LM317AHVT and Heatsink - Page 1
Forum section: electronics / beginners

author: mike_mike
author_rank: Super Contributor
timestamp: July 26, 2018, 03:51:54 pm
content: Dear Forum,I built the power supply from the attachment (with a few modifications: protection diodes and a capacitor in parallel with the potentiometer) and I mounted on one heatsink the power transistor and on another heatsink the rectifier and the LM317AHVT (I will note LM317 for LM317AHVT from now)The heatsink with LM317 is having the following dimensions: 165x80x35mm.I mounted the LM317 using a very thin piece of mica and I used thermal grease and a M3 screw and a mounting clip.The problem is that I noticed that LM317 is heating up to about 70 C while on the power supply output is connected a overload.The current through the LM317 is 0.7A and the voltage drop on the LM317 is 33V, while on the output is an overload.It is a problem if LM317 is heating up to about 70C after a few minutes of overload ?It is normal to heat up to 70 C or this temperature is too high ?

author: james_s
author_rank: Super Contributor
timestamp: July 26, 2018, 04:02:42 pm
content: That sounds pretty reasonable, with 33V on the input you are dissipating quite a lot of power. Is the heatsink getting hot too? If the IC is running much hotter than the heatsink I would look at the mounting to make sure it is properly mated. If the heatsink is getting hotter than you like then use a larger heat. There is no getting around a linear regulator dropping a large voltage at significant current getting hot, you can't fight the math.

author: Zero999
author_rank: Super Contributor
timestamp: July 26, 2018, 04:15:30 pm
content: 70oC isn't too hot.Another thing to note is the safe operating area protection is kicking in and reducing the current limit. When the voltage differential between the input and output exceeds 15V, the output current limit is reduced, to protect the pass transistor from secondary breakdown. With a voltage differential of 60V, the current is typically limited to 300mA.https://docs-emea.rs-online.com/webdocs/12cf/0900766b812cfcaa.pdfThe solution is to either route more current through the pass transistor, or add another transistor in series with the LM317HV, to reduce the voltage across it.

author: mike_mike
author_rank: Super Contributor
timestamp: July 26, 2018, 04:20:47 pm
content: Thanks for the answers.James, the heatsink is getting hot after about 30 mins of overload. The LM317 is properly mated (I think). I used a very thin piece of mica insulator (about 0.05mm thickness).The temperature was measured using a thermocouple and was the case temperature, not the junction temperature.In this case, your answers are still the same ?

author: james_s
author_rank: Super Contributor
timestamp: July 26, 2018, 04:28:31 pm
content: What is the heatsink temperature?70C case temperature doesn't sound unreasonable, it would worry me if it was above about 85-90C, but 70 is within the range that is acceptable.

author: not1xor1
author_rank: Frequent Contributor
timestamp: July 26, 2018, 04:33:29 pm
content: According to the datasheet (the first I found is from Fairchild) the worst case minimum load for stability is 12mA so R1 should be around 100 ohm and so a 2.2k potentiometer should be used for P1.In many cases LM317s would work stably with a much lower load than 12mA, but if I were you, I would test the IC on a breadboard rather than assuming to be always kissed by the blind Luck goddess.

author: mike_mike
author_rank: Super Contributor
timestamp: July 26, 2018, 04:35:39 pm
content: James, I did not measure the temperature of the heatsink, but I know that if I touched the heatsink it burned me after 2-3 seconds.not1xor1, I did not mentioned, but I used a 1.2k resistor on the board as a load. I can change this resistor to a more suitable value of resistance.

author: james_s
author_rank: Super Contributor
timestamp: July 26, 2018, 04:37:52 pm
content: Well so long as the case is not hugely hotter than the heatsink then there is not much you could do besides use a larger heatsink or forced air. Linear regulators turn the excess voltage into heat, they run hot, no way to change that.

author: schmitt trigger
author_rank: Super Contributor
timestamp: July 26, 2018, 04:51:14 pm
content: I would be far more concerned with the PNP series pass transistor overheating, as on its own it does not have thermal shutdown.Many designs thermally couple the transistor with the LM317, such that the heat generated by the former will indirectly cause the LM317's thermal shutdown to kick in.Works fine if the thermal coupling is quite tight.

author: Zero999
author_rank: Super Contributor
timestamp: July 26, 2018, 05:59:35 pm
content: Quote from: schmitt trigger on July 26, 2018, 04:51:14 pmTrue, at least this design does have current limiting, thanks to D1, D2 and R6.

author: mike_mike
author_rank: Super Contributor
timestamp: July 30, 2018, 01:58:43 pm
content: I tried to modify the schematic in order to have a minimum load current of about 10mA.I changed the 430R resistor into a 120R resistor. I used the same potentiometer (10K). I connected a 3.3K resistor in parallel with the potentiometer to have a variable output voltage from 1.25V to about 26V.I tried the schematic and it works.Are those modifications correct ?

author: mikerj
author_rank: Super Contributor
timestamp: July 30, 2018, 02:33:50 pm
content: Quote from: mike_mike on July 30, 2018, 01:58:43 pmHaving a relatively low value fixed resistor in parallel with the variable resistor gives very non-linear behaviour of the voltage control e.g. for a standard 270 degree potentiometer the first 90 degrees would change the voltage by about 17v, but the last 90 degrees would only change the voltage by about 3v.This may be acceptable (or even desired) behaviour, but if you want more linear behaviour use a lower value potentiometer and a higher value fixed resistor to set the maximum voltage (e.g. 2k7 pot, 27k resistor)

author: mike_mike
author_rank: Super Contributor
timestamp: July 30, 2018, 03:15:50 pm
content: Can this modification (from my last message) lead to unwanted power supply behavior, such as oscillations ?

author: mikerj
author_rank: Super Contributor
timestamp: July 30, 2018, 03:34:27 pm
content: Quote from: mike_mike on July 30, 2018, 03:15:50 pmNo, the modifications should not affect stability. Reducing the upper feedback resistor to 120 ohms should improve voltage regulation with small loads.

author: JS
author_rank: Frequent Contributor
timestamp: July 30, 2018, 07:14:14 pm
content: I don't know about the 1.2Ω resistor, seems a tad high, if you lower that value more current will go by the transistor and less for the regulator to take. I don't know your current requirements, at 3A (taking 0.65V for VBE) the transistor will be taking 2.6A and the reg 0.4A. If you lower it to 0.47Ω the transistor will take 2.8A and the reg 0.2A...JS

author: mike_mike
author_rank: Super Contributor
timestamp: August 07, 2018, 05:52:54 pm
content: Thanks for the replies.I have not found 120R resistor at the local electronic components shop. So I used 100R in series with 22R. I calculated that the current for 120R resistor is 1.25V/120R=0.0104A, while using the 122R the current is 1.25V/122R=0.0102A. The difference of current is very small.It should be ok if I use 100R in series with 22R ?

author: Zero999
author_rank: Super Contributor
timestamp: August 08, 2018, 04:48:50 pm
content: 122R vs 120R should make no practical difference, since it's less than the tolerance of the LM317.

author: newbrain
author_rank: Super Contributor
timestamp: August 08, 2018, 06:54:01 pm
content: Quote from: JS on July 30, 2018, 07:14:14 pmYes, I also have some doubt about the emitter resistor value; as it is, I expect the current through the transistor not to exceed about 1 A.Reasoning:We have two diodes drop (D1+D2) voltage across the resistor R6 and the BE junction of Q1.Let's say it's about 1V (as per the DS, with 3A through the diode), so 2V total (excess rounding).Across R6 we have 2 - 0.7 = 1.3V.Ohm's law gives us a current of 1.3V/1.2= ~1.1AAll the rest will be carried by the regulator.Quote from: JS on July 30, 2018, 07:14:14 pmAm I missing something? See above my calculations.

author: JS
author_rank: Frequent Contributor
timestamp: August 08, 2018, 07:14:20 pm
content: Quote from: newbrain on August 08, 2018, 06:54:01 pmYou are right, I'm ignoring the diodes, as if they were just to limit the maximum current as you said but still working on he linear mode, my mistake. Reducing the value of the emmiter resistor will send more current to the transistor even if the diodes are conducting. 1.4V at the diodes, -0.65 VBE, 0.75V/0.47Ω gives 1.6A to the tranny, but limiting the current to the reg at 140mA for resistive share between the transistorand reg. I used 0.2A at the reg resistor, which means 2V before conducting by the diodes, so using 3 diodes instead of 2 makes that work, to keep the 2 diodes the emmiter resistor needs to be even lower and the approximations of VBE and Vf are more on the limit, where you probably want to measure them at the limit conditions to know how much current will be shared.JS

author: mike_mike
author_rank: Super Contributor
timestamp: August 10, 2018, 03:19:12 pm
content: It should be a good idea to use a 1K potentiometer in series with the 10K potentiometer for fine adjustment of the output voltage ?Will the power supply work correctly if I use the 1K potentiometer in series with the 10K potentiometer ?

author: JS
author_rank: Frequent Contributor
timestamp: August 10, 2018, 04:14:32 pm
content: No problem with dual pots, you might want to use a 500Ω instead, as it might give ypu better control over the range, but 1k should be fine as well.JS

author: mike_mike
author_rank: Super Contributor
timestamp: October 07, 2018, 10:51:15 am
content: Dear Forum,I made a layout for a LM317 power supply, using the attached schematic.I don't know if I correctly made the layout. Please have a look at the layout and tell me if it is good.I want to know only if the layout is correct. The schematic that I draw in KiCAD is correct because I checked it. I also added a LED and a current limiting resistor (R4 and J3).The refdes of the components does not correspond from the schematic to the layout.I was a little bit confused by the copper pours, because I don't know if I correctly made the copper pours.

author: mike_mike
author_rank: Super Contributor
timestamp: October 18, 2018, 07:12:08 pm
content: I have a question regarding the attached schematic.In the LM317AHVT's datasheet is noted that the minimum load current can vary between 3.5mA and 12mA. When calculating the value of R1, what value 3.5mA or 12mA should I use ?The current value of R1 is 122ohms (not 120 ohms), and the minimum load current in this situation is 10.2mA. It is sufficient this value of current for stabilization or it should be higher ?

author: Zero999
author_rank: Super Contributor
timestamp: October 19, 2018, 08:12:35 am
content: Quote from: mike_mike on October 18, 2018, 07:12:08 pmWell spotted. There's a lot of variance between parts with the same number. With 10.2mA, there's the very tiny outside chance the voltage may rise slightly beyond the specification, when it's unloaded, especially at extreme operating conditions i.e. high voltages high/low temperatures.How many of these are you going to build? 99.99% of parts will work fine with 10.2mA, especially if their operating conditions aren't being pushed to extremes.

author: mike_mike
author_rank: Super Contributor
timestamp: October 19, 2018, 08:44:16 am
content: I already built 2 power supplies and I am planning to build the 3 rd one.I also planning a mini series (10 pieces) with PCB's from China.

author: mike_mike
author_rank: Super Contributor
timestamp: October 19, 2018, 11:22:33 am
content: Should I test again the modified schematic with the oscilloscope ?The only difference between the tested and the attached (current) schematic are the 3x8.2K parallel resistors (which was initially only one 3.3K resistor) and the 100R resistor (which was initially 122R resistor).And I am facing another problem. If I set the output voltage to 26.4V when I connect a 9Amps load, the voltage drops by 0.5V, reaching 25.9V. I did not find this issue last year when I tested this power supply. Why this voltage drop appears ? What should I do to correct this voltage drop ?If I set the output voltage to 19.53V and I connect a 4.74A load, then the voltage drops to 19.27V. I did not found this behavior of the power supply at previous versions.I re-tested using an older PCB and I obatined 25.9V with 7A load, and 26.0V without load. So, probably the PCB is the problem.

author: David Hess
author_rank: Super Contributor
timestamp: October 20, 2018, 01:46:54 am
content: Quote from: mike_mike on October 19, 2018, 11:22:33 amThat change should not matter.QuoteIt looks like you have a high enough minimum load on the output. Make sure that you are not falling below the dropout voltage of the combined regulator and transistor and make sure there is not a parasitic oscillation.C3 should be located at J2 and not near U1. R3 should not be required if R6 is low enough to provide the entire minimum load.QuoteIt could also be a layout problem. The lower part of the divider should terminate at a single point near J2 but the top of the divider should terminate at a single point close to U1 for best load regulation.

author: mike_mike
author_rank: Super Contributor
timestamp: November 03, 2018, 03:23:10 pm
content: What is the difference between A and B from the attached schematic, besides the direction I need to rotate in order to increase or decrease voltage ? Is there any difference between A and B ?

author: David Hess
author_rank: Super Contributor
timestamp: November 06, 2018, 08:12:20 am
content: There is no difference.

author: mike_mike
author_rank: Super Contributor
timestamp: November 15, 2018, 01:48:15 pm
content: Thank you for the reply David Hess.I tested the power supply and the results are (the schematic is the one from reply #27, with C4=2200uF/63V instead of 1000uF/63V)without load V=19.80V, with 6.17A load V=19.79Vwithout load V=26.0V, with 7.24A load V=26.0VI don't know if the wire that come from the pin 1 or another pin of LM317 to the PCB was making good connection with the screw terminal located on the PCB.If there was a bad connection, could the power supply work good and if I repair the bad connection by tightening the screw could the power supply to work bad ?

author: MarkF
author_rank: Super Contributor
timestamp: November 15, 2018, 02:12:58 pm
content: Quote from: mike_mike on October 07, 2018, 10:51:15 amI did a similar PCB which will accept either AC or DC input.

author: Zero999
author_rank: Super Contributor
timestamp: November 15, 2018, 04:18:25 pm
content: Quote from: mike_mike on November 15, 2018, 01:48:15 pmThe is no issue with those measurements. A 10mV change in output between no load and 6.17A of load is equivalent to a resistance of 10m/6.17 = 1.62mOIhms, which is very good. If there's a problem with a mechanical connection then it isn't affecting the circuit operation but it should still be resolved to improve the reliability.

author: mike_mike
author_rank: Super Contributor
timestamp: November 22, 2018, 03:21:09 pm
content: 1. Can someone explain how to calculate the voltage on the potentiometer ?It is the output voltage minus the voltage on the resistor (1.25V) ?2. If I remove the 3 8.2K resistors and the 2 potentiometers and I use instead of them only a single 2K/2W multiturn potentiometer, what can change in the power supply operation ? excluding the output voltage ?

author: MarkF
author_rank: Super Contributor
timestamp: November 22, 2018, 04:57:05 pm
content: Quote from: mike_mike on November 22, 2018, 03:21:09 pmFirst, get rid of all that resistor nonsense you have.Keep the two potentiometers if you want a coarse and fine adjustment (with the fine adj. being about 10% the value of the coarse).On the side, your capacitor values are way out of wack.Then watch this video:

author: AngraMelo
author_rank: Regular Contributor
timestamp: November 22, 2018, 05:18:44 pm
content: Im sorry to go out of the main topic here but can someone please explain to me how Vin goes to the base of the transistor but Vout comes out of the collector?

author: mike_mike
author_rank: Super Contributor
timestamp: November 22, 2018, 05:44:58 pm
content: Quote from: MarkF on November 22, 2018, 04:57:05 pmYou mean that the capacitors have a too high or too low value ?

author: MarkF
author_rank: Super Contributor
timestamp: November 22, 2018, 06:02:08 pm
content: Quote from: AngraMelo on November 22, 2018, 05:18:44 pmThe two main current paths are through the transistor and through the LM317 (green and blue). The resistor R1 (yellow) determines how much current bypasses the LM317 by setting the emitter/base voltage as a function of current. The more current required the more is bypassed.Quote from: mike_mike on November 22, 2018, 05:44:58 pmFar bigger than necessary for a DC input. Maybe a 100uF and 1uF on the input. And a 10uF and 0.1uF on the output. For me, I would just have a 100uF on the input and a 10uF on the output and let whatever is being power have it's local decoupling.

author: mike_mike
author_rank: Super Contributor
timestamp: November 22, 2018, 06:08:30 pm
content: Thank you for your help.I watched your video, and I understoon the following: If I need to calculate the power dissipated on the potentiometer, then I need to calculate the output voltage of the power supply, then decrease the value by 1.25V, then calculate the current through the potentiometer using Ohms Law and the using the formula P=I*V to calculate the power ?For example:The output voltage is 26.25V, then the voltage on the potentiometer is 26.25-1.25V=25V, the current through pot is 25V/2000R=0.0125A, and the power dissipated on the pot is 25V*0.0125A=0.3125W.Is that correct ?Please consider the attached schmatic, the DC input is unfiltered.

author: Zero999
author_rank: Super Contributor
timestamp: November 22, 2018, 07:16:25 pm
content: Yes, you've calculated the power dissipation in the potentiometer correctly.Note for future reference the maximum power rating of the potentiometer is only for the whole track, when current is passed through half of the track, the power dissipation will be halved. Thsi isn't a problem for this circuit since the current through the potentiometer is constant, but it's good to know.

author: mike_mike
author_rank: Super Contributor
timestamp: November 22, 2018, 07:39:41 pm
content: If I switch from the first schematic (with dual pots) to the second one (with 2k pot), what would change in the functionality of the power supply ?

author: MarkF
author_rank: Super Contributor
timestamp: November 22, 2018, 09:13:05 pm
content: It's NOT a matter of one or two pots.It's the 3 resistors you have in parallel with the pots. Get rid of them!Edit-Removed bad info

author: MarkF
author_rank: Super Contributor
timestamp: November 23, 2018, 05:10:36 am
content: I just noticed that you have the resistors on the emitters.They should be on the collectors for PNP transistors.Remember the outputs are being summed together andyou want to compensate for small variances of each transistor.You always want the resistors on the output side.And REMOVE R3. If you can't set R6 to meet the minimum load, then add a constant current source (not a resistor)Quote from: mike_mike on November 22, 2018, 07:39:41 pmNone with corrected two pot circuit.- You would design for the single pot to get your desired voltage range.- Then you could just use a single 3/4 turn pot- Or you could add two 3/4 turn pots with the smaller value being around 10% of the bigger depending on how fine of a control you want- Or you could use a single 10-turn pot

author: macboy
author_rank: Super Contributor
timestamp: November 23, 2018, 05:19:59 pm
content: Quote from: MarkF on November 23, 2018, 05:10:36 amNo, it is definitely correct to have the resistors on the emitter side!They act as "emitter degeneration resistors" in order to effect the current sharing. As more current flows through one of these resistors, the emitter (and therefore, base) voltage is affected in the way which reduces base current, creating a negative feedback. This enforces tight current sharing between the transistors, as changing Vbe by just a few millivolts can dramatically affect collector current. This scheme is not as effective for MOSFETs because of the much looser relationship between gate voltage and drain current.

author: mike_mike
author_rank: Super Contributor
timestamp: November 24, 2018, 08:00:14 am
content: Thank you for the replies.1. I have a 2K/2W potentiometer with the part number WXD3540. In the datatsheet it specifies "Independent Linearity". This means that the potentiometer have linear characteristics ?I don't see any other information about the linearity in datasheet.2. I made an accidentally short circuit between C and E at one of the transistors, with the multimeter probe, while I was taking some measurements and while was connected a 6-7A load at the output. After the short circuit I tested the power supply and it worked good. If the power supply works good, is there any recommendation for replacing the transistors or other components ?

author: MarkF
author_rank: Super Contributor
timestamp: November 24, 2018, 01:35:04 pm
content: If it works? - Don't fix it.

author: mike_mike
author_rank: Super Contributor
timestamp: November 26, 2018, 06:10:01 pm
content: Thank you for your help.I observed that sometimes, at random moments of time the voltage is decreasing without any reason from 26.1V to about 24.3V. This happens randomly, and it happens with load and without load at the output of the power supply.So I tested with the scope - again - and I attached some of the screenshots.I found that the voltage is sometimes coming back to the initial value if I move the PCB. I checked the connections on the PCB and they are good.The screenshots were took when the power supply had 26.1V at the output. I did not managed to check with the scope when the voltage was 24.3V.Why the voltage drops suddenly ?Are the screenshots good ?And sorry for bothering you with my noob questionsPlease help me. It is about 2 months since I started this project and I still have problems with it and I know that it is a simple project.

author: mike_mike
author_rank: Super Contributor
timestamp: November 30, 2018, 10:38:19 am
content: I found that sometimes the voltage at the output of the power supply drops suddenly from about 19V to about 17V without any reason. This happens is both situation - with load and without load.Sometimes, if I move repetitive the PCB, then the voltage comes back to 19V.What should I do to solve this problem ?Please have a look at the above message.

author: Zero999
author_rank: Super Contributor
timestamp: November 30, 2018, 11:37:06 am
content: There's no reason why that would happen, other than overheating/load. Are you sure about the connections? Try moving and shaking the PCB with the power on and no load connected.

author: mike_mike
author_rank: Super Contributor
timestamp: November 30, 2018, 11:40:15 am
content: I checked the circuit and the LM317 and the power transistors are not overheating I can hold relaxed the hand on the case of those components.The behavior is the same with and without load if I shake the PCB, sometimes the voltage drops and sometimes it comes back to the initial value.I checked the connections and they seems to be good.I used a 1.5A load at 19.87V for the last tests.I checked the voltages while the output was 19.87V and while the output dropped to 17.87V and I found that in both cases, the input voltage in LM317 and the voltage on the filter capacitors was the same. The problem is on the output of the power supply, because there the voltage dropped. I also tried with different LM317AHV and the results were the same.

author: floobydust
author_rank: Super Contributor
timestamp: November 30, 2018, 10:50:39 pm
content: It's a bit puzzling because it is an intermittent problem.I would usually suspect the parts on the ADJ pin like a noisy potentiometer or leaky C5 (rated for >50V?).The LM317 might have gotten damaged if you shorted C-E (input-output) earlier.Careful because a damaged LM317 likes to smoke and burn the potentiometer- the ADJ pin can stay at high voltage even though the pot is towards zero ohms and then it smokes. Try jumper across the pot and see if you get 1.25V out.A pass-transistor TIP36C could also get damaged with a C-E short. I would test them out of circuit.If this isn't it this stuff, then worst-case the circuit is oscillating.

author: mike_mike
author_rank: Super Contributor
timestamp: December 01, 2018, 04:30:04 pm
content: I replaced the potentiometer and now it seems to work good.I reused the old potentiometer in another power supply, made using the same schematic and the problem did not appeared.I checked with the scope on the ouptut for oscillations please have a look at reply #45. Are the results good ?I measured the voltages while the voltage at the output was low (about 17.87V) and I found that the voltage on the 100R resistor was 1.25V while the voltage on the potentiometer was lower in comparation with the moment when the voltage at the output was good (19.87V).How can I check to see what component is causing those problems ?

author: floobydust
author_rank: Super Contributor
timestamp: December 01, 2018, 09:55:09 pm
content: In electronics, you don't want to fix a problem unless you really know first what went wrong. So you go after it.My math, for 19.87V out RV2 was 1,483 ohms, and 17.87V is 1,324 ohms, a drop of 159 ohms (with R6=100R).But the potentiomenter works fine in another build. So is it the pot? It's not making sense...If the pot has a noisy spot, you can test it with a multimeter. Or connect it backwards (reverse rotation) so the wiper is somewhere else to get the same ohms but not on the noisy spot.Try to get it (PSU) to misbehave, go back to the old potentiometer or use a 1.5k resistor in place of RV2, connect a scope to the output and look for oscillations to rule them out.At these currents you will need to pay attention to wiring and single-point grounding. The end of R6 sampling output voltage should be as close as possible to theoutput terminals/binding posts."regulator to minimize line drops which effectively appear in series with the reference, thereby degrading regulation."The J2 output ground should connect to the input big filter caps' ground after J1, as well as the pot RV2. Not at the end of a long trace/wire that runs to all the little parts. All this is to get the best regulation and stability possible- you already got 10mV at 6.17A which is very good so your layout must be fine, but a pic helps. It is possible to wire a PSU like this so that it starts to oscillate under heavy load.Those scope measurements look fine. You are basically looking at noise at 2mV/div. Is the raw DC from a mains transformer and rectifier, or is it from a SMPS?

author: mike_mike
author_rank: Super Contributor
timestamp: December 02, 2018, 09:31:00 am
content: Quote from: floobydust on December 01, 2018, 09:55:09 pmI used a 50Hz transformer, with 30V secondary and 10A current and a 35A rectifier.I checked again, with another PCB and new components and now, the voltage does not drop. And I used the same potentiometer which I used when the voltage dropped.

author: Zero999
author_rank: Super Contributor
timestamp: December 02, 2018, 01:26:39 pm
content: Going from what you've said, I suspect it was the solder joint to the potentiometer. It will probably work if you put the original potentiometer back. As I said before, it's likely it was a connection problem.

author: mike_mike
author_rank: Super Contributor
timestamp: December 02, 2018, 02:07:44 pm
content: Quote from: Hero999 on December 02, 2018, 01:26:39 pmBefore I change the PCB, I tested with the wires of the potentiometer soldered on the PCB, and the result was that the voltage still dropped from about 19.87V to about 17.87V.In this case, does it still was a connection problem ?

author: Zero999
author_rank: Super Contributor
timestamp: December 02, 2018, 02:20:54 pm
content: Quote from: mike_mike on December 02, 2018, 02:07:44 pmThen it's more likely there's lose connection inside the potentiometer. Intermittent connections can be difficult to spot.My advice is don't use that suspect potentiometer in another project. Junk it. Using parts which could be faulty will just cause frustration and time wasting later on.

author: mike_mike
author_rank: Super Contributor
timestamp: December 02, 2018, 02:38:44 pm
content: It can be a lose connection inside the pot even if the pot is a multiturn pot ?I forgot to mention that I used a multiturn pot...How can I check for sure to see what component is making that problem ? I want to have a good power supply. I already made a few supplies in the last few years and they were oscillating... so I don't know what should I do... I am also very maniac in electronics and I want that those power supplies to be made by me.

author: Zero999
author_rank: Super Contributor
timestamp: December 03, 2018, 09:05:19 am
content: It's not uncommon for any potentiometer to develop an intermittent failt, multi-turn or not. The fact it works now suggests it was a problem with the pot. It's possible it was something else of course.

author: mike_mike
author_rank: Super Contributor
timestamp: December 12, 2018, 06:39:21 pm
content: Could those voltage drop problems be a result of oscillations ?If yes, then how can I find the oscillations ?I already tested with the scope on the output (please have a look at the previous posts).

author: mike_mike
author_rank: Super Contributor
timestamp: March 03, 2019, 07:51:05 pm
content: Hello,I am working at a new project. The schematic is attached in this post. The project uses an LM317T and a PNP pass transistor (TIP36C). I marked on the schematic the pins for LM317 and TIP36. LM317 and TIP36 will be connected to the board using wires.I made the layout but I am not sure if it is correct.I read a few documents about layout guidelines and I tried to respect some rules in this layout: I divided the schematic into low current and high current zones, I added copper planes for GND, I put the unpolarised capacitors near LM358, I put the bridge rectifier and 0.1R resistor in the lower part of the layout - because of the heat generated, I tried to make the traces as short as I could.The DRC does not return any error, but I want to know if the layout guidelines are respected ?If the layout guidelines are not ok, could you help me to correct the problems ?

author: mike_mike
author_rank: Super Contributor
timestamp: March 04, 2019, 01:13:42 pm
content: Quote from: mike_mike on March 03, 2019, 07:51:05 pmI am attaching updated version of the layout.

author: MrAl
author_rank: Super Contributor
timestamp: March 04, 2019, 02:00:35 pm
content: Hello,33v at 0.7 amps is more than 23 watts which is a lot. You need a good heat sink.But that's a lot of power too. The efficiency is going to be quite bad.Consider using a buck circuit which will not waste as much power.Dont know if you need it, but sometimes a small transistor is added for current limit too.

author: mike_mike
author_rank: Super Contributor
timestamp: March 04, 2019, 09:18:19 pm
content: Thank you for the replies.I built the circuit and I found that the minimum output current set by the 50k pot is about 0.45A.The designer of this circuit said that the minimum output current will be 0.15A.1. Which is the correct value ?2. How the operational amplifier commands the NPN transistor ?3. This transistor works in the linear region ?

author: MrAl
author_rank: Super Contributor
timestamp: March 05, 2019, 01:46:22 am
content: Quote from: mike_mike on March 04, 2019, 09:18:19 pmHello,I get around 0.12 to 0.14 amps minimum, however that is theoretical. When you use real components that could be off.However, if you are getting 0.45 amps then that could be because of several reasons.For one, what is your load? Remember that there will be a minimum output from the LM317 regardless of the state of the NPN transistor unless you also provide a negative voltage to the emitter so it becomes capable of reducing the output to an absolute zero.Another reason could be because of oscillation that makes a DC measurement look higher. You could check for that with a scope.Another issue though is that the LM358 probably is not fast enough for a good power supply. You could look for a faster version op amp unless you could put up with the slower speed. Slower speed means it takes longer to react to an overload.If the circuit is working properly the NPN works in the linear mode sometimes and saturated for larger current overloads.The op amp output drives the NPN, so when the current gets high enough and the op amp gain is such that it puts out roughly 0.62 volts, the transistor starts to turn on and enter the linear mode. With higher current it will turn on fully and so it will be saturated.The slew rate of the LM358 is a little slow though so it may not be fast enough for a good power supply.

author: mike_mike
author_rank: Super Contributor
timestamp: March 05, 2019, 06:05:39 am
content: Hello,The load that I used was a 1R resistor at the beginning of the tests. I also used as load a 40W/24V light bulb. In both cases the output current was 0.45A.

author: MrAl
author_rank: Super Contributor
timestamp: March 06, 2019, 02:56:13 pm
content: Quote from: mike_mike on March 05, 2019, 06:05:39 amHi,Ok well the minimum output from an LM317 without a negative power supply is about 1.25 volts, and with a 1 ohm load, that means 1.25 amps, but because the load is so low it could be competing with other resistances in the circuit so you see 0.45 amps. It could also be something wrong with the circuit itself like wiring error or something.Check it with other loads to make sure it works as you think it should without overload.Also, there is usually some input offset voltage on common op amps and that would mean an error in the measured current. You should also try changing the 0.1 ohm resistor. Try 0.2 ohms and see what you get. It should agree with the 0.1 ohm resistor except be about 1/2 the current. So 0.45 should drop to 0.225 unless there is a very bad overload and then it may be 0.45 again.

author: mike_mike
author_rank: Super Contributor
timestamp: December 11, 2019, 09:11:50 pm
content: Hello, I have got some trouble regarding the attached schematic. When I unplug the transformer (24V/100VA) from the wall socket (230V) then the output voltage from the power supply is rising for a short amount of time. Can this problem be simply solved or it needs a more complicated solution ?I will also do more experiments in the next days to see more exactly what is happening.

author: iMo
author_rank: Super Contributor
timestamp: December 11, 2019, 09:16:11 pm
content: A diode from output to input?

author: mike_mike
author_rank: Super Contributor
timestamp: December 11, 2019, 09:54:53 pm
content: I drew a more detailed schematic. Please have a look.I did not understood completely the idea behind the place were I should install the diode. Could you please be more specific ?

author: MarkF
author_rank: Super Contributor
timestamp: December 11, 2019, 10:27:49 pm
content: Diodes D4 and D5 are there if you connect something stupid to the outputs.They are basically protection in the case you connect something that has a voltage on it.- D4 is in case you connect something with a higher voltage to the output then is being output.- D5 is in case you connect a reverse polarity voltage to the output.Capacitor C3 (1000uF) on the output is extremely high. The LM317 response to transients on the load will be slowed down because of it. I would not put anything higher than 100uF. It may be causing your voltage rise when you unplug the AC.Also, you typically do not see diodes (D1 and D2) across R2. Although, they do limit the amount of bypass current.

author: mike_mike
author_rank: Super Contributor
timestamp: December 13, 2019, 09:05:17 pm
content: Can someone please recommend a schematic that sink current from the power supply ?I need it for establishing the minimum current from this power supply to about 15mA. The current sink circuit will be used as minimum load.The problem is that I did not found any current sink circuit that works starting from 1.25V ...

author: MarkF
author_rank: Super Contributor
timestamp: December 13, 2019, 11:16:24 pm
content: The current source that I can remember is the LM334. It is 1 to 40V and 10mA max.The minimum load for the LM317 is 10mA which the LM334 can satisfy.You do not need to have 15mA.

author: Zero999
author_rank: Super Contributor
timestamp: December 13, 2019, 11:28:05 pm
content: Quote from: mike_mike on December 13, 2019, 09:05:17 pmThere is no need for a current sink with this circuit, because R4 draws a steady 12.5mA from the LM317AHVT's output, which is above the worst case minimum load requirement of 12mA, stated in the data sheet.https://www.onsemi.com/pub/Collateral/LM317AHV-D.pdf

author: mike_mike
author_rank: Super Contributor
timestamp: December 14, 2019, 04:11:36 pm
content: I tested the attached schematic to see if the output voltage rise when the transformer is unplugged from the mains.The result was that the voltage does not rise when powering a load that sink a current which is below the maximum output current.If the load sink more current than the maximum output current then when unplugging the transformer from the mains then the output voltage rise.I made the above mentioned tests using LM317 AHVT from On Semiconductor. Will the behavior of the circuit remain the same if I will use LM317T instead of LM317AHVT ?

author: not1xor1
author_rank: Frequent Contributor
timestamp: December 14, 2019, 04:20:47 pm
content: Quote from: mike_mike on December 14, 2019, 04:11:36 pmYou might put a 1-2k 2W resistor in parallel to C2. That might solve the problem.

author: mike_mike
author_rank: Super Contributor
timestamp: December 14, 2019, 07:51:35 pm
content: Probably I will not use the power supply in constant current mode... I need only some protection for the PNP transistors which is available through the 2 1N5408 diodes.The next question is: if I want to use a 100VA/24V transformer for this project, how can I be sure that in all situations, the output current will not be more than 3 A ?I checked this way: the maximum current through LM317 is about 2.2A (maximum) and the current through the PNP's is about 1V/1.2R=0.833A. For 2 PNP's it is 1.66A. So the maximum output current will be 2.2A + 1.66A = 3.86A, but I am getting more than 3.86A when I have a overload at output. For example, I am getting 2.5A through LM317T ... and the total output current more than 4A...

author: David Hess
author_rank: Super Contributor
timestamp: December 14, 2019, 07:58:25 pm
content: Quote from: mike_mike on December 14, 2019, 04:11:36 pmWhat you are describing is completely normal. The LM317 and most integrated regulators include foldback current limiting which lowers the current limit when the input to output voltage is high to protect the regulator from high power dissipation and secondary breakdown of the output transistor. This has a side effect of preventing startup into a high current load.QuoteAll LM317s and most integrated regulators include foldback current limiting and will display the behavior that you describe.

author: mike_mike
author_rank: Super Contributor
timestamp: December 15, 2019, 09:53:01 am
content: I made some tests to see if the output voltage rise when the power supply is disconnected from mains.I have got the attached results:with no load at output: 0097.jpgwith about 2A load: 0112.jpgwith overload: 0124.jpgThe output voltage was 25V in all 3 cases.If I will make a higher current power supply, for example 10A (with more power transistors and a bigger transformer), then the power supply behavior will remain the same as described in this reply ?

author: David Hess
author_rank: Super Contributor
timestamp: December 15, 2019, 12:49:27 pm
content: The LM317 datasheet includes a graph of maximum output current versus input-to-output voltage as shown below. Solutions to this problem include:1. Use multiple 317s in parallel to limit current of each one to about 1/4 amp. The lab power supply design in National Semiconductor linear brief 28 uses this method to some extent.2. Use an external power pass transistor to multiple the current rating of the LM317. Note that this moves the problem to the external transistor which needs to be derated accordingly but there are many single power transistors with a secondary breakdown rating which can support more than 1.5 amps at more than 30 volts. Multiple transistors in parallel may still be desirable to handle the power dissipation.3. Use an external power pass transistor in a cascode configuration to limit the input-to-output voltage across the LM317 to a maximum of about 12 volts.In the case of 1 and 2 above, the built in LM317 current limit will be much higher than the power supply can support so it may be desirable to add a lower external current limit circuit.

author: mike_mike
author_rank: Super Contributor
timestamp: December 15, 2019, 04:50:17 pm
content: Can you please also have a look at the below reply and clarify the answer ?Quote from: mike_mike on December 15, 2019, 09:53:01 am

author: David Hess
author_rank: Super Contributor
timestamp: December 15, 2019, 07:09:53 pm
content: Quote from: mike_mike on December 15, 2019, 04:50:17 pmIt all depends on the current level and voltage across the LM317.

author: mike_mike
author_rank: Super Contributor
timestamp: December 15, 2019, 07:55:03 pm
content: This means that the output voltage is possible to rise when disconnecting the power supply from 230V mains, if the load is 9A, although the voltage didn't rise when the laod was 3A ?

author: David Hess
author_rank: Super Contributor
timestamp: December 15, 2019, 08:03:35 pm
content: Quote from: mike_mike on December 15, 2019, 07:55:03 pmYes, that is possible. When the AC input is disconnected, the input DC voltage to the regulator falls gradually. As it falls, the current limit increases so the output current increases while the input voltage is falling.

author: xavier60
author_rank: Super Contributor
timestamp: December 15, 2019, 08:27:10 pm
content: And of course the output voltage will not rise above its setting.

author: David Hess
author_rank: Super Contributor
timestamp: December 15, 2019, 10:55:18 pm
content: Quote from: xavier60 on December 15, 2019, 08:27:10 pmIt should not but some power supply designs have flaws. A bare LM317 will not have a problem.Sometimes failsafe circuits are added which clamp the drive to the output transistor during power up and power down.