman
John1397 said:
I purchased a buck regulator I apply 12 volts dc to in and get 11.5 volts out turn screw on pot makes no changes

Exactly, it won't work

jim hardy said:
Here's the root of your problem.

that is only part of the problem

the biggest problem is that there is not a big enough voltage difference between what is going in and what is wanted out

there MUST .. no exceptions ... be a minimum of 1.25 V difference between the input and output, else there will be no regulation

so if you want 11.5V, then you must have a voltage going into the regulator of at least 13V and preferably a little more than that
to take into account any voltage fluctuations of the input voltage

There are some low drop-out regulators that will go down to ~ 0.5 - 1V in/out difference

Also, if you didn't do it, it is reasonably essential to have 0.1uF capacitors from input ( and output) to ground (0V) rail

You should always refer to device datasheets to learn how to use them properly
Dave

PS ... ignore most of this post, I misread what you wrote and it doesn't fully apply to your situation

the very last sentence does tho and I have now hilited in red
it's the most important bit of advice you will likely get today

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berkeman
The OP says he wants 12v in and 1.5 to 1.6v out.

I got to start over. I purchased a buck regulator and applied 12 volts dc and it puts out 11.5 volts dc. The problem is that even if you turn the potentiometer it does not change the voltage, it could be a bad LM317. I checked the pot and it goes from 0 ohms to 4000 ohms. What I eventually want to accomplish with this device is to be able to apply 12 volts input and get out 1.5 to 1.6 volts, I do not really need any adjustment. What I want to know is if you take the 330 ohm resistor from voltage out terminal on the LM317 and connect directly to ground this seems like this would give you the smallest or lowest voltage output. If you the take the potentiometer out and put in a fixed resistor from LM317 adjustment terminal to ground. What does the value of this resistor do to the output voltage as I have seen potentiometers from 4000 ohm all the way to 10000 ohm in this type of circuit.

I got to start over.
I purchased a buck regulator and applied 12 volts dc and it puts out 11.5 volts dc.
The problem is that even if you turn the potentiometer it does not change the voltage,
it could be a bad LM317.
I checked the pot and it goes from 0 ohms to 4000 ohms.
What I eventually want to accomplish with this device is to be able to apply 12 volts input and get out 1.5 to 1.6 volts,
I do not really need any adjustment.
Now that's cleared up.
..............

What I want to know is if you take the 330 ohm resistor from voltage out terminal on the LM317 and connect directly to ground this seems like this would give you the smallest or lowest voltage output.
Have you looked at the LM317 datasheet ?
Do you understand that it tries to hold constant voltage between its Adj pin and its Vout pin ?
........................

If you the take the potentiometer out and put in a fixed resistor from LM317 adjustment terminal to ground.
What does the value of this resistor do to the output voltage
That's answered in paragraph "Basic Circuit Operation" above. However, that paragraph though should say "(see Figure 19)", instead of "(see Figure 17)"..

............................

as I have seen potentiometers from 4000 ohm all the way to 10000 ohm in this type of circuit.
Do you understand that the LM317 must be allowed to pump out some minimum current through its Vout pin ?
Will your 10K potentiometer accept that current ?

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John1397 said:
I purchased a buck regulator I apply 12 volts dc to in and get 11.5 volts out turn screw on pot makes no changes measured pot goes from 4k to zero ohms if one unhooks connection to terminal 1 adjustment to LM317 will this give you the minimum voltage of about 1.5-1.6 volts? I want 12 volts in and 1.5 to 1.6 volts out.
CWatters said:
The OP says he wants 12v in and 1.5 to 1.6v out.

OK looking at your circuit which I couldn't see earlier, it wasn't showing up for me

you have not wired the adj and fixed resistors correctly, so this part of my earlier post is still valid ....

davenn said:
You should always refer to device datasheets to learn how to use them properly

if this is the circuit of yours and is the way you wired it, then it's incorrect and it will not work

read the datasheet for the correct wiring ...

see what you did incorrectly ?

R1 should go straight to the Adj pin of the Reg, NOT to the wiper of R2Dave

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jim hardy
Detailed data sheets can be found at:
https://www.onsemi.com/pub/Collateral/LM317-D.PDF

R1 = 110 Ohms (connect between pins 1 & 2, Adjust & Output)
R2 = 27 Ohms (connect between pin 1 and GND)
This should yield about 1.56V output. Depending component tolerances, this may differ as much as 0.3V either up or down.
If you want the output lower, make R2 a lower value.

An on-line calculator can be found at: http://www.muzique.com/schem/lm317.htm

Notes:
1. The LM317 needs at least 10mA of load current, that is why the resistor values were chosen lower than the datasheet 'typical' values.
2. For minimum output voltage of 1.25V, R2 may be zero Ohms, that is pin1 Adjust connected to GND, which means the lower end of R1 is also connected to GND..

Hope this helps.
Cheers,
Tom

p.s. John, if you are comfortable calculating a resistor voltage divider, here is another way of looking at it.
The LM317 tries to keep the Output pin 1.25V higher than the Adjust pin. This means the divider output, when fed with the desired output voltage, must be 1.25V lower than what you want the output to be.

In your case you want about 1.55V output so the divider must supply 1.55-1.25= 0.3V to the Adjust pin. That defines the ratio of the resistor values. The total resistance of the two resistors in series must also be low enough for the minimum LM317 load current to flow. The minimum load current is 10milliamperes, the output voltage is 1.55V, therefore the maximum load resistance is 155 Ohms.
The resistors were chosen as the nearest standard values with the correct ratio and allowed at least the minimum current. The actual current thru the chosen resistors is 1.55/137 = 11.3mA.

jim hardy
you need a rheostat instead of that potentiometer

same thing @davenn 's saying

@Tom.G has demystified the datasheet for you..

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davenn

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John1397 said:
I got to start over. I purchased a buck regulator and applied 12 volts dc and it puts out 11.5 volts dc.
No, the LM317 is a standard dropout *Linear* regulator, not a buck DC-DC regulator. I will fix your thread title for you now.

John1397 said:

Yes it depends on the load.

The total output current must be greater than 10mA. See "Minimum load current to maintain regulation" in the data sheet. The simple way to arrange that is to make Iset more than 10mA. That way you could test the regulator with no load connected.

However that's wasteful if this is a battery powered device. If you know that you load RL will always draw IL you could reduce Iset to 10mA - IL. For example if IL is always > 6mA you can reduce Iset to 4mA.

Note that Iadj is small so the current through R2 and R2 is approximately the same. So to calculate Iset you can typically just calculate Iset = Vout/(R1+R2).

jim hardy
John1397 said:
Is this "Assume I set = 5.25 mA " based off how much load you are applying or is this an LM317 internal operation number or what would cause this number to change?

Your questions suggest that you are not familiar with circuits and Ohm's law. Is that the case ? Please advise as it helps us figure out what help you need.

Here's a picture to help out with what @CWatters just told you.
See datasheet line "Minimum Load Current to Maintain Regulation"

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## 1. What is the purpose of an LM317 linear adjustable regulator?

The LM317 is a voltage regulator that is used to provide a stable and adjustable output voltage from a varying input voltage. It is commonly used in electronic circuits to regulate power supply and protect components from overvoltage.

## 2. How do I adjust the output voltage of an LM317 regulator?

The output voltage of an LM317 can be adjusted by changing the value of the resistors connected to the ADJ pin. The output voltage is given by Vout = 1.25(1+R2/R1) where R1 is the resistor connected between the ADJ pin and the output, and R2 is the resistor connected between the ADJ pin and ground.

## 3. Why is my LM317 regulator getting hot?

The LM317 regulator can get hot due to excessive current passing through it. This could be caused by a short circuit or an overload on the output. It is important to ensure that the maximum current rating of the LM317 is not exceeded to prevent damage to the component.

## 4. How do I protect an LM317 regulator from overvoltage?

An LM317 regulator can be protected from overvoltage by using a voltage divider circuit at the input to reduce the voltage to within the operating range of the regulator. Additionally, a zener diode can be connected in series with the input to limit the voltage to a safe level.

## 5. Can I use an LM317 regulator as a current regulator?

No, the LM317 regulator is not designed to function as a current regulator. It is a voltage regulator and its output current is dependent on the load connected to it. Attempting to use it as a current regulator can result in damage to the component.