Dropping voltage with diodes.

[davelectronic]

I'm putting together a linear power supply, I expect to get 13.80 Volts at a power rating of 10 Amps. I've used a toroidal transformer with an 18 Volt secondary, the problem is I've put the AC transformer output through a bridge rectifier and filter capacitors. But now I have 27.00 Volts AC reading from the multimeter. I think this is to high to put into a linear regulator, and need to drop that voltage some what lower. So could I use rectifier diodes to achieve this ? The power supply is built now with regards hardware installed. Or is dropping voltage this way not recommended. Thanks for reading, any thoughts appreciated.

[Psi]

Keep in mind that 27V will drop back a bit once you draw some current. You're charging the capacitors up to the peak voltage but the peak doesn't have much energy.

What linear regulator are you using for 10A at 13.8V?    27V isn't that high for a regulator to handle voltage wise.
But you are going to burn off a lot of heat to take 27V down to 13.8V.   It won't be the full (27 - 13.8 ) * 10A = 130W
but it probably will be 50W or so.  We're talking a pretty huge heatsink or a fan.

Dropping voltage with diodes isn't all that practical most of the time.
It not all that accurate as the Vdrop changes with current and you need lots of diodes if the drop is large, and it doesn't save you anything because the same heat just ends up in the diodes instead of the regulator.

One option is to un-wrap the toroidal transformer and remove some turns from the secondary until you get the voltage a bit lower.
This is usually pretty easy to do with toroidal transformers, compared to other types.

[davelectronic]

Thanks for replying, yes it's a fairly big drop to 13.80 Volts. I wanted to close that gap. The linear regulator is an LM317 and I was going to use fixed value resistors to achieve the 13.80 Volts output. The transformer is a toroidal type with twin secondary outputs I have wired in parallel. I'm not sure how to remove which windings and how many turns. I have an extra toroidal transformer with the same ratings I could experiment with. Just wouldn't want to remove to many turns. I know I have to close that voltage gap to bring down the waisted power and heat. I should mention I am using current boosting transistors to obtain the 10 Amps at 13.80 Volts. Most of the time I will be drawing about 8 Amps.

[soldar]

Why not just a power resistor?

[langwadt]

One option is to un-wrap the toroidal transformer and remove some turns from the secondary until you get the voltage a bit lower.
This is usually pretty easy to do with toroidal transformers, compared to other types.

can also add windings and connect them in reverse

[Solder_Junkie]

Don’t forget to add over Voltage protection to the output, when linear regulator transistors fail it is usually short circuit.

Originally a crowbar circuit was used, more recently MOSFETs with a monitor IC, such as the LTC-4368.

SJ

[JohanH]

Why not just a power resistor?

Not a bad idea in general. An RC-filter will always lower the ripple.

[nali]

You also need to consider the *minimum* voltage troughs which need to be greater than 13.8 + the minimum dropout voltage for your reg which is 1.7V in the case of a LM317 (not considering any current bypassing scheme) so >15.5V.

What size filter caps are you using?

[Ian.M]

The voltage on-load will drop considerably, maybe as much as by a third.

There's also mains voltage variation to take into account, which will proportionately affect your output voltage - legally the UK mains voltage is 230V +10%,-6% due to EU harmonisation.  In practice none of the nom. 240V equipment was replaced, and the national grid seems to prefer to run their adjustment taps a bit high so they've got more margin as loads come on, so although in urban areas the UK mains voltage tends to be reasonably close to 240V, in lightly populated rural areas you can see significant variation.

You need to measure the supply voltage, do a full load test and do the math before you decide on any dropper, as you may have far far less voltage headroom than you think. 

Also don't forget that the RMS current in a secondary feeding a bridge rectifier + reservoir capacitor is approx. 1.5 times the average DC load current, so your transformer needs to be at least 270VA for continuous 24/7 operation.  250VA would probably be acceptable if  its never going to be loaded at the full ten amps for longer than half an hour or so, but the no-load/full-load unreg DC voltage swing will be even greater.

[langwadt]

The voltage on-load will drop considerably, maybe as much as by a third.

There's also mains voltage variation to take into account, which will proportionately affect your output voltage - legally the UK mains voltage is 230V +10%,-6% due to EU harmonisation. 

I believe the EU standard was changed to 230V +/-10% in 2009, but some suppliers might chose to have tighter limits

[iMo]

What is your DC voltage at the capacitor (wired after the rectifier) when loaded with your 10Amps (or 8Amps)??

[DavidAlfa]

I believe the EU standard was changed to 230V +/-10% in 2009, but some suppliers might chose to have tighter limits

AFAIK the standard is 240 now. There's 236V in mine!

[BeBuLamar]

The LM317 is good for 37V but yes dropping down to 13.8V does waste a lot of energy.

[soldar]

AFAIK the standard is 240 now. There's 236V in mine!

Your belief is very mistaken. All of Europe, including the UK and Spain, mandate 230V. In Spain since 2002.

Real Decreto 842/2002, de 2 de agosto, por el que se aprueba el Reglamento electrotécnico para baja tensión.
Publicado en:     «BOE» núm. 224, de 18/09/2002.

[Ian.M]

OTOH as I mentioned earlier, the various supply companies like to set the transformer taps to run the final distribution lines voltage biassed towards the high end of the tolerance range when lightly loaded, because it gives them more margin for droop at full load, and the consumers pay for power, not current, so the I²R distribution losses are reduced per KW delivered.

[DavidAlfa]

Aha, it's 230V with 7% margin so legally anywhere between 214-246V

[davelectronic]

I haven't loaded it down yet, after the rectifier and filter capacitors. It's a 300 VA toroidal transformer, rated at 18.00 Volts. I have previously used two 300 VA toroidal transformers for another power supply, they where rated at 12.00 Volts. They sagged very little with a 200 watt load. I don't have an electronic load at the moment. All I know is I need 13.80 Volts at 10 Amps maximum, although the actual current drawn will be closer to 8 Amps when running HF radio equipment. So it won't be a continuous load, more like a 50% duty cycle. When I have measured the mains where I live, many times over the years, I've never seen a reading lower than 240 Volts. This is a single very high power transistor, it's a Darlington rated at 50 Amps 300 watts. I wanted to see if I can get a linear power supply, to produce 10 Amps at 13.80 Volts with a single transistor. I'm not sure what to do with this 300VA 18.00 Volt transformer. I hoped to avoid removing windings, it never as tidy after doing that. I suppose I must find a load equivalent to 10 Amps to see what that rectified and filtered 27.00 Volts secondary drops to with a 10 Amps load. I can't use lamps unless I connect them in series, and then in parallel. Just parallel would blow them. The power resistor sounds like an interesting way of dropping power, although I imagine it would produce loads of heat. And would have to be a metal clad resistor bolted to a suitable heat sink. Now I think of it in terms of the voltage drop under a 10 Amp load, it might be ok, as you say, that 27.00 volts should drop considerably.

[iMo]

BTW. - the ham radio transmitters have around 50% efficiency, so you would need around 15-17A off the 13.8V in order to run 100Watts output TX power. I do not know what your rigs are like, but that is how it works..

Also it depends on the modes - a 100W FT8 or FM is not the same as a 100W SSB when talking the PSU's current..

PS: I've seen 100-200W linear PSUs in our local ham radio club (built in 70ties-80ties), and all what you need to be happy is a pretty large heatsink with several beefy transistors on it

[TimFox]

Have you considered adding a (much smaller) low-voltage secondary transformer in series with the primary of your high-power transformer to reduce its secondary voltage?
(Usually called a "buck transformer", popular in power distribution systems to obtain a desired line voltage.)

[davelectronic]

Now that's very interesting, I had not though of adding a smaller primary side transformer. It's something I will look closer at, although might it make the toroidal transformer secondary a bit droppy for want of a better word. I'm drawing 7.8 Amps at the moment on 11 meters FM, and ssb would be a lot more. I'm only driving the amplifier with 4 watts approximately. The PSU powering my main radio is a retail unit, it's a 30 Amp switching PSU. Yes I am aware of the power requirements for the radio equipment I have. I think I am putting out 75 to 80 watts roughly. Just a picture of the heat sink and hardware, I've still have to do heat shrink tube, don't have any that size left, so ordered some.

[soldar]

I haven't loaded it down yet, after the rectifier and filter capacitors.

You need to do that before you go any further.  Until you do that you are just spinning your wheels in search of a solution for a problem which is not known and may not exist.

[langwadt]

why linear?

[schmitt trigger]

I wonder how come nobody has asked exactly how will you deliver 10 amps from a LM317??
Are you using auxiliary series-pass transistors?

But back to your topic: if you would consider additional complexity, you could add a pre-regulator. The most efficient would be a buck topology with external switching Fets.

[davelectronic]

Yes I don't exactly know the voltage drop, but if the 300VA 12.00 Volts transformer (same manufacturer different voltage) it won't drop by much. The current is coming from an MJ11033 PNP Darlington transistor. I'm hopeful it can supply 10 Amps, as for a pre regulator, I wouldn't know where to start with something like that. And it sounds overkill for a basic linear power supply. Oh yes capacitors are 10000uf 63 volts electrolytic.

[langwadt]

Yes I don't exactly know the voltage drop, but if the 300VA 12.00 Volts transformer (same manufacturer different voltage) it won't drop by much. The current is coming from an MJ11033 PNP Darlington transistor. I'm hopeful it can supply 10 Amps, as for a pre regulator, I wouldn't know where to start with something like that. And it sounds overkill for a basic linear power supply. Oh yes capacitors are 10000uf 63 volts electrolytic.

https://www.ebay.com/itm/334836996334

[EPAIII]

18 Volt transformer for a 13.6V linear, regulated supply? WHY? You will wind up with at least 33% of the power consumed from the mains becoming HEAT. Probably more like 40% to 50%. And, of course, you will need to find a way to remove that heat from your device.

My rule of thumb when designing a linear supply is to start with a transformer secondary Voltage that is approximately equal to the desired, regulated DC output. So I would look for a transformer that can supply that 13.6 Volts at your maximum load of 10 Amps or a bit more. Then I do the detailed math and see if all is OK. An example:

13.6 VAC with a full wave rectifier:

PeakDC = (13.6 * 1.414) - 2 * DiodeDrop
PeakDC = 19.23 - 1.4
PeakDC = 17.83 Volts

Now I estimate the maximum ripple Voltage (Vpp) that I want at the full current and calculate the filter capacitors. The minimum DC input Voltage needed by the regulator (Vmin) plus that maximum ripple Voltage must equal or be less than the PeakDC calculated above. Or

PeakDC - Vmin >= Vpp

And using that I calculate the size of the filter capacitors that are needed.

C = I / (2 x f x Vpp)

So, if my regulator requires a Vmin of 14 VDC:

PeakDC - Vmin >= Vpp
17.83 Volts - 14 Volts >= 3.83 Volts

and with a 60 Hz power line frequency and your 10 Amps:

C = I / (2 x f x Vpp)
C = 10 / (2 x 60 x 3.83)
C = 0.022 F
or
C = 22,000 uF

And there is my unregulated DC power supply. I used a 0.7 V forward Voltage for the bridge rectifier, assuming standard silicone diodes. There are bridges that use Schottky diodes with lower forward Voltage ratings. If I find that the capacitor value is too high, I can go back and adjust something. While this will not produce a power supply that is as efficient as a switching supply can be, it will minimize the energy loss due to heat for a linear supply. Or, at least you can see where the wasted energy is going: transformer, diode loss, or just an unregulated Voltage that is higher than really needed by the linear regulator.

[Ian.M]

That design philosophy makes no allowance for variations in mains supply voltage.  During high line conditions it is unlikely to have enough heatsinking for the extra dissipation, and during low line conditions its likely to brownout.  To make it more robust, you have to accept higher losses, and decide what percentage supply voltage variation it should tolerate.  Do your ripple and dropout calculations at the lowest mains voltage it must work at, and the dissipation and SOA calculations at the highest.

[davelectronic]

EPAll, that's assuming exact perfect input values with no allowance for what Ian describes. I've no doubt I am probably over the limit, in terms of my input voltage being a bit high. I'd have been happier with 20.00 Volts or there abouts, with no more than a 3 Volt drop in voltage at 10 Amps. I think although not certain, the transformer will drop about 2 to 4 Volts going by the 300VA transformer I have recently used for another PSU. Same manufacturer just 12.00 Volts, I want know the exact drop underload until I test it. I still think it's going to be a bit to high really. Having used a few of these Vigatronix toroidal transformers, they seem very efficient. And I can't see from a previous post, where a buck converter would be useful, it would turn a linear power supply in to a hybrid linear plus switching power supply. I think the most practical way is to remove secondary windings, but I don't know how well that would end up. Diodes and the numbers of them needed would border on stupid. The power resistor sounds a good idea, but its probably got to be a bit of a beefy aluminium clad type resistor. Then I still have to work out the voltage drop, then calculate the value and power rating of a resistor that's suitable. Although that idea is still wasteful of power. These transformers where £42 each, I'm trying to find a way to using it with out ruining it. Ideally I would have gone for a 300VA transformer with 15.00 Volts secondarys, but there where non available at that power rating at that time.

[JohanH]

I think the most practical way is to remove secondary windings, but I don't know how well that would end up.

Instead of removing isolation and removing existing windings, you could add a few windings in reverse with regular wire.

[langwadt]

EPAll, that's assuming exact perfect input values with no allowance for what Ian describes. I've no doubt I am probably over the limit, in terms of my input voltage being a bit high. I'd have been happier with 20.00 Volts or there abouts, with no more than a 3 Volt drop in voltage at 10 Amps. I think although not certain, the transformer will drop about 2 to 4 Volts going by the 300VA transformer I have recently used for another PSU. Same manufacturer just 12.00 Volts, I want know the exact drop underload until I test it. I still think it's going to be a bit to high really. Having used a few of these Vigatronix toroidal transformers, they seem very efficient. And I can't see from a previous post, where a buck converter would be useful, it would turn a linear power supply in to a hybrid linear plus switching power supply. I think the most practical way is to remove secondary windings, but I don't know how well that would end up. Diodes and the numbers of them needed would border on stupid. The power resistor sounds a good idea, but its probably got to be a bit of a beefy aluminium clad type resistor. Then I still have to work out the voltage drop, then calculate the value and power rating of a resistor that's suitable. Although that idea is still wasteful of power. These transformers where £42 each, I'm trying to find a way to using it with out ruining it. Ideally I would have gone for a 300VA transformer with 15.00 Volts secondarys, but there where non available at that power rating at that time.

with a buck you could either skip the linear part or set the output of the buck to just about the needed voltage for the linear.

instead of removing turns fromt he secondary you can add extra turns, easy to do on a toroid, and connect them in reverse. It's probably about 3-4 turns per volt

[davelectronic]

Adding windings in reverse, and connecting them backwards. I don't understand how that works. Could you elaborate on this idea please, as I've never heard of this before.

[BeBuLamar]

why linear?

While someone already asked that if the OP would use series pass transitor to get 10A because the LM317 isn't rated for 10A. But besides that I think the OP chose linear because the cirbuit if much simpler than a switching design. Personally I wouldn't know how to build a switching power supply.
But if I need a 13.8V 10A power supply I would simply buy this
https://www.amazon.com/TekPower-TP1863-Regulated-Supply-Protection/dp/B00KXGZ2FE/ref=asc_df_B00KXGZ2FE/?tag=hyprod-20&linkCode=df0&hvadid=198066863690&hvpos=&hvnetw=g&hvrand=3951580504299729440&hvpone=&hvptwo=&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=9026816&hvtargid=pla-350892585273&psc=1

I don't know how much it cost the OP building one .

[davelectronic]

It's about the hobby, linear power supplys are, or used to be preferred over switching power supplys, because switching power supplys can induce noise in HF radio equipment. Although I'm actually using a 30 Amp switching PSU for my HF radio and amplifier right now. I wanted to see if it's possible to build a linear power supply using an LM317 voltage regulator, and a single pass transistor MJ11033 power transistor. To supply 13.80 Volts at 10 Amps. It's not cost effective, it just the the interest and fun of building it, that's it.

[langwadt]

Adding windings in reverse, and connecting them backwards. I don't understand how that works. Could you elaborate on this idea please, as I've never heard of this before.

if you add turns in the same direction you get higher voltage, if you add turns wound in the opposite direction (or connected in reverse) you get lower voltage

they add just like any series connection,  if you have "positive" 18V and put the in series with "negative" 3V you get 15V

[JohanH]

Adding windings in reverse, and connecting them backwards. I don't understand how that works. Could you elaborate on this idea please, as I've never heard of this before.

It's basic transformer theory. Imagine if you had two equally long windings on a transformer core, one wound in one direction and one in the other direction, they would essentially cancel each other out.

So you can add a few turns to the transformer and connect them in series with the existing winding. If they are wound in the same direction around the core as the existing winding, when added in series, the voltage becomes higher. If they are wound in the opposite direction, they will have an opposite force and will subtract the voltage. This works both on primary and secondary side. Primary requires more turns (usually), so it's maybe more practical on the secondary side. But if you have a split secondary, modifying the primary might be preferable.

[BeBuLamar]

It's about the hobby, linear power supplys are, or used to be preferred over switching power supplys, because switching power supplys can induce noise in HF radio equipment. Although I'm actually using a 30 Amp switching PSU for my HF radio and amplifier right now. I wanted to see if it's possible to build a linear power supply using an LM317 voltage regulator, and a single pass transistor MJ11033 power transistor. To supply 13.80 Volts at 10 Amps. It's not cost effective, it just the the interest and fun of building it, that's it.

For radio and amplifier I think they would work just fine with a non regulated power supply with a good filter capacitor.

[davelectronic]

No, I won't use an unregulated supply, to much risk of transients making it through. The transformer has twin primary and secondary windings. I would be happier doing reverse windings on the secondary. All I have here to see what that 27.00 Volts will drop to is 4 X 50 watt halogen lamps, I'm going to put each pair in series, then connect both pairs connected in parallel. See what it drops the voltage to. I'm not sure how useful this will be, as I don't have use of an ocillascope to see the drop in a waveform, peak value etc. Going back to a resistor to drop the secondary, I would put this resistor on the AC secondary side of the transformer ? I just have to work out what voltage will drop across the resistor, and the power rating that resistor needs to be. If I've got that right.

[davelectronic]

So I loaded the rectified and filtered DC with approximately 100 watts. Before connecting the load, it was 27.50 Volts. When loaded that dropped to 23.80 Volts, and a meter reading of 8.6 Amps. So going on that, I think it has to come down really ? What do you think. I know I can't look at this voltage on an ocillascope, as at the moment I don't have one. Oh, mains voltage AC was 249 Volts.

[mariush]

To some degree , you could limit the maximum voltage by the amount of capacitance.

The bridge rectifier converts your AC voltage and gives you a peak DC voltage of Vdc peak = sqrt(2) x Vac - 2 x voltage drop on diode in rectifier  =  1.414 x 18v AC -  ~ 2v  = 23.5v

That's peak voltage ... you need to add capacitors to guarantee the minimum voltage is above some threshold at all times.

You can estimate capacitance with the formula C  =  Current / [ 2 x AC Frequency x  (Vdc peak - Vdc min desired ) ]   

So for example, let's say you have a peak voltage of 23v and you want at least  14v at 10A  and you're in EU country so you have 50 Hz mains frequency, then you can put in formula :

C = 10A / [ 2 x 50 x ( 23v - 14v ) ]  =  10 / 100 x 8  = 1 / 18 = 0.055555 Farads or around 5600 uF

So at high current of 10A, the capacitors will charge up to around 14v and not get enough time to charge up higher, because they'll be discharged by the load...

You could just put a resistor in series ... voltage =  current x resistance  ... so for example at 10A of current, you'll have a voltage drop of around 1v with a 0.1 ohm resistor , and power dissipated in resistor will be P = IxI x R = 10 watts ( so you could maybe have 5  3w 0.47 ohm rated resistors in parallel)

Diodes can work, but most diodes are up to 3-5 A of current and paralleling them doesn't work perfectly. You can get diodes in bigger packages like to-220 / to-247 that you can screw to a heatsink.

[JohanH]

Going back to a resistor to drop the secondary, I would put this resistor on the AC secondary side of the transformer ? I just have to work out what voltage will drop across the resistor, and the power rating that resistor needs to be. If I've got that right.

The best place for a resistor is after the rectifier, before the capacitors. Together with the capacitors it forms a good RC-filter. This will filter ripple better than the capacitors alone. As mariush suggested, you need some beefy high power resistors in parallel so they can stand the power. Alternatively a larger one (> 10W) that can be bolted to a heat sink.

[langwadt]

To some degree , you could limit the maximum voltage by the amount of capacitance.

The bridge rectifier converts your AC voltage and gives you a peak DC voltage of Vdc peak = sqrt(2) x Vac - 2 x voltage drop on diode in rectifier  =  1.414 x 18v AC -  ~ 2v  = 23.5v

That's peak voltage ... you need to add capacitors to guarantee the minimum voltage is above some threshold at all times.

You can estimate capacitance with the formula C  =  Current / [ 2 x AC Frequency x  (Vdc peak - Vdc min desired ) ]   

So for example, let's say you have a peak voltage of 23v and you want at least  14v at 10A  and you're in EU country so you have 50 Hz mains frequency, then you can put in formula :

C = 10A / [ 2 x 50 x ( 23v - 14v ) ]  =  10 / 100 x 8  = 1 / 18 = 0.055555 Farads or around 5600 uF

So at high current of 10A, the capacitors will charge up to around 14v and not get enough time to charge up higher, because they'll be discharged by the load...

You could just put a resistor in series ... voltage =  current x resistance  ... so for example at 10A of current, you'll have a voltage drop of around 1v with a 0.1 ohm resistor , and power dissipated in resistor will be P = IxI x R = 10 watts ( so you could maybe have 5  3w 0.47 ohm rated resistors in parallel)

Diodes can work, but most diodes are up to 3-5 A of current and paralleling them doesn't work perfectly. You can get diodes in bigger packages like to-220 / to-247 that you can screw to a heatsink.

and the peak current in the rectifiers will be probably be 30-40A

[soldar]

Going back to a resistor to drop the secondary, I would put this resistor on the AC secondary side of the transformer ? I just have to work out what voltage will drop across the resistor, and the power rating that resistor needs to be. If I've got that right.

The best place for a resistor is after the rectifier, before the capacitors. Together with the capacitors it forms a good RC-filter. This will filter ripple better than the capacitors alone. As mariush suggested, you need some beefy high power resistors in parallel so they can stand the power. Alternatively a larger one (> 10W) that can be bolted to a heat sink.

You are both saying basically the same thing. Resistor in series with rectifier. Whether before or after makes no difference. They're in series.

[mariush]

and the peak current in the rectifiers will be probably be 30-40A

Nothing a couple GBU rectifiers in parallel wouldn't handle  ... or a LT4320 ideal rectifier controller and 4 mosfets.

[Zero999]

It's about the hobby, linear power supplys are, or used to be preferred over switching power supplys, because switching power supplys can induce noise in HF radio equipment. Although I'm actually using a 30 Amp switching PSU for my HF radio and amplifier right now. I wanted to see if it's possible to build a linear power supply using an LM317 voltage regulator, and a single pass transistor MJ11033 power transistor. To supply 13.80 Volts at 10 Amps. It's not cost effective, it just the the interest and fun of building it, that's it.

A properly designed, high quality switched mode regulator isn't noisy.

A basic linear regulator is often not that good. Mains capacitively couples to the secondary of the transformer. The bridge rectifier generates harmonics, when the current changes direction as they switch on/off and jellybean regulators have poorer supply rejection at high frequencies.

[nali]

I'd probably go with a buck winding to reduce secondary voltage. Maybe add a little bit of secondary resistance to limit inrush current too.

That's a nice big accessible transformer, so just loop some turns of wire to effectively make another winding using trial & error to get the voltage you need. Wire that in series with the primary; if it's anti-phase it will subtract the primary voltage & reduce the secondary voltage accordingly (it's a 50/50 trial & error which way you wire it, but simple to check).

The advantage of a buck winding is practically no wasted power.

[TimFox]

The nice thing about a toroidal transformer is there can be space for the "buck" winding.

[davelectronic]

Thank you for all your replies, reading that the peak voltage across the rectifier could be as much as 30 to 40 Amps, really. How on earth do I calculate a series resistor that could cope with such high current in the rectifier stage. How much of that 27.00 Volts should I drop ? Is calculation working out how many volts to drop across the power resistor, and it's resistance sucks up that excess power / voltage. And then working out the power rating the resistor needs to be. Would the. 50 watt aluminium clad resistors be suitable for this purpose. And it's value depending on how many volts I need to drop. I thought dropping 8 .00 volts or so would still leave enough headroom Regards voltage. I'm not entirely sure on how much voltage to drop. And would a 50 watt resistor be powerful enough to suck up the voltage I'd want to drop.

[TimFox]

The power across a resistor is the product of voltage across it and current through it.
For periodic waveforms, you need to compute a mean value over one cycle.
Important note:  the metal-clad power resistors that are bolted down require a heat sink in order to handle the rated power, just like a power transistor.
See the data sheet to determine at what case temperature that power rating applies, and how to derate it with higher temperature, then consider the heat sink temperature rise at that power.

[davelectronic]

Looks like I would need 100 watts or even more and a 1ohm resistor, I've got some 0.33 ohm 50 watt aluminium clad power resistors. Three of those to make up 1 ohm. But when the power supply is only lightly loaded, wouldn't the resistor be doing virtually nothing, until more current flows to the output. I don't know if that makes sense, but I can't see the power resistor doing anything whilst the output is not loaded, or only lightly loaded.

[TimFox]

Yes, with low current through the resistor it will do very little.
However, if you use a lower power resistor to reduce cost, it may blow up at full power.
Again, what heat sink is required for your 50 W aluminum-clad resistors?
Non-clad resistors (such as popular silicone-coated wirewound resistors) are usually rated for maximum power at a given ambient (air) temperature, but the metal resistors need a heat sink.
If you connect the resistors in series (as you suggest), then you know that the current through each of them is the same.
It is more problematic to connect low-resistance resistors in parallel, since the different wire resistances could cause the current to divide unequally between them.

[soldar]

I'm not entirely sure on how much voltage to drop.

You experiment, you tweak, that's the whole fun of it. My builds are rarely entirely "finished" as I am always tweaking here and there... or just building a replacement. That's how you learn and get the feel of the ropes.

[langwadt]

how much capacitance you have? with 10000uF 26V peak is barely enough with no resistance and 10A

[davelectronic]

Yes I guess that's what I will be doing. So long as I don't blow things up in the process. The capacitance I'm using is 2 X 10000uf 63 volt electrolytic capacitors. That should be ok for 10 Amps at 13.80 volts. I am tempted to remove secondary windings, but don't want to mess up a perfectly good toroidal transformer. If it had a single winding, I probably would have done it by now. But twin secondary windings is a bit more daunting.