Importance of input and output capacitance  for LDOs - Page 1
Forum section: electronics / beginners

author: kolonelkadat
author_rank: Regular Contributor
timestamp: October 10, 2014, 09:00:40 pm
content: I was looking in the datasheet for the ST 1117 im using and it says that the ldo is stable with 10uF on the output. Looking at the diagram, it shows 100nF on the input and 10uF on the output.The problem is that I dont have any 100nF caps. So I wonder, can I use one with higher capacitance? Specifically 100uF?

author: Simon
author_rank: Unknown
timestamp: October 10, 2014, 09:07:20 pm
content: yes and no, they are specifying 100nF because pretty much any 100nF capacitor will have very good ESR carachteristics which are required to keep the regulator stable. A 100uF capacitor may not have a sufficiently low ESR to be stable.try it, it may work, but don't assume more capacitance will automatically mean more stable.

author: kolonelkadat
author_rank: Regular Contributor
timestamp: October 10, 2014, 10:33:34 pm
content: Thanks for the input. My main concern is that it would make it more unstable over time. It seems to be working at the moment, but tomorrow, who knows.

author: SeanB
author_rank: Super Contributor
timestamp: October 11, 2014, 06:55:26 am
content: Also remember that your 100nF capacitor will most likely not be 100nF at the input voltage you have there, unless you are using a disc ceramic with leads. SM capacitors do have a very large variation with voltage, which gets worse the smaller they get. You can use a few smaller ceramics in parallel on the input right next to the regulator, which will then overall have lower ESR even if the single units do not.Often what is not disclosed on the data sheet what 100n cap they used, they might have specifies one that still is 80n at the input voltage, while your one might only be 10n at that voltage.

author: AKM
author_rank: Contributor
timestamp: October 11, 2014, 07:31:08 am
content: Almost any input capacitor >=100nF should work. Only the output capacitor usually shouldn`t have very low esr <0.1-0.5ohm depending on the ldo.

author: sleemanj
author_rank: Super Contributor
timestamp: October 11, 2014, 10:02:09 am
content: Quote from: kolonelkadat on October 10, 2014, 09:00:40 pmThe input cap shouldn't be too important, it's the output cap which is more critical for stability.The output cap needs to have a low but not very low ESR, ceramics are too low, electrolytics are too high, tantalum are about right, but people don't like tantalum for lots of good reasons.The solution is to use a ceramic capacitor, but to put it in series with a low value resistor (say 50 to 500 milliohm) to increase the ESR into the stable range.See this brief app note from TI on the subject:http://www.ti.com/lit/an/slva214a/slva214a.pdf

author: SeanB
author_rank: Super Contributor
timestamp: October 11, 2014, 11:14:22 am
content: Easier than a resistor is a "zero ohm" resistor, which will be in the right range. Not zero, not as controlled as a precision milliohm resistor but will work in most cases.

author: Kjelt
author_rank: Super Contributor
timestamp: October 11, 2014, 01:33:13 pm
content: Yes for certain LDO regulators the ESR of the output capacitor is critical BUT if so thismustbe mentioned in the datasheet with all specifications (for instance between 0,1 and 2 ohms).Now lets look at the LM1117 from TI:http://www.farnell.com/datasheets/1685500.pdfIt states clearly:QuoteNow lets look at the LD1117 from ST:http://www.farnell.com/datasheets/1776449.pdfNo mention about critical output capacitors or critical ESR requirements of this capacitor.The only mention about the output capacitor is:QuoteAlso the internal diagram of both LDO's differ slightly, so if you have the ST LD1117 unless I am very mistaken the output cap characteristics are not very critical, that said,I always make a testcircuit and look at the output voltage if it is stable under couple of temperatures and loads.

author: LukeW
author_rank: Frequent Contributor
timestamp: October 12, 2014, 12:55:08 am
content: Quote from: kolonelkadat on October 10, 2014, 09:00:40 pmI'm tempted to say the correct answer here is that you shouldget some 100nF caps. That's an extremely common value for this sort of thing, and you will want to have them next time, for your future projects, so it's worth getting some.Anyway, you've got your 10uF cap on the output, which is good... but the input capacitor is not so critical. You can probably just leave it out.Replacing a 100nF cap with a 100uF cap is a 1000x increase in the capacitance... although it probably wouldn't hurt for an LDO, as a general idea I would say that it's not an ideal replacement, in general I would try and find a closer match to the specified value in the circuit, although I understand you're only trying to use the components on hand.Do you have any other smaller ceramic (or plastic) caps in stock, eg. 1uF, 47nF, 220nF, or anything in that ballpark between 10nF and 1uF?

author: kolonelkadat
author_rank: Regular Contributor
timestamp: October 12, 2014, 01:21:31 am
content: Quote from: Kjelt on October 11, 2014, 01:33:13 pmOkay. I was worried I had either missed something or was just supposed to know. The more I looked at the datasheet, the more worried I was getting lol.Quote from: LukeW on October 12, 2014, 12:55:08 amat the moment Im using up the last scraps from my last order. This was my last 10uF cap and ive got maybe 5 more 100uF ones. At some point I will be placing a new order, but at the moment its just not in the budget.

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 03:09:32 am
content: I know this is an old topic, but pertains to what I'm wanting to know.sleemanji,that link you posted for the TI's app notes for substituting ceramic caps for the tantalum called out for in the data sheet, i'm not sure I quite understand that.Let's say I'm using an LM1117-5.0 fixed LDO voltage regulator.TI calls for a 10uF tantalum on its output, and one on it's input that is optional if far from the power supply input filter. It also says in the spec sheets that acceptable ESR ratings should be between 0.3ohm and 22ohm.Couple questions here. First off....If I'm understanding that app note correctly, I would simply use a 10uF ceramic capacitor, perhaps an X7R, (one on input rated for at LEAST my input voltage and one on the output rated for at LEAST the output voltage) and put in series a 0.3ohm - 22ohm resistor with the output and input capacitor from output of voltage regulator to ground.Or would simply putting a 100nF ceramic cap in parallel with a 10uF cap on both input and output be sufficient enough? I've looked at those cheapy 5.0v regulators sold on eBay, and that's what they are simply doing. However, I want my circuit to be done the right way as it may be drawing close to the 800mA output at times.Second...How far is FAR from the power supply filter? I will be using something like a walwart to deliver 12 volts/3A into the regulator (and another regulator also draws from that to put out another voltage). Total trace length will probably be less than 30mm or so from the walwart input to the regulator...however, I guess you'd need to add like 6-10ft for cable length of the walwart?.

author: amitchell
author_rank: Regular Contributor
timestamp: November 12, 2016, 05:45:07 am
content: Quote from: Falcon69 on November 12, 2016, 03:09:32 amYou would certainly want local bulk capacitance on your PCB, pre LDO. I am typically using 2-3 10uf X7R 1210 or 1206 50V on my 12VIN.I would be derating your ceramic capacitors by 3 times the voltage, the series resistor would count as ESR, but also voltage drop and heat especially at 800mA, does a Buck converter make more sense? How right is right here? how much output ripple is acceptable? Wallwart output can be pretty ugly, have you seen yours on a scope? all kinds of questions and answers.

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 05:54:32 am
content: I don't have a scope. This circuit just controls a few logic gates and some LED's. It's pretty basic. Nothing like High Frequency RF or anything.I guess I could just do a 6.3v 22uF tantalum on the output, but I'm worried about putting a Tantalum on the input. I've heard that any type of voltage spike (like accidentally shorting supply to ground) would make the tantalum explode or catch fire.It doesn't seem the input is that important, so maybe just keep it as a 10uF ceramic with a 100nF ceramic parallel on the input, and then just a 22uF tantalum on the output? Would that be fine?I do have a mosFET/Zener Diode protection circuit just before the voltage regulator to help protect against reverse polarity wiring if it happens, and of course a fuse.However, I just noticed also, that the other voltage regulator (LM2596) has a 220uF Electrolytic connected as well, that will be on that same input line going to the LM1117 and the LM2596, so I guess both regulators would see both the 220uF and the 10uF ceramic and the 100nF ceramic. Maybe too much?

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 06:07:18 am
content: ya, thats what I meant. Overvoltage protection. I also have the diode as you have mentioned. Tired, long day here.Okay, I'll go ahead and design it for that then. Thank You guys.But for future reference....If the data sheets says I need a Tantalum with an ESR value of 1.2ohms (for example), I can just use a Ceramic of same value (10uF for example) and just put a 1.2ohm resistor in series with it for that ESR value?

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 06:25:46 am
content: ahh, okay, understand. So I should get higher by at least 10% (if I'm using a 10% X7R capacitor) So a 10uF, I need to actually find a 10.1uF or more.Just looking at the buck converters. Cost is MUCH higher, and it looks like they will actually require the same amount of capacitors, but then would also require a resistor and an inductor, so more components, which would result in more footprint size as well as cost (not too mention the higher cost for the IC)So, the buck converters are not an option for this application. It's just too cheap and simple.

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 06:37:05 am
content: can you possibly recommend one? I need one that outputs 5volts at 800mA-1A and another that outputs 5.4volts at 2A-2.2A.Some of these aren't bad in price...like this one I'm finding...http://www.mouser.com/ds/2/115/AP65201-768589.pdfIt looks like that one uses ceramic caps. 22uF on both input and output.

author: mariush
author_rank: Super Contributor
timestamp: November 12, 2016, 06:39:29 am
content: They recommend tantalum in datasheets because (at the time those datasheets were made) they usually had higher esr compared to ceramic capacitors (around 0.5-1ohm , so they fit into that 0.3 ohm - 22 ohm window).Nowadays, electrolytic capacitors are good enough that you can find 10uF or 47uF or 100uF electrolytic capacitors with 10v-50v rating with esr below 1 ohm, in the past it was more difficult.Very low esr series like Panasonic FM or FR are borderline .. for example even 100uF 10v rated Panasonic FR is specified at 0.3 ohm impedance at 100kHz so maybe it wouldn't be smart to use themOlder series likeNichicon PWor Nichicon VR (meh, standard 85c only) have higher ESR values and are also cheaper, for example the 5mm x 11mm PW capacitors (10-47uF 25v rated for example, or 22-100uF 10v) have a specified ESR value of 0.6 ohm at 20c/100kHz (well, actually impedance, but at 100kHz impedance and esr are pretty much the same thing.. won't be the case with linear regulator but it's good enough indicato to use)Other low esr series that would most likely work would be Nichicon HE (mostly smallest sized ones in series) , Nichicon HD , Rubycon PX (85c standard), YXF , ZL , United Chemi Con LXY , Panasonic FC ...I wouldn't use tantalum capacitors in my circuits at all, if I can avoid it. I'd rather use a couple ceramic capacitors in series with a resistor if there's no vertical height for an electrolytic for example.@Falcon69 with 12v in , 5v out , I'd worry about heat dissipation especially at high currents. Remember, linear regulators waste the difference as heat, so at 800mA you'd have (12v-5v) x 0.8 = 5.6 watts. You'd definitely need a heatsink for it, and a surface mount 1117 won't cut it. I wouldn't use such a high input voltage in combination with a linear regulator for more than around 1w dissipated as heat (around 100-150mA) - 1w can be dissipated in the circuit board or in air or on a tiny heatsinkFor more than 0.5A, I'd definitely consider switching regulators .. you can find very tiny ones that only need tiny components like an surface mount inductor and maybe a diode and a couple ceramic capacitors to work, and they don't cost much.// edit : sigh... Warning - while you were typing7new replies have been posted. You may wish to review your post.guess i take a long time writing posts

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 06:45:19 am
content: okay Mariush. That makes sense about the heat. Yeah, it won't work then. I won't to try and keep everything surface mount.Blueskull. I think the most my circuit will ever see is 12volts for the input. Maybe 24volts, but doubtful. Depends on the walwart being used.So, I think I'm just going to plan for the 12volt.

author: salbayeng
author_rank: Frequent Contributor
timestamp: November 12, 2016, 06:55:21 am
content: One good way to test if your design will be stable in the future is to put in the freezer for half an hour, and see if it's stable when you first apply power (after removing from the freezer!). LM2931 is stable with 20uF ceramics at room temp, but needs 30uF to be OK at -20C.

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 06:57:12 am
content: Ya, I read that. Electrolytic caps are bad for that. They almost triple their ESR rating when -25C.

author: tatus1969
author_rank: Super Contributor
timestamp: November 12, 2016, 07:01:55 am
content: for a today's design, I would always use a regulator that is stable with "zero ohms" ceramics. Most modern LDOs support this. Pro: smaller, more reliable, better load step response, better RF rejection / EMI. And drop that old school 100nf's unless you choose different packages / ESL.

author: salbayeng
author_rank: Frequent Contributor
timestamp: November 12, 2016, 07:10:56 am
content: PTC's and Ferrite beads have very roughly 20mR(2Arated) and 40mR (5A rated) , these could be crafted into your circuit if you are using them elsewhere. They will degrade your load regulation by ~ 20mV per amp if in series with the regulator output.

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 07:19:15 am
content: thanks Blueskull. Those are nice chips, but, at 5 bucks a piece and 13 for the integrated inductor ones...abit out of my price range.Diodes Inc. seem to have alot of them for under $1.00 it seems. But kinda overwhelmed with which ones to go with.

author: mariush
author_rank: Super Contributor
timestamp: November 12, 2016, 07:31:13 am
content: And one more important note, as it seems like Falcon is a beginner.If you need something with 5v at 0.8A and 5.4v at 2.2A you don't want to size your regulators at exactly those values, you need to leave some room... for example you'd design for 5v at 0.9-1a and 5.4v at 2.4-2.5a.Also, at high currents, there will be some voltage drop on the cables between your dc regulator and your actual power consumer... so for example if you have something using 5.4v at 2A and the dc regulator is more than maybe 5-10cm away, you may want to configure the regulator to output a bit more than 5.4v, maybe 5.5v ... if you want to know how much you lose in wires alone, you figure out with the formula V = I x R ... get the wire resistance per meter , divide that by the amount of centimeters, multiply by 2 o get the total length (because you have + and - wires) and multiply by current to get your voltage drop.As linear regulators just waste the difference as heat, if you were to design a product that needs 5v at 1A and 5.5v at 2.5A , you'd need something able to provide 3.5A, regardless of voltage. The difference would be wasted as heat.So just a 12v 3A adapter wouldn't have worked.With switching regulators, everything would be more efficient. Depending on current output, most high performance switching regulators would work at around 85-95% efficiency.So let's say you use a switching regulator capable of producing 5.5v at 3.5A ... it will do this with 90% efficiency which means in order to "make" those 5.5v x 3.5A = 19.25 watts (let's just round it to 20w) , it will have to pull 20w * 100/90 = ~22 watts... at 12v input voltage, it's just a bit less than 2A... so your product may just work with a 12v 2A adapter.There are very good LDOs which only need 200-400mV above 5v to produce stable 5v from 5.5v so you get your 5v at 1A while wasting just 0.5w as heat, which can be very easily dissipated in the circuit board or just in plain air, and they take less space than a whole switching regulator.With this arrangement, 5.5v in and 5v out at 1A, your linear regulator will be about 91% efficient and a switching regulator may only be slightly better at 94-96% efficiency, so basically very little power saved. But in this case it's worth it as it's just one chip with maybe two ceramic capacitors, instead of chip+inductor+maybe diode+ceramic capacitors + maybe 2 resistors to set the output voltage .. in this situation the ldo wins because it's cheaper and takes less space and power savings are not worth going with switching regulator.

author: mariush
author_rank: Super Contributor
timestamp: November 12, 2016, 07:40:26 am
content: TPS54332 is only 2$ and can output up to 3.5A :http://www.digikey.com/product-detail/en/texas-instruments/TPS54332DDA/296-24141-5-ND/2000741ST ST1S41PHR is about 1.9$ and can do up to 4A :http://www.digikey.com/product-detail/en/stmicroelectronics/ST1S41PHR/497-13857-6-ND/4250778AP65403SP from DIodes Inc. is about 1.4$ and can do 4A as well :http://www.digikey.com/product-detail/en/diodes-incorporated/AP65403SP-13/AP65403SP-13DICT-ND/4810955and there's loads of others, just have to go to a site like Digikey or Newark/Farnell and use the filters and then also it's worth to check datasheets for "gotchas"...and it's important to note that in the case of such chips running at high frequencies layout of the parts around the switching regulator and the part choices made matters a lot. A badly laid out circuit can be unstable or be less efficient than expected.

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 07:54:17 am
content: okay, makes sense.Each regulator may not draw full power both at same time, in fact it is VERY unlikely I will even pull close to half of the rated values I stated. I haven't yet calculated it. I just calculated up everything as if it had all turned on (which is impossible, as for example, only one LED is on while the other is switched off, and vice versa).So, I should get a switching regulator that can handle about 4A (to be on the safe side), then get a low drop out voltage regulator that only has a drop-out of like 0.3-0.4 volts so it can produce the 5 volts from the 5.4 volts the switching regulator will put out?Now, these will be on the PCB, the walwart (12 volt wall plug) gets plugged into the wall, then via a barrel connector on the PCB board, it connects to the PCB supplying power to the switching regulator, then to the circuit and the LDO regulator.SO, just reading up on all these buck converters, it looks like the majority of them don't even use Tantalums. It seems almost all of them are asking for the Ceramic Caps.Heh, I was just looking at that Diodes AP65403SP one. .86 per 100 from Mouser. that's under a buck! (lol, get it, a buck?!)And yes, been reading about that frequency stuff. The Vishay Data sheets are very informative on it and explains where each of the components should go...I think if I use those same methods and apply to the Diodes chip..I should be okay...however, this is not a 4-layer board, it's only a 2-layer.http://www.mouser.com/ds/2/427/sip12116-364719.pdf<--- Page 12

author: DimitriP
author_rank: Super Contributor
timestamp: November 12, 2016, 09:10:31 am
content: Capacitor discussions are all fun and great, but for just under $11 you can get 10 of these and have some spares too...http://www.ebay.com/itm/10-PCS-LM2596-DC-DC-3A-Buck-Converter-Adjustable-Step-Down-Power-Supply-Module-/182202896686?hash=item2a6c23912e:g:AzwAAOSwAPlXhEYE

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 09:17:29 am
content: yup. I know, but I want these on the pcb board themselves with the rest of my circuit. And I'm finding out that the LM2596 is not very efficient, why I am here. Try and find a better solution, using only ceramics (no tantalums for obvious reasons, or electrolytics because of size), and also lower the cost and reduce heat from the IC chip when operating at almost full 3amps or so my circuit may draw.Those on eBay are also fakes. They do not conform to the actual specs the LM2596s is suppose to conform to. There's a video on it.

author: amitchell
author_rank: Regular Contributor
timestamp: November 12, 2016, 05:30:27 pm
content: Have you played around with TI WEBENCH? The power architect may interest you.http://www.ti.com/lsds/ti/analog/webench/overview.page?DCMP=sva_web_webdesigncntr_en&HQS=sva-web-webdesigncntr-vanity-lp-en

author: Falcon69
author_rank: Super Contributor
timestamp: November 12, 2016, 08:23:18 pm
content: yes, I played around with that for awhile. But it seems you can only use Ti's chips? Unless I'm missing something. I tried to figure out how to change it to use the Vishay or Diodes regulators, but couldn't figure out how.The chip blueskull mentioned on previous page is about $3 each when purchased in 100. That's too much for this application. I will need to use the cheaper Diodes ones at 86 cents each if I can.