I need to regulate an incoming DC supply from a wall-wart down to 5 volts. I'm experienced with designing power supplies with ordinary transformers. In this situation, the maximum voltage I can expect would be the transformer AC voltage rating * 1.4 * transformer regulation * mains tolerance (10%). This works nicely all the time I have access to the transformer's datasheet to determine the regulation %. For example, a 9V AC transformer with a 57% regulation could produce a peak voltage of 22V in the worst case. I can then calculate the virtual impedance of the transformer from its regulation and determine what the actual voltage will be at my maximum expected load - and therefore determine regulator dissipation.

Unfortunately though, when specifying a DC wall-wart, I have no control over what type of wall-wart the user will connect. The user might assume that because the input is marked "12V", they can choose any 12V wall-wart. If I'm lucky, the wall-wart chosen will be close to the maximum current drawn by the circuit - in which case the voltage will be close to the nameplate rating. But what if the user connects a wall-wart rated several times higher than the projected current draw? The wall-wart's output voltage will rise dramatically.

Exactly how much will it rise, though? From doing some measurements of a few wall-warts I have laying around, the off-load voltage varies all over the place. For a couple of 12V 1A linear wall-warts, I was seeing about 18-21V off-load. Smaller wall-warts might be worse, but I didn't have any 12V ones to check.

So, is there a rule of thumb for calculating a typical worst-case wall-wart off-load voltage? It will surely depend on the regulation percentage of the transformer used, but what's the worst typical transformer encountered in practice? Also I'm not even sure what transformer voltage a 12V wall-wart uses - since it needs to be large enough to account for transformer droop due to peak rectifier current (which could be quite large with a small, high-resistance cheaply made transformer). I would guess around 10V RMS? Probably the worst chassis-mount transformers I've come across have a regulation of around 60% - but wall-warts are so cheap and nasty I wouldn't be too surprised if it were worse. On the other hand, if the wall-wart is overrated in current, it will probably have a better regulation percentage due to the bigger transformer. So I suppose there will be a sweet spot where the transformer is big enough that it's running partially-loaded, but small enough that it has a poor regulation - giving the worst-case input voltage.

I found one source online claiming they were seeing an off-load output voltage from a switching wall-wart of twice the nameplate voltage. That seems awfully high, and perhaps not typical? But do I need to design my device to tolerate such a high input just in case such a wall-wart does come along? Are there wall-warts that could be even higher than twice?

I don't have any control over whether a switching or linear wall-wart is used - it could be either. But it's probably safe to say it won't be regulated either way.

  • \$\begingroup\$ Can you tell us more about your 5V load requirements, and what kind of efficiency minimum you are required to meet? \$\endgroup\$
    – Jim
    Commented Feb 15, 2018 at 20:30
  • \$\begingroup\$ The load current depends on the specific project I'm working on, but my current project has a worst-case draw of 230mA. I'm not too worried about efficiency, but I would like to avoid a large heatsink and lots of ventilation slots in the case. If I use a 12V wall-wart which can output twice the rated voltage at low load, then at +10% mains voltage the input could be 26V. That's nearly 5 watts in the regulator, which is uncomfortably high. But if I have to design for this worst-case, I'll probably use a switching regulator instead. \$\endgroup\$
    – Foxie
    Commented Feb 15, 2018 at 20:37
  • \$\begingroup\$ For others reading you: The tighter the transformer coupling between secondary and primary the better the transformer regulation. Loose coupling can be helpful, as in neon sign transformers where it plays an important role. But loose coupling is also an earmark of poorly built transformers, too. So to compensate, cheap wall-wart manufacturers will over-rate the secondary voltage. Unloaded, this means you will see a rather high voltage at the output. But loaded, this comes back down into the desired range. 30% or even your 60% regulation wouldn't shock me. \$\endgroup\$
    – jonk
    Commented Feb 15, 2018 at 22:09
  • 1
    \$\begingroup\$ Don't encourage the 'user' to do the power supply selection. It's an important part of the design, and should be done by the designer. \$\endgroup\$
    – Whit3rd
    Commented Feb 15, 2018 at 22:18
  • \$\begingroup\$ That's certainly one way to do it! I've seen a fair few devices where the user does have to select a power supply however. And certainly, people love to swap power supplies around under the assumption that all wall-warts are equal. The only real way to prevent that would be to use a proprietary connector. Or just rely on the thermal shutdown of the regulator to protect against user error - but that seems like a really bad design. \$\endgroup\$
    – Foxie
    Commented Feb 16, 2018 at 0:17

1 Answer 1


Well, without having much information about what constitutes a minimum load for regulation, here's a quick and dirty way to get what you want done:


simulate this circuit – Schematic created using CircuitLab

Now, as soon as the wall-wart is attached, you are at least loaded with the LED current ( higher current = better regulation, but lower efficiency of design) + the 7805 current (~5 mA internal + 5 mA for R1). I'd think this should be more than enough to get any wall-wart into its regulated 12 V, plus you have an LED to show when power is applied.

  • 1
    \$\begingroup\$ Thanks, but I'm pretty sure the lightly-loaded voltage of an unregulated wall-wart will still be close to its off-load voltage. The nameplate voltage assumes the maximum rated current is being drawn, but it rises linearly from there depending on the actual current drawn. A small load like an LED would be fine to keep a regulated wall-wart in regulation, but the wall-warts I'll be dealing with will most likely be unregulated. \$\endgroup\$
    – Foxie
    Commented Feb 15, 2018 at 22:04
  • \$\begingroup\$ Jim your circuit might work but I expect not as expected. It will probably run hot, lots of oscillations, and not well regulated. There the items of bypass capacitors, they are called out on the data sheet. Rule of thumb I use for TO220 is 65C/W junction to air so a heat sink will be required. \$\endgroup\$
    – Gil
    Commented Jul 11, 2022 at 17:11

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