Voltage overhead.

[davelectronic]

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.

[james_s]

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.

[Ian.M]

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 drop   

How many VA is the transformer and how much current do you need out at 12V?

[davelectronic]

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.

[David Hess]

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.

14.6 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.

[mariush]

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/LT4320

It'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/2396663
125c version : http://uk.farnell.com/linear-technology/lt4320hn8-1-pbf/ideal-diode-bridge-controller/dp/2406673

Digikey :

85c version :
[1] https://www.digikey.com/product-detail/en/linear-technology/LT4320IN8-1-PBF/LT4320IN8-1-PBF-ND/4693753 
[2] https://www.digikey.com/product-detail/en/linear-technology/LT4320IN8-PBF/LT4320IN8-PBF-ND/4693752

125c version : https://www.digikey.com/product-detail/en/linear-technology/LT4320HN8-PBF/LT4320HN8-PBF-ND/4754663


The 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.

[davelectronic]

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.

[Ian.M]

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.

[mariush]

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 current

link lt1083/lt1084/lt1085 : http://cds.linear.com/docs/en/datasheet/108345fh.pdf

[David Hess]

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.

[davelectronic]

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.

[dmills]

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.

[davelectronic]

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 ?

[dmills]

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.

[mariush]

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 MBR60L45CTG

max 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 FERD40U50CFP
max 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-47CTQ020PBF
Forward 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 rectifier

But 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.

[james_s]

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.

[davelectronic]

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.

[james_s]

Look at the schematic of a bridge rectifier, the diodes in it can be drawn as two pairs, each tied together at one pin.

[davelectronic]

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.

[james_s]

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.

[Zero999]

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.

Can 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 V_BE (0.7V to 1.2V) drop on top of the regulator IC's drop-out voltage.

[davelectronic]

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.

[davelectronic]

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.

Can 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 V_BE (0.7V to 1.2V) drop on top of the regulator IC's drop-out voltage.

This would be PNP transistors, so collectors on the output,and input the the emitters. And a Base resistor, also ballast resistors.

[james_s]

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.

[Ian.M]

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.