# elements and voltage

I have an American refrigerant charging cylinder with a 110v element which I wish to run on 240 volt. Would I get away with a diode in series?
I.E. run on half wave
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I presume the "element" you are talking about is a heating element as opposed to charcoal, graphite, diamonds or red phosphorous. If that is the case, running with a half-wave rectifier should cut the applied voltage to the ele,element to a bit less than 120 VRMD. Remember that there is a diode drop.
The main problem I can think of, presuming it is done correctly, is that the dc current flow could saturate a transformer if one is used to provide the line power. The old ac/dc tube radios had half-wave rectifiers, but would draw only about 200 ,A at most. A heater could draw much more.
Another possible problem is that come control circuitry might not like half-wave waveforms.
Bill
--
Private Profit; Public Poop! Avoid collateral windfall!

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Salmon Egg wrote:

Just a plain heating element and Australian 240v 50~ ac
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F Murtz wrote on 31/12/2008 :

Best practice it to use a stepdown transformer. Is it a 200 watt heater? If so Jaycar sell a Stepdown Tranny rated at 250Watts for about \$130.00.
If you must, use a 6A 1000V diode. It will take any spike you may get on the voltage. If my math is correct...have to forgive me, it's been some time since I have had to do this... if the tank heater is a 200 watt heater then it will only be putting out 180-190 watts.
I've never used a heat blanket on a bottle before... always used a heat gun. Good luck with it.
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Arlowe wrote:

Thanks
This is a dial a charger the element is built in
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I don't fully understand what it is you have. I would assume from your description it is something for charging refrigeration systems with the refrigerant. If it's something you would use occasionally and not continuously could you hire a transformer? Here in the UK any hire shop would have 240/110V transformers as, legally, you can only use 110V tools on building sites and the like
--
Stuart Winsor

For Barn dances and folk evenings in the Coventry and Warwickshire area
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Without thinking too hard on it, if you ran the heater at twice its normal voltage, it would consume 4 times the rated power. So if you fitted a diode to cut out half of the ac cycle, it should then consume twice its rated power, and run hotter.
However, if the heater has a simple thermostat fitted, you might get away with the diode solution, the heating cycle would be shorter.
regards Barry
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You are correct. My fast answer was wrong and I should have known better.
Bill
--
Private Profit; Public Poop! Avoid collateral windfall!

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Baz expressed precisely :

You wouldn't be running at twice the voltage using a half wave rectifier. The non-filtered waveform still has a substantial amount of ripple so the applied voltage is less than half the AC voltage. I believe it is a multiple of 0.45 of the AC RMS. However the evil part is determining the PRV rating of the diode to be used. I prefer to work with double the peak to peak voltage just to be safe.
But I do agree with you 100%. A stepdown transformer is the way to go.
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Arlowe wrote:

With two 6a 600v diodes in parallel it seems to take about .4amps (just tried it) The cold resistance is 314 ohms Trying to do it on the cheap can not see element markings hard to remove easily
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F Murtz wrote:

I assume that this is something that you only need to use infrequently and then for not long periods - so efficiency isn't a major issue.
You could wire it in series with a lampholder and try various wattage 240v incandescent lamps in the socket until you got the desired voltage across the heater. Start with, say, a 40W lamp and work up. If necessary, wire an additional lampholder (or more) in parallel with the first one and start with 40W lamps again. You may get lucky and one combination give you the voltage that you want.
-- Sue
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Palindrome wrote:

just measured, with the diodes ie half wave the analogue voltmeter on dc just a gnats over 100 v
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F Murtz wrote: <snip>

And what do you reckon that reading shows, the average voltage? the rms voltage? the peak voltage?
Heater elements can be very forgiving - even when dissipating twice their designed power, they can last a while.
My electric waffle iron came from the US long before they were heard of in the UK and is still surviving, running off UK mains with a series diode. It only gets used for an hour or so, each year, though. Of course, its insulation is also being subject to out of specification voltage and its thermostat contacts are being subject to out of specification opening and closing conditions - but all the exposed metalwork is well bonded to "earth" and it is supplied via an RCD - so I take the risk. I never leave it running unattended. A diode short circuit failure, thermostat contacts welding together, insulation failure aren't likely to result in more than a slightly over-cooked waffle and/or a tripped circuit. Oh, and a dead waffle iron. Replacement 240v waffle irons are cheap and easy to come by.
So, I would suggest that you give some thought to what the effect of failure would be, in your case, and whether the risk is worth taking.
-- Sue
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F Murtz was thinking very hard :

Pretty close to what I would expect. If the load voltage is 108 V and the load resistance is 314 ohms then the current should be 0.35 Amps. Very close to what you have.
If you are game, you may even install a filter, that will smooth out the ripple and raise the load voltage. You will have to use an electrolytic capacitor ,22uf 400V would be a good guess. You will need to place it parallel to the load and make sure you get the polarity right.
anyway, good luck with it.
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As others have pointed out, this will leave you running double the rated power in the element. To get close to the original rated power, you could use a triac "dimmer" type circuit to remove more than 50% of the original waveform. The actual on time you'll need will be somewhere between 25 and 50%. You can figure out the actual value mathematically (calculate the square root of the integral of the square of sin(x) over a the initial portion of a half-cycle, and determine what portion gives you a factor of 110/240 of the full-cycle voltage. Or you can just attach a RMS-reading voltmeter to the output and adjust to get the 110 V RMS value you need.
Ordinary lamp dimmers can handle up to 500 or 600 W resistive; if the heater is larger than that you'll need something bigger rated to handle it.
Dave
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F Murtz wrote:

I have nutted it out, dug out a variac tried element on 110v ac, .38Amps. With diodes, on 240v ac .346 amps dc on fluke .35 Amps dc on avo analogue. Cheapest stepdown trannie (240v to 115v) I can find \$50 and that is only 50 watts (not leaving much leeway)Next size up \$100 mark. Variac in box too big to lug around Thanks for all interesting discussion. I should have done this before I asked but I forgot I had The variac.
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