LM317: divider values and minimum load current - Page 1
Forum section: electronics / beginners

author: not1xor1
author_rank: Frequent Contributor
timestamp: June 16, 2017, 05:50:21 pm
content: just in case anybody wonders why his/her LM317 circuit doesn't work as expected...a few days ago in it.hobby.elettronica, Italian usenet group about electronics, somebody could not make his lazy LM317 behave properly because of high divider resistor valuesI was the only one to argue that the key factor was the load current while most guys sustained it was due to Iadj, so I made a simple test on a breadboard (with 1% tolerance resistors)test divider load V(Out,Adj) V(Out)a) 150+150 any 1.2513 2.5057b) 300+300 inf. 1.2513 2.5109c) 300+300 300 1.2513 2.5102d) 1k+1k inf.1.34932.7473e) 1k+1k 300 1.2513 2.5427f) 1k+1k 150 1.2513 2.5427as d), e) and f) show clearly, when there is not enough load current (just 2k\$\Omega\$ in test d) ) the reference voltage is wrong and even taking into account Iadj * 1k the output value is higher tooso to use high divider values (high value pot or trimmer) one should ensure that the load current be always more than the minimum (3.5 (typ.) - 10 mA)my English is awful, but I hope the test results speak better than me

author: Zero999
author_rank: Super Contributor
timestamp: June 16, 2017, 05:56:10 pm
content: Yes, this is a commonly known issue. Sadly the datasheet even has a schematic with 240R for R1 which just gives 5mA and is marginal.Iadj is also a factor but it can be compensated for in the calculations and is not affected by the load current.

author: Audioguru
author_rank: Super Contributor
timestamp: June 16, 2017, 06:48:49 pm
content: Quote from: Hero999 on June 16, 2017, 05:56:10 pmAlmost every circuit shown in the datasheet uses 240 ohms and the more expensive LM117. The LM317 needs 120 ohms most of the time.

author: Zero999
author_rank: Super Contributor
timestamp: June 16, 2017, 06:56:53 pm
content: Quote from: Audioguru on June 16, 2017, 06:48:49 pmSome data sheets even show the LM317 with a 240R resistor for R1.http://www.st.com/content/ccc/resource/technical/document/datasheet/group1/a0/db/e6/9b/6f/9c/45/7b/CD00000455/files/CD00000455.pdf/jcr:content/translations/en.CD00000455.pdfhttp://docs-europe.electrocomponents.com/webdocs/14e1/0900766b814e17d7.pdf

author: Zero999
author_rank: Super Contributor
timestamp: June 16, 2017, 10:09:47 pm
content: Here's a spreadsheet I did which calculates the output voltage and R2, taking the effect of IADJinto account.

author: Audioguru
author_rank: Super Contributor
timestamp: June 17, 2017, 01:28:40 am
content: National Semi invented the LM117 and LM317. ST Micro and ON Semi are just copycats.All the datasheets show the minimum load current for the LM317 at 10mA (1.2V/10mA= 120 ohms) but only National Semi explains what happens if the load current is too low because the programming resistor value is too high. The datasheets correctly show 240 ohms but it is only for the more expensive LM117 and LM217 that have a minimum load current of 5mA (1.2V/5mA= 240 ohms).If you are very picky then you can say that the LM317 needs 120 ohms only if there is no other load and when the input to output voltage is only 40V. National's datasheet shows a graph of thetypicalminimum load current at different in-out voltages but you do not know the range and you cannot buy a typical or better one.

author: Zero999
author_rank: Super Contributor
timestamp: June 17, 2017, 07:34:00 pm
content: Quote from: Audioguru on June 17, 2017, 01:28:40 amNope, some of the datasheets clearly show the LM317 with a 240R resistor.

author: Audioguru
author_rank: Super Contributor
timestamp: June 17, 2017, 10:22:24 pm
content: Quote from: Hero999 on June 17, 2017, 07:34:00 pmThey are clearly wrong and do not understand their copy of an LM317 because they have text that says the minimum load current for an LM317 with 40V from its input to its output is 10mA. Then the resistor value must be 1.2V/10mA= 120 ohms, not 240 ohms.I avoid looking at copy-cat datasheets because they are missing important information that the original manufacturer has on its datasheet.National Semi never made an LM217. The LM117 is the expensive one with a 5mA minimum load current and the LM317 is the less expensive one with 10mA.

author: rdl
author_rank: Super Contributor
timestamp: June 17, 2017, 10:25:54 pm
content: That doesn't look like a National or TI diagram. The most recent TI data sheet I have for the LM317 still uses the zig-zag style resistor symbol. However, it does show a 240 ohm resistor in the suggested circuits. The specification for minimum load current is given as 3.5 mA typical, 10 mA maximum. This data sheet is for the LM317 only, no mention of LM117.

author: Zero999
author_rank: Super Contributor
timestamp: June 18, 2017, 09:05:17 am
content: Quote from: rdl on June 17, 2017, 10:25:54 pmThe schematic is taken from one of the datasheets linked to in my second reply to the thread. There are plenty of other data sheets which show 240R and the LM317.

author: bktemp
author_rank: Super Contributor
timestamp: June 18, 2017, 09:18:05 am
content: What is wrong with 240ohms?All datasheets clearly say "Minimum load current to maintain regulation max 10mA". So if the voltage divider draws only 5mA it is the responsibility of the user to have a minimum load of 5mA.Why waste 10mA in the voltage divider if 1-5mA are enough?If you look at LM78xx datasheets, it also specifies the output voltage for a minimum load of 5mA, but none of the example circuits has a load resistor connected. Now you could argue the same way, all the datasheets are wrong, because the circuit violates the minimum load requirement.

author: not1xor1
author_rank: Frequent Contributor
timestamp: June 18, 2017, 02:13:45 pm
content: Quote from: bktemp on June 18, 2017, 09:18:05 amI just checked a few LM317 made by ST and most of them are happy with just a bit more than 1mA of load.I think that most beginners, as hobbyists, are not interested in mass production and can enjoy testing themselves the capabilities of their ICs.I think one can use one of these test circuits on a breadboardone can just compare the output voltage with a sure load (i.e. a 120resistor) and with a test load (e.g. a 270or even a 470or higher resistor) or even use a potentiometer and check at what current value the output voltage gets over the right value

author: Zero999
author_rank: Super Contributor
timestamp: June 18, 2017, 10:42:02 pm
content: Quote from: not1xor1 on June 18, 2017, 02:13:45 pmThat may be true for the small batch you tested, under normal conditions but it doesn't mean it will be the case for different manufacturers, at extremes of voltage and temperature. When designing a circuit to post on the Internet or mass produce, it's important to ensure it works well with components from other manufacturers anf under different conditions than in your lab.

author: pope
author_rank: Frequent Contributor
timestamp: February 15, 2025, 03:27:21 pm
content: Apologies for resurrecting this old topic but I think it's better that staring a new one.Can somebody explain if the minimum load requirement (i.e the R1 value) is important in situations where the load is above 10mA?In other words, if the actual load is 15mA or 150mA or 1.5A, does it matter whether R1 is 120R or 360R or anything in-between?

author: TimFox
author_rank: Super Contributor
timestamp: February 15, 2025, 03:37:51 pm
content: If the external load will always be above the minimum value, then the resistors in the voltage divider only matter to determine the effect ofIadj, similar to the effect of bias current and feedback resistors in an op amp circuit.

author: bdunham7
author_rank: Super Contributor
timestamp: February 15, 2025, 03:43:21 pm
content: Quote from: pope on February 15, 2025, 03:27:21 pmIf you have a load with a guaranteed draw of >10mA then you can use higher values for the divider resistor. However, the other factor you have to look at is IADJor the current on the adjust pin. The TI datasheet for a regular LM317 shows this to be as much as 100µA with a variation due to load or voltage changes of up to 5µA. With 360R for R1, you'll have 3.3mA and that 100µA would represent 3% of the divider current and thus a proportionate error. ThetypicalIADJcurrent and variation thereof are much lower, so that 3% would be the worst-case. For production you have to worry about worst case, for something one-off or low volume that you can 100% test then you could probably use 1k for R1 and still be fine.

author: pope
author_rank: Frequent Contributor
timestamp: February 15, 2025, 03:46:08 pm
content: Quote from: TimFox on February 15, 2025, 03:37:51 pmSo, R1 should be adjusted so as the Iadj is 100uA (max value per datasheet)?

author: pope
author_rank: Frequent Contributor
timestamp: February 15, 2025, 03:50:36 pm
content: Quote from: bdunham7 on February 15, 2025, 03:43:21 pmThank you for the detailed explanation but unfortunately due to my luck of English skills I'm having a bit difficult time to understand wht you mean.In simpler terms, using 360R is OK or not really? And if so, would the suggested 240R is better?I'm straugglin to understand this "With 360R for R1, you'll have 3.3mA and that 100µA would represent 3% of the divider current and thus a proportionate error. The typical IADJ current and variation thereof are much lower, so that 3% would be the worst-case."

author: TimFox
author_rank: Super Contributor
timestamp: February 15, 2025, 03:50:55 pm
content: Iadjis a feature of the 317.The effect of its value and variation on the regulated output voltage depends on the resistance of the voltage divider.Smaller resistors in the divider will give more stable output, but waste power in the current through the divider. This is an engineering choice.

author: bdunham7
author_rank: Super Contributor
timestamp: February 15, 2025, 03:54:44 pm
content: Quote from: pope on February 15, 2025, 03:46:08 pmNo, not at all. IADJcould be nearly zero, the 100µA is a maximum value and is a characteristic of the individual regulator and its environment (temperature, load, VIN - VOUT, etc). You select R1 and R2 to set your desired output voltage and whatever IADJhappens to be at the moment will cause an error, a deviation from your selected output voltage. The lower the values of the resistors in your divider chain, the higher the current through those resistors and that means that whatever IADJis will cause less error.You can deal with this by using low values for the divider (R1 = 120R, or even less if you like) or you can select your LM317s for low adjust current or you can adjust your divider ratio to tweak the result. But what the value of IADJactually is comes down to luck and circumstances. The datasheets don't show any graphs or relationships for the adjust current, only typical and max values for the current and its deviation.

author: pope
author_rank: Frequent Contributor
timestamp: February 15, 2025, 03:59:31 pm
content: OK so if I understand well for (let's say) 30V output a 240R/5.5k would be better than 360R/8.28k but it will draw more current (i.e the 317 will get warmer).Am I on the right track?

author: bdunham7
author_rank: Super Contributor
timestamp: February 15, 2025, 04:19:24 pm
content: Quote from: pope on February 15, 2025, 03:59:31 pmThe resistors will get warmer, but how much warmer the LM317 gets depends on the difference between the input and the output. The power dissipation will be (VIN- VOUT) * (IADJ+ IDIVIDER+ ILOAD). If ILOADis >>10mA then the other two won't matter much. For example, if VINis 40V and VOUTis 30V, then for R1 = 120R the total power dissipation in the LM317 due to the divider current is 100mW.

author: pope
author_rank: Frequent Contributor
timestamp: February 15, 2025, 04:29:51 pm
content: Great, thank you very much

author: Quarlo Klobrigney
author_rank: Super Contributor
timestamp: October 01, 2025, 09:40:55 pm
content: Meanwhile all this faffing about by Dave,https://youtu.be/ICQXqVy3Hpc?t=560I still have no idea how to measure the minimum or any current value for Iadj.I saw your .ODS spreadsheet above and converted all the .00000 values into *10^-6 etc, and still I am lost.I hate things such as "Iadj can just be ignored" and "Iadj MUST be less than 100uA"Trying to calculate the current, in desperation I turned to every version on AI where it bloviated as it does for an hour with my correction to come up with a formula where the 2 values with an answer of zero yes 0 uA.So to others here, just what is a suitable formula to measure the current because many datasheets say it is important but then say it is unimportant.I was in all this, looking for a suitable modern regulator with no ringing and be a LDO that is not from China.Why do I care, enigmas make me angry when data contradicts.

author: Zero999
author_rank: Super Contributor
timestamp: October 02, 2025, 10:00:10 am
content: The spreadsheet I posted accepts the current in µA. Using *10^-6 will produce inaccurate results.Whether IADJcan be ignored or not, depends on the application.IADJis nominally 50µA, but it can be as high as 100µA. This will typically increase the output voltage of the LM317 by 50mV per k Ohm for R2, so if R2 is 1k, then expect the output voltage to be 50mV, higher, because IADJcauses additional volage across R2.Note the reference voltage of the LM317 1.25V ±50mV, which translates to a tolerance of 4% on the output voltage. In other words, if the output were set to 12.5V, with precision resistors, taking into account IADJ, then the output voltage could be anywhere between 12V and 13V.Example. If the output voltage is 12.5V, with R1 = 111R111 & R2 = 1k then the error due to IADJis only 50mV, which is a tenth of the error due to the tolerance of the LM317's 1.25V reference, but if R1 = 1k11111 & R2 = 10k, then the error due to IADJis 500mV, which is equal to the voltage reference tolerance. The solution would be to reduce R2 to 9k57446. Obviously, you would use standard E24/E96 resistor values. I just used exact values for the purpose of the calculation.The bottom line is IADJprobably doesn't matter, if R1 is low enough to meet the LM317's minimum load requirement, otherwise it needs to be taken into account.

author: Quarlo Klobrigney
author_rank: Super Contributor
timestamp: October 02, 2025, 04:45:01 pm
content: Why would you think that to be?I'ts all amount the number of digits in the cell that is allowed.Quote from: Zero999 onTodayat 05:00:10

author: TimFox
author_rank: Super Contributor
timestamp: October 02, 2025, 04:58:08 pm
content: A simple summary:Since there is no negative power pin for the 317 (as opposed to a 7805, for example), the internal power to operate the regulator must flow through the output pin. Hence the manufacturer’s spec for minimum output current, which is the sum of the (useful) load current and the resistor divider current.TheIadjcurrent is a small bias current that flows through the junction of the resistor divider that sets the output voltage.If the load current is expected to be low, then the divider must pull enough current to meet the minimum output current spec. If a higher resistance divider is used, then the voltage error caused by the bias current and the resistors may be too high, since that current is really only subject to a specified maximum value. A quantitative question.

author: Quarlo Klobrigney
author_rank: Super Contributor
timestamp: October 02, 2025, 05:31:33 pm
content: This I know. What I want to know is how to measure the Iadj current if the load current is given as a range as up to 30mA or 10mA or 100mA.Those things are not designated in any datasheet as to its calculation.I have used my sheets for many years with the designated formulae:LM317 LM1117:Vout = (1 + R2 / R1) * 1.25 (Vref)R1 = (R2 / ((Vout / 1.25) − 1)R2 = (R1 * (Vout / 1.25) − 1))With LM431 being 2.5 for vref instead of 1.25.The LM317 is being offered in a 100mA, 500mA and 1A package and R1 specified as "must" be 470 ohms and never 240 ohms.Then there is the Vin max all over the place some down to 16V.So forgive me if I demand a standard from these manufacturers and less fiddling with experiments and time wasters.

author: TimFox
author_rank: Super Contributor
timestamp: October 02, 2025, 06:09:50 pm
content: The "standard" is a maximumvalueforIadjand a minimumrequirementforIout.If you meet the latter, then the range of the former, coupled with your choice of feedback resistors, will give you an estimate of the variation in the output voltage due to the bias current from part to part and over temperature.It is annoying that obsolete values of the minimum output current requirement seem to have persisted in the applications information in currently-published datasheets.

author: Zero999
author_rank: Super Contributor
timestamp: October 02, 2025, 07:16:02 pm
content: Quote from: Quarlo Klobrigney on October 02, 2025, 04:45:01 pmWhat's your point? Just enter IADJinto the speed-sheet in µA. It can be 0 if you want to see what the output will be, compared to the typical value of 50µA or maximum of 100µA. Type 50 and the units µA are automatically appended to it.If you don't like this and want to specify IADJin A, then change the formula, delete *0.000001 from the formulae in U11 and U21 and remove µ from the formatting of E6.Quote from: Quarlo Klobrigney on October 02, 2025, 05:31:33 pmAs mentioned above, IADJdoesn't depend on the load current.Like any current, it can be measured with a multimeter set to current mode.QuoteThe TL431 is a different part.No data sheet says R1 must be 470 Ohms and never 240 Ohms. A wide range of resistor values can be used for R1. If the circuit connected to the output always draws more than the minimum load, then R1 can be higher.I've not seen an LM317 data sheet which specifies a maximum input voltage of 16V. There are certain conditions i.e. high output current, when the voltage difference between the input and output voltages, can't be over a certain value, due to the safe operating protection. For example, from memory, the LM317 has a maximum output current of 500mA, when VIN-VOUT> 15V. This means that at higher voltages, the current limit is reduced, to protect the output transistor.QuoteGood luck with that. Read the data sheet carefully and design the circuit so it will work with parts from a wide range of manufacturers. This is normally not too difficult. Don't rely on a super minimum load current, keep VIN-VOUT> 3V and <40V and if you need the full 1.5A, VIN-VOUTshould be <15V

author: Quarlo Klobrigney
author_rank: Super Contributor
timestamp: October 12, 2025, 04:54:57 pm
content: I'm getting better in bash and I created a "little" terminal app to calculate all that I need to know without xls or ods sheets asking me to save before exiting!Further testing of all permutations and accuracy is necessary to confirm its accuracy due to the long nights.Here is a snapshot:

author: gbaddeley
author_rank: Regular Contributor
timestamp: October 13, 2025, 12:39:03 am
content: I have always used R1 = 100, 110or 120 with LM317/337. Then calculate R2 for the desired output voltage, ignoring the insignificant current through the C pin.Bypass R2 with 10 - 47 uF to improve output ripple / noise.

author: Quarlo Klobrigney
author_rank: Super Contributor
timestamp: October 13, 2025, 10:31:26 pm
content: I know about the bypass caps but for the newbies of course.I am not.Here below are 3 panels for the computations from the design for all 3 regulators...Comments are welcomed.