LM317 Power supply questions. - Page 1
Forum section: electronics / beginners

author: rohan-04
author_rank: Contributor
timestamp: May 31, 2013, 12:43:10 pm
content: Some of you may know that I planned to make a power supply with an LM317 and a couple of switchs. I've now decided to use the conventional ten-turn pot and a moving coil meter to make the power supply.The limits for the new power supply are :-a)1.5v-15V - 1A variable voltage.b) 5V -1A output to a usb female.The schematic is attached.The promblem im have trouble getting around is, when the LM317 is turned to output a low Voltage out (in this case 4V>) the power rating at 1A is at 23Watts. I plan to use a 240VAC Primary - 24VAC Secondary - 2A Transformer. I think I could have also made a very silly error in my calculations, so if you guys could help me again.Thank you.

author: rohan-04
author_rank: Contributor
timestamp: May 31, 2013, 01:09:47 pm
content: Exactly! so what can I do to reduce this? My first guess was to cut the out put current to 0.8ma. That would bring it down conciderably.

author: mariush
author_rank: Super Contributor
timestamp: May 31, 2013, 01:14:57 pm
content: The LM317 can only dissipate about 15w of heat.A 24v AC transformer would be rated as RMS voltage. The bridge rectifier would convert that to DC with a peak voltage of 1.414 x 24 = ~ 33v , minus the voltage drop on two diodes of the bridge rectifier. So at the input, you'd have about 31-32v.To get 1.25v at 1A, the LM317 would have to dissipate (32v - 1.25v ) x 1A = ~ 31.75w, which is not possible.You need to always have this voltage difference between input and output as small as possible.My advice is to get a transformer with center tap or two independent secondary windings. There are transformers ready made with 2x9 volts, 2x10v or 2x12v.The LM317 for the USB can always be powered from one winding, while the adjustable one can switch between using one winding, or both windings depending on the output voltage.With a 2x9v transformer, the rectifier DC voltage would be about 11v (after bridge rectifier diode drops), for 12v windings you'd have about 15v.With 2x9v or 2x10v you can make your adjustable power supply have two ranges : 1.25-9v and 9v-18v and put a switch on the front which basically commutes the AC1 in your design between using one winding, or both.With 2x12v, the peak DC of about 15v for a winding is big enough to do 1.25-12v directly. If it's center tap the second winding is pointless. If they're independent windings, you can put them in parallel to get twice the current amount.

author: digsys
author_rank: Unknown
timestamp: May 31, 2013, 01:29:45 pm
content: First observation - put both Leds across their respective outputs. The way you have them, you're ONLY proving the switch works :-)Use the TO3 version of the LM317 and IF you can't have a multi-tap transformer, use the OP Voltage to drive 1 or 2 Linear PNPPre-Regulators, also TO3. That way you wash off most the heat on dummy devices. Alternatively, you can use the PNP Transistors asparallel current boosters, with the LM317 just providing control current. Check the application notes.

author: rohan-04
author_rank: Contributor
timestamp: May 31, 2013, 01:32:03 pm
content: Quote from: mariush on May 31, 2013, 01:14:57 pmThats a good idea ill look into it now. So basically If i break it into two segments of voltage being 1.25v-7.5V and 7.5V-15. I'll need to have transformer (C.T) that gives 2x9V? and togther 18v? to supply the 15V? Also what about the LEDS? do we need to account for their voltage drop?

author: rohan-04
author_rank: Contributor
timestamp: May 31, 2013, 01:42:10 pm
content: Quote from: digsys on May 31, 2013, 01:29:45 pmIm gonna have to research that answer and get back to you in a day...

author: Paul Price
author_rank: Super Contributor
timestamp: May 31, 2013, 01:46:31 pm
content: Go with a switcher, less than a watt or two for heat wasted and can put out 3 times the current, so easy to build.You can use the non HV adj. version of the switching chip as your voltage from your transformer and bridge rectifier will be max about 33V, much less than 55V input, but so what, the 24V from your bridge rectifier does not require the HV version chip in the circuit shown, so it is even cheaper to build.You do not need 55V of high voltage into the switching regulator that you see in this schematic, your 24V transformer circuit will work fine.This switching circuit by itself, is clean enough at it's output to use with just about any circuit you will work with. And all this is done without using a wasteful linear regulator, and this circuit, even in a little tiny way will help save the planet from global warming.Here's some ideas to play with:If you really want a very,very, quiet smooth output, change the voltage adj. potentiometer to a dual-ganged version and you can make the switcher always track at a voltage a little above the input voltage to the LM317 linear regulator so it always has just the right min. required input voltage to work perfect.You see, you use the switcher circuit to supply the voltage going into the input of your LM317 regulator, to a voltage that is just about a few volts higher than the output of your LM317 regulator circuit, so the switcher can function either as a high-current supply alone (connected to jacks on the front panel of your power supply), or as apre-regulatorthat so greatly reduces the amount of heat wasted in the linear regulator.Maybe a simple voltage raising circuit using two LED's connected in series or a zener diode can make the setting voltage of the switcher always be set a few volts higher then the LM317 regulator, and then, by just using one pot, you can vary the output voltage of both regulators. That is because the voltage across a zener or a one or two LEDs when on and lighting up is quite well regulated and does just right to serve to set the needs-not-to-be-so-closely-accurate voltage output of the pre-regulator a little higher. Schematic available on request.

author: rohan-04
author_rank: Contributor
timestamp: May 31, 2013, 02:14:03 pm
content: Ok so now Im doubting the LM317, should i cuck this circut and go for a SMPS build instead? An SMPS is a better way of making a power supply right? as it handles the heat better?

author: mariush
author_rank: Super Contributor
timestamp: May 31, 2013, 02:32:55 pm
content: Quote from: rohan-04 on May 31, 2013, 01:32:03 pmTransformers with center tap will be listed in catalogues or online stores as total voltage, with a mention saying they're of center tap variety... for example you may see "18V CT" - that's 18 v AC rms, center tap , two secondary windings of 9v AC rms linked together in the middle.So 24 AC CT will be two 12 v windings linked in the middle.Center tap transformers only have three wires in the secondary side.. some allow you to separate the center wire in two to have two separate secondary windings, some don't. You have to visually see the terminals of the transformer to determine that.Transformers with independent secondary windings have just that, two separate windings. They'll be listed as full voltage, or 2x half voltage.You can link those in parallel to get more current, or in series to get more voltage, or you can link the middle two terminals to create a three terminal center tap transformer from it.So again, most transformers are advertised as Voltage RMS - when you convert that to DC voltage, you multiply that value by 1.414.So a 12v AC rms transformer will have a peak of 1.414 x 12 = 16.9 volts.But when you use a bridge rectifier to convert to DC, there's a voltage drop on the diodes in the bridge rectifier. A diode drops about 0.4-1.1v depending on how big the bridge rectifier is and how much current goes through it, and in a bridge rectifier two diodes are always on, so you have to always keep in mind, you lose about 2 times the voltage drop on a diode inside the rectifier, about 0.8-2.2v.So your 16.9 peak voltage becomes 16.9 - 2.2v = 14.7 volts. We're always using the worst case scenario values, you might have better bridge rectifier but we just have to be sure it would work in all cases.Now, remember this is a PEAK voltage. The rectified voltage will not be a straight, smooth, DC line, it will vary between this peak of 14.7 volts and go down close to 0v for very short amounts of time. This is where the capacitor after the bridge rectifier comes in: it fills up when you get close to peak voltage and then when the voltage from transformer goes down, it fills the gap.With an infinite size capacitor, you will get a perfect straight DC of about 14.7v but infinite capacitors are impossible. With more realistic values of something like 3300uF-4700uF, the DC voltage will fluctuate by about 1-2 volts, maybe a bit more depending on how much current your circuit uses.So with this in mind, assuming in worst case scenario the voltage will vary by 2 volts in the capacitor, the linear regulator will at some points only receive about 14.7-2v = 12.7 volts.The linear regulator itself needs about 2 volts above the output voltage to regulate the output properly, so that's what limits your output voltage ... 12.7v input - 2v drop on regulator = 10.7v ... so a range of 1.25-10v is adequate and safe.Now with these values, 12.7v DC input , 1.25-10v @ max 1A your maximum wasted power will be (12.7v-1.25v )x1A = 11.45w , the lowest would be (12.7v-10v) x 1A = 2.7w.Both are below the maximum of about 15 watts, so as long as you have a good heatsink on the chip, it's perfectly safe and good.One more thing you have to keep in mind... transformers don't always output exactly 12v ac rms (if they're rated for that) ... at low amounts of current, they may output up to 10-15% more.Let's say you only use 100mA ... in this case the transformer may output 10% more, or about 13v AC rms ... that converts into over 18 volts and even with the drops in the bridge rectifier you may have a voltage after the bridge rectifier of over 16v.If you don't know this, you may say "well, 14.7v dc is less than 16v so i'll use 16v capacitors to smooth out the rectified voltage" but you really have to use capacitors rated for at least 25v, just for such cases.LM317 is a good linear regulator. It's easy. Just use a transformer with smaller voltage like I've explained above.SMPS are harder to make and not work right on prototyping boards so you'll have to solder them on pcb, get adequate inductors, low voltage drop diodes, it's much harder to get them right.If you're a beginner like you seem to be, it's a good project and learning experience to make one of these linear power supplies first. You'll learn from everything a lot.Regarding LEDs... leds are current devices, they don't care about voltage that much as long as it's over a certain value. You need to use a resistor to limit the current going into the led, and the resistor will also drop the voltage on it.The formula is simple ... let's say your led is rated for 3v 10mA ... at 10mA it would be super bright, so use 5mA to make it light less bright and live longer.You're going to power it from the input voltage as that won't vary... if your input is about 12v as I did the math above, you have the formula V = I x R so 12v = 0.005 x R so R = 12 / 0.005 = 2400 ohm. You round up to a standard value like 2700 ohm or you can put two 4700 ohm resistors in parallel to get 2350 ohm. It's not a critical value, if it varies by a few hundred ohms, you get 4 or 6 mA of current in the led, still below the maximum of 10mA.

author: rohan-04
author_rank: Contributor
timestamp: May 31, 2013, 02:42:23 pm
content: Alright, So tomorrow im gonna go hunt for a transformer. And im also gonna trim down the requirements and make it just a 12V system, and seee how it goes.Thanks alot, you solved alot of my questions without me even asking them

author: Paul Price
author_rank: Super Contributor
timestamp: May 31, 2013, 02:46:39 pm
content: It is a myth that you always need a PCB to make a circuit like this.I have built many well-working switchers like I show in the schematic, some putting out at least 3 to 5 Amps or so, just on perfboard, no PCB is really required, just try to keep leads short, but kinda neat looking, use reasonably larger wire for ground and power carrying connections, but use no wire so big it is clumsy to work with.I have even built this circuit without any board at all just hanging together in mid-air, and it works very very well.(of course, the power supplied into the switcher circuit was coming from a discarded 18.5V 3.5Amp wall wart lying on the table that once was used to power a Dell laptop and so I didn't have a big transformer and capacitor and bridge suspended in mid-air!)Both types of power supplies have their building and operation pros and cons, but both are fun to work with.

author: Paul Price
author_rank: Super Contributor
timestamp: May 31, 2013, 02:48:05 pm
content: You could just abandon the LM317 and make a switcher, so much easier to build or use my suggestion above to use both technologies and you will learn about switchers and linear power supplies.You can get all the low-voltage drop diodes, maybe a small heat sink, and a large toroid inductor, maybe just about all the parts you need, (except for the LM2976, LM2975, etc regulator) for free to build this supply from salvaging the parts form a junk PC computer power supply.I don't think, after making so many of these switchers in every PCB and Perfboard way, that SMPS are harder to make and not work right on prototyping boards, that you'll have to solder them on pcb, or it is hard to find or get adequate inductors, low voltage drop diodes, that it's much harder to get them right.

author: JackOfVA
author_rank: Unknown
timestamp: May 31, 2013, 03:06:26 pm
content: If you use an LM-317 note that they have a minimum current rating as well as a maximum current rating. TI, for example, spec's the LM-317 it manufactures as 3.5mA "typical" minimum current requirement and 10 mA "maximum" minimum current in order to maintain regulation.The voltage divider that you use to set the output voltage may draw sufficient current to meet this requirement.

author: Paul Price
author_rank: Super Contributor
timestamp: May 31, 2013, 03:10:29 pm
content: In your original schematic:The 240 ohm resistor connecting from the output pin to the Adj. pin of the LM317 already ensures the min. current requirement for regulation is already met.Did you know that you can also use a 7805 or 7812 regulator to create a variable output regulator without buying a LM317?With a small negative supply for bias(LM555 circuit to make), they can also have a min. regulated output voltage down to zero volts. The 240 ohm resistor in the original schematic has to be increased to 3.3K to get it to work and the center ground pin become the adj. terminal.

author: JackOfVA
author_rank: Unknown
timestamp: May 31, 2013, 03:37:24 pm
content: Drop between OUT and ADJ is 1.25 volts, and there's a 240 ohm resistor between these two points.1.25V / 240 ohms = 5.2 mA. Meets typical but not worst case minimum current.It's very likely adequate 99% of the time since TI is usually conservative in its worst case specs, but it is a bit short of the worst case spec sheet value, so if one were to design to meet the worst case specification, something closer to 120 ohms would be required.Or am I missing something?

author: Paul Price
author_rank: Super Contributor
timestamp: May 31, 2013, 03:53:55 pm
content: For the 24V circuit he is using, the 240 ohm is more than adequate. See the attached graph at the point Vin<30V the requirement is less than 3mA. See attached graph from NatSemi Spec Sheet.I also show some very blurry photos(cell phone camera not so good) of theentire switching regulator built on a 1.5 x 3-in approx perf board.The black sausage-like object at the bottom(top view) is the inductor, the switcher chip is on right side with a small heat sink and there is a slightly off-board connector to fit the PC power supply plug for power coming in.

author: JackOfVA
author_rank: Unknown
timestamp: May 31, 2013, 04:23:44 pm
content: Quote from: Paul Price on May 31, 2013, 03:53:55 pmI think the plot of Imin versus voltage differential is of "typical" Imin, not worst case. ON Semi, for example, has pretty much the same curve, but still specs Imin at 3.5 mA typical; 10 mA worst case. Fairchild data sheet has no similar plot, nor does STM for their LM317 offerings.I agree, though, that 5 mA in the original circuit is adequate for almost all cases, but have a nagging concern about how often the "worst case" condition will surface.

author: Paul Price
author_rank: Super Contributor
timestamp: May 31, 2013, 04:27:49 pm
content: Not to split the current hair, but the graph is titled "Minimum Operating Current", and I would assume "operating" means that a regulator is operating as a regulator and the 240-ohm supplies current n x s.

author: Bored@Work
author_rank: Super Contributor
timestamp: May 31, 2013, 04:32:25 pm
content: Quote from: JackOfVA on May 31, 2013, 03:37:24 pmNo, you are not.What people typically miss in the LM317 datasheets is that all the circuits in them are typically for theLM117, not the LM317. And guess what, the LM117 has a max (worst case) Imin of 5 mA, vs the 10 mA of the LM317.The 240 Ohm resistor one typically finds in the datasheet schematics is for the LM117. It typically also works for the LM317, but if you want to be on the save side, change it to 120 Ohm, and half the other resistor of the divider, too.This is an issue with LM317 datasheets since forty or so years. And there must be millions of devices with LM317th out there with a too weak Imin. Apparently most of them work most of the time, so it isn't such a big thing.

author: Paul Price
author_rank: Super Contributor
timestamp: May 31, 2013, 04:45:49 pm
content: My hare-brained mind tells me to quit yakking and eat my lunch, but the LM117 maybe is the one much more critical with 240 ohm min current setting than the LM317 because it must meet spec from -85 to +150 deg C.I always use a 220-ohm, because it is easier to find, standard 5% value that is very common in use.

author: rohan-04
author_rank: Contributor
timestamp: May 31, 2013, 05:02:58 pm
content: Right so now in the morning im gonna go out an buy a 240VAC to 12VAC transformer, and hopefully build up the circuit to give out 10V. What value cap should i use on C1? Is 1000uf a fair amount? or should i up it to 3300 or 4700uf?. About the value of R1, I initially planed for a 15V circuit so i thought i would use 180Ohm plus a 2k ten turn pot to get 15V out of the equation. Now to get the pot to give 10V what value will I need 300 odd Ohms, will that be too much?

author: Paul Price
author_rank: Super Contributor
timestamp: May 31, 2013, 06:39:21 pm
content: 4700 uF is better. Use the same pot resistance values, change the resistors the pot connected to, used with.Use the formulas! Have you read the spec sheet? The part about output voltage setting is not too technical.

author: c4757p
author_rank: Super Contributor
timestamp: May 31, 2013, 06:42:17 pm
content: Quote from: Bored@Work on May 31, 2013, 04:32:25 pmEureka!I was wondering why that resistor is always too high in the LM317 datasheets - ignoring the example circuits, I've always come up with a value around 82-120. I've never used LM117, so I never realized it had a lower minimum current.

author: codeboy2k
author_rank: Super Contributor
timestamp: May 31, 2013, 08:02:40 pm
content: sonofabitch!I've always wondered why every schematic I see with the LM317 uses that ubiquitous 240 ohm resistor almost religiously. Although I am pushing 50yrs old now, I have never worked with the LM117 so a reason never came to me.When I do a design with the LM317 I always end up with 100-120 ohms there (usually 100 since it's often already part of the BOM). That's what I've always used, based on worst case design.In fact, every single design I make is based on worst case everything, including possible failure modes. I will design for that as a worst case sometimes too, and increase values to provide protection in case of that failure.I find that there is too much stuff on the web that is not really engineered, just slapped together and/or copied from somewhere else.

author: Bored@Work
author_rank: Super Contributor
timestamp: May 31, 2013, 08:33:23 pm
content: Many companies do no longer offer the LM117 at all, and over time have "updated" their datasheets by meticulously replacing every LM117 label in their schematics with LM317 - of course keeping the old magic 240 Ohm resistor. Probably having the interns done it. Why invest in checking 40 year old schematics in a datasheet for a jelly bean IC?And unfortunately with every datasheet revision they seem to remove application examples. Of course, some stuff is outdated, e.g. how to do a switching regulator using a linear regulator, but if you have to repair or revise old stuff it helps if you can go back to where the original developer got his idea from.

author: rohan-04
author_rank: Contributor
timestamp: May 31, 2013, 08:44:44 pm
content: Sorry to be doing this guys, but I got side tracked with all the replies that i forgot that 12V is crucial, Im building this system for 12VDC mainly. Anyhow so I've now worked this out.T1p:240VACT1s:14VAC=>14x1.414 = 19.7V DC.=>19.7-12= 7.7 V ( Vin-Vout)=>7.7-2.2=5.5V (diode)=>5.5 - 2 = 3.5v (lm317)P= Min = 14.2P=Max = 7.5WSo will using a 14V transformer work? Do 14V transformers even exist?

author: c4757p
author_rank: Super Contributor
timestamp: May 31, 2013, 08:53:26 pm
content: Quote from: rohan-04 on May 31, 2013, 08:44:44 pm2V is the diode drop and capacitor ripple, right? Then you're going to need more than 1.5V overhead for an LM317. (And somebigcapacitors, if you expect only about 0.6V ripple.)QuoteThey do indeed; Mouser has 17 (not all in stock).

author: rohan-04
author_rank: Contributor
timestamp: May 31, 2013, 09:36:36 pm
content: Yeah sorry there were a few errors in that, fixed it now but it still looks wrong to me. Im now using a 4700uf as C1.

author: mariush
author_rank: Super Contributor
timestamp: May 31, 2013, 09:41:17 pm
content: Since you're in India, chances are you'll find 12v AC transformers much faster than 14v AC.12v x 1.414 = 16.9vInstead of a ready made bridge rectifier or four regular diodes, you could use 4 Schottky diodes, these have lower voltage drop.For example, 1N5818, 1N5819 should be fairly easy to find even in Indian local stores ... they're 1A diodes with a maximum of 0.5v drop:http://www.farnell.com/datasheets/1449529.pdfSee page 5, you'll see that at lower currents the voltage drop will be even smaller than 0.5v.1A is OK, since at any point two diodes will share the current and you only want to output up to 1A. 1N5817 is not adequate since it tops out at 20v and your transformer may have pulses getting close to 20v. 5818 handles 30v, 5819 does 40v, so they're OK.But if you want, you can search for 1N5821 or 1N5822, these can do up to 3A and have only slightly higher maximum voltage drop, about 0.55v.So if you make the bridge rectifier with these schottky diodes, you'll have about 1v in total drop on the rectifier, which leaves you with about 15.9 volts.To get 12v out of a LM317, you need a buffer of about 2.5v at 1A :http://www.fairchildsemi.com/ds/LM/LM317.pdfsee graph on page 4, figure 4. (Sidenote: maybe I picked the wrong LM317 datasheet, there's 10s of versions and I'm sure a lot of them say only 2 volts are required, but still better safe than fail and let's just say 2.5v)So you have to keep the voltage above 14.5v all the time. Let's round the 15.9v to about 15.5v to be extra cautious, that means the capacitors you use after the bridge rectifier are allowed to go down only a maximum of 1v.You can use this formula to determine capacitance :capacitance = ( Current x Duty Capacitor [0.7 usually] ) / ( Vripple x frequency [2x mains frequency])Current is 1A in your case, Vripple is 1 volt, frequency is 50 Hz in India... so C = 1 x 0.7 / 1x100 = 0.7/100 = 0.007 Farads ... that's 7000 uFSo if you want the capacitors to go as much as 1v below what they receive, you need at least 7000uF of capacitance. You could use 3 x 2700-3300uF 25v capacitors which should be even better.So it should be doable with a 12v AC transformer and large capacitors at the input and schottky diodes instead of regular diodes.

author: c4757p
author_rank: Super Contributor
timestamp: May 31, 2013, 09:45:50 pm
content: ...this seems to be a recurring theme, mariush. I type out a long response with all the calculations and everything, and when I hit 'preview' I see you've already replied, usually better than mine!

author: rohan-04
author_rank: Contributor
timestamp: June 01, 2013, 08:58:34 pm
content: Ok, Im no it. 12VAC and schotty's + massive caps.Thank you everyone for your help....

author: jackeagle
author_rank: Newbie
timestamp: August 06, 2013, 10:00:50 am
content: hello... I have a small doubt can i use 24V 5A transformer has I/p for LM317T to get a adjustable 1.25-22v??

author: kizzap
author_rank: Unknown
timestamp: August 06, 2013, 10:25:58 am
content: you should be able to, however you will need to take heat dissipation into consideration, especially at low voltages with higher current.say you are drawing 1A at 1.5V, and assuming that you are not using a centre tapped transformer or anything of the sort, you will get roughly 33V DC out of the rectified and filtered AC, in which case you will need to drop 31.5V at the 1A, or 31.5 watts of power, which will need an impressive heatsink to be able to cope, if it can do it at all.-kizzap

author: Dave
author_rank: Super Contributor
timestamp: August 06, 2013, 03:59:52 pm
content: Massive caps are an idiotic idea, your transformer will be pushed into saturation all the time.The bigger the capacitance, the higher your tdischarge:tchargeratio will be, but due to conservation of charge, your transformer will still need to deliver as much charge as the load will pull from the cap. So, you havelesstime to delivermorecharge to the capacitor bank, therefore you will needhugecurrent spikes to get the job done. Huge current spikes -> transformer core saturation. They aren't good for the capacitors either.Pick a reasonable capacitance (that would be somewhere 2000-4000uF per ampere).

author: jackeagle
author_rank: Newbie
timestamp: August 07, 2013, 02:19:09 pm
content: Quote from: kizzap on August 06, 2013, 10:25:58 amThank U for u r reply..i m using center tapped transformer 24-0-24V 5A the heat sink is quite large..

author: kizzap
author_rank: Unknown
timestamp: August 09, 2013, 03:59:39 am
content: If you are only planning on limiting yourself to a single supply, I would honestly consider using a 9-0-9 transformer, which will give you 25V rectified with the outer taps and you can use a relay to switch to the center tap to give 12.5V.Otherwise, if you use just a single tap of the transformer you are looking at using, you will have to Dissipate 150 Watts of power if you plan to use it down the lowest voltage (1.25) at 5A, whereas my solution will limit it to roughly 50-60W, which is more manageable.-kizzap

author: megajocke
author_rank: Regular Contributor
timestamp: August 10, 2013, 03:54:38 am
content: Quote from: Dave on August 06, 2013, 03:59:52 pmThis may be a bit off topic, but where does this myth that drawing a high current from a transformer will cause it to saturate really come from? I see it repeated all the time by a lot of people, although in reality flux is primarily related to voltage per turn and frequency as the primary drivers, with load current having only a secondary effect. And even then, this secondary effect is in the opposite direction of what this myth suggests; loading the transformer causes a voltage drop in the primary winding resistance, whichreducesthe flux rather than increases it.It is increasing the voltage or lowering the frequency that will saturate a transformer.And on the topic of rectifiers, changing the capacitor value in a typical transformer fed rectifier will not actually change the current waveform appreciably because the charge phase is dominated by transformer resistance (and/or possibly inductance) unless the transformer is absolutely huge compared to the load on the rectifier.Simulating an unrealistically stiff transformer will however show that the current gets peakier for a larger capacitor value. Low power SMPS without PFC fed from the mains is one application where this actually happens because the mains has very low impedance in relation to the current draw of the SMPS.QuoteThat sounds reasonable at around the 10-40 V range or so, giving about 1.5 - 4 V pk-pk of ripple, but it may be quite excessive at higher voltages.

author: megajocke
author_rank: Regular Contributor
timestamp: August 10, 2013, 05:20:21 am
content: Quote from: mariush on May 31, 2013, 09:41:17 pmI wrote a reply, but it seems I lost it.Anyways, it seems a 12 V transformer is not going to be enough at even 5% low mains voltage regardless of how large smoothing caps one uses even if using schottky diodes. A 15 V 30 VA (2 A secondary current) unit seems more suitable, allowing for 10% low mains voltage with 3300uF of smoothing and normal diodes.The attached LTspice simulation shows such a configuration where transformer regulation is assumed to be dominated by winding resistance, neglecting transformer leakage inductance. This is likely to be a good approximation for toroidal core transformers but may not be suitable for split bobbin EI types which can have considerable leakage inductance. Secondary RMS current is 1.8 A so a 15 V 30 VA transformer does seem to be a suitable size.No load voltage could be, accounting for 10 % high mains and 10 % transformer regulation, up to 25.7 V before the rectifier voltage drop.

author: c4757p
author_rank: Super Contributor
timestamp: August 10, 2013, 05:22:28 am
content: What's with the thick plot lines?? I was trying to figure out why there was HF riding on top of those signals...

author: megajocke
author_rank: Regular Contributor
timestamp: August 10, 2013, 05:30:07 am
content: There's a setting in LT spice found inTools (menu) -> Control Panel -> Waveforms -> Plot data with thick lines (checkbox).I remember now that I turned it on previously when exporting some waveform picture. It's not on by default.

author: Dave
author_rank: Super Contributor
timestamp: August 10, 2013, 09:00:21 am
content: Quote from: megajocke on August 10, 2013, 03:54:38 amIf everything you have written was true, and the only real limiting factor was the winding resistance, have you ever thought why huge linear power supplies use massive transformers?According to you, they should only need a transformer cores that could accomodate wires of the appropriate diameter.The key here is density of flux (B).

author: jahonen
author_rank: Super Contributor
timestamp: August 10, 2013, 11:14:30 am
content: Quote from: Dave on August 10, 2013, 09:00:21 amNo, for transformers (except those in flyback switch mode converters or similar "bi-phase" style supplies) flux density is fixed by the number of volt-seconds applied by the primary winding waveform along with the core characteristics. Load current is not a factor in flux density since primary and secondary windings "push" against each other while maintaining the balance of energy transferred.It really boils down to that you can't fit enough sufficiently thick wire in a small transformer core, thus requiring larger core size. Thus core has certain power capability in practice.Regards,Janne

author: megajocke
author_rank: Regular Contributor
timestamp: August 11, 2013, 07:13:59 am
content: jahonen, that's a good explanation.How then, does size relate to power rating? I'll try to explain. If it were possible to keep the winding current density and core flux density constant independent of core size, scaling the size of a core would change transferrable power as the fourth power of the length scaling factor. This is because the voltage per turn scales as the core cross sectional area while the winding wire size (or number of turns) scales as the area of the "window" in the core where the winding passes through.However, such a scaling would increase the power dissipation of the transformer as the cube of the length scale (because the power dissipation density in both winding and core is constant) while the cooling surface area only increases as the square of the length scale. Therefore flux density and/or current density needs to be lowered as the core size is increased, to keep the same temperature rise.Practically, therefore, the power scaling ends up being proportional to roughly the cube of the length scaling factor, or in other words the volume/weight of the transformer.

author: Dave
author_rank: Super Contributor
timestamp: August 11, 2013, 04:00:32 pm
content: I read up on transformers and it turns out you are right. My bad.

author: rohan-04
author_rank: Contributor
timestamp: October 20, 2013, 09:50:22 pm
content: Sorry for not replying wasnt aware of the thread being active.I decided to trash the old plan of using it to as 1.25-12v, and decided to use it at 9v max. I attatched a schematic below with the reworked details. Will the schematic attatched work? also will it be possible to get 1.5A from the system? That would total to 18watts, anything that needs to be changed?

author: Jon86
author_rank: Frequent Contributor
timestamp: October 20, 2013, 09:55:25 pm
content: I know I'm a bit late to the party, but just as a side-note, if you tie the USB data lines together and don't connect them to 5v, the device sees it as a charger and will pull the full current that it wantsThat's if you want to use it for charging usb devices, of course

author: rohan-04
author_rank: Contributor
timestamp: October 20, 2013, 10:14:04 pm
content: Quote from: Jon86 on October 20, 2013, 09:55:25 pmIs that so? so there is actually no need for the resistors and all that jazz?

author: jorgemef
author_rank: Regular Contributor
timestamp: September 09, 2022, 10:16:15 am
content: I too have a power supply with LM317 TO3 chinese kit. Works well when with 1,4A as long as VI-VO<15V.My transformer is 21V3A which after regulation goes about 30V. So I can only get 400mA when voltage is below 14v.I was wondering how to limit voltage and maintain the 1.4A current over the range. First considered using a lamp before LM317 which has non-linear resistance.But now I thought again and will perhaps limit the input voltage to the LM317 with a 7815/7818 before the LM317 which I can bypass with an external switch. So I have the option to have the full range covered with full range current.I also found this, which avoids external switch. Seems a better option, but maybe more expensive:https://www.eevblog.com/forum/projects/lm317-pre-regulator-for-wide-output-voltage-range/Cheers,Jorge

author: tunk
author_rank: Super Contributor
timestamp: September 09, 2022, 10:43:42 am
content: Probably wrong, but it could be that your heatsink/cooling is insufficient.

author: ledtester
author_rank: Super Contributor
timestamp: September 09, 2022, 10:49:17 am
content: Quote from: jorgemef on September 09, 2022, 10:16:15 amYou will still be dissipating a lot of heat at low voltages.Instead of a linear device like a 7815 try inserting a buck regulator (called a "preregulator") before the LM317. That will drop the voltage but dissipate a fraction of the heat. Even the cheap buck regulators you can get on ebay/aliexpress should work. There are trade-offs, such as adding more noise to the output.A more sophisticated design would have the output voltage of the preregulator set to the desired output voltage of the LM317 plus a couple of volts -- such an arrangement is called a tracking preregulator.

author: rdl
author_rank: Super Contributor
timestamp: September 09, 2022, 10:51:32 am
content: As a simple solution, you could just use some chunky power resistors in front of the regulator to drop the voltage input. You could even set up some kind of switching arrangement so you could still get the higher voltage if you need it. There's still gonna be some heat dissipated though.

author: MrAl
author_rank: Super Contributor
timestamp: September 10, 2022, 11:32:41 am
content: Quote from: rohan-04 on May 31, 2013, 12:43:10 pmA lot of replies so just a quick note...You can use a switcher, a buck, or if you really need a linear output then use a buck switcher front end with the linear LM317 backend. You get low power dissipation and good regulation with cleaner output.

author: geggi1
author_rank: Frequent Contributor
timestamp: September 10, 2022, 11:53:20 am
content: If you got issues finding a transformer with 15V secondary you can always use a shunt transistor capable of handling the higher dissipated power.See page 17 in the datasheet.https://www.ti.com/lit/ds/symlink/lm317.pdf?ts=1662791888714&ref_url=https%253A%252F%252Fwww.google.com%252F

author: jorgemef
author_rank: Regular Contributor
timestamp: September 15, 2022, 01:04:23 pm
content: Quote from: geggi1 on September 10, 2022, 11:53:20 amBut 15v transformer will be the same as if using outside transistor plus 15v zener to limit voltage drop on LDO at 15v right? And that configuration will benefit from integrated protection from the LM317, but with limited voltage range. Tested it and with 1,5A with outside transistor + zener, voltage drop on transistor stage is around 4V. This on page 17 is interesting, will give more few volts range, but then there is no current limitation for the high power transistor in case of short in the output. Could this configuration be combined with configuration on page 16 with a transistor pulling down LM317 adjust when current increases? Not much interested in DC/DC pre-stage as already have two other DC/DC power supplies and the EMI is a beast and goes through everything downstream.

author: jorgemef
author_rank: Regular Contributor
timestamp: September 22, 2022, 09:16:31 pm
content: DC/DC pre regulator is not so bad. With a buck converter we don't get so much EMI. I was used to flyback EMI which convert 310v to 12v and those do generate much EMI. But a buck to reduce from 30v to 10v or 15v generate very less EMI. Will incorporate one of those with a switch so I can have the option for higher current in the low voltage range.