Help me design a 13.8v 5A low noise linear cheap power supply

Hi, although there are circuits on the net, I want to build a 13.8v 5A low noise linear cheap power supply.
The power supply must be simple so few components that can handle the power.
The continuous current that will be drawn from this PSU will be 3-3.5A for many hours.
To avoid large heatsinks (even better passive cooling?) or fans (fans are unacceptable in this design) I think that much more powerful regulators should be used (or even more than one) so as for them to run cool, in conjunction to low voltage difference from their input to their output.
In terms or ripple, I won't be too extreme, but it would be used in communications equipment so it has to be acceptable.
I would like to have a "capacitance multiplier" circuit to reduce ripple and followed by a linear regulator capable of keeping the voltage stable.
But there may be better circuits and more simple.

I can custom order any voltage/current toroid for it, so you name it. But take into account the low heat generation, so not a much larger voltage at the input of the regulator/multiplier compared to the output.
So any suggestions/simulations/etc?

A 12V 5A transformer seems like a good start. Add full diode bridge and smoothing capacitor (say 10,000 uF). This gives you 17 VDC no load. As you draw 3-4 Amperes it brings down the voltage to 12-13V. It's stable if your house voltage is stable.

In days when transformers were cheap this was the basis for myriad power supplies. If you're lucky you can do without regulation or heat sinks or capacitance multiplier.

HI,

neazoi said:

To avoid large heatsinks (even better passive cooling?) or fans (fans are unacceptable in this design) I think that much more powerful regulators should be used (or even more than one) so as for them to run cool, in conjunction to low voltage difference from their input to their output.

Click to expand...

The generated heat of a linear circuit does not depend on which parts are used. Neither bipolar, nor NPN, nor PNP, nor Mosfet, nor size, nor resistor, nor diodes.
All generate the same heat. And it is always Ptot = I x Vdrop.
So the only way to reduce the generated heat is to reduce the voltage drop.
But for a system to be able to regulate the output voltage ... it needs headroom (voltage).
The input needs to be higher ... in any case ... as the output.
The smaller the headroom, the smaller the generated heat. But the higher the risk for output ripple.

The regulation needs to withstand worst case high input voltage: when mains input is high, when transfomer ratio is high, when output load is low.
And it needs to be able to generate ripple free output voltage at worst case low input conditions: low mains voltage, high output current, nominal capacitor ripple voltage (with ideal capacitance), but also increased capacitor ripple voltage due to low capacitance (aging, production)...
There are "stiff" transfomers and more weak transfomers. A stiff one (toroid) suffer from high inrush current and high peak current when caoacitors are charged (overtones).
A weak transfomer suffers from having big difference between no load output voltage and full load output voltage.
Depending on transformer you may expect 120% no load output voltage. So a 12V transformer may output 14.4V, causing more than 20V peaks, so that the capacitor voltage may go up close to these peaks (reduced by diode voltage drop).
Mind that for 5A average output current the peak transformer current may easily go up to 20A. .. reducing the peak plateau...reducing the "full load capacitor voltage".

The necessary headroom depends on circuit. A high side NPN will need at least 2-3V (depends on type and base drive circuit).
A more sophisticated high side Mosfet P-Ch circuit may work down to 0.1V .... but the risk for an unexpected output ripple is high.

A lot to consider...a lot to calculate and estimate. It always will be a compromise.

Klaus

Klaus is giving good advice. the size and type of linear regulator makes no difference to the amount of heat it has to dissipate. That is down to plain ordinary Ohms Law (Voltage in - voltage out) / current out.

For a simple regulator that suits your requirement I would suggest a linear 3-pin device like the LM317 with a wrap-around bypass transistor. It's simple, effective and gives short circuit and overheat protection. Like this: https://www.circuitstoday.com/12v-15a-voltage-regulator
For 5A output you can omit two of the bypass transistors and the associated resistors and for variable voltage so you can get 13.8V use an adjustable regulator.

How it works is quite simple: the regulator IC does all the work at low currents, as the load current increases, so does the voltage drop across the resistor at it's input pin. When the voltage across the resistor exceeds the transistors B-E turn on voltage it starts to share the work and eventually passes most of the load current. You have to pick a suitable resistor value to start the transistor conducting before the IC maxes out. The great advantage of this simple design is that if the ICs thermal shutdown operates, it stops drawing current so the transistor shuts down too. It is wise to put them on the same heatsink so they protect each other.

I would also add a diode from the output to the input (top of the reservoir capacitor) wired non-conducting to protect against residual voltage from the load when the AC is turned off and also add a 1nF capacitor across each of the bridge rectifier diodes to minimize switching noise.

Brian.

betwixt said:

Klaus is giving good advice. the size and type of linear regulator makes no difference to the amount of heat it has to dissipate. That is down to plain ordinary Ohms Law (Voltage in - voltage out) / current out.

For a simple regulator that suits your requirement I would suggest a linear 3-pin device like the LM317 with a wrap-around bypass transistor. It's simple, effective and gives short circuit and overheat protection. Like this: https://www.circuitstoday.com/12v-15a-voltage-regulator
For 5A output you can omit two of the bypass transistors and the associated resistors and for variable voltage so you can get 13.8V use an adjustable regulator.

How it works is quite simple: the regulator IC does all the work at low currents, as the load current increases, so does the voltage drop across the resistor at it's input pin. When the voltage across the resistor exceeds the transistors B-E turn on voltage it starts to share the work and eventually passes most of the load current. You have to pick a suitable resistor value to start the transistor conducting before the IC maxes out. The great advantage of this simple design is that if the ICs thermal shutdown operates, it stops drawing current so the transistor shuts down too. It is wise to put them on the same heatsink so they protect each other.

I would also add a diode from the output to the input (top of the reservoir capacitor) wired non-conducting to protect against residual voltage from the load when the AC is turned off and also add a 1nF capacitor across each of the bridge rectifier diodes to minimize switching noise.

Brian.

Click to expand...

Thank you all for the replies!
I am thinking that maybe it will be easier (and cheaper) to avoid all the hassle use a regulator such as theLT1083? LT1083/LT1084/LT1085 - 7.5A, 5A, 3A Low Dropout Positive Adjustable Regulators (analog.com)
There are even ready kits in amazingly low prices.

What if one wants also over current protection? Can it be done with this simple LT1083 (or any other device that comes into your mind)

They already have short circuit protection in them. Adding an adjustable current limit would be a bit more complicated and would have to be done before the regulator input pin as it has no external feedback.

Note that many of the packages are now obsolete and may be hard to get.

Be wary of Ebay kits, they may be cheap but read the small print - at least one I spotted claims 10A constant output yet shows a PCB clearly fitted with 1N4005 rectifiers. Another says "The IC is old. Used", most have heat sinks that might work if you fit a large fan and operate them inside a freezer.

Brian.

neazoi said:

I am thinking that maybe it will be easier (and cheaper) to avoid all the hassle use a regulator such as theLT1083? LT1083/LT1084/LT1085 - 7.5A, 5A, 3A Low Dropout Positive Adjustable Regulators (analog.com)
There are even ready kits in amazingly low prices.

Click to expand...

You have already got lots of excellent advice! But you also need to pay attention to the backend: the transformer, rectifier and filter set.

Remember that the transformer must be nominally rated for the load + losses (in the regulator) + margin (lowest mains voltage + highest load current)

Because your power demands are modest, you can go for a toroidal transformer (it may be cheaper locally - but it all depends). Hope the fuse will not blow!

With reasonable care you should be able to get 50% efficiency and perhaps without a fan.

Somehow I prefer center tapped transformers but the decision is more subjective than objective.

How well the regulator works depends much on how you have conditioned the backend.

For 14V output, you can need about 18-19V input to the regulator.

c_mitra said:

For 14V output, you can need about 18-19V input to the regulator.

Click to expand...

and for 12.5-13v output? 17v transformer (unloaded)?

betwixt said:

They already have short circuit protection in them.

Click to expand...

The more I see this LT1083 the more I like it. It will operate at more than half it's limit. Let me ask, If say the transformer is rated for 5A and the lt1085 something like 7-8a, what will happen if I try to draw accidentally 40A out of this PSU? Will it just shutdown?

Note that for a rectifier-capacitor filter power supply you need to de-rate the transformer's RMS current rating about 50% from the desired DC output, due to the high peak current drawn by the rectifier-filter, and the consequent high I squared R loss from the transformer winding resistances.
If you try to draw a DC current equal to the transformer RMS current rating, the transformer will seriously overheat and likely burn up.

crutschow said:

Note that for a rectifier-capacitor filter power supply you need to de-rate the transformer's RMS current rating about 50% from the desired DC output, due to the high peak current drawn by the rectifier-filter, and the consequent high I squared R loss from the transformer winding resistances.
If you try to draw a DC current equal to the transformer RMS current rating, the transformer will seriously overheat and likely burn up.

Click to expand...

So I need a transformer current rating double that of the output current of the PSU?

neazoi said:

So I need a transformer current rating double that of the output current of the PSU?

Click to expand...

Yes, for good margin.
Minimum 1.6 times.

crutschow said:

Yes, for good margin.
Minimum 1.6 times.

Click to expand...

Good hint!
and for 12.5-13v output assuming a LT1083 used? 17-18v transformer (unloaded)?

The other option to over-rating the Tx markedly - is to use a split bobbin Tx, pri on one half, sec on the other - these have about 8-10% leakage L and provide a very much better current waveshape in the pri/sec, you still need to over-rate, with about 18-20V no load, and 7 amp current capability for 5ADC out ...

neazoi said:

and for 12.5-13v output? 17v transformer (unloaded)?

Click to expand...

At full load, the regulator may need a min drop of 1.5V; so the input must be greater than 14.5V

At full load, the ripple at the input can be significant; if the ripple is 3V, than the min input to the regulator may be 17.5V.

With this config, most of the power is coming from the filter caps and the transformer is only topping them up twice a period.

At 5A, diode drops may be more than 1V (no, not 0.6V) and with a bridge rectifier, you will have 2V total diode drop.

I would suggest you go for a ready made transformer 15-0-15V with rated continuous current of 4A. They will be easy to find and cost will be reasonable.

After rectification and with a pi filter in the output, you will be getting 15x1.4-1=20V; more than sufficient for your purpose.

a pi filter ( C - L -C ) will give you the average voltage at full load minus losses, 15Vrms has an average of 15/1.111 = 13.5 v, minus say 1.6V for the bridge rect = 11.9, less IR losses in the sec of the Tx ...

correction - a pi filter ( C- L - C ) will or can give higher than the average voltage if the 1st C is quite large - as it acts to capture the peak volts - the price for this is higher peak and therefore rms currents in the Tx ...

Easy peasy said:

a pi filter ( C- L - C ) will or can give higher than the average voltage if the 1st C is quite large - as it acts to capture the peak volts - the price for this is higher peak and therefore rms currents in the Tx ...

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That is why I like the center-tapped winding. Each half is only half-active and that keeps the RMS current low. You can therefore use thinner wire (but double the number of turns) and the power lost the diodes is reduced.

c_mitra said:

That is why I like the center-tapped winding. Each half is only half-active and that keeps the RMS current low. You can therefore use thinner wire (but double the number of turns) and the power lost the diodes is reduced.

Click to expand...

Interesting point, I thought there was no difference from a single output transformer. The transformer will be custom made anyway, so I will order a 15-0-15V with rated continuous current of 4A then.
The rest would be easy with a LT1083 in place and it would also switch off by itself if I accidentally try to draw more current than it can give I guess.

There is only a point I now consider, the LT1083 is more powerful (8A if I remember right), so If I try to draw more current from it it will shut down.
But the transformer will be rated at 4A. So what happens if I try to draw more say 8A out of this PSU, that has this 4A transformer?
Will the transformer burn out? Will the LT1083 shut down because of the voltage drop in the transformer, due to the higher current asked from it?

neazoi said:

There is only a point I now consider, the LT1083 is more powerful (8A if I remember right), so If I try to draw more current from it it will shut down.
But the transformer will be rated at 4A. So what happens if I try to draw more say 8A out of this PSU, that has this 4A transformer?
Will the transformer burn out? Will the LT1083 shut down because of the voltage drop in the transformer, due to the higher current asked from it?

Click to expand...

Your transformer will be designed for 30V*4A=120VA conservatively and you can assume that you will get >100W of continuous power.

The voltage at the first filter capacitor will be the peak voltage, i.e, 15*1.41=21V minus one diode drop; say about 1V-20V

This filter capacitor is the main source of power and the transformer tops it up every 10ms. You need to size it well.

If you draw 8A from the transformer, the capacitor will drop the below the regulator minimum voltage and the regulator will shut down.

The transformer will not burn down because it is supplying power to the filter cap only during 16-20V; the voltage at the filter cap will not drop below 16V (in the steady state; because the regulator will shut down at lower voltage and no current will be taken) and will be limited to 20V on the max side.

The transformer will supply power to the filter cap only in small pulses (between 16-20V; approx duty cycle 20%). The current can be high during this period.

Your power requirement is 14Vx5A=70W; losses at the regulator will be (18-14)x5A=20W; losses at the rectifier will be 5W (2.5W at each diode).

I guess that the filter may drop about 1V or so.

So, your transformer will stay healthy; you will need to size the filter capacitor suitably.

There are free capacitor calculators on the net or you can work out yourself.

Thank you very much for the explanation.
Ok I think I will order this toroidal 15-0-15V with rated continuous current of 4A then based on your suggestions.
Then two diodes (types suggestions?) one in each arm of the transformer output, with their anodes connected together. The center tap of the transformer connected to the ground. Then a big electrolytic (starting value point?) and then the LT1083 in a configuration maybe similar to this Power Supply Regulated 7.5A @ 1.25 To 30Volt - LT1083 - Xtronic . I am not sure if I really need a pi L/C filter before the LT1083, I think this is too overkill as most commercial receivers should cope with a little rip. I do not know.