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 110v@60Hz Steam Mop from the US, which I'm trying to get to work on the 240v@50Hz 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 110v@60Hz and 240v@50Hz, 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 240v@50Hz version is about 9% 'more' than the full-wave 110v@60Hz version, so that's approximately the equivalent of 120v@60Hz, which is what the heater element is rated to anyway.
Can anyone explain why this wouldn't work?
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 110v@60Hz Steam Mop from the US, which I'm trying to get to work on the 240v@50Hz 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 110v@60Hz and 240v@50Hz, 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 240v@50Hz version is about 9% 'more' than the full-wave 110v@60Hz version, so that's approximately the equivalent of 120v@60Hz, which is what the heater element is rated to anyway.
Can anyone explain why this wouldn't work?