# How to test a LM317T Voltage Regulator

Discussion in 'Power Electronics' started by Binary Buddha, Sep 24, 2016.

1. ### Binary Buddha Thread Starter Member

Sep 24, 2016
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• I know nothing about electronics except how to put batteries in. I'm a total noob. Be gentle. I'm here to learn.
• Why? I heard a rumor about counterfeits and that they should be tested. I just need a simple test to verify it's doing what it's supposed to be doing.
• I'm armed with Eagle CAD and I don't know how to properly use it.
• Yes, I've read the datasheets. But, my lack of knowledge and understanding the basics keep me comprehending it. It's like reading Klingon. http://www.mouser.com/ds/2/389/lm217-974117.pdf
So, I'm trying to figure out a simple test to see if the all the regulators I got in are counterfeit or not. Would the below diagram be a suitable test? I'm assuming I set my multimeter to DCV with the red probe on VOUT and the black probe on GND.

2. ### crutschow Expert

Mar 14, 2008
13,488
3,372
That would tell if it's basically working.
The output should be about 1.25V with 1.25A through the 1Ω load.
You only want to do that test for a second or two without a heatsink since the device will be dissipating about 7.75W.

If you want to test whether the voltage is adjustable you would need to add a pot and resistor to the circuit, along with a higher resistance load resistor.

Note that your 9V source has to supply 1.25A to test the posted circuit.

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3. ### Binary Buddha Thread Starter Member

Sep 24, 2016
30
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Thanks for responding!

I couldn't find a 1Ω so I used a 560Ω and got 1.249V. Is that about right? As for the power, it's a 9V battery at 0.83mA... I'm guessing adding a second 9V battery wouldn't work.

4. ### #12 Expert

Nov 30, 2010
16,676
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1.249V is within specs.
Your little 9V batteries aren't strong enough to test the limits of a good LM regulator chip.
If you want to push the limits, you will need more than a couple of amps.
What you tested was the reference voltage. There is a lot more to learn about self-limiting current operation, safe area of operation limits, transient response speed, regulation under load, etc.

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5. ### Binary Buddha Thread Starter Member

Sep 24, 2016
30
0

I know. It's the whole reason I'm here. Would one of those 12V square flashlight batteries do the trick?

Part of the reason I need a working regulator is to build this.

http://www.instructables.com/id/Variable-ATX-bench-powersupply-FabLab-edition/

With that, getting proper amps and voltage for supply won't be much of an issue. ... At least I hope so...

6. ### Binary Buddha Thread Starter Member

Sep 24, 2016
30
0
Is this right? Should result in about 11.946V

Last edited: Sep 24, 2016
7. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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Adj pin should not be grounded.
Check the datasheet for proper diagram

8. ### #12 Expert

Nov 30, 2010
16,676
7,320
Yes. Those massive cubes should be able to throw 10 amps for a little while, and an amp, all day.
Use 120 ohms for R1 to satisfy the minimum load of 10 ma.
Change R2 to fit the 12V requirement. 1033 ohms is the calculated value.

9. ### Binary Buddha Thread Starter Member

Sep 24, 2016
30
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Can you clarify? I'm apparently not interpreting the datasheet correctly. Another forum said that VOUT is the regulated voltage. If that's true, then what does ADJ connect to? In this diagram from the datasheet, I thought that thick black line under R2 meant ground.

10. ### #12 Expert

Nov 30, 2010
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The black line under R2 does mean ground, but that doesn't mean attach the adjustment pin to ground. Your drawing in post #9 looks good.

11. ### Sinus23 Member

Sep 7, 2013
178
465
The black line does indeed...

#12 beat me to it

But in post #6 the adj was shorted straight to the ground

12. ### Binary Buddha Thread Starter Member

Sep 24, 2016
30
0
120Ω or 1200Ω? R=V/I = 12/0.01=1200 ... No?

Change R2 to fit the 12V requirement? I'm assuming I use Io(max) on the datasheet, but not sure which number to use where since it's a potentiometer. 1033Ω for R2? I=V/R = 12/1033=~12ma ... kind of confused...

13. ### Binary Buddha Thread Starter Member

Sep 24, 2016
30
0

So what does the wiper pin on the POT connect to? POT pin 1 is input from R1... POT pin 3 is GND... POT pin 2????

Last edited: Sep 24, 2016
14. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,789
771
The pot needs to be around 5KΩ
I cannot make out the POT value from the diagram.
And connect pin 1 and pin 2 of pot together .
The POT can be varied to adjust the output to your desired voltage

15. ### dl324 Distinguished Member

Mar 30, 2015
3,377
651
It's good that you're aware of the issue with counterfeit parts, but testing is probably beyond the capabilities of most hobbyists.

Basically you'd need to test all of the parameters that the manufacturers check and it's unlikely you'll have the appropriate equipment. In addition to reference accuracy, you'd need to check: max current, thermal protection, load regulation, line regulation, ripple rejection, drift, ....

If you're determined to do this, you'll need to build some test circuits and have an oscilloscope.

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16. ### dl324 Distinguished Member

Mar 30, 2015
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651
The pot should be 1K. The output voltage can only be adjusted to around 9-10V and 1K will give a 10V adjust range.

17. ### dl324 Distinguished Member

Mar 30, 2015
3,377
651
As mentioned, connect terminals 1 and 2 together. That will give you increasing voltage with clockwise rotation.

Also, learn style from other (well drawn) schematics you see. Current flow should be primarily from left to right and top to bottom. Wires should be drawn using multiples of 90 degrees; with the exception of wires entering/leaving buses.

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18. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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I use 5K.
I can adjust from 1.25V to 20V with a 24V input.
Data says 5K.

19. ### dl324 Distinguished Member

Mar 30, 2015
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That's fine, but my point was a 1K will give the OP better resolution because he has a 12V input voltage.

And, since he is learning, he should learn how to select appropriate component values.

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20. ### #12 Expert

Nov 30, 2010
16,676
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R1 always has 1.249 volts across it. That is a constant for this chip.

John Berry likes this.