Is a 6.3 V AC transformer secondary enough to produce accurate regulated 5 V DC? - Page 1
Forum section: electronics / beginners

author: SharpEars
author_rank: Unknown
timestamp: October 17, 2014, 03:21:36 pm
content: The subject pretty much says it all. If I have a transformer that has a 6.3 V secondary, post bridge conversion to DC and with a linear regulator in the circuit, is it possible to produce an accurate 5 V fixed DC output?Or is 6.3 V not enough to do that with all of the voltage drops in the bridge (to DC conversion) and linear power regulation portions.The reason I am asking is because 6.3 V seems to be a very common secondary and using a 12.6 V secondary will cause a linear regulator to potentially produce way too much heat due to the 12.6 -> 5 V differential.

author: Simon
author_rank: Unknown
timestamp: October 17, 2014, 03:26:24 pm
content: You voltage from the rectifier will peak at around 7.5V so if you have a large smoothing capacitor and your taking a small amount of current you will get away with it but otherwise 6.3V - 1.4 rectifier = 4.9V although you can get diodes with less voltage drop to give a bit more headroom. With careful component selection and an LDO regulator you should just manage it.

author: Richard Crowley
author_rank: Super Contributor
timestamp: October 17, 2014, 03:31:05 pm
content: In theory, a 6.3V AC transformer output will produce 8.8 V DC which should be adequate for your garden-variety 3-terminal regulator (such as 7805, etc.)

author: SharpEars
author_rank: Unknown
timestamp: October 17, 2014, 03:35:39 pm
content: OK, so let's say we go with a 12.6 V transformer secondary instead to avoid these issues. Isn't it going to be difficult to design or use an existing linear regulator efficiently, since there is a large disparity between the input voltage (12.6-1.4=11.2V post bridge rectifier) to the linear regulator and the output voltage of 5 V - a difference of 6.2 V.

author: SharpEars
author_rank: Unknown
timestamp: October 17, 2014, 03:36:22 pm
content: Quote from: Richard Crowley on October 17, 2014, 03:31:05 pmI don't understand how the DC voltage can be higher than the AC voltage, especially when 1.4 V will be subtracted by the bridge rectifier alone in the conversion from AC to DC.

author: Monkeh
author_rank: Super Contributor
timestamp: October 17, 2014, 03:38:24 pm
content: Quote from: SharpEars on October 17, 2014, 03:36:22 pmAC voltage is generally (in this case, is.) specified RMS. Vpk is greater than 6.3V, so with adequate smoothing capacitance you get a higher DC voltage than AC.So long as you apply enough capacitance and don't attempt to reach the full power rating of the transformer, you will easily achieve >7VDC, allowing a normal LDO or even a crappy old 7800 series regulator to operate normally.

author: SharpEars
author_rank: Unknown
timestamp: October 17, 2014, 03:43:58 pm
content: Quote from: Monkeh on October 17, 2014, 03:38:24 pmWith 1.4 V taken out by bridge rectification pre-regulation and an adequate capacitor, we are at 7.5 V DC. I thought the 7800 series would need a Vin of 8 V (i.e., Vout+3V)?Correction:Vdrop is 2 V, so the math works out...

author: Simon
author_rank: Unknown
timestamp: October 17, 2014, 03:47:04 pm
content: You AC voltage is specified in RMS the peak is 1.4x the RMS so peak voltage out of your rectifier is (AC RMS x 1.4) - Bridge voltage drop, that voltage will be the maximum available on the capacitor. But if you fully load the output the voltage will be dragged down to VRMS - Vbridge

author: SharpEars
author_rank: Unknown
timestamp: October 17, 2014, 03:48:37 pm
content: Quote from: Simon on October 17, 2014, 03:47:04 pmSo, given a 10 amp secondary rating, how close can I get to 10 amps, before I don't have the 7 V I need?

author: Monkeh
author_rank: Super Contributor
timestamp: October 17, 2014, 03:49:09 pm
content: Quote from: SharpEars on October 17, 2014, 03:43:58 pmIt's not a fixed voltage. But okay.6.3 * 1.414 = 8.9V, - a generous 2V = 6.9V. More than adequate for an LDO.

author: SharpEars
author_rank: Unknown
timestamp: October 17, 2014, 03:50:06 pm
content: Quote from: Monkeh on October 17, 2014, 03:49:09 pmAre there low part count LDOs that can accurately do 10 amps at 5 V (i.e., within 0.01%)? Hopefully a regulator IC...

author: Monkeh
author_rank: Super Contributor
timestamp: October 17, 2014, 03:51:08 pm
content: Quote from: SharpEars on October 17, 2014, 03:50:06 pmYou can make one. But if you want 10A, use a switcher.Also, don't constrain your regulator choices to the 78xx range. They're ancient and crap. Even the 1117 is a more attractive choice, and it's ancient and crap, too.

author: SharpEars
author_rank: Unknown
timestamp: October 17, 2014, 03:51:58 pm
content: Quote from: Monkeh on October 17, 2014, 03:51:08 pmI want low (ripple) noise and efficiency is not an issue, so it must be linear...

author: Monkeh
author_rank: Super Contributor
timestamp: October 17, 2014, 03:53:43 pm
content: Quote from: SharpEars on October 17, 2014, 03:51:58 pmYou can get low noise with a switcher.10A is going to require discrete transistors.

author: mariush
author_rank: Super Contributor
timestamp: October 18, 2014, 04:12:01 am
content: So let's recap.You have an AC voltage. The basic idea is to convert it to a DC output using a bridge rectifier (or four independent diodes), then use a capacitor to smooth out what comes out from the bridge rectifier, and then you can use a voltage regulator to get exactly 5v or whatever you want.Whenever you rectify the AC to DC, the energy goes through two diodes, so there's always going to be a loss equal to 2 times the voltage drop on an individual diode in the bridge rectifier.For low currents (let's say 100-300 mA), you could use some Schottky diodes which can have a voltage drop as little as 0.1-0.3v. Bridge rectifiers made for high currents can have a drop of about 1-1.2v per diode.Then, depending on the capacitor size, the amount of voltage you will get will vary. A simple formula that approximates how much capacitance you need is : C = Current / ( 2 x AC Frequency x Vripple) where Vripple is how much you're willing to let the voltage sag.Then, you have all kinds of linear regulators, or switching regulators... some are better than others. Linear regulators typically need to have the input voltage higher than the output voltage by some amount, in the case of a classic chip like 7805 you're looking at about 2v.Other classic regulators like 1117 only need about 0.8-1v to work properly, but they can only handle a lower amount of current, in the case of 1117 it's about 1A compared to 1.5-2A for the 7805.There are also other linear regulators like (for example)LT1085 / LT1084 / LT1083which can handle up to 3A / 5A / 7.5A with just 1V drop but when you go above around 2-3A, it's not a good idea to use linear regulators because of the power dissipated in the regulator which would force you to use a big heatsink.With linear regulators, everything between the input voltage and output voltage is converted to heat and must go somewhere... for example if you have 8v at input and want 5v at the output at 2A, that means the regulator will produce (8v - 5v) x 2A = 6w of heatSo let's look at an example... you have a 6.3v AC transformer and you want 5v @ 0.5A .This means the peak DC output before the losses in the bridge rectifier will be equal to 6.3v x 1.414 = 8.9v DC.Now, I could use a bridge rectifier but considering it's only 9v peak and 0.5A, I'm looking at some diodes like the classic1n5817-1n5819and I'll go with the middle one, 1n5817 which can handle up to 20v and 1A and have a typical voltage drop of about 450mV at 1A (scroll down in datasheet, value is there).The voltage drop is important.. but note it's specified at 1A and it's "typical" so depending on your luck, the actual drop may be a bit higher, or a bit lower, at 1A. As our maximum current is only 0.5A, we're obviously not going to get close to 450mV drop @ 1A, so the voltage drop on the diode will be smaller than that figure of 450mV. But just to keep things simple, let's say the voltage drop per diode is 500mV.With this in mind, we now know that the peak DC output after the bridge rectifier made out of 4 x 1n5817 diodes will be 6.3v x 1.414 - 2 x 0.5v = 7.9vNow you need to pick a capacitor to smooth out this dc output... the more capacitance, the most time the output voltage will be closer to that peak of 7.9v .. BUT there's no point using a very large (and expensive) capacitor to keep the voltage high, if the linear regulator only needs a bit of voltage above the 5v it needs to output. So you have to pick your linear regulator, determine the minimum input voltage that regulator needs to output 5v properly, and then choose a smoothing capacitor accordingly.In this case, as I have a maximum of 5v @ 0.5A, I can use aLM1117linear regulator, which is designed to output up to 0.8A of current with only about 1.25v drop at 0.5A (it's all in those tables in the datasheet).So, the regulator will need 5v + 1.25v = 6.25v at minimum to output 5v properly, and we have a dc voltage with a peak of about 7.9v. To be very safe, let's choose the smoothing capacitor to make sure there's always at least 6.5v DC to the linear regulator, which means we can afford a ripple voltage of 7.9v - 6.5v = 1.4vNow we can use that formula C = Current / ( 2 x AC Frequency x Vripple) = 0.5 A / ( 2 x 60 Hz x 1.4v) = 0.5 / 168 = 0.002976 Farads or 2976 uF. The next standard value would be 3300uF, so a 3300uF 16v rated or better would be perfect.--For 5v and 10A, if you really want linear regulators I suppose you could use two LT1083 regulators in parallel (see page 14 in datasheet), but you'd still need to feel them with at least 6.5v (due to the dropout voltage of 1.5v) and you'd need a bridge rectifier capable of at least 10A which means the voltage drop inside the bridge rectifier will be much larger) and also you'll need much much larger smoothing capacitors.So the best idea would be to use a transformer designed for more than 6.3v AC, for example let's say 9v AC.In that case, you'd have a peak dc voltage of about 9v x 1.414 = 12.7v , minus about 1.2v x 2 = 2.4v in the bridge rectifier , so about 10v. Now determine capacitance size to make sure there's always 6.5v even at 10a... so C = 10A / [ 2 x 60Hz x (10v - 6.5v) ] = 10 / 120x3.5 = 0.02380 Farads or 23800 uF so something bigger than that would work, like 27000 uF or 2 x 15000 uF for example.

author: IconicPCB
author_rank: Super Contributor
timestamp: October 18, 2014, 05:00:52 am
content: A lot will depend on how the secondary voltage is specified; at rated secondary load or no load condition.If no load then a lot will depend on transformer design ( regulation ) and how the secondary voltage reduces with applied load.Suck it and see. If it does not provide the necessary voltage then go to a higher rated transformer.

author: chickenHeadKnob
author_rank: Super Contributor
timestamp: October 18, 2014, 07:12:49 am
content: Quote from: SharpEars on October 17, 2014, 03:36:22 pmIf you need more voltage headroom and cost isn't a concern linear technologies makes something they call an "Ideal diode bridge" .link to LT4320:http://www.linear.com/product/LT4320It switches 4 mosfets that you supply to work as diodes. The drop across the two fets is as good as the Rds(On) of the fets. Less drop less heat.search the forum for the thread on this part number. A member is using them in production.

author: Zero999
author_rank: Super Contributor
timestamp: October 18, 2014, 08:14:34 am
content: What about using the LM431 with a MOSFET?The LM431 can be biased to a higher voltage using a voltage doubler. It needs to be a logic level MOSFET (one that will pass the full current with neglidgable voltage drop given a gate voltage of 5V).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 18, 2014, 09:00:27 am
content: ^ That will work, though I'd note, be weary of oscillation; the TL431 is an op-amp, not a "programmable zener", despite what marketing typically calls it.OP: what the heck do you need 0.01% at 5V, 10A for? Not only is there no application that actually *needs* that, there is no application that can use that: the contact resistance of the best gold-plated banana jacks drops several orders of magnitude more voltage than such a tolerance!Or to put it another way, if you can show us the application that seems to require such precise regulation, we may be able to suggest accommodations to relax that requirement.Tim

author: kxenos
author_rank: Frequent Contributor
timestamp: October 18, 2014, 09:15:31 am
content: An other thing that you have to take into consideration is the regulation of the transformer and the voltage input variation. For example, if you have a 230V/6,3V transformer, for a 10% drop in line voltage you will have a 10% drop in the output voltage. So if you want input 230V±10% Then your lowest voltage will be 5.67V. About transformer regulation, trasformer manufacturers will specify in their datasheet the regulation of their transformers, typically about 80%. That means that in the maximum load the real output will be 80% of the nominal output. For example If you have a 230V/6.3V 3VA transformer with specified regulation of 80% that means that for the maximum output current of 3VA/6,3V = about 0.5A the output voltage will be 0.8*6.3 = 5.04V. And for the same 10% drop in mains the lowest voltage out of the transformer will be about 4.5V.So, I think that you should look for a higher voltage one where the only drawback is the power loss on the regulator.Keep also in mind that for power calculations you will have to calculate for the maximum accepted line voltage for example 230*1.1 = 253V

author: kxenos
author_rank: Frequent Contributor
timestamp: October 18, 2014, 09:25:39 am
content: Now I saw that you need 10A. I think you should start looking for a AC-DC SMPS with an output of 9V and a linear post-regulation stage with 2 x LM338K in parallel.

author: Simon
author_rank: Unknown
timestamp: October 18, 2014, 09:29:45 am
content: the Op has not posted in a while and we still don't know what transformer he has and what he is trying to power, that would help greatly in coming up with a solution.Also to note because of the horrendous power factor correction of a rectifier and capacitor a transformer of more than 10A capacity is recommended - No engineering is never that simple, but that's what makes it fun.

author: Zero999
author_rank: Super Contributor
timestamp: October 18, 2014, 01:13:18 pm
content: Quote from: T3sl4co1l on October 18, 2014, 09:00:27 amI see what you mean but I suspect it has a fairly low risk of oscillation because the MOSFET is configured as a source follower, the gate capacitance will be lower than expected.Quote from: kxenos on October 18, 2014, 09:15:31 amAre you sure? In my experience it seems to be the other way round. Transformers are specified under 100% load conditions, so when there is no load connected, the output voltage will be higher than the rated value, thus a 6.3V transformer with 80% regulation will have an unloaded output voltage of 6.3/0.8 = 7.875V.

author: Dave
author_rank: Super Contributor
timestamp: October 18, 2014, 01:41:30 pm
content: Quote from: Hero999 on October 18, 2014, 01:13:18 pmCorrect, unloaded transformers will have a higher output voltage.

author: kxenos
author_rank: Frequent Contributor
timestamp: October 18, 2014, 02:08:41 pm
content: Quote from: Dave on October 18, 2014, 01:41:30 pmYou are both right..