Voltage overhead. - Page 1
Forum section: electronics / beginners

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 12:13:55 am
content: Wandering if this is achievable with a low drop out linear voltage regulator. So I've got a transformer with a secondary of 14.60 Volts AC unloaded reading. With a 12 Volts linear regulator LDV of 0.5 Volts, and taking a 1 Volt drop across a full wave bridge rectifier, would it be possible to get at least a 12 Volts output from the above. Any thoughts appreciated.

author: james_s
author_rank: Super Contributor
timestamp: May 10, 2017, 12:22:40 am
content: I would think so, but it depends on how much the output of the transformer sags under load. When you rectify and smooth the AC from the transformer you will end up with around 1.4x the AC voltage and that should be more than enough headroom even once you factor in diode drops.

author: Ian.M
author_rank: Super Contributor
timestamp: May 10, 2017, 12:44:26 am
content: At low currents, with a big enough reservoir cap its pretty easy, with up to 20V into the regulator, but assuming the transformer has 10% regulation the mean output voltage will drop to under 11V as you approach the transformer's full RMS load current (worse if the mains supply is lower than usual). You can buy yourself a little more margin by using high current Schottky diodes in the bridge rectifier to reduce its total dropHow many VA is the transformer and how much current do you need out at 12V?

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 01:32:51 am
content: Thank for the replys, now the iffy part...It's a rewound 700 watt MOT with high temperature silicone wire secondary. The cable is rated for 45 Amps continuous current. I'm only looking for 15 - 20 Amps output, by way of series pass transistors. I'm in no doubt the transformer can deliver 15 Amps continous and 20 Amps 50% duty cycle. If I can keep the voltage regulation up between 11.40 Volts minimum, and ideally 12 Volts or just over. I know a MOT is poor in efficiency, but just trying it for the first time. I got as many turns of this high temperature silicone wire on as I could. I could have dropped a wire size for more turns, but would have lost some current ability. Not sure how it looks now you see its a MOT. It will be forced air cooled all the time it's on.

author: David Hess
author_rank: Super Contributor
timestamp: May 10, 2017, 01:53:29 am
content: Quote from: davelectronic on May 10, 2017, 12:13:55 am14.6 volts AC is 20.6 volts peak (multiply by 1.414). Subtracting two diodes is pessimistically (1) another 2 volts leaving 18.6 volts. A high voltage drop regulator is 3 volts leaving 15.6 volts. Half of the 3.6 volts remaining is a reasonable amount of input capacitor ripple.So what is the problem?(1) The power factor is low and the peak current is high for a rectifier capacitor input so 1 volt is actually reasonable for silicon rectifiers. Using schottky rectifiers will halve it but that is not really necessary here.

author: mariush
author_rank: Super Contributor
timestamp: May 10, 2017, 01:58:36 am
content: For such high currents, it may be worth going with a bridge rectifier made with the LT4320 and four mosfets with low rdson :http://www.linear.com/product/LT4320It's not THAT expensive and you can buy it even in DIP package so you could pair it with 4 to-220 or some other through hole mosfets and you won't waste 20-30 watts into a bridge rectifier.Farnell :85c version :http://uk.farnell.com/linear-technology/lt4320in8-pbf/diode-bridge-controller-9v-72v/dp/2396663125c version :http://uk.farnell.com/linear-technology/lt4320hn8-1-pbf/ideal-diode-bridge-controller/dp/2406673Digikey :85c version :https://www.digikey.com/product-detail/en/linear-technology/LT4320IN8-1-PBF/LT4320IN8-1-PBF-ND/4693753https://www.digikey.com/product-detail/en/linear-technology/LT4320IN8-PBF/LT4320IN8-PBF-ND/4693752125c version :https://www.digikey.com/product-detail/en/linear-technology/LT4320HN8-PBF/LT4320HN8-PBF-ND/4754663The datasheet suggests some mosfets but also has examples and formulas to tell you what mosfet parameters to look for when picking the mosfets by yourself if the ones in the datasheet can't be found.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 02:13:31 am
content: From what David says it looks promising, I've got some Motorola 7812K TO3 regulators. I was looking at the LM2940CT LDR regulator, and had considered shottky diodes for the rectifier. I really don't fancy fets for a bridge rectifier configuration. So I was just concerned about the drop, even with a large filter capacitance. But if you think 3 Volts for the regulator and a, SAY 1 - 1.5 Volt drop for a rectifier is achievable, I'll give it a go, and commit to drilling case holes for TO3 heatsinks.

author: Ian.M
author_rank: Super Contributor
timestamp: May 10, 2017, 02:25:11 am
content: The DC mean output current of a bridge rectifier is typically 62% of the RMS input current. Load the transformer secondary to about 30A with a resistive load and re-measure the voltage. For a suitable load, try a coil of steel wire in a bucket of water - you want about 0.44 ohms.That will give us enough data to tell you if its achievable.

author: mariush
author_rank: Super Contributor
timestamp: May 10, 2017, 02:48:26 am
content: If you're thinking of paralelling multiple ldos, have a look at LM1084 which should do 5A with 1v drop.The Linear made version LT1084 even has example in datasheet showing you can parallel them by using a small value resistor for balancing the currentlink lt1083/lt1084/lt1085 :http://cds.linear.com/docs/en/datasheet/108345fh.pdf

author: David Hess
author_rank: Super Contributor
timestamp: May 10, 2017, 03:12:17 am
content: I did not realize that the current requirements would be so high. That will eat into the voltage margin and you probably want to support as high a ripple voltage on the capacitor as possible so use schottky rectifiers or the mentioned synchronous rectifier and use a low dropout regulator. That will lower the requirements for the input capacitance.If the allowable input ripple is 2 volts, then the input capacitance needs to be 8200uF per amp/volt so 8200uF * (20 amps / 2 volts) = 82,000uF which may seem like a lot but is typical at these kinds of currents. That is about $10 worth of new snap-in style capacitors but you might be able to find some big surplus screw terminal capacitors.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 09:33:55 am
content: Yes this was my concern, being so close to the drop out voltage of a standard linear regulator. And a quite high current demand, i knew i might be to close for comfort. I have a load of 50 watt halogen lamps to test AC voltage of the transformer to see its voltage drop before rectification and filtering. I'm probably not going to go standard linear regulator, but opt for a LDV regulator. Also try and source a suitable shottky diode package and type. Thanks for all the replys and help, cheers.

author: dmills
author_rank: Super Contributor
timestamp: May 10, 2017, 09:56:45 am
content: You did remember to knock the magnetic shunts out of that MOT core....Otherwise it will droop badly as these are usually designed with deliberately very high leakage inductance.I would normally have wound for a few more volts, just to deal with a low mains condition (You generally want to size transformer output voltage to stay in regulation with the input 10% low, and size the regulator thermal design for mains 10% high).One thing that can help a lot with MOTs is to fit a small back transformer to lower the primary voltage a bit, takes the core further away from saturation.Regards, Dan.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 10:54:43 am
content: Yes i did drift out the magnetic shunts, i didn't want anything affecting maximum current I'm looking for. That being 15 -20 Amps, not sure on which shottky key diodes to go for, I've found some TO220 package 3 lead devices. But it looks like each one drops 0.6 Volts for each package, so 1.2 Volts if i used a pair for a full bridge rectifier. As i understand it, its about 1.4 Volts drop for a conventional bridge rectifier. Does any one have any suggestions for high current axial lead, low FWD voltage drop Schottky diodes ?

author: dmills
author_rank: Super Contributor
timestamp: May 10, 2017, 12:20:12 pm
content: Those current levels say 'Ideal diode' made of 4 mosfets and some control logic to me, Rds(on) can easily be a few milliohms and 4 D2Pack devices will easily handle the current while saving you maybe 20W-30W or so of heat at 20A.Most of the usual suspects (TI, Linear) have easy to use 'ideal diode controllers' suitable for bridge rectifiers, and the saving in heatsinking will usually pay for the extra parts, some of these components are even available in thru hole.

author: mariush
author_rank: Super Contributor
timestamp: May 10, 2017, 02:22:29 pm
content: Indeed.If you insist on using plain diodes to create a bridge rectifier, one sort of a hack would be to use the common cathode dual diode packages you commonly find in older ATX power suppplies, those can have a forward voltage lower than 0.5v at high currents. (You can just not use one of the two diodes in the package)For example :1.7$ON Semiconductor MBR60L45CTGmax 0.55v drop at 30A per diode but around 0.4v at 10A per diode and you have two in the package, so if you connect them together in parallel then you'd have ~ 0.35v-0.4v per element.1.6$STMicroelectronics FERD40U50CFPmax 0.5v voltage drop at 20A per diode max 50v ... about 0.35v at 10a per diode, so with two diodes in parallel you'd have less than 10a per diode, so low voltage drop.2.8$Vishay Semiconductor Diodes Division VS-47CTQ020PBFForward voltage : 450mV @ 20A per diode , maximum 20v input voltage(note: all these obviously need to be heatsinked but only some of them - like FERD40U50VCFP for example - are in insulated packages, so if you choose one that's not insulated, you'd need to use separate heatsinks or you'd need to use some insulator between them and the heatsink as the metal is connected to cathode most of the times, so without insulator, you won't have a rectifierBut seriously, if you spend 8$ on four of these, you may just as well spend 6-7$ to buy the LT4320 i suggested and four 0.5$ to 1$ mosfets, and you may not even need a heatsink.. let's say 10mOhm rds(on) x 20a = 0.2w dissipated power on a to-252 / dpaq / to-220 / to-whatever through hole package would be nothing.

author: james_s
author_rank: Super Contributor
timestamp: May 10, 2017, 03:33:36 pm
content: For that amount of current at 12V, I would just get a surplus hot swap server power supply. They cost peanuts on ebay and most will do 50+ amps without breaking a sweat.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 05:43:34 pm
content: Thank you for the shottky diode links. Yes i did wander how you go about connecting them up when in a bridge configuration. As the anodes are tied at one pin. But i see it if only using one diode per package, yes that could work out expensive. I've not looked in to the fet option before, but will have a look and see what it involves. I've a lot of converted server power supplys, its something i do for fun. The MOT is a result of our microwave going pop, and the opportunity to use the MOT out of it. So again its just for fun, and see if i can get a result out of it.

author: james_s
author_rank: Super Contributor
timestamp: May 10, 2017, 06:05:59 pm
content: Look at the schematic of a bridge rectifier, the diodes in it can be drawn as two pairs, each tied together at one pin.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 06:11:30 pm
content: As the anodes are tied together and, and each device having a pair of Diodes, I can only utilise one diode per package. Think that right, as each device has tied anodes I couldn't just use two T0220 packages, as I see it in need four, each devise as a single diode. Bit wasteful if thats the case, and these things ain't cheap lol.

author: james_s
author_rank: Super Contributor
timestamp: May 10, 2017, 06:21:10 pm
content: Can you get one that has the cathodes tied together and use one of each? Many dual diode parts are available in 2 or 3 different configurations.

author: Zero999
author_rank: Super Contributor
timestamp: May 10, 2017, 06:30:08 pm
content: Quote from: davelectronic on May 10, 2017, 01:32:51 amCan you post a schematic of how you intend to connect the series pass transistors?Normally adding series pass transistors to the output incurs an extra VBE(0.7V to 1.2V) drop on top of the regulator IC's drop-out voltage.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 06:36:17 pm
content: I don't know, but at a guess what I've been looking at is common anode, and I need common cathode. Common not meaning 0 Volts, just that they share a pin. What I do know is the ones with anodes tied together are like £6.00 plus postage per devise. That's just not workable for a fun project. If I'm only dropping 0.5 Volts across the regulator, I might get away with 1.4 Volts for the drop across the bridge rectifier. If my capacitance is high enough.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 06:40:17 pm
content: Quote from: Hero999 on May 10, 2017, 06:30:08 pmThis would be PNP transistors, so collectors on the output,and input the the emitters. And a Base resistor, also ballast resistors.

author: james_s
author_rank: Super Contributor
timestamp: May 10, 2017, 06:41:22 pm
content: You should be able to get them cheaper than that, just look on digikey at Schottky diodes, they are available in a whole range of configurations, single, dual, common anode, common cathode, series, IIRC they're not terribly expensive.

author: Ian.M
author_rank: Super Contributor
timestamp: May 10, 2017, 06:48:56 pm
content: Higher capacitance means higher peak current and more voltage drop across the diodes.Have you measured the AC output voltage with a load of about 30A yet? If it droops too much this project may be dead before you even go to a distributor's parametric search looking for diodes.You'll find common cathode Schottkys with enough current and voltage rating in most PC power supplies feeding the +12V rail. Look for a dead 800W or higher one.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 07:43:44 pm
content: Just done some testing on the transformer with 50 watt lamps. The readings are from 1 - 6 lamps. No lamps 14.7V, One lamp 14.5V, Two lamps 14.4V, Three lamps 14.2V, Four lamps 14.1V, Five lamps 14.0V, Six lamps 13.9V, I was quite impressed with that. Normally I use the the lamps for testing server modded power supplies, whilst 6 lamps is 300 watts, it comes out less on a meter. But I'm sure that at 6 lamps I was in the high twentys in number of Amps. My highest meter only goes to 20 Amps. But those reading above are one lamp added up to a maximum of 6 lamps. I could have gone past 500 watts in lamps I have, but by lamp 6 I didn't see the point. So it might be workable with an LDV regulator and a standard square package bridge rectifier ? What do you think.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 08:01:37 pm
content: Yes I might have overlooked the transistors drop across them....I'm hopping a reasonable, maybe not to much voltage drop on AC test will be exceptable when the filtering is added. In other words as the AC didn't sag to badly, it should be ok with capacitance after rectification.

author: Ian.M
author_rank: Super Contributor
timestamp: May 10, 2017, 08:06:34 pm
content: What was the current with four lamps? If it was under 16A, then try to get a current measurement with five lamps. Due to the burden voltage of the meter's 20A range, please read the transformer voltage while the ammeter is in circuit.That should give enough data to determine the equivalent source impedance and run a SPICE sim to see how much ripple to expect at 20A DC load for various capacitance values so you can see how much voltage drop you can tolerate in the bridge and how that impacts the regulator headroom.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 08:42:23 pm
content: With 5 lamps it's taken my meters 20 Amps range into over range, but the voltage was 14.0V, with 4 lamps it's reading 17.67 Amps at a voltage of 14.1V steady. Not sure if this helps, but that's as much as my meter will read with 50 watt lamps.

author: Ian.M
author_rank: Super Contributor
timestamp: May 10, 2017, 09:43:17 pm
content: That's great.No load was 14.7V, so the voltage drop @17.67A is 14.7-14.1=0.6V. This is consistent with the 0.3V drop for two lamps, and in the ballpark with the 0.8V drop for 6 lamps.0.6V/17.67A is 0.034 Ohms and that's the equivalent impedance of the secondary. (I say equivalent impedance because some of its transformed from the primary side - its *NOT* simply the DC resistance.)To a first approximation the transformer can be modelled as a sinusoidal voltage source 14.7V RMS with 0.034 Ohms series resistance.I simulated it with LTspice, with a 100,000uF reservoir capacitor ans 20A load current and found that with ideal diodes you'd have 2.6V headroom for the regulator, however with the only 25A Schottky LTspice had a model for (MBRB2545CT) there was only 1V headroom. At under 35,000uF reservoir capacitance the 25A Schottky bridge couldn't keep the ripple troughs above 12V.A controlled MOSFET bridge with individual MOSFET Rds_on of 10 milliOhms, and 100,000uF reservoir capacitor would give you a much more respectable 1.8V headroom, considerably easing your regulator design challenge.The last digit of the equivalent source impedance is pretty dodgy. If you take it as 0.04 Ohms series resistance, it reduces the headroom with the above 25A Schottky to only 0.8V. I suggest re-measuring the loaded and open circuit voltages to 2 decimal places so it can be calculated more accurately. You should also measure the mains voltage - even a 5% droop will make this supply impractical.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 10:01:50 pm
content: Thanks for running that in simulation, I can lower my expectations to 12 Amps continuous and 15 Amps 50% duty cycle. This pertains to running an RF linear amplifier. If this transformer can run 150 Watt linear amplifier I'd be happy with that. I could drop a wire size from currently 12AWG to 14AWG, but I was hopping I could just make the grade with that many turns of cable on the secondary. I don't know what your attachment is, my tablet office app won't open it. I might be able to view it on my pc.So my question is at 12 Amps, 12 Volts, could I get away with a standard bridge rectifier, and an LM2940-12 LDR regulator with a Volts drop of 0.5 Volts for the regulator. And 1.4 Volts for the bridge rectifier, or do I have to go with schottky Diodes, or the IC fets option.

author: Ian.M
author_rank: Super Contributor
timestamp: May 10, 2017, 10:08:19 pm
content: The attachment is aLTspiceschematic. I'd drop a size on the wire to get more turns in. otherwise even with the MOSFET controlled rectifier option you'll have to spend far too much on caps to be worth it..

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 10:09:10 pm
content: Forgot, although I'd like to stay as close to 12 Volts as possible, I can still operate the amplifier down to 11.40 Volts minimum. But the closer to 12 Volts would be better. And I can scrape by with 12 Amps continuous current.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 10:15:05 pm
content: Dam, dropping a wire size drives the cost up, I can only get this high quality silicone wire in 5 or 10 meter lengths. And the wire I unwind is £8.50 down the drain. If I go with schottky Diodes that's even more expense. Surely I can keep it above 11.40 Volts at the lower 12 Amps. Just asking what you think.

author: Zero999
author_rank: Super Contributor
timestamp: May 10, 2017, 10:23:31 pm
content: Quote from: davelectronic on May 10, 2017, 06:40:17 pmLike this but with a low drop-out regulator rather than the LM317?https://www.eevblog.com/forum/beginners/linear-regulated-power-supply-with-lm317t-mosfet-50n06/msg656109/#msg656109If so then, yes you lose another 0.7V to 1.2V, across R3 in the above schematic.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 10:30:24 pm
content: Yes that configuration, I can drop to a working voltage of 11.40 Volts and a continuous current of 12 Amps. I really didn't want to rewind it with thinner wire gauge if possible. With the current wire it's 12AWG, and I couldn't get another turn on it. Kick in the teeth if I've got to drop to 14AWG to get the turns count on there.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 10:35:30 pm
content: So do you think it possible to stay above 11.40 Volts with the windings I have on there now, that using a standard rectifier, and LDR regulator of 0.5 Volts.

author: Ian.M
author_rank: Super Contributor
timestamp: May 10, 2017, 10:53:58 pm
content: Dropping to 12A max eases the problem considerably. However you'll still need a lot of capacitance if you are using an ordinary silicon bridge rectifier and you want enough headroom to regulate. Unless either you can give us a part number for the bridge you plan to use and it has a SPICE model available, or you are prepared to measure its voltage drop at a range of currents up to 20A (or its max rating, whichever is lower) so we can fit a model to it, we cant do a lot better than say "suck it and see!".For those playing along at home here are a couple of high current Silicon diode SPICE models:Code:***********************************************SRC=1N1183;DN1183;Diodes;Rectifier >5A;50V 35A.model 1N1183 D (IS=23.4N RS=1.75M N=1.61 BV=66.6 IBV=4M+ CJO=4.4N VJ=.75 M=.333 TT=144U)* Motorola 50 Volt 20 Amp 100 us Si Diode 11-23-1990***********************************************SRC=1N1200;DN1200;Diodes;Rectifier >5A;100V 12A.model 1N1200 D (IS=29.7N RS=10.6M N=1.70 BV=133 IBV=4.5M+ CJO=494P VJ=.75 M=.333 TT=5.76U)* Motorola 100 Volt 12 Amp 60 us Si Diode 11-23-1990***********************************************SRC=MDA2500;MDA2500;Diodes;Bridge;50V 25A, Full Wave* single diode from MDA2500.model 1D_MDA2500 D (IS=1.52E-08 N=1.59 BV=5.00E+01 IBV=5.00E-06+ RS=1.79E-03 CJO=4.69E-10 VJ=.29 M=0.4 TT=4.32E-06)* Motorola 50 Volt 25.00 Amp 3.75 us Si Diode Bridge 07-08-1990I've revised the simulation to add them, drop the load current to 12A, and the capacitor to 27,000uF.Personally I'd strip some dead PC PSUs for high current Schottkys. Even if you have to use two separate packages for the negative side of the bridge because you are unlikely to find common anode diodes, I reckon it will be worth it to have the extra headroom and cooler running diodes,and not need to spend so much on capacitors.

author: David Hess
author_rank: Super Contributor
timestamp: May 10, 2017, 11:00:13 pm
content: 30 amp dual schottky diodes in TO-220 packages are less than $1 each at Mouser and 40 amp ones can be had for less than $2.When I built a big linear power supply like this, I used a really big surplus transformer which had a 25 volt 30+ amp center tapped secondary. So instead of a full wave bridge, I used a full wave center tap which only requires 2 diodes and therefore has half of the voltage drop. The diodes I used were stud mount 1N3900s on a big heat sink which are not ideal but I have a drawer full of them from the original power supplies that the transformer came out of. In retrospect, using half of the original massive bridge rectifier would have been better.If you could rewind your transformer for twice the voltage and a center tap, then that will halve the losses. It would be fascinating to make a full wave center tap rectifier with a pair of n-channel MOSFETs and comparators.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 10, 2017, 11:13:52 pm
content: I could do shottky diodes, but had the bridge rectifier KBPC3510-G data sheet below. That drops 1.1 Volts. And the LM2940-12 LDV linear regulator drops 0.5 Volts. That's what I'd hope to use. The transistors are the MJ11015 high power Darlington type transistors. Volts drop across them on diode check is 0.6 approx from Base to emitter and Base to collector.

author: james_s
author_rank: Super Contributor
timestamp: May 11, 2017, 12:21:17 am
content: All you can really do is try it, rig something up using the parts you have and then modify as needed.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 12:33:51 am
content: Quote from: james_s on May 11, 2017, 12:21:17 amI think your right James, I think, only think mind. Under load with the above components, I'm going to get 11.80 - 11.90 Volts worst outcome. I could be totally wrong. But a drop of 1.1V for the rectifier, 0.5V for the regulator, and 1.2V across the resistor transistor network. If I got 11.40 Volts hopefully a little more, at 12 Amps I'd be happy with that.

author: Ian.M
author_rank: Super Contributor
timestamp: May 11, 2017, 06:46:18 am
content: Simming it with a silicon bridge rectifier, it doesn't look too bad with enough good low ESR capacitance but that bridge will kill your headroom. Due to the supply current being concentrated near the peaks of the voltage waveform, I'm seeing about 45A peak through the bridge for 12A average DC current. That's only 20.5A RMS so well within your bridge's ratings. However, if you look at Fig.3 of its datasheet, on the 35A curve, the instantaneous forward voltage dropPER DIODEis 1.2V. That means you are loosing 2.4V that you can ill afford in the bridge. For comparison, the sim says the Motorola MUR2500 50 Volt 25.00 Amp bridge I have a model for drops 0.9V per diode. A good set of Schottkys can save you about a volt of total drop in the bridge, headroom you badly need if you want to push above 12A, or when the mains voltage drops when half the country puts the kettle on in the commercial breaks in Corrie or at the end of East Enders.Its looking like you need 20mF minimum capacitance to keep the ripple trough above 12.5V so you've got enough headroom to regulate. However I'm seeing 16.7A RMS ripple current in that cap for a 12A load so you are going to need a bank of them to split the ripple current up to something more reasonable so they don't puke their guts at full 12A load current. The best deal (price per mF) on decent quality caps I could find at Mouser washttp://www.mouser.co.uk/ProductDetail/United-Chemi-Con/EKYB250ELL682ML40S, (25V 6800uF, 4.22A ripple current), and you'll need at least six in parallel to meet the ripple current spec. Eight would give you some margin, but you might as well go the whole hog and use a bank of 10 as there's a price break at qty 10.Recalculating with 68mF capacitance, using 1/10 the worst case ESR from the above cap, and assuming 0.05R wiring resistance between the bridge and the caps, it looks like you can maintain 1V headroom at 15A load current. The ripple current is 24.6A RMS, well within their rating.Unless you've got some server grade 25V caps hanging around, it looks like it will cost you £13.50 for the caps. They are going to need decent busbars as well - probably the easiest option without getting a pcb made would be to mill a break in the copper of a piece of single sided copper clad, drill it for all the caps and sweat solder two heavy gauge solid coper wires to it to improve its current carrying capability.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 09:35:11 am
content: Thanks for the info Ian, I've never used simulation, so no idea how that looks. But having watched a few videos on it, so have some idea. I can see how close it is, and yes i got the bridge rectifier wrong, it says 1.1Fv per element, so that makes sense. I might just rewind it, as much as it pains me to waste cable. The cable on there now is 12AWG with an outer diameter of 4.5mm, a drop in cable size to 14AWG is 3.6mm outer diameter. That would allow the more turns count, but i would definitely have to drop my expectation to 12 - 15 Amps then, not much of a big deal.As its just a matter of removing bolts, I'm tempted to build it, try it, then rewind the transformer if its not executable with the 12 AWG cable. Might as well see.

author: David Hess
author_rank: Super Contributor
timestamp: May 11, 2017, 09:42:36 am
content: Quote from: Ian.M on May 11, 2017, 06:46:18 amI built a couple of power supplies like I described above and the stranded copper wire between the transformer and rectifiers was so large that it supported the flat heat sink that the stud package rectifiers were mounted on; that is not a good construction practice but it worked out.The screw terminal capacitors were so large that the regulator board including the heat sinks mounted to the input capacitors instead of the reverse or having the input capacitors separate.The regulator circuit was the one Hero999 shows but with an LM337 negative regulator instead of an LM317 positive regulator so I could use 2N3055 NPN transistors; it worked fine but if I did it again, I would use the configuration shown below from National Semiconductor's Voltage Regulator Handbook.I added a fan directly above the pass transistor heat sink and made a simple temperature control circuit using an operational amplifier and the -2mV/C temperature coefficient of a diode epoxied to the heat sink. This lowered the thermal resistance of the heat sink by an order of magnitude which is no surprise. Since the error amplifier was an integrator, the voltage driving the fan was roughly proportional to power dissipated (the temperature sensor operated at a constant temperature) although I never took advantage of that; at nominal output current, the fan was completely silent.The only issue I ever had was with the emitter ballast resistors which were not rated for enough power and not cooled by the fan. At an unrealistically high output current for an extended time, they would glow dull red which was more alarming than a problem.If I did a high power regulator like this again, I would add external controlled foldback current limiting to the protection already built into the integrated regulator. I might make a delta Vbe based temperature control for the fan so no adjustments are needed for the offset which varies between diodes. An SCR crowbar on the output would be a good idea to protect whatever is being powered from a pass transistor short; this requires a fuse which is probably located between the input capacitor and the regulator.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 11:16:07 am
content: My circuit is pretty much the same, with out that diode on the regulators input. I've built a few now, all with multiple pnp pass transistors. The last one was a 13.20 Volt psu which can do 15 Amps, by way of a 7812K and 4 x MJ4502 transistors, its in use now on a cb radio linear amplifier, and power a couple of radio scanners. That one has a 300VA toroidal transformer in it with a secondary of 15 Volts. But reads AC 17.6V no load. Its 22 volts DC rectified and filtered, and 19 Volts at the bridge under a 12 Amp load. What I'm going to try first with this MOT and its 14.7V AC output, is the 35 Amp diode bridge, the LM2940- 12 LDV regulator, and 2 x MJ11015 transistors. Capacitors I'm trying with this will be snap in radial electrolytics 10000 each x 6 capacitors in a bank. If that can manage above 11.40 Volts at 12 Amps i will be happy with that. But closer to 12 Volts for the same load would be better. If it drops under 11.40 Volts with the above, then i will wind a slightly lower wire diameter on the MOT. Last one i built below.

author: Ian.M
author_rank: Super Contributor
timestamp: May 11, 2017, 11:31:44 am
content: While you are gold plating it, what about using a DC to DC converter or an auxiliary transformer or PSU to provide a boosted*supply for the control circuit? Then you can use a LM723 with a simple NPN Quasi-darlington#pass transistor so you can regulate right down to 0.5V input-output voltage differential!The LM723 makes it easy to implement foldback current limiting and load voltage sensing. Protect it against sense disconnect by shunting each sense terminal to the corresponding power terminal with a 1N4001.* You only need 5V extra headroom - a 5V 1A USB supply, riding on top of Vunreg would do nicely.# Using multiple output transistors, with emitter resistors for current sharing, and feeding the driver transistor from the boosted rail

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 11:47:37 am
content: Now you really got my mind in overdrive...I don't know about a dc to dc converter, but its got me thinking. So has an additional transformer for regulation with higher voltage for the control transformer. I'm thinking first which one to try, and a parallel configuration maybe. But how to implement the control transformer, so it regulates only, and all power is taken from the MOT.I never gold plate btw, its to expensive. But you've sure got me thinking.

author: Ian.M
author_rank: Super Contributor
timestamp: May 11, 2017, 12:03:12 pm
content: That's fairly simple. The boosted rail feeds *everything* except the fan and the collectors of the paralleled output (pass) transistors. I You need 1A so you've got plenty of base drive for the output transistors (forced HFEof 20) 2A would be better if you are using crappy low gain output transistors. It will only draw what it needs from the boost rail till the unreg rail collapses to within about 0.5V of the output, at which point the boost rail current will start to rapidly increase due to the driver transistor trying to keep the load voltage up. It might be a good idea to detect the excessive increase in driver collector current by adding a sense resistor and use it to turn on a PNP to inject current into the LM723 CL pin (via a resistor) and trigger the current limiting.Run the fan off Vunreg via a LM317 set to 14V to prevent it being over-voltaged. You can even add a thermistor on the heatsink to vary the fan voltage according to temperature so its not too noisy on light load.I've just figured out how to get the boost rail essentially for freeNo extra mains wiring, no extra SMPSU boards or DC to DC converters, just 4 components (2 if you use a common cathode Schottky, and a single large cap).Use a pair of 2A Schottky diodes off the AC terminals of the bridge to feed their own reservoir capacitor. 10mF will give you a boost rail that stays above 16.5V when loaded with 1A - either use two of the same caps as the main output or something cheaper as it doesn't need the high ripple current rating. Revised LTspice schematic attached

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 12:10:25 pm
content: As intersting as it would be to build a control circuit for the regulator, i feel it would start getting to busy under the hood. Compared with different cable diameter it would work out more expensive.I've took the plunge and ordered the same quality high temperature silicon wire, but in 14AWG instead of the 12AWG that's on it now. The 14AWG wire has an outer diameter of 3.6mm and conductor diameter of 2.1mm, and I've ordered 10 meters of it. I should easily get 18 Volts plus on it with that. But i will aim for 20 Volts AC so i can use components i already have. Cable cost £10.58, if the MOT had a slightly bigger core i wreckon I'd have got away with it.

author: David Hess
author_rank: Super Contributor
timestamp: May 11, 2017, 12:25:49 pm
content: Quote from: davelectronic on May 11, 2017, 11:47:37 amThe control transformer does not need to be higher voltage. Since its output is floating, the secondaries can be stacked so you have a high current main supply and a slightly higher voltage low current supply for the control circuits. The control supply only has to be large enough to supply the base drive to the transistors.This configuration is how the big Astron linear power supplies work although they wind the control transformer winding as part of the main transformer winding.

author: Zero999
author_rank: Super Contributor
timestamp: May 11, 2017, 03:50:41 pm
content: Quote from: davelectronic on May 10, 2017, 11:13:52 pmThe volt drop on the base-emitter using the diode test function on your meter will not be indicative of real life conditions. The diode test function uses a tiny current, compared to what you'll have when your regulator is fully loaded. It's a Darlington pair so the voltage drop will be higher than a single BJT. When fully loaded, VBEwill be somewhere between 2V and 3V, giving a total drop-out voltage of 2.5V to 3.5V.

author: Ian.M
author_rank: Super Contributor
timestamp: May 11, 2017, 05:42:29 pm
content: Quote from: David Hess on May 11, 2017, 12:25:49 pmIts obviously better to stack another supply but if you are really stuck with the transformer you've got, and the separate peak rectifier I posted earlier doesn't give you enough headroom, or has too much ripple, you can use a capacitively coupled bridge to provide the control supply.Here's one with a regulated 17.2V control supply, for 5V headroom for the control circuits at up to 2A.The reason for the discrete Zener + Sziklai pair control voltage regulator rather than an integrated regulator is to let it act as an amplified capacitor below the dropout voltage so its better behaved when heavily loaded.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 06:25:56 pm
content: This might have been an option, but I've removed the secondary as of this morning. Some 14AWG is on its way. The bit I'm struggling to understand is how the control transformer doesn't interfier with the collector output of the MOT. Surely the control transformer has to feed the input of the regulator, oh hang on, I confused slightly. If the control transformer only feeds the regulator, does the power to the transistor ballast resistors come from the MOT. And the only purpose of the MOT is to feed the emitter to collector current ?I have looked at the schematic Ian, I'm still trying to get my head round the two transformers relationship in the circuit. In other words, so the circuit doesn't try to take its high current from the control transformer. Hope that makes sense. Although I will rewind the MOT now with greater turns, this concept could be useful for the future.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 06:30:58 pm
content: Quote from: Hero999 on May 11, 2017, 03:50:41 pmHopefully I will get a greater turns count with the next gauge down wire I've ordered, with greater turns count.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 08:34:35 pm
content: Quote from: Ian.M on May 11, 2017, 12:03:12 pmWon't the second voltage from the shottky diodes to the regulator get dragged down, when the the output is loaded heavy ? As it's the same source. I can't see a schematic in your post Ian, just two bridge pdf files.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 08:47:44 pm
content: Yes I think I can see it, where the two schottky Diodes feed another separate filter capacitor. And the regulator voltage would stay above the drop out voltage. So that does nothing but feed the input rail to the regulator ? Think I've got that right. And there's nothing there to drag that rail low. Great idea ! Wished I'd seen it before stripping the core again.

author: richard.cs
author_rank: Super Contributor
timestamp: May 11, 2017, 09:43:59 pm
content: Quote from: davelectronic on May 11, 2017, 12:10:25 pmThe silicone wire is part of the problem here, if you built it with enamelled wire (as is usual for transformers) then the outer diameter is around 0.1 mm more than the conductor diameter. You could have 12 awg solid core enamelled with 2.05 mm conductor diameter and an overall diameter around 2.2 mm, and fit nearly twice the number of turns on as with 3.6 mm diameter silicone insulated wire.

author: james_s
author_rank: Super Contributor
timestamp: May 11, 2017, 09:48:07 pm
content: You can even get square enameled wire for when space is a premium. Insulation rating up to 200C which is hotter than I'd want any transformer I was using to run.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 10:07:06 pm
content: Quote from: james_s on May 11, 2017, 09:48:07 pmI know, I've wound a MOT after splitting the E I sections. On reassembly I bonded the core back together with super strong epoxy. It drew more current from the primary than it did before separation. Meaning just the primary, secondary open circuit. I didn't have a welder to mechanically reconnect the core. They say just touching is enough for mutual inductance, I believe the separation, and glue bonding hindered the processes of the transformers mutual inductance. I'm going nowhere near 200°C I'd be surprised if I reached 85°C with my application. It only requires 12 - 15 Amps, and most of that is at a 50% duty cycle. Coupled with forced air cooling at two speeds, series resistor for idle cooling, thermal switch shorting out the resistor for full speed fan cooling.

author: james_s
author_rank: Super Contributor
timestamp: May 11, 2017, 10:29:10 pm
content: Large hose clamps are another effective way of attaching core halves. I've also used cut C cores from Alphacore to wind custom transformers, a hose clamp works well for assembling those.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 11, 2017, 11:16:49 pm
content: Quote from: james_s on May 11, 2017, 10:29:10 pmThis is a great idea, just wish I'd thought of it back then when I did separate that MOT core. I'm struggling with some health issues, so everything is super slow. If I can avoid a triathlon I will. That last psu might take the average guy a weekend, two at the most. Took me 5 months, just over lol. Do a little here and there, the day I throw my hobby in is the day I throw in the towel.Just want to say thanks to everyone for the help, I've got a lot of food for thought from all your posts. When I get a better voltage on the secondary I will post the result.

author: Ian.M
author_rank: Super Contributor
timestamp: May 12, 2017, 02:14:40 am
content: I've been playing around with a LM723 sim - if you hadn't torn it down I *think* you could have got 12V at up to 15A out. Zipped LTspice sim attached.However it would run quite a bit cooler if you put a 6V RMS 1A auxiliary winding on for the boost rail, and connect it to the auxillary bridge in place of the 1500uF caps coupling the auxillary bridge to the main secondary, and it would be nice to have *FRACTIONALLY* more headroom so winding the main secondary to deliver about 1V more at full load would be a good thing.

author: floobydust
author_rank: Super Contributor
timestamp: May 12, 2017, 02:48:41 am
content: Here is an old Elektor magazine LM723 circuit and they used a voltage-doubler similar to your idea, but oddly enough not to drive the triple.I'd add current limiting resistor on VC or Q3 base as things will go bad with up to 18V drive into Q4-Q6. Q3 or U1 will cook.I always include a protection diode from output to the main filter cap. These designs blow up if you ever connect a battery to it and the power supply is off, or if the output ever sees more voltage than the setpoint.

author: Ian.M
author_rank: Super Contributor
timestamp: May 12, 2017, 07:50:04 am
content: Quote from: floobydust on May 12, 2017, 02:48:41 amGood point, bit in this circuit its self-limiting over periods longer than a few us. Considering the capacitively coupled bridge as a charge pump, it can only deliver 14V*1500uF*100=2.1A. Over that, the boost rail drops below the unreg rail and Q4-6 pick up the load. However, transiently the 3.3mF bulk capacitor on the boost rail can supply a lot of current, and Q3 is under-rated for driving 3x 2N3055, so adding 100R in series with U1:Vc and swapping Q3 for a 5A Ic_max part would be a good idea.If the CC bridge is replaced with a proper PSU for the boost rail, it should either be current limited or a current sense resistor added in Q3's collector circuit with a PNP to pullup U1:Cs via a resistor to trip the main current limit.Quote from: floobydust on May 12, 2017, 02:48:41 amYes, and a 20A fast fuse on the output would be a good idea to improve survivability if some idiot hooks up the battery backwards!

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 12, 2017, 09:25:58 am
content: Thanks for posting the schematics, and all the circuit suggestions. For this one I'm going to just use the LM317K or the 7812K, using the MOT was just a cheap and chearful means to an end. The main reason being logevity of the MOT in a psu environment, if it didn't survive long i won't lose a lot if the psu failed. I am planning on using an LM723 in the future, but would make that with a purchased transformer, and knowing its going to have good longevity with a proper mains transformer. The MOT was just an opportunity as our microwaves magnetron went pop, and not chance of repair or parts. Before it went in the bin, i thought i would knock up a cheap but effective psu with the rewound MOT. If the MOT fails early, not to much circuit investment would be lost. More in terms of hours building as opposed to money. Everything takes me double or longer than the average guy. But i certainly hope to try the LM723 in the future, with all the bells and whistles added.

author: Zero999
author_rank: Super Contributor
timestamp: May 12, 2017, 11:37:22 am
content: A voltage doubler is a good idea to bias the input to the output transistors above the output voltage.What about the TL431 plus a MOSFET? The transient response when the load is suddenly applied isn't great but other than that it seems to be quite stable and will have a very low drop-out voltage.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 12, 2017, 06:19:00 pm
content: Maybe I shouldn't have unwound the 12AWG off the core, it looks like there are a lot of options avaliable for a borderline input voltage from the transformer. I really like some of these ideas, I thought it would be a headache and might not work with a modified input, but all the options look good.

author: Zero999
author_rank: Super Contributor
timestamp: May 12, 2017, 08:09:29 pm
content: Another thing you could consider is replacing the diodes in the bridge rectifier with MOSFETs and use the LT4320 ideal bridge rectifier controller. The size of the filter capacitor can then be reduce3d slightly. You'll need to use MOSFETs with a very low on resistance because the peak current will be around 50A. 5mOhm MOSFETs would give a total voltage drop of 0.5V across the bridge.http://cds.linear.com/docs/en/datasheet/4320fb.pdfHave you considered a switched mode power supply on the secondary side of the MOT? It would be a good learning experience. You could even make it isolated and power factor corrected - the sorts of things normally done with an SMPS running from the mains, except you don't have to worry about dangerous voltages.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 12, 2017, 10:00:43 pm
content: It's something to think about, I've heard of the fets bridge in the past, but never considered it. Only because I've always had plenty of voltage headroom. Space is tight in the case, I didn't think the MOT would swallow up much room.With the last psu I mounted the circuit board above the transformer, using very long M4 bolts as standard offs. This time the MOT is a bit higher than the toroidal transformer I used in the last psu. At the moment I've only used copper strip board as crazy as it sound, making large heavy solder tracks when load current demands it. It's the only down side with big power and heavy current, everything needs to be large wire diameters, or multiple twisted pairs. A smps secondary, not something I've heard of before, coming out of a linear transformers secondary. Sounds a bit past my ability at this stage. Maybe if I get as far as an LM723 based power supply, I'd have more confidence to try that sort of thing.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 16, 2017, 07:05:27 pm
content: Just an update...The 14AWG turned out to be poor quality, thin walled. Not seen in the same stuff that was 12AWG. So with a control circuit for the regulators input, what would be the minimum voltage I could wind the MOT with to get the current from it. Going on my regulator feed from a separate transformer, or use a dc to dc converter step up for the regulator. So if I want to keep the current up at 12 Volts regulated output, is 12 Volts AC unloaded on the MOT high enough if the regulators input needs are met. Thanks for reading, any help appreciated.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 16, 2017, 07:24:09 pm
content: I've found a reasonable steps up converter for the regulators input from the MOT. Is 12 Volts AC target unloaded on the MOT usable, after rectification and filtering. I realise the peak voltage is important for the current delivery at 12 Volts regulated output. But if the regulators input is achieved, could the MOT supply the current at 12 Volts AC and 16.92 Volts after filtering, less the rectifier and pass transistor voltage drop.

author: Ian.M
author_rank: Super Contributor
timestamp: May 16, 2017, 07:48:26 pm
content: No. You need to shoot for about 10% more RMS voltage AT FULL LOAD than the desired DC output + enough extra on top of that for the diode drop and the regulator headroom.To a certain extent you can trade off secondary voltage against reservoir capacitance, but increasing the capacitance rapidly becomes expensive as it has a knock-on effect on the specification of the diodes required.

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 16, 2017, 08:28:34 pm
content: Ok, might be a step backwards then. I've rewound the first good quality 40 Amp cable, for a maximum unloaded output of 11.40 Volts. A higher voltage means buying more cable. So if using a step up converter for the regulator input I think will regulate properly. The bit I'm unsure about is the 11.40V x 1.41 = 16.074V , from the MOT to the input to the transistor ballast resistors emitter. I'm thinking input from the dc to dc step up converter to the regulators input, and the 16.074V in to the emitter ballast resistors. But that's an unloaded voltage. The MOT loaded to 200 Watts drops to 11.00V exactly. So thinking 15.51V through the ballast resistors, although the actual voltage from emitter to collector is probably going to be less. I know the peak is different, but is that enough voltage from the MOT ?

author: Ian.M
author_rank: Super Contributor
timestamp: May 16, 2017, 09:20:27 pm
content: The problem is the lowest voltage of the ripple. Ignoring the source impedance of the transformer winding, forward resistance and voltage drop of the diodes etc. plot two half cycles of a full wave rectified sinewave, with amplitude 11*sqrt(2)V as accurately as yo reasonably can on graph paper. Rule a horizontal line across at the output voltage you want, then above it a parallel line adding the total bridge Vf drop and the regulator dropout voltage. Draw a straight line tangent to the previous peak of the half sinewave to the point where the upper line intersects the rising slope of the next half cycle. That line + the half sinewave from the intersect point to the tangent point is the ripple waveform, and the slope of the line that constructed it, together with the max load current gives you the required capacitance byC=I/(dV/dt)which is derived from Q=CV. You can also calculate how much current must flow to recharge the capacitor, but that gets a bit trickier. You can approximate it from the change in charge and the time between the intercept and tangent points, with the slope at the intercept point giving you an upper limit.Alternatively, simulate it in LTspice or your other favourite Spice package, which lets you model the transformer secondary source impedance and see the effect of non-ideal diodes.SPICE also has the benefit of letting you easily calculate the capacitor RMS ripple current, which if you exceed the cap's max rating will kill it fairly quickly

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 16, 2017, 10:14:40 pm
content: Thanks for taking the time with the expansion, I know roughly what you mean, but couldn't see it unless I put it on paper. Maths is not my strong point, visualisation to product is. My best option I think is to use a simulation program. I'm a Linux user, so will have to look in to what's going to work on that system. I should have known better than to get cheap cable off ebay. I'm going to have one more go with the 25 Amp cable, same as the 40 Amp stuff I already have. Just can't get the turns count on this 700 watt transformer core with the 40 Amp cable. I'm starting to question cost now, it's only that I want to see the finished viability of using a MOT in a low voltage power supply. Another £16.00 on cable coming up, I must be mad...

author: Ian.M
author_rank: Super Contributor
timestamp: May 16, 2017, 10:30:52 pm
content: LTspice generally runs pretty well under WINE on Linux

author: davelectronic
author_rank: Frequent Contributor
timestamp: May 16, 2017, 11:25:40 pm
content: I will have a look, give it a go, cheers Ian.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 12, 2017, 10:46:35 am
content: I've not managed to get LTSpice working on linux, what I've got now is the transformer in the chassis and the 35 Amp bridge rectifier in circuit. This drop 1.58 Volts from AC 14.7 Volts down to 13.12 Volts. I'm using a low drop out voltage regulator with minimum input of 13.60 Volts. The circuit has two MJ11015 darlington high power transistors, what I'm hoping is to use some snap in 25 Volt 6800uf capacitors x 9 capacitors, so 61200uf of capacitance to keep the voltage up for regulation.These capacitors are reasonable price for that value. I was wandering if that is a high enough capacitance to regulate 12 Volts at a maximum of 12 Amps current, i can fit 10000uf capacitors x 9 so 90000uf total. But these capacitors are nearly twice the price of the 6800uf capacitors. No other change in components. So I'm looking to find out if 61200uf of capacitance for 12 Volts 12 Amps is high enough. Any help appreciated.

author: mariush
author_rank: Super Contributor
timestamp: June 12, 2017, 01:51:10 pm
content: Capacitance (Farads) = Current / [ 2 x Mains Frequency x ( Vdc peak - Vdc min) ]Use LT4320 and four mosfets like Hero99 (and i think me) recommended to get smaller losses in the bridge rectifier - you'd have around 15w dissipated in the bridge rectifier right now.It's not THAT expensive and you may be able to get by with just 3-4 x 10k uF 25v instead of 9 6800 or whatever.If you don't have the bridge rectifier making heat there, you may even be able to use cheaper 85c rated capacitorsFor example, you can get Samwha 22000uF 35v from TME.eu for 2 uk pounds each :http://www.tme.eu/gb/details/hc1v229m35045ha/85c-snap-in-electrolytic-capacitors/samwha/Or 33000 uF 35v for 2.8 uk pounds (3000h@ 85c) :http://www.tme.eu/gb/details/hc1v339m40050ha/85c-snap-in-electrolytic-capacitors/samwha/In 105c you have 10000uF 50v rated for 1.6 uk pounds :http://www.tme.eu/gb/details/he1h109m30050ha/105c-snap-in-electrolytic-capacitors/samwha/

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 12, 2017, 06:28:53 pm
content: Thanks for the reply, I've committed hardware to the case now for the bridge rectifier. There's precious little room inside the case as it is, I'm not sure what else is needed in the fets rectifier circuit. My two boards are stacked, one for the capacitors, the other the control board. I could always just go with a separate regulator input from the transformers AC for the regulator. That shouldn't sag with a less than 1 Amp regulator load.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 13, 2017, 05:07:19 pm
content: So I've mounted the transformer and bridge rectifier, added 10000uf just as a test, getting 21 Volts no load. Might be ok yet with a moderate level of capacitance.

author: Ian.M
author_rank: Super Contributor
timestamp: June 13, 2017, 06:34:57 pm
content: Put a load on it, (ideally at least 2/3 the final max load) and scope it. Measure the slope of the ripple between halfwaves and the minimum and maximum voltage. That should give you enough data to determine if its practical and how much more capacitance it will need.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 13, 2017, 07:40:34 pm
content: It's the ideal thing to do for sure, but I've not got the use of a scope at the moment. I'm going to take a chance on 60000uf, I could go to 90000uf but that would cost double what I can get 60000uf in 6800uf snap in high ripple current capacitors. Should be ok up to 12 Amps...

author: Ian.M
author_rank: Super Contributor
timestamp: June 14, 2017, 05:08:34 am
content: Quote from: davelectronic on June 13, 2017, 07:40:34 pmThat's a pity. You *really* need that data to avoid throwing more money away.What I am concerned about is whether the diode drop becomes excessive at your required load current - I expect the average output voltage to be under 13V and that adding more capacitance to maintain a higher voltage will be ineffective due to the increased peak current in the diodes increasing their voltage drop.The discharge rate of the bulk capacitor can be calculated so you don't have to actually measure the slope, but it would be worth setting up to measure the peak and trough of the ripple under load using a diode and a capacitor(Peak reading voltmeter is easy. For the trough reverse the diode and use a 1Meg pullup to a higher voltage. Calibrate out the diode drop using known DC input voltages)Also measure the mean DC voltage, and ripple directly using the capacitor for DC blocking as a cross-check. You'll need to know whether the DMM is true RMS or half or full wave average (try both polarities!), calibrated as RMS equivalent for a sinewave, as you need to convert the ripple to a pk-pk voltage.Another option would be to use an Arduino (or other MCU with ADC and UART) as a 'pooor man's' oscilloscope.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 14, 2017, 09:03:14 pm
content: Wish I had a scope..Probably just use 9 x 10000uf capacitors. When think how many times I've rewound the MOT etc, it's all bolted in now with 14 AWG on it. I've used the same case as my previous project psu (now in use) This one's just a bit of fun really. I can't run it 24/7 it would rack up my electricity bill. Last MOT primary I tested was 700 watt unit, it drew 3 Amps no load. I know there wasteful in the power department, it's an occasional use DX HF radio psu for high power linear amplifier.I've thought about building it so the MOT only powers up on HF linear use, and a more modest transformer runs long term RX radio state. But it was only a thought. I've put so much in to it I might as well do it properly and see it to the end.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 26, 2017, 10:31:26 pm
content: That's 90000uf 9 x 10000uf capacitors, it's giving 21 Volts. Let you know the end voltage result when I've built the control board. It's got the low drop out voltage regulator under the small heatsink and 40mm fan. Should be interesting to see the loaded voltage.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 26, 2017, 10:32:36 pm
content: Capacitor bank, only one picture posted in first try.

author: davelectronic
author_rank: Frequent Contributor
timestamp: July 14, 2017, 12:24:33 am
content: Finally finished the MOT project, as I only had 14.70 Volts secondary voltage to work with, I lowered my expectations to 10 - 12 Amps at 12 Volts. In practice I've got that easily, and fairly sure I could push 15 Amps at a 50% duty cycle out of it. It's to power a HF linear amplifier, so duty cycle anyway. Just wanted to say thanks for all the help, much appreciated.PS. Front posts, rear sockets unloaded voltage reading.And 150 Watt halogen lamps load. Current and voltage reading. After 90 minutes continuous use, the transformer reach 76°C the transistors 52°C and rectifier 48°C. A total capacitance of 94700uf was used.