Trying to figure out transformer's max current rating - Page 1
Forum section: electronics / beginners

author: Frankentronics
author_rank: Regular Contributor
timestamp: April 30, 2020, 09:11:04 pm
content: Greeitngs,Is there a way to figure out the maximum current that a transformer can handle?I tried to Google a datasheet and was unable to find one.The transformer has a company logo ETM and the following number, CY0906VN.The output of the transformer is roughly 12 V AC.Is there a test that I can do to figure out the maximum amp rating?Thanks...

author: David Hess
author_rank: Super Contributor
timestamp: April 30, 2020, 10:13:29 pm
content: I usually get an estimate by measuring the cross section of the wire used for each winding.

author: Frankentronics
author_rank: Regular Contributor
timestamp: April 30, 2020, 10:47:52 pm
content: Quote from: David Hess on April 30, 2020, 10:13:29 pmAn estimate is good enough.It's hard to get to the windings, but I was able to measure .15mm for the primary winding and I can kind of estimate the thickness on the secondary winding to be around .30mm. As you can see in the photo, I am holding a strand of known thickness, 0.2mm, next to the exposed secondary winding and I can see that the wire of the winding is a bit thicker than the known .20mm strand I am holding against.From a practical standpoint, I want to connect the rectified output of this transformer to three voltage regulators, 7812, 7809 and 7805. I think it's best to connect the voltage regulators in parallel. I don't even anticipate all the outputs to be used simultaneously, but if I even use two of them simultaneously, is it reasonable to assume that 500mA of current through each regulator, simultaneously, would be fine?Oh, also, should I use 1N4004 diodes for the bridge rectifier?Thanks...

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 30, 2020, 10:58:52 pm
content: Quote from: Adinol on April 30, 2020, 09:11:04 pmCompare the size of the transformer that you have with similar sized transformer on Digikey or Mouser. This will give you an idea of the VA rating.Divide the VA rating by the voltage to get the approximate current.Regards,Jay_Diddy_B

author: Ian.M
author_rank: Super Contributor
timestamp: April 30, 2020, 11:36:17 pm
content: I doubt a 12V transformer will be satisfactory to feed a 7812 regulator, as unless its a very large one, it wont have enough voltage headroom to avoid ripple breakthrough at significant load current. Also, note that for a full wave bridge feeding a reservoir capacitor, the required RMS secondary current rating is approximately half as much again as the DC load current.A photo of the transformer, dimensions of the core and its winding apertures*, weight, and also the primary and secondary winding resistances (measured as accurately as possible#) would be helpful in figuring out the transformer's capabilities. We'd also need to know the turns ratio, which is more problematic, although it can be approximated from the unloaded primary to secondary voltage ratio. So-far assuming 120V in and 12V unloaded out, it would appear the turns ratio is 10:1. Exact unloaded voltage measurements would be helpful. Also, repeating the voltage ratio measurement in reverse with the nominal primary unloaded and a line frequency low voltage (<12V) AC source connected to the nominal secondary would help further refine the exact turns ratio.If its possible to add a temporary ten turn winding of fine magnet wire round the center limb of the core, without damaging the other windings or insulation, then apply mains to the primary and measure the primary voltage and the voltages across the temporary winding and original secondary, you can calculate the primary and original secondary turns counts with a fair degree of accuracy, and (making some assumptions about the purity of the copper, grade of iron used for the core and a wild-ass guess at the temperature rating of the insulation on the windings) from that and the other data gathered, characterise the transformer fairly accurately.However if you don't mind a potentially destructive test, after confirming its physically big enough to be likely to have an adequate rating, it may be simpler to load the transformer with a 1.5A load, check the secondary voltage remains above 12V and 'smoke test' it for half an hour at that load current then check the transformer for excessive temperature rise. Fuse the primary circuit at no more than 1A to prevent excessive excitement if it fails!*... or if toroidial, inner and outer diameters and height.#You need a four wire ohmmeter to get an accurate measurement of the secondary resistance - If you don't have one, improvise with a 100mA DC current source checking and noting the exact current under load. Depending on the capabilities of your meter you may also need a x10 or even x100 OPAMP amplifier to boost the developed voltage to a significant proportion of the FSD of the lowest range of your meter.

author: Manul
author_rank: Super Contributor
timestamp: April 30, 2020, 11:45:43 pm
content: If it is 12V AC, I'm not sure if it will give enough headroom for a typical 7812. Because standard ones require ~2V headroom, you need too choose a low dropout one (LDO). Some voltage is lost in bridge rectifier. Maybe you will need schottky bridge, to save some voltage. Also keep in mind, that there will still be ripple after rectification and filtering. That ripple must be included in headroom for LDO, so the lowest ripple point needs to be above regulator dropout, not average value (which DMM will show).About transformer current. More or less you can estimate by windings resistance. If you draw some current, there will be voltage drop in the windings (and power dissipation I^2*R), so you will start to loose considerable amount of output voltage and when votage is too low to be useful it is the limit (or too much heat from power dissipation).Quick, out of head estimation that if secondary winding is above 1 ohm, it will not be a very good transformer for 1.5A DC.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 01, 2020, 01:05:54 am
content: Wow. Lot's of useful information in your replies.Thank you.My first concern is that this transformer will not be adequate for a 7812 regulator. So, I need to move on to a different transformer. I looked for another and found a slightly bigger transformer with 15.28V AC output. So, I conclude it's a better choice.I will do some of the precision measurements on this 15V transformer tomorrow and post the info here.What I'm actually hoping to do is to make a capacitance multiplier, as Dave explains in this video......then connect the three regulators (7812, 7809 and 7805 in parallel) and have a very stable 3 value voltage source to test small audio amp circuits. I want to build battery powered amps so 12V and 9V are needed values (the 5V is just a bonus while I'm at it).I'm thinking the capacitance multiplier will reduce the ripple to such minimum that this power supply will nicely simulate a battery power source.Again, I'll post more info on the 15V transformer tomorrow so that you guys can tell me if you think it's suitable for this project.Many thanks to all...

author: srb1954
author_rank: Super Contributor
timestamp: May 01, 2020, 01:15:29 am
content: I don't think a transformer this size is going to be very suitable for your application even if you run only one regulator at a time at 500mA load. Looking at the size of the transformer I would estimate it has a rating of 5-7VA (volt-amps)The transformer needs to have a suitable VA rating which you can very roughly estimate at VA= 1.8 x Vdc x Idc for a bridge rectified AC into a capacitor smoothing filter. As an example, taking just a single 7809 regulator (which will require at least 3V input to output voltage differential if it is to maintain voltage regulation) you will need a transformer rated at least VA = 1.8 x (9V + 3V) x 0.5A = 10.8VA, which is significantly higher than my estimated rating of the example transformer. ForThis transformer seems even less suitable for the 12V and 5V regulators. The rectified output voltage is likely to be to low to operate the 7812 except at very light loads. With the 7805 regulator the DC input voltage to the regulator is likely to be too high and the regulator will dissipate excessive heat.

author: Manul
author_rank: Super Contributor
timestamp: May 01, 2020, 01:20:23 am
content: If you will use capacitance multiplier, you may again go into trouble about 7812 even with 15V transformer, because capacitance multiplier will drop some 2 or 3V. Needs to be tried, a lot of variables involved, and depends a lot about the transformer performance and load current.

author: james_s
author_rank: Super Contributor
timestamp: May 01, 2020, 01:32:57 am
content: You can load it with a progressively heavier load while keeping tabs on the temperature. When it starts to feel uncomfortably warm you've probably gone a bit too far. Give it an hour or so to stabilize after each increase.Going off the physical size should get you in the right ballpark.You can also measure the DC resistance of the secondary and calculate the I^2R losses from that.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 01, 2020, 01:48:53 am
content: Hm, from all the follow up replies it seems that I should go back to step 1 and rethink this project. I thought this was going to be a simple and quick project.Basically I need to use a beefier transformer with about 20V output, correct?But I guess the ideal transformer for this would be a transformer with three separate secondary windings, correct?I do have a bunch of salvaged transformers, so I'll look through them all tomorrow and see what I got. Some of them do have 2 or 3 output windings, but what are the odds that they will be correct specs...?Thank you all...

author: exe
author_rank: Unknown
timestamp: May 01, 2020, 09:28:00 am
content: Output voltage changes with the load. Smaller transformers have worse regulation, that is, they have higher output voltage under light load. So, a small 12V tranny can output, say 16V under no load. After rectification it can go quite high. Also diode dropout voltage is smaller under smaller currents. Also take into account mains variation...

author: AVGresponding
author_rank: Super Contributor
timestamp: May 01, 2020, 09:35:51 am
content: For a 500mA current at 12V you need a 6VA transformer. That one looks to be at least that, and maybe as big as 15VA or so.Full wave rectification will give you the nominal VAC of the secondary x1.414, which is 16.96VDC, though assuming a ~470uF smoothing cap you'll probably see something like a fraction over 16VDC, maybe 16.5VDC or so, off load.The dropout voltage for a standard 7812 is no more than 2.5V or so, so anything over 14.5V will be fine.Incidentally the nominal voltage of a transformer may be given for no load, partial load, or full load; read the datasheet, as YMMV. Since you can't find a datasheet for yours, I suggest you use the opportunity to test it, and find your own data.If you're getting a measured 12VAC out of your transformer off load, chances are it'll be fine at full load, as it'll probably not drop more than 10% or so, which will still leave you above the dropout voltage.

author: exe
author_rank: Unknown
timestamp: May 01, 2020, 09:46:21 am
content: Quote from: AVGresponding on May 01, 2020, 09:35:51 amIt's normally suggested to have 40% overprovisioning, as nominal transformer VA rating is meant for resistive load, while a diode bridge conducts only a fraction of time, creating a pulsing load. There is also a voltage drop on the diode bridge.

author: AVGresponding
author_rank: Super Contributor
timestamp: May 01, 2020, 10:05:41 am
content: Quote from: exe on May 01, 2020, 09:46:21 amDuty cycle is a factor too (operational duty cycle, not waveform). You can over-engineer it to the point of absurdity if you really want to, but unless you're delivering always-on, mission critical equipment, I'd not bother.The drop over the bridge shouldn't be significant enough to put it in the drop zone, but ofc that depends on the specs of the diodes/bridge being used, and the loading...Another factor to consider is the OP mentioned simulating 12V and 9V for battery devices. Using normal primary cells, the fresh, no-load voltage for those nominal ratings are 13.2V and 9.9V respectively.Then there's the low voltage cutoff, which might be anything from 6.4V to 8.8V for the nominal 12V system, and 4.8V to 6.6V for the nominal 9V system.How far down the rabbit hole shall we go today?EDIT: Edited to clean up a maths f**k up on the nominal 12V low cutoff values

author: Ian.M
author_rank: Super Contributor
timestamp: May 01, 2020, 10:39:12 am
content: Quote from: AVGresponding on May 01, 2020, 09:35:51 amThat's true as far as it goes, but are you being deliberately misleading?e.g. "For a 500mARMScurrent at 12VAC RMSyou need a 6VA transformer" (my clarification added inred) is true, however that does *NOT* mean you can get 12V DC out at 500mA DC from a 6VA transformer supplying a rectifier and linear regulator.A full wave rectified capacitor input load has a much higher RMS input current than its DC load current. Hammond Mfg (transformer division) recommends*the DC load current does not exceed 62% of the RMS current capability.Quote from: AVGresponding on May 01, 2020, 09:35:51 amAgain, true as far as it goes, but lets see what happens at 500mA DC load current. The O.P. is in North America, with a 60Hz mains supply, so there is 8.33ms between peaks. The bridge rectifier only conducts close to the peaks, so for most of that time the load current must be supplied by the capacitor. In 8ms that's 4 millicoulombs of charge, which (if not intercepted by the rising side of the next half cycle) is a voltage drop of 8.5V! Its clear that 470uF is grossly insufficient for the expected load current. Increase it to 1000uF and you can expect a bit under 4V peak to peak ripple. Further increasing the capacitance has diminishing returns, as the peak current increases dramatically and the winding resistance (both, referred to the secondary) and increased diode Vf drop nibble away at the peak output voltage, pushing the whole ripple waveform bodily downwards.That peak to peak ripple adds to the headroom required above the output voltage as does the bridge rectifier's forward voltage drop, and the regulator's minimum input to output voltage drop at the required load current. For a 78xx regulator at 500mA that's likely to be over 2V. Therefore 16.5V off load on the resevoir cap is going to be insufficient for more than 10V out, and will be marginal even for that as no allowance has been made for mains supply voltage variation, which in rural areas with a significant industrial/agricultural load on the same feeder could be as much as 10% low at peak load.'exe' proposes 40% overprovisioning (presumably of secondary RMS current capability vs DC load current) - that will result in an overloaded transformer which may be tolerable for short period operation with a low enough duty cycle that the transformer can cool thoroughly between overloads. Earlier I mentioned "approximately half as much again" i.e. 50% overprovisioning - that's still an overload - so if you want it to run 24/7 at full load current you need more margin. I recommend following Hammond's formula from the design guide below.*See Hammond'sDesign Guide for Rectifier Use. I highly recommend anyone interested in linear PSUs and other line frequency power transformer applications save a copy.

author: AVGresponding
author_rank: Super Contributor
timestamp: May 01, 2020, 11:22:05 am
content: Quote from: Ian.M on May 01, 2020, 10:39:12 amThis is just wrong. The only time that would be applicable is when there is no load, and the only current flowing would be to maintain the leakage from the capacitor and the quiescent on the linear reg and DUT.The diodes/bridge will conduct when the voltage difference between the transformer secondary and the smoothing capacitor exceeds the forward voltage of the diodes, 0.2V-~1V per diode, depending on type and loading.1uF per mA of loading is a general rule of thumb for linear supplies, unless you have a special use case.EDIT: To restate what I said earlier; different manufacturers quote their VA ratings differently, and given the absence of a datasheet on this particular device, the only way to know what it's capable of is to characterise it yourself.

author: exe
author_rank: Unknown
timestamp: May 01, 2020, 11:31:04 am
content: Quote from: Ian.M on May 01, 2020, 10:39:12 amThe reason 40% works for me is that nominal values are meant for absolute worst scenario, like 75C ambient temperature (but this needs to be mentioned in the datasheet) and 125C core temp. Heating is what limits tranny's performance. With better cooling it's possible to get a bit more. Although, I of course, do not encourage pushing parts beyond maximum limited defined by the manufacturer.It's also often not possible to get exactly 50% margin: it's either a bit less than that, or much more as trannies usually go in, say, 60VA, 90VA, 120VA ratings, but nothing in between. Then there is a dilemma: to buy a beefier part, or to try luck. Again, not encouraging anyone to go with smaller parts.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 01, 2020, 11:54:19 am
content: This got a bit too technical for my level of understanding.But to clarify my needs. When I said 500mA I was just guessing what my maximum current at any given time in the future might be if I really had to use this power supply. The main objective was to use up one of the small transformers I've had laying in my salvage parts bin for years and make some use of it. I did see what the size might be what I'd expect to be inside of a wall wart so I figured I'd be able to draw up to 500mA.If my main use of this small power supply is to simulate a 9V battery, so that I don't have to use up batteries for breadboard tests, I think I have to think the average maximum current would be in a range that gives me decent battery life. Perhaps a good guess would be that I wouldn't want the current to exceed 20mA.I do have bigger transformers in my salvage bin and perhaps I should really just consider using a bigger one, since I don't have any other uses for any of them anyway, at this time.But I am inclined to save the bigger transformers for more important future projects, so I have to make a decision.I was really thinking of this power supply as a quick little project, just so I can have an additional power supply from time to time.Hope this makes sense.Thanks...

author: Ian.M
author_rank: Super Contributor
timestamp: May 01, 2020, 12:39:55 pm
content: Y'all may find this LTspice sim educational if you haven't done a lot of linear PSU design. Its a fullwave rectified 12V RMS 60Hz line frequency (so 120Hz after rectification) voltage source feeding a 470uf reservoir capacitor via an ideal diode with a 500mA load current (as AVGresponding proposed).The peak to peak ripple is 6V, not too far off the ballpark 8V my back of the envelope calculation gave. The trough of the ripple is below 11V. The RMS current from the source is 983mA, nearly twice the load current.The nasty 'gotya' is the ripple current in the 470uF reservoir capacitor of 533mA RMS. That's a very large current for a small capacitor, but is somewhat unrealistic due to the ideal diode and no source resistance or capacitor ESRYou can make the sim more realistic by adding a resistance in series with the source (secondary resistance + primary circuit impedance transformed down by the square of the turns ratio), and replacing D1 with a realistic diode model e.g a 1A silicon or Schottky diode chosen from the available models, with an instance parameterN=2added (Ctrl right click and set Value2 and make it visible) to model the double diode Vf drop of a bridge rectifier. Also try increasing C1, and note its influence on the ripple trough voltage, but do keep an eye on the RMS ripple current through the cap, and compare it with the ratings of commercially available electrolytics.

author: Ian.M
author_rank: Super Contributor
timestamp: May 01, 2020, 12:55:07 pm
content: @Adinol,For breadboarding, the power supply needs to be short-circuit proof. Unfortunately smaller transformers will overload before the short-circuit protection of your chosen regulators cuts in, so it would be advisable to add a polyfuse, appropriate to the estimated RMS secondary current rating for your transformer, in series with the winding.(Give us as much data as poosible for your chosen transformer and ask us to estimate that.)If you decide to run the transformer at a significant overload factor, fit the polyfuse in close thermal contact with the winding surface so the heat from the winding will reduce its trip current if the overload is long duration.Consider using a LM337 adjustable regulator which will give you a greater range of output voltages and let you simulate a discharging battery. Also don't forget the internal impedance of a battery typically increases as it discharges - check battery manufacturer datasheets and add an appropriate low ohm resistor, externally in series with the output when breadboarding with your PSU or you may get a nasty surprise (e.g. motorboating or other instability or distortion) when you attempt to use your completed amplifier for an extended period powered by real batteries.

author: srb1954
author_rank: Super Contributor
timestamp: May 01, 2020, 01:15:18 pm
content: QuoteA 6VA rated transformer would be hopelessly undersized for this application. To get a regulated 12VDC at 500mA, using a 7812 regulator, you need something like a 10-12VA rated transformer. A transformer with approximately 15V no load output voltage and a current rating of at least 800mA RMS is required as a minimum. This is barely adequate for this application as it doesn't provide any margin for optimum regulator operation under low mains input voltage conditions. The transformer secondary voltage may also need to be fine tuned depending on the voltage regulation of the transformer under load.The peak voltage quoted above of 16.96VDC for a 12V transformer only applies under zero DC current draw from the rectifier/smoothing capacitor. The DC output of a rectifier/smoothing capacitor drops considerably under load, typically down to about 70-80% of the no load peak voltage, depending on the smoothing capacitor value and the effects of voltage regulation of a typical transformer.A smoothing capacitor value of 470uF is too low for this application. I would recommend at least 1000uF as a bare minimum. Increasing to 2200uF will reduce the ripple voltage and give the regulator a little more usable input voltage when the mains voltage dips a little. Also, you need to be aware of the capacitor ripple current rating to make sure your electrolytic capacitors last the distance. In this case the ripple current rating required will be around 600mARMS for the DC output current of 500mA.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 01, 2020, 01:53:55 pm
content: I appreciate all the useful information and I'm glad I asked about the transformer, in the first place.I still have some transformers to go through, but I think I already found a good candidate. It's a much larger transformer (also no datasheet available) and it outputs over 18V. My main use will actually be using the 9V output from my power supply to design practice guitar amps powered by a 9V battery. I can't imagine how that large 18V transformer would not be sufficient for this application.I think it's a good point that batteries are not actually providing the voltage that is written on the label, so now I do think that I should use an LM317 and design the circuit so that the output can be adjusted to simulate the range starting from what a brand new battery might provide, down to the cutoff of what a drained battery might provide.Perhaps others have already designed battery simulator power supplies and I can just use an existing circuit to complete the project.Thanks...

author: Emo
author_rank: Regular Contributor
timestamp: May 01, 2020, 02:13:23 pm
content: Ok,some rough estimates based on dimensions and wire thicknessThe maximum current is approx 2.5 - 3.5 A per mm2 copper wire(depending on winding geometry etc)So in this case 0.3 mm dia = 1/4 * Pi * 0.3*0.3 = 0,07 mm2max current is approx 3 * 0,07 = 0,2 A(0,18 - 0,25)0.4 mm wire does hold max 0,5 A which does better match the estimate powerRule of thumb for the number of windings per Volt is 45 divided by the surface area of the middle leg.(again only for laminated iron core at 50 Hz)Max power is approx 1.5 times surface area middle leg times length of the middle leg for laminated iron core in cm2.(from the photo I estimate 1.5 * (1 * 1.5 * 2.5) = 5.6 WAll rule of thumb approximations.So try to measure the physical sizes more exactly and you will have a better idea of the transformer propertiesEric

author: exe
author_rank: Unknown
timestamp: May 01, 2020, 02:13:41 pm
content: Quote from: Ian.M on May 01, 2020, 12:55:07 pmI think small trannies around 5VA and below are "inherently short-circuit proof". Although, again, ideally this should be mentioned in the datasheet.BTW, on my power supply I use RS 671-8934 tranny, which is 2x12V 1.25A. I think I managed to get continuous 2x15V@1A from it with temperature inside enclosure about 60-65C (at front panel, tranny is installed at the back, so not sure what is the temperature there). Not that I recommend this... I checked temperature of individual parts, including the tranny it was fine. If one wants, I can repeat the measurements to see how well it is doing under full load.

author: Ian.M
author_rank: Super Contributor
timestamp: May 01, 2020, 02:44:22 pm
content: I wouldn't bet on small transformers being short-circuit proof unless the datasheet or label explicitly say so (e.g. may say 'impedance protected'). Over my years in the repair trade I've seen many transformers of approx 5VA, often in linear wall-warts, with open-circuit primaries due to the typical buried thermal fuse in the windings tripping. Some could be salvaged if the thermal fuse was sufficiently accessible to permit replacement, but many had signs of 'cooked' insulation, had developed shorted turns, or had the thermal fuse too deeply buried.One 'trick' to estimate how badly overloaded a transformer is, is to measure the primary resistance when stone cold, (and note the ambient temperature), and again when hot immediately after disconnecting the mains supply then use the resistance ratio and tables for the resistivity of copper vs temperature to estimate the average winding temperature.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 01, 2020, 03:26:16 pm
content: I think there's a chance for me to learn useful skills here.How about if I post all the relevant date for the small 15V transformer and you show me how you do the calculations for that one. Then I do calculations for the bigger 18V transformer and you guys check if I made any mistakes.(We can forget the smaller 12V transformer that I initially asked about).So, here is some date for the small 15V transformer. I used a variable transformer to do the tests.When the input peeks at 117.7V the output peeks at 15.18V.When I reverse it and try to adjust the variable transformer 120V for the output, I took a reading of 120.4V when it peeks. It took 16.49V from the variable transformer to produce 120.4V output.To do those readings I set my DMM to capture the MAX reading for a time period of 1 minute of it being hooked up.The resistance readings are as follows.primary - 118.6 Ohmssecondary - 1.8 OhmsI am including some photos that show how I took those voltage readings.Thanks...

author: Ian.M
author_rank: Super Contributor
timestamp: May 01, 2020, 03:57:29 pm
content: When reversed, you *MUST* keep the input voltage under its normal output voltage as its important to keep it out of saturation to get a good measurement for calculating the turns ratio. *DON'T* try for 120V out, - the output voltage with approx 14V in will be far more useful.Also it would be helpful if you can edit the post to add the core dimensions.It would be best to keep all data for one transformer in one post to prevent mixups, then those of us replying with calculated ratings can link to or quote the specific reply containing our source data.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 01, 2020, 08:16:47 pm
content: Oh, I see.So here are the correct specs for the small 15V transformer.Primary winding resistance - 118.6 OhmSecondary winding resistance - 1.8 OhmPowered up forward:In - 117.7 VOut - 15.18 VPowered backwards:Out - 117.7 VIn - 15.71 VThank you very much...

author: srb1954
author_rank: Super Contributor
timestamp: May 02, 2020, 08:51:48 am
content: A quick simulation using these transformer parameters shows that you should be able to manage 450mADC output from a 7812 regulator when using a 2200uF input smoothing capacitor. The smoothing capacitor needs to be rated at 600mA RMS ripple current or better.You can reduce the smoothing capacitor to 1000uF but then the available output current will drop to around 380mA DC before the regulator has trouble maintaining full output regulation due to the excessive ripple on its input supply voltage. The residual ripple on the output of the regulator will also be considerably increased with the smaller smoothing capacitor.The above figures are based on your 117VAC input so there is not much safety margin for drops in the mains input voltage. Also the available current will drop as the transformer heats up and the winding resistance increases. The transformer will get moderately warm when operated at this load.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 02, 2020, 02:43:49 pm
content: Thank you for running the simulation, although I thought we were going to just do manual math, so I learn how to calculate the same for the large transformer. But perhaps there's just too much math involved to do it manually.In any case, this is all interesting and useful information.So, I was thinking of using 2 or 3 parallel capacitors, as Dave pointed out some advantages of doing that, in one of his tutorials. So, if I use 3x 1000uF it should be good, correct?I don't understand what it means that the smoothing capacitor needs to be rated at 600mA RMS ripple current or better. The only parameters that are written on these caps are the capacitance and voltage. I also have a component tester that shows ESR, but I guess I would have to look up the data sheet for each separate cap to see what the RMS ripple current rating is. One issue is that I like to use salvaged components, but perhaps this is not a practical approach in this case.Also, since I am just experimenting and learning I'd also like to breadboard a capacitance multiplier, using a BD139 transistor, a 1K resistor, and a 470uF cap and see what happens. I do understand that this will be a problem for the 7812 but I'm still curious to see what happens. Perhaps I can still make a small power supply that bypasses the capacitance multiplier for the 7812 and still uses the multiplier for the 7809 and 7805. Of course, the real project will be using that larger 18V transformer.Would you mind running a simulation with a multiplier? I'll do the same on a breadboard and see what happens.Thanks...

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 02, 2020, 04:02:00 pm
content: So, I just ran some tests on the breadboard.First I calculated that to get 500mA I would need a 24 Ohm resistor rated at 6W. I don't have one, so I gabbed the following values: 36R, 100R, 180R, 270R and 3K9.I was surprised to see that non of my 7812's actually output 12V, at no load. Some were as low as 11.77V, so I used the one that went up to 11.97V. This was with the small 15V transformer that brings the rectified voltage to 18V and change and using a 2200uF cap. I got the same result, with no load, with or without the capacitance multiplier.So, with the loads this is what I measured (voltage across load and current):36 R - 10.37V - 300mA100 R - 11.95V - 118.6mA180 R - 11.95V - 66.1mA270 R - 11.96V - 43.7mA9.9 K - 11.96V - 3.0mAI didn't do Ohm's Law calculations to check if the numbers match, but I'm sure they do.Is my conclusion correct that if I build this circuit to power up loads up to about 50mA, it should work just fine.Thanks...

author: Ian.M
author_rank: Super Contributor
timestamp: May 02, 2020, 05:27:22 pm
content: It looks like it 'falls off the cliff' somewhere above 120mA. Try paralleling load resistors to get intermediate values e.g. add 1K resistors in parallel with the 100R one to go up in approx 12mA steps to determine its current capability more accurately. The 1K resistors will each dissipate a small fraction of a watt so can be ordinary 1/4W resistors stuffed in the breadboard.If you have a scope, use it, AC coupled to check the capacitance multiplier emitter and regulator output for signs of ripple breakthrough. If not, assuming your DMM has a reasonably sensitive AC voltage range, put a 1uF non-electrolytic capacitor in series with your multimeter to block DC and use it to check the ripple.N.B. solderless breadboarding with power resistors is always a dodgy business - far too many of us who have seriously used breadboards have one marred by melted or even burnt impressions of overheated parts! *Never* trust breadboard contacts to carry more than a couple of hundred mA, and any time you are expecting to dissipate enough power for a part to get hot, its probably best mounted off-board, in a soldered or 'choc-bloc' screw terminal block lashup with any closely associated components that *MUST* be physically close to it for stability, then connected to the breadboard with solid core jumper wire.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 02, 2020, 07:46:04 pm
content: I really like this methodical and analytical building of a circuit and I'm really learning a lot. Thank you.For now the solderless breadboard is my only practical option, but thank you for letting me know about the potential problems. I'll have to look into other options to set up higher current tests int the future.I breadboarded the circuit to easily add 1k resistors and I logged the results, starting again with 100 Ohm.100 R ................ 11.9V - 118.6mA100 R + 1x1k ..... 11.9V - 130.2mA100 R + 2x1k ..... 11.9V - 114.8mA100 R + 3x1k ..... 11.9V - 153.2mA100 R + 4x1k ..... 11.9V - 164.5mA100 R + 5x1k ..... 11.9V - 176.0mA100 R + 6x1k ..... 11.9V - 187.1mA100 R + 7x1k ..... 11.9V - 197.7mAHere is where I had to switch to my 10A input on my DMM and here is also where the voltage across the load started dropping.100 R + 8x1k ..... 11.7V - .22A100 R + 9x1k ..... 11.5V - .23A100 R + 10x1k ... 11.4V - .23AIf I understand correctly, this power supply, at 12V, can handle up to about 60 Ohm load, which would make the current just shy of 200mAIs my conclusion correct?And of course, if I were to make a rating for this 12V power supply I should round the figure down to 150mA or less rating, at 12V.I will actually build a bigger supply with the bigger transformer, but out of curiosity, does that sound like an adequate power supply to be used to test battery powered 12V guitar practice amp circuits? Of course, one of the objectives is long battery life.Thank you...

author: Ian.M
author_rank: Super Contributor
timestamp: May 02, 2020, 09:07:27 pm
content: You haven't mentioned checking for ripple break-through yet.Apart from that, it does so-far look like its good up to 200mA. Next check the secondary AC voltage, first with no DC load and then while drawing as close to 200mA as you can get, to see how much the secondary voltage is dropping. It would also be worth checking the voltage in to out across the regulator at 200mA load to see how much headroom there is.Make sure the regulator heatsink is adequate for its expected dissipation then try running it for half an hour at 200mA load, checking that nothing gets excessively hot, especially the transformer, transistor and regulator.Personally, if the ripple's low enough, I'd clap a LM317 in there so you can adjust the output between 1.2 and 12V (or a bit more at low loads), add a 350mA polyfuse in series with the secondary and box it up with the heatsink on the rear of the case and call it good.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 02, 2020, 11:44:09 pm
content: I had to dig out the old oscilloscope from storage. I still have to learn how to use it.It was worth going through the trouble because I am very pleased with the results of the test.If my interpretation of the scope screen is correct, the peek to peek voltage of the remaining ripple, after the capacitance multiplier, is just shy of 3mV. I set the Y axis to 5mV per division and I had my probe on 1x (I'm attaching a photo of the scope screen in case you want to check).By comparison, the ripple before the multiplier is 520mV peek to peek.Here are some other specs.I built the bridge rectifier from four 1N4004 diodes, so the rectifier does not have a heat sink. But it is not warming up at all at 200mA load, so I guess it doesn't need a heat sink. I still have to run a half hour test under load, though.The load I used was 58 Ohm and the DMM measured .20A of current. Voltage across load was 11.8V.AC voltage across secondary winding was 14.09V AC under load, and 15.17V AC with no load on.IN and OUT voltages across 7812 regulator as as follows:Without load:IN - 18.34VOUT - 11.96VWith load:IN - 13.54VOUT - 11.75VWhen my kid goes to bed I'll have time to run it for half an hour while I watch it.Again, I am really enjoying the process and I appreciate all the help and instructions.Thank you very much...

author: Ian.M
author_rank: Super Contributor
timestamp: May 03, 2020, 12:03:06 am
content: That's pretty good for cleaning up the ripple. The regulator will reduce it further by a couple of orders of magnitude, so yes, it should be clean enough to replace a battery supply in audio applications.The AC voltage is dropping less than 10% so its almost certain the transformer is well within its ratings. However it looks like its dropping out of regulation at 200mA. That's not surprising with only 1.79V of headroom. You'll have to derate it a bit or if you want a full 200mA @12V, squeeze another half volt or so's headroom out of the bridge by replacing the 1N4004 ordinary silicon diodes with Schottky diodes. Any 1A to 3A, 30 to 40V ones should be fine.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 03, 2020, 03:32:23 am
content: OK, I did the half hour test. There's not building up of heat on on any of the components, running the 200mA load. The only component that got kind of warmish was the 100 Ohm resistor.Nevertheless, I am going to heat sink all the 78xx regulators as well as the BD139 transistor.Tomorrow I will build the boards, which will include a new rectifier board.However, I would like to add short circuit protection using a 12V relay, as seen in this video.I happen to have a salvaged relay like that and like the idea of building this, unless there are any technical issues I am unaware of.Should this circuit breaker be placed between the rectifier and the capacitance multiplier?Thanks...

author: srb1954
author_rank: Super Contributor
timestamp: May 03, 2020, 04:11:42 am
content: I just did a simulation with LTSpice as this is much easier and quicker than manual calculation. The maths for a highly non-linear circuit like a bridge rectifier and capacitor filter is horrendously difficult and takes many hours of work to calculate. With a circuit simulation you can get an answer in a few minutes and can quickly change the component values and get updated answers in seconds.I can do a simulation if you post a drawing of the full circuit you are using. However, I suggest you download the free LTSpice simulator yourself and learn how to use it. It is a very valuable tool to understand what is going on inside a circuit although you still ultimately need to check out the final circuit on a breadboard. A circuit simulator doesn't always give a 100% accurate representation of the circuit performance as it is only simulating an approximate model of the behaviour of real components.On the capacitor ripple current rating you need to look up the manufacturer's datasheets to obtain this figure. The manufacturer determines this ripple current figure from the type of capacitor, its value, physical size and internal construction to provide an acceptable operational life of the capacitor. Using a salvaged electrolytic capacitor is perhaps not the best idea as you have no idea what conditions it has previously been subject to and how much of its life has already been used up.

author: Ian.M
author_rank: Super Contributor
timestamp: May 03, 2020, 06:34:39 am
content: That video from reply #37 isn't actually short circuit protection. Its a no-volts trip and only works with a current limited supply. When you short the output the supply voltage collapses below the relay's minimum holding voltage, so it drops out, breaking the circuit to the output and the diode feeding its coil. If you do decide you want to use it, it needs to go after the regulator so the regulator can limit the current into the short circuit. For higher current applications, its also worth adding a NC pushbutton in series with the relay coil so the same relay can be used for manual output on/off switchingIf you want actual circuit breaker action, you need to sense the current. Usually this involves sensing the increased voltage drop across a low-ohm resistor. However all the simple transistor circuits need one Vbe drop to trigger so you'd want to put them on the input side of the regulator and as you don't have enough voltage headroom, that wouldn't be a good idea. You could make a simple real over-current trip that has minimal voltage drop by winding a current sensing coil around a form B (NC) or form C (changeover) reed switch, and experimentally adjusting the number of turns so it operates at the desired current, then using it to break the coil circuit of the no-volts trip, but its a royal PITA to adjust the current for each individual reed switch by adding/removing coil turns.+1 for downloading and learning LTspice. Half an hour with it can replace a day of manual calculations, burnt parts and fingers finding out an idea wont work, or refining component choices for a workable idea.Capacitor ripple current rating is closely dependent on the capacitor's ability to dissipate heat ad thus its surface area. If your design needs a high ripple current rating it can be easier to get it by using two or more smaller value caps in parallel, mounted sufficiently far apart that airflow between them isn't restricted.

author: exe
author_rank: Unknown
timestamp: May 03, 2020, 08:30:31 am
content: Yesterday I measured temperature of my tranny in the power supply load.Setup: 2x12V@1.25A 30VA tranny in a linear power supply sourcing 2A current. After ~30 or so minutes the hotest spot is the bolt that connects it to the enclosure. It measures ~85C (ambient temperature is 23C). The output voltage noticeable sagged as it wormed up, approx 0.5-0.7V less than at start. The tranny is rated to work up to 105C (thermal rating of the output leads, windings can withstand more). So, I can imagine this can be exceeded during a hot day.I'll post a video later, it's not a high-quality content, but may be entertaining.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 03, 2020, 09:09:22 pm
content: Quote from: srb1954 on May 03, 2020, 04:11:42 amI've looked into options a few times over the years and I never really found anything that worked out for me on my Mac. I remember trying to use the DigiKey online app and also another online app that was also tied to some merchant, so you could easily order parts. I also remember downloading some apps but there was always something that didn't quit work for me so I ended up resorting to pen and paper, or Adobe Illustrator.I'll give it another try to see what is currently available for the Mac.Quote from: srb1954 on May 03, 2020, 04:11:42 amOh, I totally get that. I actually use a combination of new and salvaged components and I know electrolytic caps are the last component I should be using. I try to salvage good brands just to have different values in my salvage bins.This works for now but I'll definitely order replacement caps once I am compiling an order for other components.Quote from: Ian.M on May 03, 2020, 06:34:39 amThis makes a lot of sense, now that you explain it. In fact, I was having trouble understanding how it was a breaker.I like your idea of adding an NC button. In fact, I think it makes more sense to have the NC button and to put the LED on the other pin of the relay, so that the LED is on when the circuit is also on. I like the idea of having push buttons to turn on and off the 3 supplies, 12, 9 and 5 Volts, and the LED woudl tell me when they are on. Then, if it trips I know I messes up something on the breadboard and I can troubleshoot.Quote from: Ian.M on May 03, 2020, 06:34:39 amPerhaps that's something I should consider for the bigger battery simulator supply. It would be a good learning experience.Quote from: Ian.M on May 03, 2020, 06:34:39 amI built a new rectifier board, using Schottky diodes, and I ended up using three 1000uF caps, rated at 35V. I didn't spread them apart enough, I know, but I only had that small board to spare and this will be placed inside of an enclosure that has a fan blowing at the board.Now that it's all built on the final boards I can't detect any ripple, whatsoever. I used the actual same components that were on the breadboard, the only thing that changed is the rectifier and the fact that I also added 7809 and 7805 in parallel and each of those has caps on ins and outs.I'm not complaining, but I find it interesting that there's no ripple that my scope can detect, with or without a 100 Ohm load.I really appreciate all the expert information. It really helped me out.Tomorrow I plan to put it all inside the enclosure.

author: srb1954
author_rank: Super Contributor
timestamp: May 04, 2020, 03:06:27 am
content: LTSpice is available for the Mac.I have installed it on my Mac but I find the user interface difficult to use compared to the PC version. I tend to use the PC version on my old Windows machine although I suppose you install a Windows emulator on your Mac and run the PC version of LTSpice on that.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 04, 2020, 03:13:16 am
content: Quote from: srb1954 on May 04, 2020, 03:06:27 amOh, I see. Somehow I ended up on this sitehttps://ltspice-iv.en.lo4d.com/windowsthe first time and I didn't look too closely. Sorry...I'll give it a try.Thanks...

author: exe
author_rank: Unknown
timestamp: May 07, 2020, 06:00:02 pm
content: My power supply:https://www.youtube.com/watch?v=tX_r-sxKpAg&feature=youtu.be. Again, the video is amusement only, it's not meant to prove anything. Okay, it meant, but it failed.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 08, 2020, 02:10:38 am
content: Thank you for posting your video.I think if you place your circuits into a larger cabinet and also install a fan it would make a big difference. It would be worth taking the time to redo this part of your build.I am actually finishing up my power supply and I hope to be able to post pictures of the final product within a day or two. I thought I'd be dine by now but it's just amazing how much time things take to build.This little supply discussed in this thread is actually going to be inside the same cabinet as a much larger power supply that I am just finishing up.

author: rdl
author_rank: Super Contributor
timestamp: May 08, 2020, 03:24:41 am
content: Quote from: exe on May 07, 2020, 06:00:02 pmAll you need is some ventilation and it'll be fine. Provide some openings so air can flow through, preferably from bottom to top, past the hottest parts. A fan is probably not necessary.

author: Frankentronics
author_rank: Regular Contributor
timestamp: May 09, 2020, 01:41:50 pm
content: Here is a quick update on my project.So, this little power supply discussed in this thread is actually being added to a large adjustable power supply that I am still finishing up.I am attaching a picture of the work done so far.The large power supply utilizes a large salvaged transformer (not shown in this picture) and a salvaged bridge rectifier outputting +/-20V as well as +/-45V. The +/-20V output is connected to the adjustable supply and it all works fine.There was room on the chassis to add a small isolated supply with the three fixed voltages. You can see, in the picture, the small transformer, the small rectifier (with the 3 1000uF caps) and the heat sink with the three 78xx regulators.Now I am building the cabinet and figuring out the internal wiring connections that will use some salvaged ribbon jumpers.Once this is al finished I will post images of the final product.For now, many thanks for all the help, here.Thanks...