LM317 keeps blowing up - Page 1
Forum section: electronics / beginners

author: jorenca
author_rank: Contributor
timestamp: January 27, 2016, 06:45:43 pm
content: Hi guys,I'm making a regulated lab power supply somewhat similar to the EEVBlog power supply.However, my LM317 keeps dying on me.I have relatively big heatsing attached to it.I have no idea what I'm doing wrong.When I measure a blown LM317, it has it's ADJ pin almost (10 ohm or so) shorted to the OUT pin (if this matters at all).I'm attaching a shematic of how I use the regulator.Any help appreciated, I'm desperate now after 5 regulators destroyed.

author: AF6LJ
author_rank: Unknown
timestamp: January 27, 2016, 06:56:07 pm
content: Show us your circuit.

author: Andy Watson
author_rank: Super Contributor
timestamp: January 27, 2016, 07:01:14 pm
content: Are they genuine LM317 ?

author: wblock
author_rank: Regular Contributor
timestamp: January 27, 2016, 07:11:11 pm
content: The left pin on an LM317 is the adjust, the middle pin is the output, and the right pin is the input. If you have it hooked up like that diagram, we have probably found the problem.

author: mikerj
author_rank: Super Contributor
timestamp: January 27, 2016, 07:13:23 pm
content: Why do you have a switch on the reference pin? Leaving the reference pin open circuit is not a valid way to use the regulator.Tell us what you are trying to achieve with this circuit.

author: jitter
author_rank: Frequent Contributor
timestamp: January 27, 2016, 07:27:24 pm
content: Where are your resistors that set the output voltage (R1 + R2 in the typical application below of a LM317)? Why are there switches in those pins at all?

author: jorenca
author_rank: Contributor
timestamp: January 27, 2016, 07:33:38 pm
content: They are not genuine LM317 I guess.I have checked the pinout of the regulator several times before I connected it.The switches are actually jumper cables that I will connect to a PWM adc.However, currently im connecting the ADJ pin to ground to get 1.25V and I do get it.When I start playing around with the jumper cables, however, first my load resistor burns and then, when I disconnect the load, the LM317 heats over 150C and I disconnect the power in.My initial tought was that when I unplug the ADJ pin jumper, this results in high voltage output from the regulator (so I measured) and then I overload the regulator with the10 ohm load resistor, which results in my LM317 dying and even with disconnected load to overheat anyway.But doesn't LM317 have some protection?At the moment my guess is that I'm shorting out something while I'm probing my circuit, but I dont think I am.Heres a more descriptive schematic, the 1 ohm resistor will be used for current sensing.

author: tautech
author_rank: Super Contributor
timestamp: January 27, 2016, 07:34:31 pm
content: Quote from: jorenca on January 27, 2016, 06:45:43 pmUse member jitter's circuit along with protection diodes as shown in most 317 datasheets.

author: KL27x
author_rank: Super Contributor
timestamp: January 27, 2016, 07:38:01 pm
content: Well, if you go by jitter's post, he has R1 and R2 mislabeled/swapped.Either that or you would have to switch R1 and R2 in the formula. The first half anyway. I'm not sure which resistor actually belongs in the second half.Jorenca, what voltage out are you trying to get? Are you omitting a ton of parts from your schematic? (Namely R1 and R2?)

author: jitter
author_rank: Frequent Contributor
timestamp: January 27, 2016, 07:42:31 pm
content: Hah, indeed...Here's thedatasheetand page one shows the typical setup and formula correctly.

author: jorenca
author_rank: Contributor
timestamp: January 27, 2016, 07:51:34 pm
content: I'm trying to get 1.25V to 11.25V out.Basically, I want to use the EEVBlog power supply schematic, replacing LT3080 with LM317 (because of availability issues).I am aware that the constant current mode will be impaired because of LM317 output being minimum 1.25V.I have no resistors connected between the OUT and ADJ pin.No parts related to the regulator are ommited in my previous schematic.Can disconnecting the ADJ pin while having a 10 ohm resistor connected between OUT and ground be the cause?Even so, shouldn't the LM317 protection kick in before the regulator gets fried?The schematic from the datasheet I cannot use, because I want to drive the ADJ pin with 0 to 10V(with a PWM adc and opamp gain of 2) and get OUT = ADJ_pin_voltage + 1.25

author: KL27x
author_rank: Super Contributor
timestamp: January 27, 2016, 07:52:10 pm
content: Trivia:You can use any basically any fixed linear regulator the same way as an LM317, using the same formula, by using the ground pin as the adjust pin. Just replace the 1.25V with w/e the voltage of the regulator. The resistor ladder basically changes the gain. The regulator is trying to keep the difference between the output pin and the ground/adjust pin at the internal reference voltage. In the case of an LM317, that is 1.25V. For an LM7805, that is 5V.E.g., with a 5V regulator you can get any output voltage 5V or higher, by adding R1 and R2 and using the formula Vout = 5V*(1+ R2/R1)The LM78xx regulators have the heatsink on the ground pin... so when you do this, you can't connect that heatsink to ground anymore. The voltage on the heatsink will now be a stupid Vout-5V. The LM317 is meant to be adjustable, so the heatsink is placed on Vout which makes more sense.

author: macboy
author_rank: Super Contributor
timestamp: January 27, 2016, 07:56:51 pm
content: You need input and output capacitors very near the LM317 or it will oscillate and die a hot smoky death. Currently you have a capacitor located near the load (not near the LM317) and it is disconnected along with the load. Do not disconnect the capacitor. You need an input cap too, close to the LM317. You alsomusthave a minimum load or again, you have trouble. The minimum load is 10 mA, so if you are set up for a fixed 1.25 V output, then you must have a 120 ohm resistor from out to ground at all times (not to be disconnected by your relay). The basic adjustable LM317 circuit has a 120 ohm between Out and Adj, and a variable R from Adj to Ground. There is always 1.25 V between Out and Adj, so always ~ 10 mA flowing through that 120 ohm resistor, satisfying the minimum load requirement. (Refer to schematic posted by jitter ... note that R2 is 240 ohm in that case, this is only allowed if using a version of LM317 that has a 5 mA minimum load requirement).

author: KL27x
author_rank: Super Contributor
timestamp: January 27, 2016, 08:09:59 pm
content: QuoteYes, that would work fine. When grounding the adjust pin, you are basically turning the LM317 into an "LM7801.25" 1.25V fixed regulator.When you disconnect the adjust pin, I'm not sure where that pin is going to be pinned and/or if it will just float. I suspect the output would just jump to ~Vin -1.5V, but you're the one with the circuit and a multimeter. That would be drawing 1.7A with your Vin and your 10R load resistor. That's 2.55W, and a TO-220 can't do that even with a small heatsink, at least not for more than something on the order of seconds or minutes.Just as bad when you GROUND the pin and get Vout of 1.25V. That will pull only 125mA through the load. But that will drop 18.25V through the regulator. Which results in 2.25W. Almost the same. *edit, sorry, seeing your Vin is only 15V, so adjust accordingly.If you go by the max current of 1.5A for this part in a TO-220 package at 70C using the minimum voltage drop of 1.5V, that puts max dissipation for this package at 2.25W. And that will only work as long as the part is kept below 70C, or so. And unless using a very good heatsink and a fan, you probably won't get that. I'd say with a dinky heatsink, you ought to reduce that by 50%.You should have input and output caps, yeah, but I doubt oscillation will be causing the device to die, since your load is just a dummy resistor. Personally, I have never seen an LM317 perform a thermal shutdown. At least not in a way that it continues to work after it cools down. But then I'm in the same boat as you. I bought mine off eBay many years ago, and I can't be sure they're real.FWIW, my own LM317 bench supply has a large aluminum heatsink with integral fan and a 20V input like yours. I have never really documented the limits, but a 10 ohm load would be pushing it at any Vout.

author: jorenca
author_rank: Contributor
timestamp: January 27, 2016, 08:38:33 pm
content: just a small note that the input voltage of the LM317 is about 15V (measured 15.4V) after the 7815.Indeed with ADJ disconnected the output is ~13V.When I ground the ADJ and get 1.25V out, with the 10R resistor the heatsink gets to about 90C after a few minutes.Is 70C on the heatsink the maximum reasonable temperature that the regulators should operate at?I will add the caps asap.Can I consider the floating ADJ pin while having 10R load as the reason for all my dead regulators?p.s.: My goal is about 200mA max output at 1.25V,0.8A at >5V

author: KL27x
author_rank: Super Contributor
timestamp: January 27, 2016, 09:26:10 pm
content: This is just my opinion:QuoteFor long term reliability, I would personally want to see something around 70-85C, but 90c might be ok, too. It might appear to work fine at even 120C.... but I wouldn't rely on it long-term.QuoteThe difference between Vin and Vout is being dissipated by the regulator. So at lower Vout, you can't draw as much current.Using Vin of 15V, the difference between Vout 13V vs 1.25V with a 10 ohm load is going to be the difference between 2.6W (at 13V/1.3A) and 1.72W @ 1.25V/125mA. So yeah, that might be the difference. With your current heatsink it looks like your regulator is topping out somewhere between 1.72W and 2.6W. If your temperature reading is accurate, it is probably closer to 1.72W.

author: MrSlack
author_rank: Frequent Contributor
timestamp: January 28, 2016, 09:16:14 am
content: That cap is probably discharging into the LM317 when you kill the power. Once this happens the behaviour is undefined and it will probably explode when you reapply power.Read the datasheetthen stick a couple of diodes in the circuit as described. Also stick a heatsink on it as it needs somewhere to dump all the energy.If you'd got a properly configured LM317, it is very difficult to blow it up.Also get a genuine LM317T from ST or somewhere. The clones will explode as well.

author: AlxDroidDev
author_rank: Frequent Contributor
timestamp: January 28, 2016, 11:48:33 am
content: Another thing: if you're going to have an open/close switch, it's better to have it on the input side, so the LM317 is not consuming any power when it's not needed.

author: AF6LJ
author_rank: Unknown
timestamp: January 28, 2016, 01:17:30 pm
content: 70C for a heatsink tempriture is a bit high. unless the ambient temp is in the 50s...I try to keep the device temp around 10 to 20 degrees above ambient.

author: jorenca
author_rank: Contributor
timestamp: January 28, 2016, 03:13:29 pm
content: well, today I added input and output capacitors (1uf aluminum foil).I checked the resistance between the LM317 pins (with the LM317 desoldered) on my board and all seemed good.I grounded the ADJ pin and had no load connected to the OUT pin (only the output cap was connected to OUT).As a result, the regulator got damaged again! I did not even disconnect the ADJ pin from ground.Could not having a load really damage the LM317?Isn't the load only required for stabilizing the output voltage?

author: AF6LJ
author_rank: Unknown
timestamp: January 28, 2016, 03:28:32 pm
content: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&sqi=2&ved=0ahUKEwjWtLah58zKAhUS72MKHbv8D0oQFggcMAA&url=http%3A%2F%2Fwww.ti.com%2Flit%2Fds%2Fsymlink%2Flm317.pdf&usg=AFQjCNGQYAR2K_GvDLW8Yv0iC96N2stKVQ&sig2=1bMlAB2BcmX2cQc2Z7ps8Q&bvm=bv.112766941,d.cGc&cad=rjaHere is the datasheet and buried within is the power supply schematic you are looking for.

author: mikerj
author_rank: Super Contributor
timestamp: January 28, 2016, 03:28:40 pm
content: Quote from: jorenca on January 28, 2016, 03:13:29 pmAre you still powering up the regulator with the adjust pin floating? If so why would you do that, especially when you've already been told it's a bad idea?Follow the advice in the datasheet, and the numerous examples they give.

author: AF6LJ
author_rank: Unknown
timestamp: January 28, 2016, 03:30:30 pm
content: Grounding it directly is also a bad idea...

author: jorenca
author_rank: Contributor
timestamp: January 28, 2016, 03:50:14 pm
content: I'm not powering the regulator with the ADJ pin floating.I tought I should make a 10mA current sink with another LM317 to attach to my power supply output just to have the minimum load satisfied.I saw from the LM317 datasheet that I should use 120ohms to get about 10ma current.I'm attaching a photo of my configuration.However, when I measured if it was working (amps from ADJ to ground) it got fried even in this set up!!I'm desperate now.

author: rdl
author_rank: Super Contributor
timestamp: January 28, 2016, 04:12:01 pm
content: Try building the basic standard regulator circuit and see if you can get that to work. Once you do that you can try adding modifications. You know the old saying, "First you make it work, then you make it good".

author: wblock
author_rank: Regular Contributor
timestamp: January 28, 2016, 04:28:29 pm
content: Where is the ground connection in that picture?

author: KL27x
author_rank: Super Contributor
timestamp: January 28, 2016, 05:18:15 pm
content: On second thought, I find it curious that the LM317 is dying, but the LM7815 is not. Are these the same package size?The LM317 is dropping less than half the voltage at 13V out as the LM7815.

author: rdl
author_rank: Super Contributor
timestamp: January 28, 2016, 05:29:45 pm
content: It's kind of hard to guess what's going on. I read the entire thread and still I'm not sure what the circuit is supposed to be doing. I suspect the diagram provided is not an accurate representation of what has been built. That's why I suggest getting the basic regulator circuit to work first.

author: jorenca
author_rank: Contributor
timestamp: January 28, 2016, 05:33:54 pm
content: same package, same heatsink.Anyway, I was recording how the constant current configuration fails, and it magically did not!I have now connected another LM317@10mA as a load and a 20R load at 1.25V and all seems good.I guess I somehow short out the output when I'm measuring (because I have a tiny multimeter with tiny leads that are hard to use).

author: rich
author_rank: Frequent Contributor
timestamp: January 28, 2016, 05:42:03 pm
content: worth noting that the heatsink tab is electrically connected to Vout, so perhaps shorted Vout tab to Adj when trying to measure current in your breadboard circuit.Quote from: jorenca on January 28, 2016, 05:33:54 pmDo you mean the sametypeof heatsink, or are they sharing one physical heatsink?

author: jorenca
author_rank: Contributor
timestamp: January 28, 2016, 05:54:11 pm
content: same type, heatsinks are both isolated.

author: macboy
author_rank: Super Contributor
timestamp: January 28, 2016, 07:26:47 pm
content: Quote from: jorenca on January 28, 2016, 03:50:14 pmWhere is the heatsink?Where is the input capacitor?Where is the output capacitor?Are you even listening?

author: KL27x
author_rank: Super Contributor
timestamp: January 28, 2016, 07:35:20 pm
content: Quote1. Why are you trying to measure current between ADJ and ground?2. How are you measuring current? Do you know how to use a DMM to do this? (There are a couple ways to do this and one way to blow fuses/circuits).

author: rdl
author_rank: Super Contributor
timestamp: January 28, 2016, 07:41:33 pm
content: Try building the basic circuit I posted. It's nearly fool-proof, so if it doesn't work then you've probably wired it wrong or you may have bad parts. Use 120 ohms for R1 and a 1000 ohms (1K) pot for R2. Don't worry about measuring any currents, just check the voltage between ground and +V out. You should get about 1.25 - 11.5 volts.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: January 28, 2016, 07:48:54 pm
content: Build this:the 240ohm resistor is so that it will always draw at least 10mA, if you are going to use anything higher, use a 120R [or less] resistor between the ADJ pin to the OUT pin.Cadj is optioal, for testing you dont need it.the voltage divder the diode and the capacitors are a must.keep everything as neat and clean as possible, close as possible to the LM317 and use large heatsink, if the heatsink gets over 60C put a fan on just for testing, you dont want to burn yourself.DO NOT ground anything, only use your multimeter set on DC volt and measure directly the output, no load. see if you can get the output voltage to change by changing R2.a picture of your setup would be nice.good luck!

author: Audioguru
author_rank: Super Contributor
timestamp: January 29, 2016, 01:59:37 am
content: Quote from: ZeTeX on January 28, 2016, 07:48:54 pmOhm's Law says that 1.25V/240 ohms = 5.2mA, not 10mA. The schematic is wrong and shows the 240 ohm resistor that is used for the more expensive and less current needed LM117.Most schematics on the datasheet from the inventor (National Semi) show the LM117 and 240 ohm resistor.You might use a low input voltage and be lucky and get an LM317 that needs a low output current. Then it will work with a 240 ohm resistor. Luck is gambling.

author: jitter
author_rank: Frequent Contributor
timestamp: January 29, 2016, 06:16:18 am
content: Yeah, that's why you often see R1 halved to 120 Ohms (obviously R2 must be halved too) in real world applications that draw very little current. That way there's no need for an extra load resistor on the output.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: January 29, 2016, 09:22:29 am
content: Quote from: Audioguru on January 29, 2016, 01:59:37 amoh yes, my mistake. better use 120ohm resistor.