Bad calculation...

Externet

Well-Known Member
Hi.
To make the fan motor inside a 120VAC household heater work with 240VAC...



Measured the mains with the fan ON = 124VAC
Measured the voltage across a 10 Ω resistor placed in series with the fan motor = 4VAC
That means 120V across the fan motor with current of 0.4 A trough both. That means the motor impedance is 300 Ω

A plain 300 Ω resistance ( as a light bulb of 120V 48 W ) in series will allow to work on 240VAC

240V-------------------------120Vlamp48W---------------------------Fanmotor-----------------------neutral

Can it be a 300 Ω resistance from a 0.48 Watt 0.04A 12 V lamp instead ? As an automotive incandescent light bulb ?
Where is the mental fart ? Wattage must be the same ? Bulb has to be able to work on the 120VAC drop voltage ?

240V-------------------------12Vlamp300Ω---------------------------Fanmotor-----------------------neutral

 
That type of light bulb changes resistance with temperature of the filament. Measuring the bulb cold, with an ohm meter will not give you a good reading.
Do not put a 12V bulb in this project.
Do you have two fans. Wire them in series.
 
Thanks. All incandescent lamp bulbs change resistance from cold to hot. I did not measure any lamp bulb resistance for post #1.
I have seven heaters, but not able to work in pairs. They are in different rooms.
Question was, what other parameter for the series resistor ? 300 Ω is 300 Ω from this, that or the other, but clearly will not work. Where is the mental fart ?
 
The heating elements are basically resistors, so you could work out what resistance to use in series. However, it is a really bad idea because the resistance in series will generate just as much heat as the heating element. Also, a lot of heaters have various power settings, and the resistance is different for each power setting, so you would need to change the series resistor.

The fan will be highly inductive, so although the impedance is 300 Ohms, you don't know what it's resistance and inductance are, just that they give 300 Ohms when in series. Inductive impedance and resistance don't add up like resistors in series. The total impedance is the square root of the sum of the square of the inductive impedance and the square of the resistive impedance.

You would need somewhere between 300 Ohms and 520 Ohms, depending on how inductive the motor is.

If you measure the DC resistance of the motor, the resistance needed in series could be calculated more accurately.

You could use a transformer.

Putting a 12 V lightbulb in series is a terrible idea, because the resistance changes so much, and because you could easily have far more than 12 V across the lightbulb, so it will burn out almost immediately.

If you want to run the heaters, it'll be far cheaper to buy 240 V heaters as they are much cheaper than the resistors (and probably heat sinks) that you would need to go in series with any 120 V heaters.
 
it is a really bad idea because the resistance in series will generate just as much heat as the heating element.
Thanks, Diver.
The fan motor is the only item to change into running on 240V on this thread. Any heat generated by a series resistor to the motor is fine, as the appliance is a heater.

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Irrelevant but just to explain : there is two heating elements in parallel inside, that on 120VAC yield 1500W and would be wired in series for 240V operation. I believe is the norm for many heaters, factory wired in series for sale in 240V markets. Those should have 240VAC motors; cannot buy in U.S.
 
Does your romex supply to the unit have 4-wires (bare ground plus, black (hot 1), red (hot 2), neutral (white) and bare ground)?

If so, you can simply connect your heater from hot-hot for 230vAC and hot-neutral for a 115v supply.
 
Can it be a 300 Ω resistance from a 0.48 Watt 0.04A 12 V lamp instead ? As an automotive incandescent light bulb ?
Where is the mental fart ?
The fart is that the 300Ω is at 12V and 0.04A.
When put in series with the 0.4A fan, the voltage will go way above 12V and you will see a nice, momentary, very bright light from the bulb as it blows.

A 50W, 120V lamp in series, may possibly work, but the starting current of the motor may blow it.

An alternate is to connect a small 50W space heater (example) in series with the fan.

A third alternate, that dissipates no power, is a series non-polarized capacitor, such as a motor run (not start) capacitor.
A size of about 6µF should be in the ball park.
 
If you have a transformer rated at 50 VA or more that has a 120/240 V primary, you can just wire the motor to the 120 V part and the mains to the 240 V part. Leave the secondary disconnected.

A lot of transformers have two 120 V primary windings that are put in series for 240 V and in parallel for 120 V. Just put the windings in series, connect the mains across the series combination like you would if using the transformer normally, and connect the motor across either winding.

There will be about 0.2 A flowing in each winding, so each winding is handling about 24 VA, so a 50 VA transformer will be fine as the current in each winding will be what it would normally be at full load, and there will be no secondary current, so the transformer will be cooler than it would normally be at full load. However it might be mounted inside a heater.
 
Crutschow suggestion of a capacitor ~8uF is something I had forgotten to consider. It is small to fit in, does its job and cheap. Thank you.
[ Am in U.S. and cannot buy that heater unit for 240VAC. Only available for 120VAC. Distributor does not sell the fan motor as a spare part for 240VAC ]
 

The resistance of the bulb of 300 Ohms would only be at 12 volts.
For other voltages:
At 1v: maybe around 30 to 40 Ohms.
At 12v: 300 Ohms.
At 120v: infinite resistance because it blew out.

This is fairly easy to imagine.
If you want to drop 120v, then whatever you put between the wall socket and the fan must be able to handle 120v. A 12v bulb can not handle 120v or even 24v without blowing out rather quickly.

On the other hand, if you want to use it for a flash bulb for taking pictures in a dark room then maybe you are on to something

You might be able to get this to work with a 120vac 50 watt bulb, but the startup current will be a lot more than 0.4 amps so you'd have to make sure the fan does not get damaged.
A 300 Ohm resistor would have to be rated at least 100 watts to handle this without getting exceptionally hot. Even then it will be as hot as a 50 watt heater. Not that much for a heater but still could be significant in a cooling application.
Maybe a 50 watt soldering iron
 
Thanks, Al. A soldering iron heating element. Good one too, cheaper than the capacitor and contributes with heat !.

View attachment 147167

Hi,

Well I should add that the soldering iron core will get VERY VERY hot, no kidding, and I'm sure you know why.
Unfortunately, that means making a special housing.

One idea to reduce the temperature *might* be to make 2 sets of 2 in parallel, then put those in series. Those four elements will (or should) make up the same resistance but have 4 times the power rating and thus will not get as hot as a normal single soldering iron core. Of course you must test it carefully to make sure it does not get too hot for your placement and that it still drops the right amount of voltage.
Fans can easily take a little less voltage but not sure how they can handle a higher than nominal voltage.
 
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