# Modifying 110v device to run on 240v

Discussion in 'General Electronics Discussion' started by Samon, Mar 14, 2014.

1. ### Samon

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Mar 14, 2014
Hi all,

I have a reasonable amount of electronics experience with DC systems, but have never really done much involving AC, so I'm probably missing something that makes this clear... hence I'm here asking for some help!

I have a [email protected] Steam Mop from the US, which I'm trying to get to work on the [email protected] power we have here in Australia. I've been discussing this with a mate over the last few weeks and we can't agree on whether this would work or not...

Most of the internals (pump, controller, etc) are all DC, and the power supply is fine for 110-240v, so that's no problem, the only issue is the actual heater element, which runs directly from the 110v mains and is stamped '120v;.

It's been suggested that I could simply throw a diode in series with the heater, performing a very rough half-wave rectification by chopping the 240v in half, driving the heater element with 240vrms for half the time, rather than the 110/120vrms the whole time. I'm not so sure this would work, as if I remember correctly back to my highschool electronics days, it's all about the "area under the curve", simply cutting the waveform in half doesn't actually cut the total power in half.

Taking a purely mathematical approach, I've calculated the 'area under the curve' for one cycle for both [email protected] and [email protected], then halved the 240v one to eliminate the half of the wave lost via the diode (but the time factor is untouched, it's just 'off' for 50% of the time).

110vac (155peak) @ 60Hz:

Area under the graph for one full cycle = 1.6552
Area under the graph over 1 second = 1.6552 * 60 = 99.312

240vac (339peak) @ 50Hz:

Area under the graph for one full cycle = 4.3163
Area under the graph for half of one full cycle = 2.1581
Area under the graph over 1 second = 2.1581 * 50 = 107.907

From this I figure the half-wave [email protected] version is about 9% 'more' than the full-wave [email protected] version, so that's approximately the equivalent of [email protected], which is what the heater element is rated to anyway.

Can anyone explain why this wouldn't work?

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Try it again assuming the resistance of the element is fixed and graph the power consumed by the element over a 240V half cycle.

3. ### Samon

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Mar 14, 2014
Hmmm... maybe it's been too long, but how would I graph that?

So it's a 1450w element...

V=I*R, therefore I=V/R
P=I*V, but since I=V/R, P=(V^2)/R and therefore R=(V^2)/P
R=(110^2)/1450=12100/1450=8.34ohms (tested with a multimeter, showing closer to 10ohms, but that is still going through the thermal fuse and a the entire power lead, which I assume would affect the reading)

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Assume 10 ohms.

Then calculate the power at 110V RMS (110V RMS has the same heating effect as 110V DC)

Now do the same for 240V RMS. Then divide that by two to simulate the effect of using a diode to power the element for alternate half cycles.

Do you note a problem?

5. ### shumifan50

551
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Jan 16, 2014
Forgive my simplistic approach.

By inserting the diode, no voltage will be over the element for half the cycle. This will allow the element to dissipate the heat into the water, before the current flows again. So I would guess that even if you had too high a voltage(within reason - not 240V) the element should survive as it is not just the rated wattage but also the rate of dissipation of the heat.
Ensure a diode that can handle the current.

6. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Samon, you've already worked out that P = V^2 / R. Assuming R is roughly constant, P is proportional to V squared.

That means that doubling the voltage will quadruple the power. This happens because doubling the voltage (for a fixed resistance) also doubles the current, and power is voltage multiplied by current.

So without the diode, using a 120V-rated element at 230V will double the current (which is probably a problem in itself) and quadruple the power. If you add the diode, the mean power will be twice what it should be.

7. ### shumifan50

551
56
Jan 16, 2014
As a matter of interest, a friend of mine South Africa(230V) put a big capacitor over the supply to his US TV etc and it worked for him. So maybe that is another solution.

8. ### duke37

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Jan 9, 2011
You would need a very large and expensive capacitor for this power. Use a transformer, also large, expensive and heavy.

You could use a series resistance but this will dissipate the same power as the device unless the diode bodge is also added.

9. ### Samon

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Mar 14, 2014
This is basically the original calculation I did which is why I believed it would not work in the first place.

P=(V^2)/R
P=(120^2)/10=14400/10=1440W
P=((240^2)/10)/2=(57600/10)/2=5760/2=2880W

But this where I'm not certain that my lack of AC knowledge is affecting my logic... Is putting a diode in series and chopping out half of the power supply really the same as just halving the power? I guess I'm wondering if something along the lines of shumifan50 is the case... that the 'pulsed' nature of the power, with equally-spaced areas of zero power 50 times per second would allow the element to cool enough to prevent it from burning out (this is the argument my friend who suggested this would work is using).

10. ### shumifan50

551
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Jan 16, 2014
The trouble is that we don't know the efficiency of heat dissipation of the element to the water at the different water temps, so calculation is not possible. I also noted that the calculation of power changes with different wave forms/phases, so maybe the calculations are not correct.
For the better electronics guys here it is a bit like calculating heatsinks - unfortunately I am no good at this and do it by trial and error.

The mop is useless unless it works in your area. The diode idea is fairly cheap and, unless it is a very expensive mop, worth trying. There won;t be a market for it where you are.

As a final note: In remote controls the IR transmitter LEDs are driven with up to 250ma (rated at 45ma) for short bursts to get the required range and left to cool between bursts. They are typically off for 25ms between bursts.

Good luck and let us know what happens. This is a fairly common problem I have seen asked a few times before, but never answered or reported back.

Last edited: Mar 14, 2014
11. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Excellent -- that was a correct assumption.

That's correct.

Yes. Over a period greater than 1/f sec (technically a multiple of this -- f = frequency) the power dissipation does average out that way.

No, shumifan is not correct. The power already pulses on and off. The time it takes for the element to heat or cool is far larger than the period between half cycles.

The issue (as pointed out by Kris I think) is whether the element would be able to handle the higher current. However this is mooted by the fact that your total average power dissipation would still exceed the rated value by 100% and that would likely cause problems first.

Totally irrelevant in this case.

The heat transfer characteristics are not being changed. The waveform is known, and the time constants involved are very large compared to the frequency.

That's also a fairly simple thing, especially for nearly constant power which we have in this case.

The problem is that even if it works, the power drawn form the supply is twice the rated amount and is unsafe.

If it ever got to a steady state condition, it *might* work, with the on/off cycles shorter than previously. However in getting to this situation you are likely to overstress things and also overtemp the unit. This could result in thermal fuses blowing, or the element going open circuit. In the worst case you could have a more pyrotechnic failure (think fire).

12. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
I agree with Steve. You would be dissipating twice as much power, which is a VERY significant factor, and your peak current would be twice as high, which would stress everything in the system.

Remember that the 110VAC system already operates at twice the current of a 230VAC system (for a given amount of power), which is why fuses and wiring in 110VAC systems have higher current ratings than the 230VAC system in your house.

The instantaneous peak power that element will draw (at the voltage peaks) at 230VAC is FOUR times what the element, and other parts of the unit, are designed to handle.

So I'm afraid you've lost this battle. But you've learned a few things!

If you decide to continue, do plenty of testing outdoors, using an industrial RCD/ELCB/GFI (whatever you call it) and keep it away from anything flammable.

Edit: Also, read your fire insurance documentation. It will definitely not cover you if you use the diode idea; it may specifically mention running equipment designed for a different mains voltage as an exclusion to the policy, since it is a known cause of fires (and at low power levels as well)!

Last edited: Mar 15, 2014
13. ### shumifan50

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Jan 16, 2014
@Steve @Kris

Thanks for the explanations, I again learned a lot and appreciate it.

The only bit I am still not clear on is the heat dissipation. How do we know that the thermal efficiency of the transfer between the element and the water is at its limit. If it is not, it might be able to dissipate more heat than it does at 110V. I would guess that, as it is a home appliance, there is a significant safety factor. What is true regardless, is that the unit will, if the element does not blow, produce steam at a much larger volume than designed and could become a 'steam bomb'. Hopefully the thermal fuses will protect against this or some safety valves.

From all the discussion, it seems the best solution would be to replace the mop with a local model. If the same model is sold locally, you might be able to get a properly rated element and replace it, but this will most likely cost as much as replacing the mop and other components might still fail.

14. ### Samon

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Mar 14, 2014
Thanks everyone for taking the time to explain this in terms I can finally understand!

The same model is available locally, however at approximately twice the price (about \$200 compared to the \$110 I paid for the 110v version - priced per volt? ) - if you add the cost of the wrong version as well then this becomes a \$310 steam mop

I tried calling the manufacturer to see if I could order just the 240v heater element - apparently even they can't get *any* replacement parts - if it's within warranty they replace the whole unit, otherwise it's a case of throw it out and buy a new one!

After trawling the internet for part numbers, etc I have found that the heater element is the exact same part number as in a \$130 handheld steamer, so in theory I could buy one of those and swap it out, but then I've spent \$240 and only end up with the equivalent of something I can purchase locally for \$200!

On a side note, anyone on [email protected] power interested in purchasing a never-used once-dissassembled steam mop?

15. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
I don't think that makes sense. Given that nearly all the power drawn by the element is converted into heat energy, and that most of that energy can't go anywhere except into the water being heated, I don't see what "thermal efficiency" even means in this context.

I tried to explain why, but I accidentally closed the tab and lost the explanation. Probably just as well, because I don't know much about the physics of heat. My barely informed opinion was that doubling the power dissipation in the heating wire (or whatever it was) would increase the temperature difference across the insulating material within the element tube and possibly damage it, and that the tube temperature would also increase due to the thermal resistance between the tube and the water due to the bubbles of steam.

No doubt there are folks here that really understand this stuff and can give an accurate explanation.

16. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Hmmm, let me try to explain.

The thermal coupling between the element and the water to be heated isn't perfect. This means that the element gets hotter than the water.

Heat flows very much like water or electricity. It flows from a place that is hotter to a place that is cooler. That path has some resistance to the flow of heat, so for a given temperature differential, heat will flow at a certain rate.

If you maintain the same resistance, but increase the amount of heat you are pumping in at one end, more heat will flow, but the temperature differential will become higher.

That means, if you operate the heating element at twice the power, the differential between the water temperature and the element temperature will double (roughly, because there are other factors).

If the element was 50C higher than the water temperature, it might reach 150C when the water was boiling. Now it might reach 200C when the water boils.

It is quite reasonable to expect that materials which are stable at 150C might be far less so at 200C.

It is true also that if you are placing heat into the water at a higher rate, there exists the potential to produce steam faster or at higher pressure. However I think this is less of a concern because the device will almost certainly have some way of regulating the pressure and venting excess. However the risk does remain.

17. ### shumifan50

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Jan 16, 2014
Thanks for the explanations.

18. ### duke37

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Jan 9, 2011
One way of reducing the power of the element would be to use phase angle control, like a light dimmer but MUCH fatter.

The peak current will be very high, the radio interference could be severe and fuses/trips may object.

19. ### Kiwi

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Jan 28, 2013
Is it possible to get a local electrical shop to wind a new element on the original former?