Curious about differences in LDO Voltage Regulator - Page 1
Forum section: electronics / projects

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 03, 2023, 05:45:35 pm
content: So, I built a circuit that sums up voltages in various configurations. I've been working on this in four different iterations, but the fourth one ran into an interesting issue.Part of my board uses LDO voltage regulators to generate +5V and -5V from +10V and -10V, respectively. In my third version of the board, I used the NCP1117 5-volt regulator, and it worked like a charm. However, due to supply issues, I used the LDL1117 regulator as a replacement. However, the problem was the output. The LDL1117 was outputting ~-0.75 volts instead upon start-up. I checked the LDO input voltage, and it was set at 10 volts. As a quick fix, I used an extra NCP1117 component in place of the LDL1117 and it works fine now.Troubleshooting the issue, I took off the LDL1117, attached some wires, and hooked it up to a DC power supply. At 10 volts no-load, the output was 5 volts. Odds are, the load on the board couldn't be handled by the LDL1117. The thing is, the load between versions 3 and 4 of the board is exactly the same (just some op-amps, capacitors for the op-amps, and a voltage divider). It didn't seem too heavy for the NCP1117, so I'm curious as to why the LDL1117 is having trouble.The NCP1117 has a dropout voltage of around 1.07 volts at an output of 0.8 A while the LDL1117 has it at 0.6 volts at 1.2 amps. The input voltage at 10 volts satisfies these requirements. Both regulators can output ~2 amps during a short circuit at worst. With the input voltage at 10 volts, that would give a dropout of 5 volts (VIN-VOUT), and the graphs on both regulators show that at that dropout voltage, each of them could handle ~1.5 amps. I also tried increasing the input voltage to 12 volts, but that didn't change anything. What else would cause such a discrepancy between the two voltage regulators that would have one work at start up while the other does not, under the same circumstances?

author: ArdWar
author_rank: Frequent Contributor
timestamp: August 03, 2023, 05:57:20 pm
content: Does it just locking-up or does it maybe actually oscillating?The differing dropout voltage might offer a clue here, the lower it gets (within a similar design) the weirder a LDO can behave under edge cases.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 03, 2023, 06:11:51 pm
content: I think it may have been locking up. At one point in my attempt to troubleshoot the thing, the LDL1117 showed an output of 5 volts at start up. However, after turning off the power and turning it back on again, it showed ~-0.75 volts (perhaps it may have been oscillating at a very low level, but I didn't notice it). How odd. You hope that the problem would be solved, but after turning it off and turning it back on again, the problem persists despite the good fortune.With what you said about the differing dropout voltage, I can see how it being manufactured to work at a low level can maybe cause instability, but that's manufacturing issues. Wouldn't the actual dropout voltage be large enough to bypass that?

author: ArdWar
author_rank: Frequent Contributor
timestamp: August 03, 2023, 06:53:19 pm
content: Yeah, that was just a wild guess. If nothing else was significantly changed in the design, a significantly lower *design* dropout voltage usually has to come with a compromise somewhere that might cause it to behave differently to a "normal" 1117.Another thing to consider is if your negative LDO start up faster and causing brief negative bias at 1117 output, preventing it to work properly (and that ST part may happen to be more sensitive to it than others)Anyway, when did 1117 even come with UVLO? That combined with dropout voltage difference should be enough reason to consider that particular 1117 *not* a jellybean 1117 anymore...Also, weird pairing between 1117 with LT1175 huh?

author: magic
author_rank: Super Contributor
timestamp: August 03, 2023, 06:56:18 pm
content: ST LDL1117 looks like a completely different chip internally with completely different regulation loop, it's not a traditional LM1117 clone and may be not quite 100% compatible.Check if your capacitors meet the recommendations, it may be unstable.The datasheet also mentions some SOA protection, maybe it's kicking in due to the high input-output differential voltage and high inrush current?Maybe your circuitry draws abnormally high current at low supply voltage for some weird reasons?(That being said, start with the capacitors).editOr maybe look for some other 1117 regulator which is more true to the original? There is like a million manufacturers of those things.Or simply replace with good old LM317 or LM7805. You have absolutely no need for low dropout here, with all the capacitor selection drama it tends to bring.LD1117 appears to be ST's traditional 1117.

author: wraper
author_rank: Unknown
timestamp: August 03, 2023, 07:04:27 pm
content: You did not provide any info about heatsink (or copper polygon on PCB) and actual current consumption of the circuit. The difference in could be because overheat protection in particular specimen of LDL1117 kicks in at somewhat lower temperature.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 03, 2023, 07:15:28 pm
content: You brought up an interesting point. The datasheet for NCP1117 doesn't show UVLO while the LDL1117 does. Unfortunately, it doesn't really say anything about it other than the turn-on threshold is 2.3 volts. Looking up the idea, perhaps it was stuck in an undefined behavior during power-up caused by UVLO? But, why wouldn't it work properly when the optimal/min input voltage was achieved?Odd pairing? Why would you say that? Though, funny, the LT1175 works fine. My previous design used a different part (NCP1117 and MC79M05), but I had to change it due to space considerations and the availability of parts (especially this one). I tried to go with an alternative that was smaller and one that could provide the same output current as the old one.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 03, 2023, 07:16:48 pm
content: Ideally, current consumption (in theory) was found to be 300 mA in simulation from both regulators. There is no heatsink. The component was soldered to pads on the board, and the ground pour that's connected to the ground pin on the regulator is a solid ground plane.

author: magic
author_rank: Super Contributor
timestamp: August 03, 2023, 07:20:36 pm
content: Quote from: ArdWar on August 03, 2023, 06:53:19 pmWait, is the output really 0.75Vnegative?This likely means the regulator is completely turned off, and perhaps you are right that it goes crazy because of that negative voltage.I don't think original 1117 has this problem, and 317 surely doesn't because they are used for dual-rail circuits all the time.Quote from: ArdWar on August 03, 2023, 06:53:19 pmLook at the "block diagram" in the datasheet - it is a completely different chip. MOSFET rather than bipolar, P-ch rather than NPN (hence the lower dropout), a different regulation loop.In this case, I suspect the P-ch MOSFET is simply turned off and current only flows from ground through protection diodes. OP could try to verify this by measuring Vin and GND pin currents.Maybe the regulation opamp gets phase reversal when the output pin goes below ground?editAbsolute maximum rating on output voltage: -0.3V.Perhaps not the best choice for a dual-rail application.Or you could add a Schottky diode which prevents the 5V rail from going more than 0.3V negative and see if it helps.

author: mariush
author_rank: Super Contributor
timestamp: August 03, 2023, 07:47:03 pm
content: It's not clear from the schematic if you're using ceramic capacitors or something else.A lot of 1117 regulators require some output capacitor with higher esr, like at least 0.1 ohm ESR, some require at least 0.3 ohm ESR. If you only use ceramic capacitors, that could be causing issues.There's some minimum load required to regulate as well, a few mA, which you probably have, but it wouldn't hurt to add some "standby" led consuming a couple mA.Lots of better regulators you could have used, like for exampleMIC5219-5.0 :https://www.digikey.com/en/products/detail/microchip-technology/MIC5219-5-0YM5-TR/1030586... would be nice, but it can't dissipate a lot of heat, so I doubt you could do 10v in, 5v out at 300mA ... you could maybe have a couple diodes in front to drop some voltage.TLV76750 is stable with ceramic capacitorshttps://www.digikey.com/en/products/detail/texas-instruments/TLV76750DRVR/10820367LD29150 in PPAK footprint would be able to radiate the heat easier :https://www.digikey.com/en/products/detail/stmicroelectronics/LD29150PTR/725294- but this one also requires high esr capacitors on output, 1 ohm or higher is suggestedLF50 alsohttps://www.digikey.com/en/products/detail/stmicroelectronics/LF50CDT-TR/591636- recommended more than 0.1 ohm

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 03, 2023, 07:53:49 pm
content: Thanks, magic. I appreciate your reply. Interesting point you brought up regarding the internal workings of the regulation loop. I see what you mean by it being a different topology. That was something I didn't consider.With regards to SOA protection, wouldn't the high, actual dropout voltage (5 volts) be enough to handle the 'short-circuit' at startup? At ambient temperature, Figure 16 starts at ~5.5 dropout voltage allowing for ~2 amps of short circuit current. That seems to drop as the voltage gets larger. But, even starting at something like 12 volts (which increases the dropout voltage to 7 volts), the regulator still outputs the incorrect, low voltage.With regards to the capacitors, I was using ceramic capacitors (X5R and X7R) for the regulator outputs while the op-amps used both ceramic (0.01u, 0.1u, 10u), and tantalum (6.8 uF) following the datasheet. The regulators used 0.1 uF and 10 uF as per the NCP1117 datasheet. I thought it would work as a simple replacement, so I didn't follow the recommendation of the LDL1117 calling for 4.7 uF in the typical application diagram.Yeah, I'll have to revamp my parts selection for the next batch. At least this is just a prototype. I don't quite remember why I chose what I did, but if low dropout voltage is causing the issues, then I see no harm in going for larger ones. But, I think one big consideration was space and part availability. That's why I couldn't use the parts that I did before...so as I was typing this, I see your new post. Yeah, the output was definitely negative. How could the output of the negative LDO affect the positive one? The only ways they are connected are through the op-amp and the caps at the op-amp that are tied to +5 and -5 on either side of the cap.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 03, 2023, 07:56:29 pm
content: The caps were ceramic and tantalum. I used ceramics for mostly all of them (0.01uF, 0.1 uF, and 10 uF) while the tantalum caps were only for the op-amp (6.8 uF). With regards to my choice of parts, admittedly, I couldn't remember why I chose what I did, but I distinctly remember that there were space issues and issues with part availability.

author: magic
author_rank: Super Contributor
timestamp: August 03, 2023, 09:02:04 pm
content: Quote from: LoveLaika on August 03, 2023, 07:53:49 pmVDROPon fig. 16 and 17 goes up to 18V. I think it's not the difference between the rated output of the chip (5V) and the input, but the difference between the actual output (-0.75V) and the input. So you should be looking at 10.75V. However, even then the chip is supposed to still provide over 0.5A current, which is more than the 300mA you said your opamps consume.Power dissipation with 10.75V difference and 300mA current is a fairly substantial 3W, the regulator should be getting really hot if it's working. Is it getting hot? If not, chances are that 5V rail current really comes from ground through protection diodes, either in the regulator or in the opamps (if some of their pins are grounded, perhaps IN+ in inverting stages).Quote from: LoveLaika on August 03, 2023, 07:53:49 pmThe opamps pull their quiescent current from the positive rail and pull it down, until something starts to supply current to keep the rail up. If the positive regulator fails to keep up for any reason, the rail may go negative until it is supplied by some diode somewhere. And if the negative rail voltage makes the regulator go crazy, it may stay that way indefinitely. It is a possibility.Get some 1N5818 from your junk box and connect it between +5V and GND. If this helps, you know the LDO is to blame.I think you should be able to find lots of alternatives in SOT223. Many genuine 1117 variants are available in this package - for those, look for parts with ~1.2V dropout and adjustable option available.

author: wraper
author_rank: Unknown
timestamp: August 03, 2023, 09:11:57 pm
content: Quote from: magic on August 03, 2023, 09:02:04 pmNo, if this happens, it means that circuit is abusing LDO at startup, and one LDO being more resilient against this condition than other does not mean that LDO is at fault and circuit is fine.

author: magic
author_rank: Super Contributor
timestamp: August 03, 2023, 09:23:16 pm
content: Actually, permanently adding such schottkys may be a good idea anyway, to prevent damage to protection diodes in ICs if the circuit operates with missing positive or negative rail for prolonged time. It's a known issue of all dual supply circuits.However, if the LDO has -0.3V absolute maximum rating, even a schottky won't really protect it (their Vf is often 0.3~0.4V). And if it misbehaves when rail voltage goes beyond absolute ratings, you really are screwed. Definitely a better LDO is in order.

author: ArdWar
author_rank: Frequent Contributor
timestamp: August 03, 2023, 09:31:49 pm
content: Easy way to test is to power up the positive rail first before the negative rail and see if it works reliably. LT1175 got output reversal protection so the negative rail is still expected to work even if its output is pulled positive by 1117.Then test the other way around. You'll know if the problem is something to do with LDL1117's response to output reversal if the first one is okay but second one always fail.Quote from: wraper on August 03, 2023, 09:11:57 pmIt's at "fault" in the sense of a part that normally you trust to "just work 100% of the time" ends up giving you an exercise of power supply sequencing and timing headache.UVLO for example would be a welcomed feature on most single supply application, especially battery powered one. But a unwelcomed one if it ends up holding the output in a high impedance state just a tad bit longer than a completely linear part would, allowing it to be pulled into undesired level causing undesired mode of (in)operation.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 03, 2023, 09:40:42 pm
content: In my design, I didn't use Schottky diodes, but I used 1N4148 diodes. D27 and D28 are diodes that connect from the regulator's output to ground in the appropriate fashion. When initially powered during testing, I got that bad output while -5 volts was coming from the negative LDO. Actually, now that I think about it, the way D27 was connected to the output, the voltages would have caused the diode to be foward-biased turning it on. Still though, why -0.75 volts rather than something closer to ground/0?Yeah, my plan now is to look for new regulators and try them out with this board. Once this prototype is acceptable, I can mass-produce it (hopefully with no issues).

author: TimFox
author_rank: Super Contributor
timestamp: August 03, 2023, 09:57:24 pm
content: It is possible that the -0.75 V unwanted output is an effect of "latch up", caused by a temporary condition where a pin of the IC is temporarily pulled negative with respect to the substrate, which is ordinarily the most negative node of the part in normal operation.Usually, a reverse voltage in excess of a PN diode drop is required to latch up a device, so a Schottky diode between the output (or whatever) and the substrate suffices to stop the latch-up condition when the negative supply comes up first and induces a negative voltage in the wrong place. A PN diode (1N4148, etc.) will not work in this location.I ran into this problem with a mixed-signal CMOS device that used split rails (+/- 6 V), where both rails were provided by 3-terminal linear regulators (7806/7906).If one raw supply (in my case, lab supplies during testing) comes up before the other, the one that is not yet powered looks like a very high resistance and currents through the device from the other supply caused the "ground" pin (which was actually an input, the substrate was the negative rail) to float out of the range between rails.The only way to reset that device was to carefully sequence the power, but sometimes it overheated during latch up and was destroyed.After the factory engineer explained that to me, small Schottky diodes from the power rails to ground/common fixed the problem.edit: Thinking back, the "ground" pin on that IC was externally connected directly to circuit ground/power supply common.If the positive supply came up first, while the negative supply was still off, the high output resistance of the OFF regulator to circuit ground was connected to the negative power pin of the IC.The ground pin was only an input to the device, so the current from the positive supply could only flow out the negative power pin, which then floated positive, and the "ground" input pin was then negative with respect to the negative power pin, initiating the latch up (parasitic SCR).

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 03, 2023, 10:34:34 pm
content: Thanks for your explanation. I think I kind of understand it, if not fully. Would simply replacing the PN diodes that I mentioned before with an appropriate Schottky (in the same orientation) enough, or is this something to just add on alongside that? The pins you speak of in your case, it sounds like the output pin of the regulator is more negative than its "ground" pin (would that be its substrate)? If that's the case, then, current would flow from ground to the output pin via the diode, but how would a Schottky diode mitigate such an effect as compared to a PN diode? Is that due to a Schottky's lower forward voltage, so it can "activate" faster than a PN diode would?

author: TimFox
author_rank: Super Contributor
timestamp: August 03, 2023, 10:42:05 pm
content: As I understand it, to induce a latch-up you need a voltage in the wrong place that is at least a P-N drop (0.6 to 0.7 V, roughly).If you place a Schottky in that location, reverse-biased in normal operation (as are your P-N diodes), it should keep the reverse voltage to only 0.3 V, insufficient to cause a latch up.For example. a Schottky from the negative regulated rail to ground, that will conduct if that rail goes positive with respect to ground.I don't remember if I used an 1N5711 signal-level, or a 1 A rectifier Schottky diode.For a positive fixed linear regulator, the "ground" pin should be the substrate, but the "input" pin may be the substrate for a negative regulator, since it's the most negative point in the circuit.

author: wraper
author_rank: Unknown
timestamp: August 04, 2023, 09:40:11 am
content: Quote from: ArdWar on August 03, 2023, 09:31:49 pmIt's not part that giving you a headache. It's that you are violating safe operation conditions and one part is more forgiving than another. Does not mean that the "good" part won't fail in future as it's not the part itself that is the root problem. Not to say negative voltage on the output can cause trouble not only for LDO but also for circuit it's powering.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: August 04, 2023, 02:20:35 pm
content: This is an issue that I would have never considered until the problem actually occurred. I sure hope that this little accident didn't harm the op-amps and that the board is still usable with fixes. Even if the power supply sequencing is the cause of the issue, it seems cheaper (and less time-consuming) to find a new positive regulator and Schottky diode replacements that can withstand the lock-up. The alternative would be to redesign the power supply that I'm using (not bench-top). I can't exactly simulate this in software, so trial and error will have to do.For what it is worth, I'm glad this issue was caught before mass production. Truly, thanks to all of you for helping make sense of the situation.

author: AnalogTodd
author_rank: Regular Contributor
timestamp: August 04, 2023, 05:01:13 pm
content: Quote from: LoveLaika on August 04, 2023, 02:20:35 pmI designed linear regulators for a major semiconductor company for over 25+ years and this situation you describe is one that has vexed many system designers for a long time. Supply sequencing and loads that go to opposing supplies have caused many people issues for decades. Many regulators cannot stand having their output pulled more than a diode drop in the opposing direction relative to ground (i.e. a positive regulator cannot have their output go more than a diode drop negative and vice versa for a negative regulator). Fixes for this include reverse-biased schottky diodes on each output or to find a regulator that can handle having its output taken to opposing voltages and still start properly. If you're already using the LT1175, look at the LT1763 or LT1963, those will start even with their outputs many volts negative (yes, I'm biased towards those parts, if you want to know why you can ask).You have likely not damaged your op amps, they ran at a lower voltage and biased on the inputs to different voltages than you wanted. They are probably the reason you found this issue. Because they run between V+ and V-, when V- comes up first you're pulling current out of V+ and dragging it negative. What happens inside a chip that isn't designed to handle this is that an N-type tub connected to that output gets pulled more than a diode below the P-type substrate and turns on. A parasitic NPN is formed with the N-tub (the one connected to the output) becoming an emitter, the P-substrate becoming the base, and every other N-tub on the chip acting as collectors. It's not a transistor with great beta or anything, but it often doesn't need to be to prevent the chip from starting properly. In order to actually avoid this issue a chip needs designed from the beginning to handle it.

author: magic
author_rank: Super Contributor
timestamp: August 04, 2023, 07:16:03 pm
content: Quote from: wraper on August 04, 2023, 09:40:11 amIt is the part. Good parts won't fail because they are designed to handle this condition, which is essentially unavoidable in dual rail circuits.Aforementioned LT regulators work with a few volts towards the opposite rail, so a simple PN diode to ground protects them easily. LM317/337/1117 type chips can deal with ADJ going one diode below/above the output, or more if enough series resistance exists between ADJ and ground (the usual case). I think it may be good idea to use schottkys with them, because "one diode drop" becomes less than 0.6V if the regulator happens to be hot for any reason. You also need to think about other ICs on that rail which may have some inputs tied to ground.Dunno about LM78/79 and too lazy to try to analyze them at this time, but they are widely used in dual rail circuits, often without any protection, and there doesn't seem to be much complaints about problems with them.editLM79 datasheets shows dual rail supply examples with simple silicon diodes (1N4001) for protection.Quote from: LoveLaika on August 03, 2023, 09:40:42 pmBecause 0.75V is the forward voltage of your 1N4148 (a small diode with not so low forward voltage), or the internal diode in the regulator, or the combination of both.Also, because the regulator is malfunctioning and apparently doing nothing to actively pull the rail up.

author: TimFox
author_rank: Super Contributor
timestamp: August 04, 2023, 08:12:20 pm
content: In my somewhat different case, it was not the 78/79 regulators that latched up, it was the CMOS device connected between +6 and -6 V rails.Schottky diodes from each rail to ground/power-supply common fixed that problem, but PN diodes did not.Different devices latch up in different ways, but usually stay latched up until power is removed (or they melt due to high current).

author: AnalogTodd
author_rank: Regular Contributor
timestamp: August 04, 2023, 08:21:25 pm
content: Quote from: magic on August 04, 2023, 07:16:03 pmIf you want to dig into technicalities of it all, it is the part itself combined with the fact that using it in a dual rail circuit shows supply sequencing issues allows the output to be pulled negative w.r.t ground and violates the absolute maximum rating of the part (VOUTrated from -0.3V to VIN+0.3V). As you mentioned, the LT regulators work rather well because they are designed to handle this exact situation (and parts are tested to make sure they can as well, VOUTis rated +/-20V).Quote from: magic on August 04, 2023, 07:16:03 pmThis is why I recommended schottky diodes instead of something like a 1N4148 or 1N4001; their forward voltage is comparatively much lower than that of the internal p-n junction and prevents the parasitic NPN from turning on.