help with wattage conversion!

Should work fine as long as the pins are right for the appliance.

I don't think it will, if I believe the numbers from the cell phone charger. The 50 watt converter rating probably refers to heating load, so it's the same as 50 volt-amps. At 230 volts, that's only 0.22 amps. Obviously the charger has a lousy power factor to take 69VA in and put only 2VA out without turning into a smoking ruin.

You're probably better off to get a US style charger for that phone when your friend arrives. If it's a brand sold in the US, I'm almost certain it will have the same power connection.

Mike

| Hi. I probably learned this in my physics class back in high school, but it | has slipped away. My friend from Europe is coming to visit, and I want to | buy her a Europe to US appliance converter (converting the US 110-120V | coming from my wall outlet to her 230-240V appliance). I found a "reverse | converter" for sale online for this purpose here: |

| | I see that it is up to 50 watts. I know that watts is the same as "VA", but | that's all I know about that. She will only use it to recharge her cell | phone, not for a hairdryer or something. I asked for the numbers on her | recharger, and she gave me this: | | 230 V - 0.3 A 50/60Hz | 5V -0.4A | | I hope someone can help me out to determine if she can use this 50 watt | converter.

230 V * 0.3 A = 69 watts. You'll need something a bit bigger.

There's obviously something strange going on here. If you put 69 watts into something the size of a cell phone charger it's going to get damned hot. Touch a 60 watt light bulb lately?

The input is likely 0.03A judging from a charger that I have rated 120/240V

0.07A 50/60Hz with output 1.5V 0.8A

Maybe the charger is a switcher. In any event, it matters where/how the energy is dissipated, not how much is "put in".

The nominal output is 400 mA at 5 volts, or a couple of watts. There's no way that thing is going to take in 69 watts and return 2 without getting a tad warm. Have you touched a 60 watt bulb lately? Even if it's as big as a 3 by 3 cube, that's 54 square inches of surface. Normal convective losses are something the order of 120 degrees C rise for every watt pushed through a square inch. This is looking like a watt per square inch, and I'm saying you'd blister your fingers if you touched it before it melted. Obviously the ratings are a tad conservative for steady state.

Thanks for all your responses. Well I asked my friend to take some photos. Maybe that would help.

Here's the sticker:

And the part of the charger cord which goes into the wall:

Since they don't make the Siemens MC60 (her cell phone model) here in the US, I can't buy a US version of the charger cord. I have found cords for similar models which I am considering. Siemens also makes a dual voltage charger for the MC60. No idea why they packaged the single voltage charger with her phone instead, since it is a tri-band world phone. I guess it's just something you pay extra for. She can't find this in any store. Oh well.

Thanks again,

~mioi

do you have a dryer plug? they're 240 volts. sammmm

You assume it will run 69 watts input, steady state. It won't - that's obviously a max rating. How could a supply be so inefficient that it needs ~70 watts in to produce ~ 2 watts out?

But that's not the point I was trying to make. My objection is to the statement "if you put 69 watts into something the size of a cell phone charger its going to get damned hot". That is simply not true. You have to look at where the energy is dissipated. If it is not dissipated in that "something", whatever it is, that is the size of a cell phone charger, that thing is not going to get hot. Take your 60 watt bulb example in a table lamp. You put 60 watts in to the plug at the end of the lamp cord - but the plug doesn't get hot like the bulb does. There is only a very small fraction of the energy disspated in the plug/line cord/switch - the large majority is dissipated in the bulb. The plug is even smaller than the size of a cell phone charger, but it doesn't get hot because very little energy is dissipated there.

Obviously we're not on the same wavelength here. I maintain if you take a blsck box into which you put 70 watts, and out of which you get two watts, the reamining 60 odd watts are going to turn to heat. In a 3 by 3 by 3 box sitting on a table somewhere, it means you're stuffing about 1 watt per sqaure inch into ambient. If the mode of heat transfer is convective, it's going to get about 150 degrees C above amibient.

I do agree the rating on the label mis states the steady state power input, but the point I am making is that power in has got to equal power out in the steady state (obviously if within the box there's a chemical conversion going on while a battery is being charged my version of steady state doesn't start until the battery is fully charged). In a box with about 56 square inches of surface, without forced cooling, it's going to get very hot.

I doubt you're disputing that, so I probably misunderstood the point you were making.

But in this example, we know only 2 watts are leaving the device as electrical energy, therefore if the label represented steady state usage, 67 watts are being dissipated IN the converter.

Since this is obviously NOT happening, the rating is probably accounting for a startup surge and a 50 Watt step-up transformer would be fine.

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