Dropping voltage with diodes.

[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.