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Is it safe to use a passive plug adapter rated for 110 V in a 220 V mains country?

With passive plug adapter I mean something like this:

random pic from Amazon

Those bad boys are also known as "travel adapter(s)", and sometimes contain a status LED that is lit when the mains voltage is present.


This question was inspired by an identical question on the travel stackexchange:

Adapter voltage different to both mains power and appliance expectations

I had a bit of a discussion with a member of that stack, and I just wanted to check with this community.

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    \$\begingroup\$ Do you have a link to the 110 V adapter? BTW the one in your picture will be rated at 230 V because it is a UK type. \$\endgroup\$
    – Andy aka
    Apr 29, 2017 at 9:36
  • \$\begingroup\$ No, it's just a generic electrical safety question \$\endgroup\$ Apr 29, 2017 at 9:50
  • \$\begingroup\$ LOL then I think you know the answer! \$\endgroup\$
    – Andy aka
    Apr 29, 2017 at 9:58
  • \$\begingroup\$ I do, I just wanted to check w/ the community and don't have time to answer right now \$\endgroup\$ Apr 29, 2017 at 10:19
  • \$\begingroup\$ I would like to express skepticism that the premise of this question can actually happen in the real world. If anybody ever designed and marketed an adaptor with a nominal input voltage rating of 120V that can be inserted into a 220 or 240V mains receptacle from another country, that person should be given a stern talking-to. Adding an LED just makes the offense that much worse. But there are other types of indicators such as gas discharge lamps. Perhaps what you are calling an LED is actually a gas discharge lamp. \$\endgroup\$
    – user57037
    Apr 29, 2017 at 16:31

1 Answer 1

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A direct mechanical adaptor such as this, with NO LED indicator, if designed anything like properly, will be safe on 110 VAC or 230 VAC.

An included LED which is designed to operate on 110 VAC will PROBABLY survive on 230 VAC but may not. Such a LED probably uses a series resistor.
PER mA of LED current a series resistor dissipates about 60 mW on 110 VAC and 120 mW on 230 VAC per mA of LED current, but when the voltage is doubled the cureent will about double, so energy dissipation in the resistor will rise by about a factor of 4 and in the LED by 2+ times*.
(It will have a fwd series diode or reverse parallel diode across the LED so power will be consumed only during a positive half cycle relative to LED -ve.)

As the resistor dissipation double on 230 VAC relative to 110 VAC the resistor power dissipation may cause failure on 230 VAC. Also, there is a chance that a resistor that operates OK on 110 VAC will fail from voltage breakdown (regardless of dissipation) on the higher voltage.

The LED MAY use a series capacitor. The voltage rating of this will determine 230 VAC survival.

Any series or parallel diodes used need to withstand peak 230 VAC voltage.

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Resistor dissipation:

  • If LED voltage is say 3V at X mA and 4V at 2X mA then ~~~=:

Say I_LED is designed at 2 mA at 110 VAC.
I LED will vary with AC waveform point. Assume Veffective is 110 VDC but it probably won't be.
R = V/I = (110VDC -3V_LED)/2 mA = 53.5k
Very non critical - say 56K used.
Power_R = V^2/R = (110-3) ^2 / 56k = 193 mW. This is for a full cycle so 1/2 cycle power = about 100 mW

At 230 VAC if VDC mean = 230 VDC then I~+ 230/56K = 4.1 mA.
As Vf_LED at 4.1V ~= Vf at 4V.
I actual = (230-4)/56k = 4 mA.
ie the LED voltage has minimal effect on current BUT the current doubling MAY damage the LED.
Power-56K is now about 230V x 4mA x 50% duty = 460 mW.
Ideally a 1W + resistor with say 400 VDC rating should be used. If a very small SMD resistor was used that usually didn't fail at 110 VAC it may well fail at 230 VAC.
The above figures are for 2 mA LED current. Scale approximately linearly with current for other currents.

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  • \$\begingroup\$ Assuming a simple current limiting resistor, it seems like resistor power dissipation is proportional to V^2, not V. Please let me know if I am missing something. \$\endgroup\$
    – user57037
    Apr 29, 2017 at 16:22
  • \$\begingroup\$ @mkeith No. Thanks. Will edit. I (almost) can't imagine why I put that. Maube the back of brain said "constant current" - which is of course not the case. \$\endgroup\$
    – Russell McMahon
    May 1, 2017 at 12:24
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    \$\begingroup\$ @mkeith That got to be a long edit :-( - I see why Idid what I did. I said that resistopr dissipation double "per mA" which is correct, but then failed to factor in the fact that current also~= doubles :-(. \$\endgroup\$
    – Russell McMahon
    May 1, 2017 at 12:39

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