Calculating for Adj Voltage Regulator - Page 1
Forum section: electronics / beginners

author: Falcon69
author_rank: Super Contributor
timestamp: October 12, 2014, 05:52:00 am
content: Okay, My first attempt at this, so be gentle.I am looking at purchasing thishttp://www.st.com/web/en/resource/technical/document/datasheet/CD00001883.pdfto achieve a 5.2v output.I'm confused abit on the voltage reference. I am guessing it is the R1 120 ohm resistor they have in the picture fig. 3.If it is, then the calculation should be:5.2v = 120 * (1+R2/120)5.2/120 = 1 + R2/120(5.2/120) - 1 = R2/120[(5.2/120)-1]/120 = R2R2 = 125.2 ohmDid I do that correctly?Also, 10uF capacitors on input and output, and particular kind? Just 50v Ceramics?

author: Falcon69
author_rank: Super Contributor
timestamp: October 12, 2014, 06:01:05 am
content: Oh, I also wanted to ask, It's a 3A regulator, those resistors, what wattage? 1/2, 1/4, 1/10?If they are just used for reference so the voltage regulator knows what voltage to put on its output, Then there shouldn't be any current, or very little, flowing through them, correct? In that case, I can use 1/10w.

author: sleemanj
author_rank: Super Contributor
timestamp: October 12, 2014, 06:08:33 am
content: VRef is a reference VOLTAGE ("V", you see), generated internally to the device, 1.2v, you can see in the diagram that VRef is seen measured across the resistor indicated, this is what causes the regulator to regulate, it "does what it needs to" to make the VRef inside match the voltage across R1.

author: sleemanj
author_rank: Super Contributor
timestamp: October 12, 2014, 06:13:47 am
content: Quote from: Falcon69 on October 12, 2014, 06:01:05 amUse Ohm's law. If you know the output voltage of your regulator, and you know the resistance of those resistors, you know the maximum current that must be flowing through them to ground.

author: Falcon69
author_rank: Super Contributor
timestamp: October 12, 2014, 06:18:22 am
content: sleemanjI don't see where you are seeing this 1.2v.If you scroll down to look at the revision changes, revision number 14 it says they removed the 1.2v ref. So how do we know what the Vref is for this chip to figure out the resistor value?

author: Falcon69
author_rank: Super Contributor
timestamp: October 12, 2014, 06:26:09 am
content: ohms law says 5.2v*3A = 15..6 watts! I don't think that is right. I've seen these voltage regulators with small resistors on them. 15w resistors are pretty huge.

author: free_electron
author_rank: Super Contributor
timestamp: October 12, 2014, 06:38:36 am
content: THe chip wants to see 1.2 volts across the 120 ohm resistor.so it is ging to adjust its output to try to get that.1.2 volt across 120 gives us 10 milliampere. that is the current that needs to flow in that branch.5.2 volts / 10 milliampere = 520 ohms. we already have 120 ohms. so the remaining resistor is 400 ohms.in reality the chip really wants to see 1.25 volts.1.25/120 gives you the current .5.2 volts / current gives you total resistance . subtract 120 and you are there.

author: Falcon69
author_rank: Super Contributor
timestamp: October 12, 2014, 06:59:50 am
content: ahh, that makes sense. thank you free_electronSo I would need a 379.2 ohm resistor for R2 to get 5.2 volts and a 10watt resistor (0603) is more than enough, in fact, nearly double then what is needed.Thanks!

author: sleemanj
author_rank: Super Contributor
timestamp: October 12, 2014, 07:21:22 am
content: Quote from: Falcon69 on October 12, 2014, 06:26:09 amTable 7 on page 11 details it (well, as FE says, it's more like 1.25v, but anyway, 50mv is neither here nor there)The revision history does not mean they "removed a 1.2v Reference", it (probably) means there used to be a 1.2V fixed output version of the regulator, which they no longer produce.Quote from: Falcon69 on October 12, 2014, 06:26:09 amErr, no, how on earth...I don't even.If you use a 400R resistor for R2, the formula is thereby....1.2 * ( 1 + (400/120))Work that out, and what do you get... tada, 5.2vYou can apply Ohms law then...5.2 = (120+400) * I5.2 / 520 = II = 0.01A, 10 milliamps flow through the resistors R1 and R2Then you can calculate power for each resistor, I * I* R = P120R * 0.01*0.01 = 0.012W, 12 milliwatts400*0.01*0.01 = 0.040W, 40 milliwatts

author: Falcon69
author_rank: Super Contributor
timestamp: October 12, 2014, 07:34:53 am
content: Ya, I still dont see it. Very confused. Table 11 is on page 21. Page 7 has the electrical characteristics, Table 3 on it. But I don't see any Vref on it.On the first page, it says Dropout at 3A is guaranteed at maximum 1.2V. Is that the Vref? But, then above, it says Dropout Typical at 1.3V at 3A. I'm so confused. Is this just a poorly written data sheet? How can you have guaranteed maximum dropout at 1.2v, then say it's typical at 1.3v for same amperage?

author: Falcon69
author_rank: Super Contributor
timestamp: October 12, 2014, 07:39:32 am
content: FOUND IT!I had the page and table numbers reversedThanksEDIT: I see you edited your post, so maybe YOU had the numbers swapped. LOL. Sneaky

author: sleemanj
author_rank: Super Contributor
timestamp: October 12, 2014, 07:42:53 am
content: Quote from: Falcon69 on October 12, 2014, 07:34:53 amDo not simply believe the broad statements in datasheet summaries.

author: IanB
author_rank: Super Contributor
timestamp: October 12, 2014, 07:51:20 am
content: The information you need is in the diagram on page 6 and the table on page 11.In particular note the formula:Vo = Vref (1 + R2/R1)where R1 = 120 ohm (per diagram) and Vref = 1.25 V (per table on page 11).If you want Vo = 5.2 V, then:5.2 = 1.25 (1 + R2/120)Therefore:R2 = 120 (5.2/1.25 - 1) = 379.2 ohmsRounding up, you need a 380 ohm resistor.

author: David Hess
author_rank: Super Contributor
timestamp: October 12, 2014, 02:23:53 pm
content: The resistor issue is pretty well covered above so I will limit myself to the capacitor issue.The LD1085 is obviously a clone of the Linear Technology LT1085 so you might want to peruse the Linear Technology datasheet for further edification:http://www.linear.com/docs/3741The LT datasheet does not quite entirely cover the issue but does make recommendations for the capacitors which do *not* include film or ceramic capacitors. This is because the low ESR of these capacitors will lower stability and may in some cases cause problems. With some cleverness though, you can probably get away with using them anyway. I would just follow the recommendations on page 9 of the LT datasheet which agree with the ST recommendations such as they are.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: October 12, 2014, 10:41:40 pm
content: QuoteGot a link?, they might come in handy.

author: Falcon69
author_rank: Super Contributor
timestamp: October 13, 2014, 07:03:18 am
content: Thanks David. I will look into it.And HackedFridgeMagnet, I meant to say 1/10W, my bad.

author: Falcon69
author_rank: Super Contributor
timestamp: October 13, 2014, 07:15:43 am
content: Looks like I will need to change the output to a 50uF Aluminum one for the cap instead of the 10uF ceramic I had. Aluminum much cheaper then a Tantalum. That data Sheet you Linked Davis says a 10uF Tantalum or a 50uF aluminum is good.

author: Falcon69
author_rank: Super Contributor
timestamp: October 13, 2014, 07:23:35 am
content: Okay, Can't find an SMD package for a 50v 50uF Aluminum Capacitor. Only can find 47uF. And what should the Ripple Current range I should look for?Any other ideas? Maybe a Ceramic and a 1ohm resistor in series?

author: sdg
author_rank: Regular Contributor
timestamp: October 13, 2014, 07:33:16 am
content: Quote from: Falcon69 on October 13, 2014, 07:23:35 amLook for a bigger one, like 56 µF or more.I think Nichicon makes 56µ/50v SMDs

author: Falcon69
author_rank: Super Contributor
timestamp: October 13, 2014, 07:33:49 am
content: oh, it's okay to go bigger then the 50v they recommend?

author: sdg
author_rank: Regular Contributor
timestamp: October 13, 2014, 07:53:19 am
content: The average electrolytic cap is +-20-30% anyway, so 56 or 47 µF instead of 50 µF shouldn't make any difference in this situation.And if your PS outputs up to 5 volts, perhaps a 20v cap on the output side should be enough. The input cap ratings depend on the peak input voltage your regulator will see.I use electrolytic caps rated for twice the peak voltage they will ever see, but that's a rule out of thin air and I have no technical basis for that (other than I hate it when electrolyte drips on the carpet).

author: Falcon69
author_rank: Super Contributor
timestamp: October 13, 2014, 08:06:10 am
content: Okay, cool. If I don't need but maybe twice my output (5.2v) then maybe this one will work? It's a 1206 casehttp://www.mouser.com/ProductDetail/AVX/TAJA106K010RNJ/?qs=%2fha2pyFaduhD94bTpKsWaWa9qnAGCC4jTqAM%252bZo1%2fgomwsac%2fMNfrw%3d%3d

author: David Hess
author_rank: Super Contributor
timestamp: October 13, 2014, 11:56:13 am
content: Quote from: Falcon69 on October 13, 2014, 07:15:43 amA 10uF ceramic output capacitor is one of the configurations I would expect to work poorly causing the regulator to oscillate because of excessively low output ESR. Some regulators especially including low dropout ones will definitely fail in this way.The usual solutions to this are to either add a small value resistor in series with the large value ceramic capacitor to control its ESR or to use a small ceramic capacitor, say 0.1 uF, in parallel with the larger aluminum electrolytic capacitor as an alternative to a more expensive solid tantalum capacitor.That ratio of about 4 to 1 of aluminum electrolytic capacitor to solid tantalum capacitor is no accident. For a given capacitance, good aluminum electrolytic capacitors have an ESR about 4 times higher than the same value of solid tantalum so if you want the same ESR which has a direct effect on output ripple, the aluminum electrolytic capacitor needs to have a value which is about 4 times larger.This ignores that aluminum electrolytic capacitors perform worse at high frequencies than solid tantalum capacitors which is why a smaller valued ceramic capacitor is often used in parallel with them.