Firstly I am not at all experienced in electrical matters so forgive any wrong terminology.

Here's the situation: I recently moved out of US to a country that uses 220v. We have a single phase step-down transformer that provides 120v to the entire house. There are a few appliances that we would like to order with the option of either 110 or 220 versions. (220 versions are more expensive)

What I would like to know is, is there an increase in power consumption if I am to run an appliance through the step-down system vs the direct 220 sockets?

eg. 300w food dehydrator 110v version run through transformer vs 300w 220v version plugged into native wall socket

Another way to ask the question would be; Is the total power consumption of a 600w machine run through a step-down transformer 600w? Or is it more?

Thanks for any insight you can provide on this!

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    \$\begingroup\$ Yes. More transformers = more losses. Higher currents = more losses. Lower voltage, same power = higher current. \$\endgroup\$ – Tom Carpenter Feb 19 '17 at 18:34
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    \$\begingroup\$ I'm curious: what does a food dehydrator do, technically? Is it just a fan? Is it a fan with a heater? Is it just a heater? \$\endgroup\$ – Marcus Müller Feb 19 '17 at 20:06
  • \$\begingroup\$ @MarcusMüller fan + heating element + thermostat + timer w/ drying racks \$\endgroup\$ – mFabien Feb 19 '17 at 20:10
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    \$\begingroup\$ For all the computer/entertainment/electronics, have a good look at the external power supply: many, if not most, have switch-mode power supplies that just work on both 110V and 220V. For example, I've yet to buy a single laptop whose charger would need more than a plug adapter to work in both the EU and US. \$\endgroup\$ – Marcus Müller Feb 19 '17 at 20:18

For an ideal transformer, power in equals power out: it's 100% efficient.

For a realistic AC line step-down transformer, power in is a little bit more than power out: it's not quite 100% efficient, but it should be close.

Thus, if you plug a 300W load into a step-down transformer (assuming the transformer is rated for more than 300W), expect it to draw a little more, perhaps 325W - 375W depending on quality of construction.

  • \$\begingroup\$ Thanks so much for your reply. Just for a little more info. What would you suggest as a good ballpark % to add in calculating the expected power draw? My particular interest is for a 600w machine & I'm trying to figure out whether it's worth paying more for the 220 version based on expected use. \$\endgroup\$ – mFabien Feb 19 '17 at 18:42
  • \$\begingroup\$ I'm not exactly sure, because I had a hard time finding a spec sheet for similar devices. However, I would expect 80-90% efficiency. Also, see @jonk's excellent answer regarding frequency /breaker limitations and make sure that your device will operate as expected. \$\endgroup\$ – uint128_t Feb 19 '17 at 19:09

You didn't bring up something else -- frequency. You don't mention the country, but I suspect that if they are supplying a mains voltage of \$220\:\textrm{V}_\textrm{AC}\$ then it's probably also \$50\:\textrm{Hz}\$. Just using a transformer alone to drop the voltage to US levels doesn't change the frequency and not all devices work properly (or well) on the wrong AC frequency. (I'm assuming here that you are using a transformer designed for the operating frequency of your country's main supply.) So that will be one consideration. You need to make sure that if you buy the cheaper US rated devices that they will still do okay on the different frequency. Electric water and space heaters are probably just fine (though I can't speak for the fancy ones with electronic panels in them.) But just be aware that a vacuum cleaner's motor might not work the same or as well, even with the voltage right.

Regarding efficiency, all transformers are imperfect and will sustain some losses. Also, the lower voltage implies a higher current for the same power. And this implies the need for more expensive wire. Perhaps a lot more expensive. Keep in mind that losses in the wire are equal to \$I^2 R_{wire}\$ and that if you double the current (as required for the same power at the device) that the losses go up by a factor of 4 (if you leave the wire alone.) Also, the voltage drop across the wire is \$I R_{wire}\$ and this means more of a drop, too (again if you leave the wire alone.) All this together means that you need a lower \$R_{wire}\$ value for similar results and that if you truly want the exact same losses in the wire then you need wire with \$\frac{1}{4}\$ the resistance since the current will be twice as high.

Just be thoughtful about each choice and make sure you have adequate wiring for the US voltage devices. Consider the implications of frequency, of wiring and transformer losses, wiring voltage drop to the device, and what wiring you have installed to get to the device, when making a choice between the two options.

Also, you might consider limitations of your breakers and US devices, generally. For example, a skillet in the US will top out at about \$1500-1800\: \textrm{W}\$ because most US breakers are set for \$15\:\textrm{A}\$ rating. However, that's not true for many European skillets, which have the advantage of higher voltages. There are times when you actually would like to have a higher wattage skillet for cooking larger quantities of food.

So think about all the details. But you've got options, it seems. So I think that puts you in a better circumstance than not having the US voltage available. You can pick and choose, at least. Just be aware while doing so.


The power consumption already factors in the voltage, so 300W are always 300W regardless of the voltage. The lower the voltage, the higher the current, but same power consumption overall.

HOWEVER... Be aware that the transformer itself will have some losses that will add up to the power consumption of the appliances you connect to it.

There's no such thing as a free lunch, sorry.


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