Lm317 — Parallax Forums

# Lm317

Posts: 2
edited 2022-04-15 21:12

Hello,
I’d like to ask, how is possible that my “improvised Voltage supply” made with lm317T is only showing 6volts with any load on output(without load i can easily regulate all 16 volts i have on input-DC). I’m using 10k potentiomer in typical scheme. But its really pointless to have 16v input, while i can regulate not even half of it. Thanks for responses.

• Posts: 1,187

Can you post the schematic? Also, remember there is a certain minimum load needed. If you dont meet the spec, it will act quite strange. From memory its something like 20 mA measured at the output terminal.

• Posts: 23,514

Also, you need to check the data sheet for the dropout voltage. It's well above zero for the LM317 under load.

-Phil

• Posts: 23,514
edited 2022-04-16 03:49

Here's a "typical" circuit for the LM317, used as a voltage regulator:

Is this what you're using? What are the values in your circuit of the fixed and variable resistors? For example, if your fixed resistor (R1) is 240R, as in the example circuit, and your variable resistor (R2) 720R, you can expect no more than

VO = VREF (1 + R2 / R1) = 1.25V x (1 + 720 / 240) = 5V

This is regardless of how high the input voltage is. To increase the output range at the high end, you need to increase the value of R2. But you'll never get 16V out with a 16V input, due to the dropout voltage, which can range from 1.5V to 2.25V with a load at room temperature.

Also, you'll never get less than the reference voltage, which is 1.25V.

Another thing you need to be wary of is heat dissipation. For example, with a 16V input and a 5V output, drawing 500mA, the regulator must dissipate

(16V - 5V) x 0.5A = 5.5W

It'll get HOT without a proper heatsink, and its internal thermal overload protection will cause it to switch off.

Also, you did remember the input and output caps, right? They're vitally important.

-Phil

• Posts: 15,012

@"Phil Pilgrim (PhiPi)" said:
Is this what you're using? What are the values in your circuit of the fixed and variable resistors? For example, if your fixed resistor (R1) is 240R, as in the example circuit, and your variable resistor (R2) 720R, you can expect no more than

VO = VREF (1 + R2 / R1) = 1.25V x (1 + 720 / 240) = 5V

Yes, unless you have a noisy pot, with wiper contact bounce issues. Then, you can find much more than 5V spikes. as you adjust the supply.
If you are designing a pot-variable supply, it's safer to think about open-wiper possibility, and choose the circuit accordingly.

• Posts: 23,514

Yes, unless you have a noisy pot, with wiper contact bounce issues. Then, you can find much more than 5V spikes. as you adjust the supply.

Only true if the wiper isn't connected to one end of the pot. Otherwise, the spikes in the above example would be limited to 5V.

-Phil

• Posts: 2

@"Phil Pilgrim (PhiPi)" said:
Here's a "typical" circuit for the LM317, used as a voltage regulator:

Is this what you're using? What are the values in your circuit of the fixed and variable resistors? For example, if your fixed resistor (R1) is 240R, as in the example circuit, and your variable resistor (R2) 720R, you can expect no more than

VO = VREF (1 + R2 / R1) = 1.25V x (1 + 720 / 240) = 5V

This is regardless of how high the input voltage is. To increase the output range at the high end, you need to increase the value of R2. But you'll never get 16V out with a 16V input, due to the dropout voltage, which can range from 1.5V to 2.25V with a load at room temperature.

Also, you'll never get less than the reference voltage, which is 1.25V.

Another thing you need to be wary of is heat dissipation. For example, with a 16V input and a 5V output, drawing 500mA, the regulator must dissipate

(16V - 5V) x 0.5A = 5.5W

It'll get HOT without a proper heatsink, and its internal thermal overload protection will cause it to switch off.

Also, you did remember the input and output caps, right? They're vitally important.

-Phil

Yea, i have exactly this circuit, i understand that there is voltage dropout voltage, i tried various resistors on R1 position , but it’s just
changing “sensitivity or regulation” either way i cant reach ~6v on output with any load, i would like to know if is possible to regulate loaded output with even more volts.

• Posts: 15,012

@mattejp said:
Yea, i have exactly this circuit, i understand that there is voltage dropout voltage, i tried various resistors on R1 position , but it’s just
changing “sensitivity or regulation” either way i cant reach ~6v on output with any load, i would like to know if is possible to regulate loaded output with even more volts.

but you also said
" I’m using 10k potentiomer in typical scheme."
which is not 'exactly that circuit' ?

You should tabulate which loads and ranges where it 'works as you expect', and what loads where you 'can't reach 6v',and what VIN is in that case.

There is a maximum current rating and maximum temperature rating on these parts.

• Posts: 1,187
edited 2022-04-16 12:57

Suggest you build the above circuit exactly as Phil has shown it. Verify that you get the expected voltages. This eliminates any possible issue with having a bad part. Once you get that, then start swapping resistor values to get the output voltage you want.

• Posts: 1,524
edited 2022-04-16 14:25

In case you're still using the basic circuit, as depicted in posts #4 and #7, there are chances the regulator has been accidentally damaged, by shorting its INPUT terminal to GND, while the output capacitor still has any residual charge left into it, or even if some kind of "battery" is left connected to the output (I once did it, at the 80's, while trying to craft a motorcycle battery charger; lessons learned).

In order to rule that possibility out, you'll need to test if the LM317 is still functional, or has suffered any damage, and should be replaced, before any further proceedings.

Start by adding the two protection diodes, as depicted in figure 9 (page 11) of the following TI datasheet, and also remember that, as depicted, the 240 Ohm resistor will not ensure a minimum 10 mA consumption by itself, in order to ensure proper regulation at any arbitrary load condition; better go down to ~120 Ohm (<= (1.25 V / 10 mA)).

The Cadj capacitor is usefull to mitigate minute variations introduced by the variable adjust resistor wiper, while moving it (an analog voltimeter will help you notice how "noisy" it can be, while trying to reach the proper output level).

Always remember: LM317 is kind of an old school guy (as I am, by the way), so it'll not be happy with "any" or "some" kind of capacitors (in, adj, and out), and really can start to react weirdly (aka: oscillate the output voltage), when the output current is suddently pushed hard, even whitin its design limits.

If this happens, it's time to find the culprit, by replacing one at a time with another type of capacitor. Keeping them very close to regulator pins (almost "tacked to") is a good start point, since it'll also avoid excessive inductive coupling at such areas.

Fingers crossed here; good luck!

Henrique

P.S. Just reccaled: In the case of the battery charger, I resorted to a 1N5402, at D1's position... just in case...

• Posts: 1,370

Take a look at the TI LM317 datasheet. Most examples are in there.

• Posts: 749

@mattejp said:

Yea, i have exactly this circuit, i understand that there is voltage dropout voltage, i tried various resistors on R1 position , but it’s just
changing “sensitivity or regulation” either way i cant reach ~6v on output with any load, i would like to know if is possible to regulate loaded output with even more volts.

Do you have a heatsink attached to this? and is the heatsink isolated, as in not connected to ground?

• Posts: 1,370

Maybe your LM317 got spiked, overheated and its now bad. Try another known good one.

• Posts: 1,697

mattejp,

• Posts: 4,044

...just in case, check the pin out for the LM317; it's not the same as other TO-220 style regulators.

Looking at the face of the 317 (the side with the device ID), the left pin (1) is the ADJUST, the middle pin (2) is the OUTPUT, and the right pin (3) is the INPUT pin.

• Posts: 20,222

Microsoft support suggests you turn the device off, wait 10 seconds then power up again to resolve the problem.

Apple support suggests you buy a new device for \$1600.

JK, I couldn't resist. I can't add to the numerous great suggestions here already. But I will say that the venerable LM317T is very inefficient, it dissipates power as heat, exactly as a big power resistor would. More current=more heat. More voltage drop= more heat. Very wasteful, especially for a battery-powered device. These days, high-efficiency switching regulators are dirt cheap. As are 5V booster modules, which can take 2-4 volts up to 5V. I have a new favorite, these tiny (0.4" square) 3-pin modules: https://www.amazon.com/dp/B07L76KLRY

• Posts: 2,346

No kidding, I got some of those 3.3V to +/- 12V....can't fault them