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I watched an interesting video in which the caster reverse engineers a modern 850W computer PSU. From this teardown one can see that 90% of the bulk of the PSU is devoted to synchronous conversion of the 120V input to the ultimate output which is a 12V flyback transformer. Once this 12V DC source is obtained, the outputs are generated using either direct output from the transformer or the output from two different synchronous buck regulators (one for 5.5V and 3.3V). In general, once the 12V is obtained, the only things downstream in the power supply are the following:

  1. 5V buck regulator
  2. 3.3 buck regulator
  3. voltage sensing circuit board (sends "power good" signal to motherboard)
  4. -12V inverter, which is another buck regulator (non-synchronous)

These components compose less than 10% of the mass of the PSU.

So, what I don't understand is, given the relatively small size of these components, why are DC-DC ATX converters so big? For example, PowerStream sells a 650W DC-DC ATX PSU which is just as big as the 850W PSU in the teardown video. But since the PowerStream is taking a 12V input, why does it need to be so big? Since it would only seem to need the 4 components listed above, couldn't it be much smaller?

One possible clue is that there is a reference to 1500V AC "isolation" in the specs. So, does that mean they are converting the input to 1500V AC, then downconverting again? If so, that would explain the bulk. If this "isolation" is indeed the reason, why would they need to do that? If the PSU is getting 12V power from a battery, the power will be smooth, so why go to all the trouble to take to 1500V AC and back again to 12V DC?

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    \$\begingroup\$ Why are 600W and 900W PSUs the same size? Because they have to fit the same standardized format. \$\endgroup\$ Commented Nov 1, 2020 at 17:36
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    \$\begingroup\$ Note that the "12V" input converter has in input range of 9-18V. The 12V output has a tolerance requirement of +/-5% and so requires regulation. Ultimately the device has to do pretty much the same as the 12V AC input version (except for AC rectification). \$\endgroup\$ Commented Nov 1, 2020 at 17:54
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    \$\begingroup\$ ATX supplies are that big because they wouldn't be ATX supplies otherwise. If you needs something smaller, try an SFX (SFF ATX) or TFX. \$\endgroup\$
    – Mast
    Commented Nov 2, 2020 at 14:55
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    \$\begingroup\$ Standard ATX PS/2 is 150x86x140, ATX PS/3 (incorrectly known as micro ATX) is 150x86x100 (all in mm), etc. \$\endgroup\$
    – Mast
    Commented Nov 2, 2020 at 14:57
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    \$\begingroup\$ @Fax ATX very much specifies the size. It's like, a section in that wikipedia article: en.wikipedia.org/wiki/ATX#Physical_characteristics \$\endgroup\$ Commented Nov 2, 2020 at 15:29

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Even though it is DC/DC rather than AC/DC, all of the outputs are fully isolated from the input so 100% of the total energy has to go through transformers, the largest component type in the device. To do that the input DC has to be chopped, just like the input AC in the other supply after it has been rectified and filtered. So the only real savings in volume are the possible elimination or reduction of the large input capacitors. The main switcher is low-volts-to-low-volts rather than high-volts-to-low-volts, but the total energy is the same and in round numbers the watts-per-cubic-inch is the same, thus the overall volume is the same.

And, because the input current is approx 10 times greater in the DC/DC supply, the transformer primary windings will be fatter, as will any noise suppression chokes. One could argue that (for equivalent output power levels) the DC/DC supply should be larger than the AC/DC supply, not smaller

The supply is fully isolated because the manufacturer cannot know the "quality" of the DC source out in the field. There are many reasons why you would not want the 12 Vdc input and the various DC outputs to share a common ground. 1500 V is the breakdown voltage rating of the insulation between the transformer primaries and secondaries. It does not refer to any actual internal voltage level.

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    \$\begingroup\$ Well, you are right. The DC-DC PSUs are actually a little bigger on average than the equivalent 120V PSUs. Why do the outputs need to be isolated? Can't the buck converters just be supplied 12V directly from a battery, assuming the battery is standalone (not in a car or something)? \$\endgroup\$ Commented Nov 1, 2020 at 17:58
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    \$\begingroup\$ "... assuming the battery is standalone ..." would be exactly the problem. \$\endgroup\$
    – Transistor
    Commented Nov 1, 2020 at 19:26
  • \$\begingroup\$ In a specific system with a limited and controlled DC source - maybe; even probably. But in a generic product for an unknown markets and uncontrolled environments - no. \$\endgroup\$
    – AnalogKid
    Commented Nov 1, 2020 at 19:58
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    \$\begingroup\$ Isolation breaks any ground loops. For example, when you have current flowing in the 12V wire, the resistance of wire will cause a voltage difference between them, so if these are unisolated, there would be a voltage difference between a computer running at 600W and computer running at 500W. Also, if you connect the two computers together with say USB cable, there might be current flowing in the USB cable to balance the PC voltages. Now, imagine if one of the computers suddenly have a loose negative battery connection, the poor USB cable would pass about 50A and blow up. \$\endgroup\$
    – Justme
    Commented Nov 1, 2020 at 21:23
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Aside from matching form factor, power supply volume is more or less constant with watts for a given technology level. The power level is the same, so probably they're using the same topology (a forward converter). The transformer will have fewer turns in the primary but made of larger cross-section wire(s) to handle the ~20x higher current with similar I^2R losses.

As to why you would want isolation- it's a REALLY good idea to avoid ground loops. You don't want enormous currents flowing through your USB cable shells and such like. If the minus lead got a little loose you could burn up your mombo really easily.

Some space is saved by not requiring the mains reservoir capacitors but the rest is going to be similar, and the DC-input supplies are probably not nearly as optimized as their high-volume cousins. Keep in mind that 850W out at 85% efficiency means 1kW input, so it will be drawing 83A (100A at 10V in), which means fat wires or fat really thick traces.

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  • \$\begingroup\$ What is a USB cable shell? I guess I don't see the advantage of the isolation if the 12V input power is relatively pure (like a standalone solar-energy type battery). If it was car battery, sure I can see isolating that, but for a standalone battery, why add all the extra transformers/capacitors and suffer all that energy loss? \$\endgroup\$ Commented Nov 1, 2020 at 18:02
  • \$\begingroup\$ USB cable shell is the metal part of the USB cable, which is connected to ground and also supply common on a PC. I don't see any "extra" losses. In fact if you tried to make a non-isolated 12V supply from a 12V nominal input you would need a complex buck-boost converter that might be less efficient than a simple forward converter. \$\endgroup\$ Commented Nov 1, 2020 at 18:05
  • \$\begingroup\$ Ok, how about using TWO 12V batteries in series, so the source would be 24V. So, that would give about 250 amp-hours of clean 24V DC power and normal buck regulators could be used to supply the various outputs. \$\endgroup\$ Commented Nov 2, 2020 at 5:14
  • \$\begingroup\$ @TylerDurden that would be a different design. It would rule out using that PSU in 12V vehicles, which is probably the most common use case. 2x12=24V nominal battery is used in many UPSs, not just those that go up to mains voltage but those built in to ATX supplies . My car laptop PSU actually does something like 11-30V anyway. It's only 100W admittedly, and maybe 50% bigger than a 100W laptop supply, plus in a metal case for heatsinking. \$\endgroup\$
    – Chris H
    Commented Nov 2, 2020 at 11:22
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    \$\begingroup\$ Many things are possible, not so many are economically viable as products. It could also be made much smaller using higher frequencies without compromising isolation. Cost is usually a significant factor. In niche products the cost of design may be significant as well as cost of production. \$\endgroup\$ Commented Nov 2, 2020 at 11:25
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Isolation is certainly something that requires a certain board layout, AND it makes transformer windings larger.

But that's not the point. The limiting factors for size are

  1. thermal transport ("cooling")
  2. size of transformers / inductors
  3. size of capacitors
  4. size of power semiconductors

For 3., voltage makes a big difference – a 220 µF capacitor that can withstand 6 V fits on the smallest of your fingernails, whereas a 220 µF capacitor that withstands surges of say 500 V is actually big. That's basically because the layers that isolate the two electrodes simply need to be thicker, this making the same size of capacitor have less capacitance.

But you're right, that's a "primary side problem", i.e. it affects few capacitors in the PSU.

Now, 2. requires one to think about three things:

  • isolation
  • ampacity of the wire
  • stored energy in the magnetic field.

If you need to deal with a high voltage winding, well, that's going to need thicker / safer isolation around its wires.

If you decide to push all power through the same secondary winding to step-down from that to other voltages, well, that winding will carry a lot of current and hence need high-diameter conductors.

And lastly, if you can't design your switch-mode supply with a high switching frequency, then you need to store a lot of energy in your inductors / cores while the primary feeds in power. That makes the whole thing bigger.

On the other hand, you don't want to switch too fast - that increases the losses in the switching transistors quadratically, and then these need to be larger and produce more heat, for which you need heat sinking – you get the idea, that makes things bigger.

Now, there's a few things that are generally a bit harder than other things:

Think about the energy you need to store to be able to give out stable voltages even if the input drops (which it very well might do, especially in 12 V or 24 V vehicle power!). You'll find out that it's easier to go 210–260 V -> 12 V (because a little bit of capacitance stores a lot of energy, and NO WAY the input will ever drop below the output) than to go from 8V – 16V to 12 V. You'll need to make sure that somewhere there's enough energy stored to make this work! You'll typically find switch-mode architectures in there that have multiple coils / transformers to make sure there's some intermediate storage of energy from which the output can always get power. That of course is larger in components, and means more losses, which means more cooling, which means more size and weight.

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The input side of the AC-DC will be largely the same as the input of the DC-DC convertor.

The AC-DC convertor first rectifies the mains input to DC and does some basic filtering on it. This DC voltage then gets chopped and downconverted using a isolated transformer.

The DC-DC converter does some basic filtering and then chopped and passes it through an isolating transformer.

So on the whole the major components are present in both, input filtering, isolating transformer and output regulation.

One possible clue is that there is a reference to 1500V AC "isolation" in the specs.

That doesn't mean what you think it means. That refers to being able to put a one pole on the input terminals and the other pole on the output terminals and then put 1500V on it with no current leaking through. This requires the isolation transformer I mentioned above.

If the PSU is getting 12V power from a battery, the power will be smooth

No it isn't, one common source of 12V power is from a vehicle where a bunch of other components are switching on and off, a set of sparkplugs are firing a few thousand times a second and the dynamo is constantly changing speed. This voltage cannot be considered stable by ATX standards. Therefore they need to do quite a bit of filtering. At least as much as you would do when getting your power from mains powergrid.

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Mini-Box picoPSU-120

You can get very small DC-to-ATX power supplies - like the 120 watt 'picoPSU-120' pictured. Takes a 12v input and outputs +12v, -12v, +5v and +3.3v right on the ATX connector.

Of course, if you need more than 120 watts, or if you want isolation, you'll need something proportionally larger. But while 120W may not run a big gaming rig, the latest USB-C-powered Macbook Pro consumes less than 100W even when charging.

The converter you linked is the size of a standard PSU so it fits in the space in your case for a standard-sized PSU.

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  • \$\begingroup\$ That is not ATX. ATX power supplies need to have a +12, -12, +5.5, +3.3 outputs at a MINIMUM, and usually have SATA outputs. Also the wattage for a PC is generally at least 400 Watts. The pico supply shown just has a single 12V output. \$\endgroup\$ Commented Nov 3, 2020 at 17:12
  • \$\begingroup\$ @TylerDurden Read the product listing again - it supplies +12, -12, +5.5 and +3.3v and it has output connectors for 3 drives. And while 100W wouldn't power a gaming rig, any laptop powered by USB-C - the latest Macbook Pro for example - runs on less than 100W. \$\endgroup\$
    – mjt
    Commented Nov 3, 2020 at 17:53
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The units shown are sold as "industrial grade", and certainly at industrial grade prices - such equipment will be built more sturdily and with more safety and reliability margins designed in, since (especially with 48V equipment, which is commonly used in serious telecomms installations) it is not unlikely to end up in applications where YEARS of 24/7 operation will be expected of the equipment. So these should be compared to server power supplies (which are sometimes made surprisingly small* - but then, VERY dense and heavy and using small high RPM counterrotating fans which would be FAR too noisy for an office), not office computer power supplies.

Also, fuses, breakers, switches, connectors, relays for high-wattage, low-voltage DC circuits need to be extremely heavy duty (due to the way a DC arc will NOT self extinguish, and due to how heat dissipation in conductors goes up with the current squared), and are hard to miniaturize - a switch that can handle 50 amps DC is very different from what you need for 16 amps AC.

*Thinking of units like the Supermicro PWS-741P-1R here - surprisingly small, but heavier than an equivalently sized brick, and noisy.

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