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
Private Profit; Public Poop! Avoid collateral windfall!
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
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
Good luck with it.
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
For Barn dances and folk evenings in the Coventry and Warwickshire area
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.
So your answer is No! Use a step down transformer.
However, if the heater has a simple thermostat fitted, you might get away
with the diode solution, the heating cycle would be shorter.
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.
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.
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.
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.
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
I have nutted it out, dug out a variac tried element on 110v ac,
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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