LM317 dead or something else? - Page 1
Forum section: electronics / repair

author: ETITsynthesizer
author_rank: Contributor
timestamp: July 03, 2022, 07:44:31 pm
content: I have a custom circuit that I designed. there is a 24VDC input going into an LM317 set to 3.3v output with fixed resistors 1.2K + 2K to ground. I destroyed all my 74HC IC that were supposed to be getting 3.3v. the board was pulling > 1A with the 3v3 rail at 6.5V. one of my hex inverter IC was oscillating at 800Hz making the relays go crazy. I think this was the result of not having a minimum load of 10mA. big mistake. I pulled all the socketed IC for now. the LM317 is at 21.7V with no load. 21.63V with 330R. 21.61 with 150R. when I disconnect power I tested the top of the lm317 resistor divider 1.2K at 1.4 Ohms in both directions. I don't think I cooked the 1K2 resistor. Maybe I did? so I think it went something like this, no load, no regulation caused the regulator to output 21V which caused the 74HC IC's to short. the shorted 74HC IC's then created a load big enough to again keep the LM317 out of regulation. did I cook the 16V cap on the 3.3v rail? what do I replace first?

author: Audiorepair
author_rank: Frequent Contributor
timestamp: July 03, 2022, 08:22:44 pm
content: I really don't think it is wise to supply a voltage regulator with 24v, when you only want to get 3.3v out of it.It is likely to get very hot.5v to 9v would be the usual ballpark to aim for.

author: MikeK
author_rank: Super Contributor
timestamp: July 03, 2022, 08:41:23 pm
content: 24V - 3.3V = 20.7V20.7V x 1A = 20.7Watts, i.e. Good Gawd!

author: Zero999
author_rank: Super Contributor
timestamp: July 04, 2022, 12:20:02 pm
content: I suspect it's the LM317 was bad first, which caused the output voltage to be too high and damaging the rest of the circuit. This could be caused by the input voltage being too high, or the wrong polarity.Shorted ICs should reduce the output voltage not increase it. The LM317 has current limiting and thermal protection, so is unlikely a fault on the output will damage it.The lack of a minimum load shouldn't damage the lm317, but it can make the output higher than normal.How much current does the circuit use in normal operation? I take it that it's not designed to draw 1A and only did so because the voltage was too high.I would replace the LM317 first. What sort of capacitor is it? If it's aluminum, leave it along, it's unlikely to have been damaged. If it's tantalum, the replace it, as that sort of abuse might have damaged it, which could cause it to go into meltdown later on.

author: rkabz
author_rank: Contributor
timestamp: July 04, 2022, 01:59:12 pm
content: Suggest you have a good read ofhttps://intranet.ctism.ufsm.br/gsec/Applications/lm317.pdfand note the value of R1/R2 to provide minimum load for stability, especially at 3.3 volts output. R1 is typically 240 ohms or 120 ohms in this app note. No problem running from 24 volt input, but you need to be aware of the power dissipation limits imposed by the load current. Don't forget that the R1/R2 are contributing to the load. Also take note of decoupling capacitor values and and positioning. Hope that helps.

author: BrokenYugo
author_rank: Super Contributor
timestamp: July 05, 2022, 08:30:30 pm
content: Sometimes linear regulators just short their output transistor for fun ("fun" being a stand in for whatever silicon wizardry actually happens), not something you hear about often but it sucks when it happens to you. I'm unlucky with such things and have had two fail that way, running reasonably cool in production equipment.When in doubt, install a crowbar or other overvoltage protection on the regulated rail.https://en.wikipedia.org/wiki/Crowbar_(circuit)

author: Neomys Sapiens
author_rank: Super Contributor
timestamp: July 05, 2022, 10:49:52 pm
content: What they do not tell explicitely about the LMx17 (or it's more powerful siblings, the LMx50 and the LMx38) is that the comprehensive protection features (..protected against Over-current, Over-Temperature..), which make the device sound like an almost fail-safe solution for any linear regulation needs, are only able to protect the chip if the condition arises in a benign, gradual way AND THEY WORK ONLY AS INTENDED ONCE THE REGULATOR HAS REACHED STEADY-STATE OPERATION. Especially at high input-output differentials, all of the mentioned devices depend absolutely on MASSIVE heatsinking (and even that does not prevent everything).I made that experience in the most painful (mainly money-wise, but including some burned fingertips) way possible, i.e. working with the LM138/338 in the mid-80s. When this regulator is starting up at about 80% of its rated Vi-o differentioal into a load of about 50% of it's rating, it will die like flies if not well fitted to a large heatsink.At high I-O differentials, the heatsink needs to be larger that the maximum steady-state dissipation, as it must cover a part of the possible transient dissipation too. The possible alternative is to introduce a limiting impedance before the regulator, which amounts to willfully degrading its performance.Furthermore, the startup behaviour of this series of regulators is not very well described except in secondary analyses of it's circuit. This means that depending on the I-O voltage differential and the set output value, it can fry your connected load by overshooting even if it is not destroyed itself.One possible way to go is to soft-start the device, which might or might not work well with a certain type of load. In the case of digital circuits, I suspect the latter.Of course it is possible to build a nearly undestructable power supply around those regulators. But in order to achieve that, you have to go way beyond the 5 or 7 component minimum appnote circuit.