I'm working on making my own bench power supply, I want up to 7-8 amps. The switched mode power supplies are much cheaper than the transformers. Why would one choose a transformer instead of switched mode, if they're more expensive and heavier? Is longevity really the only reason? Is a 15v, 8 amp transformer better at pushing its 8 amps than a 15v, 8 amp switched-mode power supply?

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    \$\begingroup\$ In trying to think of a really good reason for "heavy iron" in the context of today's tools and ICs, and assuming comparisons between high quality designs in both cases. The best I can come up with is things like operating near a dental office using an x-ray machine. Those machines inject very NASTY spikes back into the building wiring. The heavy iron transformer will make it very difficult for the spike to get across into your power supply circuit. But those spikes might very well cross over using a switched mode supply system. Massive iron is hard for spikes to cross. \$\endgroup\$
    – jonk
    Commented Oct 22, 2019 at 4:44
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    \$\begingroup\$ If you want to perform better than 8 bit ADC measurements, (5 volts / 256 = 20 millivolt quanta), and you want to MAKE YOUR OWN SUPPLY, I'd go with the big-iron. \$\endgroup\$ Commented Oct 22, 2019 at 10:39
  • \$\begingroup\$ @jonk I am curious, dont these modern SMPS are equipped with heavy filtering to neutralize these heavy spikes? I opened one and they had various inductors to filter out these spikes of various frequencies. \$\endgroup\$ Commented May 28, 2020 at 12:46

1 Answer 1


In engineering, it's usually fairly straightforward to get the wanted, specified, behaviour. The spec says 15v 8A, you can usually get 15v 8A.

The more difficult bit is to tradeoff the different costs that come with any given approach.

One cost you mention in your question, longevity, or fear of a short life, is quality. While high price and a brand name aren't necessarily guarrantees of good quality, a price too low to be true and absence of a data sheet are often guarantees of bad quality. A SMPS can be built for a few dollars by almost anyone, big iron OTOH is unlikely to get produced like that. Are you going to spend the time on research and money on quality when you could buy a cheap knockoff?

There are other costs to the different approaches. Each sends back rubbish into the mains, just different rubbish. The transformer, if feeding a standard diode/capacitor rectifier takes current in big, 3rd harmonic producing pulses, a problem for the supply if on a large scale, but generally no problem at all at this level. The SMPS sends back swathes of RF, usually only suppressed enough to meet EMI regulations, if you're lucky. If you want to do sensitive measurements in those bands, you may have to search to find better gear.

The SMPS is much lighter. Maybe not much of a difference today, but when your shelves are groaning, and when you're old and wizzened and you're groaning (how would I know about this?) then lighter instruments can be beneficial.

Draw up a list of all the costs and benefits of each approach, and see if there's a killer reason for one or the other, for you. If not, flip a coin.

As you're making a bench power supply, that's presumably going to be adjustable. One way to adjust the output is with a linear regulator, the other is with SMPS. The linear will be quieter, but much less efficient than the SMPS. One common way to get the best of both worlds is to use a linear regulator as the adjustable output stage, preceding that with an SMPS regulated to keep the minimum practical drop across the linear stage.

The adjustable SMPS will be preceded by some form of transformer stage, both for isolation from the mains, and to step the voltage down to something appropriate for the output. That can be either big iron + recitifiers + capacitor, or can be SMPS to fixed output, or if it has enough control range, to variable output to drive the last linear cleanup stage. It does actually make sense to use a big iron + rectifier + caps that's then followed by a non-isolated buck converter, as you can use much smaller caps in the rectifier than would normally be the case, making best use of the varying capacitor voltage.

  • \$\begingroup\$ +1 This is exactly the type of answer I was looking for! Very nicely explained. I wanted to know which if SMPS are more energy efficient or Transformer type. You elaborated this by stating they throw out "different kind of" rubbish. So in terms of energy efficiency, can a comparison be drawn say for similar watt output for computer power supply? I am really fascinated how iMac's PSU are so compact and has no cooling fan so I assume they dont throw out much heat and energy efficient. \$\endgroup\$ Commented May 28, 2020 at 12:42
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    \$\begingroup\$ Both a big iron transformer, and an SMPS, are both 'nearly' 100% efficient. However, you'd often pair a big iron transformer with linear regulators, and those are inefficient, but at least the inefficiency of those is very easy to calculate. Successive generations of faster/lower switching losses FETs have enabled computer PSUs to push the efficcncy from 80% to 85% to 90%, halving the losses means twice the power in the same space. Is the variable bit, say 0-30 V, going to be done by a linear regulator, or adjustable SMPS? Do you have output noise requirements? \$\endgroup\$
    – Neil_UK
    Commented May 28, 2020 at 13:04
  • \$\begingroup\$ Allright, makes sense. But doesnt conventional big iron transformer almost always use linear resistive regulators to control the voltage, so inherently less efficient? To answer your question I believe the variable bit as you mentioned is adjusted by SMPS and not linear regulator in iMac PSU Since these machines are quite sensitive I am not sure what output noise tolerances they have. \$\endgroup\$ Commented May 28, 2020 at 13:24
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    \$\begingroup\$ The inefficieny is in the linear regulators, not the transformer. Just because they're used together doesn't mean you have to. I've updated the question to include transformer + rectifier + capacitors driving an adjustable non-isolated buck which gets your voltage down to perhaps Vout + 2 V, thne a linear stage which cleans up the last 2 V. \$\endgroup\$
    – Neil_UK
    Commented May 28, 2020 at 15:31
  • \$\begingroup\$ Well so you are saying that the linear stage is positioned last in the line when the voltages are low to mitigate the losses. Thats something really smart thing to do indeed. Thanks a lot for knowledge intensive yet intuitive contribution! \$\endgroup\$ Commented May 28, 2020 at 15:39

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