Linear Power Supply Design - Calculating Transformer Power and Capacitor Values - Page 1
Forum section: electronics / beginners

author: paul_g_787
author_rank: Frequent Contributor
timestamp: October 17, 2021, 02:22:18 pm
content: I am looking at building a linear power supply for a project I am working on and thought this would be a great learning opportunity.I have a transformer here I bought some years ago to tinker with when I was learning about transformers which I am hoping to use.https://cpc.farnell.com/block/vb3-2-2-18/transformer-3-2va-2-x-18v/dp/TF01254It is: 3.2VAPrimary: 230VSecondary: split 2x18V)I have attached a schematic of my proposed design.I have assumed a min/max input voltage of 216V - 253V due to the statutory limits.Giving my a min output voltage of 33.8V and max output voltage of 39.6V.This is how I calculated the voltage of the smoothing capacitors and the rectifier diodes. Is this correct?However I do not know if this transformer will be powerful enough for my project so I need to calculate it and this is where I am a little stuck and need a little mathematical assistance.I essentially have four problems: How would I calculate the following?The maximum output current of this power supply.The input current at assuming the maximum output current.The value (µF) for the Class X capacitor across the primary coil.The value (µF)of the two smoothing capacitors C1 and C2.Edit:I have identified my first mistake. I did not take into account the correct voltages after rectification from AC to DC.The DC voltage will be the peak voltage of the AC signal (not the RMS voltage).So assuming a maximum 253V (RMS) on the primary, the secondary will be 39.6V AC (RMS).This is (according to the below link) 56V Peak. (2x28V).https://www.allaboutcircuits.com/tools/peak-voltage-calculator/1N4001 has a peak reverse voltage of 50V AC, so this is a good choice of diode.The 25V capacitors are too small.Even with the 2V drop from the rectifier (leaving us with 26V) so I have specified 35V.

author: TimFox
author_rank: Super Contributor
timestamp: October 17, 2021, 03:39:39 pm
content: For conventional transformer/rectifier/filter circuits, I recommend using "PSUD II" freeware from Duncan Labs.https://psu-designer-ii.software.informer.com/2.2/Note that when using this software, you must specify the transformer in terms of no-load voltage and equivalent resistance at the secondary. The information from Duncan includes how to estimate these parameters from the usual specification of voltage with specific load current. The load can be specified as a resistance, constant current, or stepped current (to estimate response to transient load current). Besides output voltage and ripple, it also gives results for filter capacitor ripple current and rectifier currents.

author: Ian.M
author_rank: Super Contributor
timestamp: October 17, 2021, 04:23:52 pm
content: 1N4001 has a grossly inadequate reverse voltage rating for your application. See Vishay's1N400n seriesdatasheet, max. ratings section.The transformer's no load voltage, is (according to itsdatasheet, page 3) 1.57 times its nominal, fully loaded voltage. Assuming the mains supply is 253V (i.e 230V+10%), we can also expect the secondary voltage to be +10%, so at no load, there can be (18V+18V)*1.1*1.57 = 62.2V RMS across the whole secondary widing. Neglecting Vf drop, at no load, the capacitors charge to the peak voltage, of 87.9V between the +VDC and -VDC rails, which is the peak reverse voltage 'seen' by the diodes. 1N4002 with 100V PIV would do the job with a little margin, but 200V PIV 1N4003 would be a better choice. If you have higher rated 1N400x series diodes in stock you might as well use them.Similarly, the capacitors need at least a 50V rating, and 63V would be more advisable.I don't know what you are planning on using this for but the transformer is only rated for 88.8mA RMS secondary current, which with your chosen rectifier and reservoir capacitor topology, only gives you 55mA DC current. The high no load voltage will also make it difficult to use simple three terminal regulators with it.

author: mariush
author_rank: Super Contributor
timestamp: October 17, 2021, 04:37:18 pm
content: The transformer is too small to be usable for a linear power supply. Well, you can do a max 100mA power supply, something like that.Its specifications are 3.2VA 2 x 18v AC windings.You can either use the windings separately, or you can use it as a single 36v AC secondary winding.With 36v AC secondary, 3.2VA means the maximum AC current Iac peak = 3.2 VA / 36V = 0.088A or around 90mAWith 18v AC secondaries you'll have 1.6VA on each winding, and still around 90mA on each : 1.6VA / 18V = 0.088AIf the secondaries are completely independent, you could parallel them to get lower secondary voltage but double the current, so you could have 18v AC and 0.177A AC current.When you rectify the AC voltage to DC using a bridge rectifier, you get a DC voltage with a peak value that can be estimated with formula:Vdc peak = sqrt(2) x Vac - 2 x voltage drop on diode inside bridge rectifier.Since you have a small transformer with low output current, you can assume the voltage drop on diodes inside the rectifier will be small, around 0.7v. If you had higher current like 3A or more, you go with 1v.So, let's go with assuming you're gonna parallel the secondary windings to get lower voltage and more current.Vac = 18v , Iac = 0.177AThen, Vdc peak = 1.414 x 18 - 2 x 0.7 = 25.4 - 1.4 = 24vWe can also use the formula Idc = 0.62 x Iac = 0.109A or 110mA to approximate how much current you can do on DC.The 0.62 value is a constant that works reasonably well with bigger transformers, let's say in the 25-200 VA range but it's fairly closeSo... so far you estimate Vdc peak = 24v and Idc = 0.11AHere you have to keep in mind two things:* at very low output currents, or with minimal load, a transformer will often output more than what it's designed. This is even more pronounced with very low VA value transformers.A 25-50VA transformer's voltage may be 5-10% higher than nominal (ex a 12v AC transformer may output 13v AC if there's nothing consuming power) but a <10VA transformer may have even 15-20% higher voltage without a load.So, your transformer may output 20v AC instead of 18v AC without any load on it.edit:See Ian.M's message above and the datasheet at page 3... seems this particular transformer is even worse, with 1.57 factor or 57% more than nominal voltage. May be worth adding a led and a current limiting resistor right after the bridge rectifier as a minimal load / power on indicator ... configure it for 5-10mA or thereabouts, to tame this transformer a bit.* your AC mains voltage is rarely exactly 230v - depending on how loaded the AC grid is, your voltage may be 215-220v in busy periods, or may be 240-250v in the middle of the night when few people consume power - the AC can be -5%..+10%It's important to keep this in mind because you can't expect the transformer to output exactly 18v AC and you have to allow for some variation when you do the math... account for higher voltage not only due to no load but also for high AC voltage.Going back to the calculation above, I determined Idc peak = 24v and Iac = 0.11A assuming 230v AC and some load.So as an exercise let's go with worst case scenario where you have 250v AC and you have 10% extra due to no load.Your transformer ratio is 230v : 18v = 12.77:1 so with 250v AC you'll have 250/12.77 = 19.577v ACAssuming a 10% variation for no load, you have 19.58x1.1 = 21.54v AC (worst case scenario 250v AC 10% no load extra)Now if we go in the formula for Vdc peak, you get Vdc peak = 1.414 x 21.54 - 2x0.7 = 29v peak DCLet's also do the other worst case scenario, where the AC voltage is low (215v) and you take 100% of current, so transformer doesn't give an excess voltage215v / 12.77 = 16.84v ACVdc peak = 1.414 x 16.84 - 2x0.7 = 22.4vNow, this is peak DC voltage ... and 100-120 times a second the voltage is gonna go down to 0v and back to peak DC voltage, unless you add some capacitance to smooth the output voltage.You can estimate how much capacitance you need using this formula : C = Current / [ 2 x AC Frequency x ( Vdc peak - Vdc minimum desired ) ]You estimated that the transformer can't give you more than 0.11A, so let's use 0.15A as a value for current.You're using a 230v transformer, so you're probably in a country that uses 50Hz mains frequency.Your peak DC voltage will be between 22.4v (worst case estimation with 215v AC) and 29v peak max but you can't rely on max, so let's be conservative and go with 22.5vNow the minimum voltage desired .. it's basically a trade-off between how much capacitance you're willing to introduce (big capacitors are expensive) and how much you're willing to let the voltage "sag"Let's say in this scenario, we're OK with a minimum DC voltage of 20v, which would allow us to use a linear regulator to get 1.25v..18v output.We can put the numbers in formula and get C = 0.15A / [ 2 x 50 x (22.5v - 20v) ] = 0.15 / 250 = 0.0006 Farads or 600uF minimum capacitance to stay above 20v all the time.Of course, you can go with 680uF or 820uF or 1000uF ... it's only gonna be smoother dc voltage, closer to 22-24v DC... but you know for this little current (0.15A) it would be pointless to go with more capacitance.Because you know you may have peak dc voltages of 29v or close to that amount, you also know now that you should use capacitors rated for at least 35v, ideally 50v.Had you used just the standard calculation for 230v, you may had assumed 24v dc peak means you can use 25v rated capacitors and over time, you may have damaged the capacitors and damage your power supply.

author: paul_g_787
author_rank: Frequent Contributor
timestamp: October 17, 2021, 04:43:00 pm
content: Quote from: Ian.M on October 17, 2021, 04:23:52 pmOK thank you for this information!! I just looked into the datasheet and the figures you provided and this is starting to make sense to me now!I just re-calculated, factoring in the un-loaded voltage and came up with the same figures as you did.

author: exe
author_rank: Unknown
timestamp: October 17, 2021, 05:09:03 pm
content: > The maximum output current of this power supply.In ideal case it's VA / V, so 3.2/18 = 0.177A (total from both channels). Then there is a derating (usually 50%), but you tranny is short-circuit proof (as says in the datasheet), so I'd say it doesn't need any derating. But it will have very poor regulation (=voltage will sag significantly under full load comparing to no-load voltage).BTW, in the datasheet, regulation is specified as 1.57. That means, no-load voltage is 18V*1.57V = 28Vac. After rectification that will be 28*1.44=~40V (all assuming 230Vac in). That's kinda close to maximum reverse voltage for your diodes. I'd propose 1N4002-1N4007.> The input current at assuming the maximum output current.Why do you need this one?> The value (µF) for the Class X capacitor across the primary coil.I don't think it is needed.> The value (µF)of the two smoothing capacitors C1 and C2.Usually for full-bridge rectifier it anywhere between 2000-3000uF per amp. In your case, I think, you need to double that because capacitors are only charged (but you better check that by, e.g., simulation). If you need a number, then I'd say 820-1000uF will do the job.PS whoops, I didn't notice other replies in the thread. Probably most of stuff I wrote was already mentioned

author: Terry Bites
author_rank: Super Contributor
timestamp: October 20, 2021, 07:48:42 am
content: This little calculator is nicehttps://www.changpuak.ch/electronics/power_supply_design.phpStep by step designhttp://www.skillbank.co.uk/psu/index.htm

author: Ian.M
author_rank: Super Contributor
timestamp: October 20, 2021, 02:38:07 pm
content: The trouble with web based calculators is archiving the design process so you can reproduce the results when you need to revisit the project several years later. Therefore I strongly prefer a SPICE simulation or spreadsheet, calculated and stored locally on my PC which can simply be archived with the other design files.You can make a rough sim of your transformer + rectifier topology inLTspice. The attached sim is parameterized so it can easily be modified for different transformers and has various .measure statements to report critical results like the expected ripple etc.Assumptions:The transformer's regulation can be modelled as a DC resistance referred to the secondary side, and the secondaries can then be treated as voltage sources.Worst case, the reservoir capacitors will be 20% under nominal value.You can find a 'near enough' SPICE model for the diodes in the bridge rectifierObviously, those assumptions aren't totally realistic, but are certainly good enough to get your design 'in the ballpark'.