thermal knife power supply (part 2)

Question

This is "part 2" because my original question has only brought up more queries that i am unable to solve. The first part can be seen here for any who are interested: "Hot-knife"/Thermal cutter power supply design

I apologize for making two posts for the same project but my other has seemingly died :(

currently i am trying to construct a thermal knife to cut and seal ropes, both to avoid expensive store bought ones and as an educational project. i have bought a blade online to use that is supposed to be 60W, it can be seen here:

==================No longer relevant, see update======================== To obtain 60W of power im going to use a 60VA transformer to transform 240VAC, to 4V 15A output. so now my speedbumps:

1. I understand 15A is a very large current so will a voltage of 4V be low enough to be safe?
2. Before building, I wanted to test these conditions using a benchtop supply. When i connect the blade across the two terminals however i cannot get the voltage to go above 0.6V, even pushing the current to the supplies max of 20A. This would only dissipate 12W of power not the 60W i had wanted to test. My theory is that this is due to the extremely low resistance of the blade which according to V=IR would require a very large current to produce a moddest voltage. Any thoughts on another way to test before committing to the transformer?
3. I have been told a ceiling fan speed control is a good addition to fine tune the current going to the transformer. using Australian mains of 240VAC, would something like this be appropriate? http://www.ebay.com.au/itm/AC-250V-50-60HZ-Fan-Speed-Rate-Control-Rotary-Button-Wall-Panel-Switch-/301422113787?pt=LH_DefaultDomain_15&hash=item462e2873fb

Thank you for any and all help :)

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Edit - Update 1

Request was made for blade datasheet, it can be found here: http://www.saferight.com.au/_literature_167996/Hot_Knife_Datasheet Am i correct in assuming a 60W power draw for the blade? It is possible that some of this power is used by other components in the tool, in which case i think my only option would be a guess and check method as I have been unable to find any other useful datasheets.

I have done some measurements to deduce the resistance of the blade (far too small to measure on a multimeter). Set the power supply to constant current (CC) and then measured the voltage across the blade at different intervals. tabulated V/I using excel and I believe the resistance is 8.39mOhm

Using P=VI and V=IR with the data P=60W and R=8.39mOhm yields values of I=84.57A (im going to need some thicker cable) and V=0.7095V

Being on mains of 240VAC, this would indicate an effective transformer ratio of 338 (using N=240/0.7095) i should be able to achieve this by modifying a transformer as shown in the link provided by Bruce Abbott.

Thank you all, The advice (and scolding) have been a great help thus far, please tell me if this new theory is on the correct path before i start asking around for bigger power supplies to test with.

Answer 1

0.6V at 20A indicates a resistance of 0.03Ω. P=V²R, so for 60W you need to supply 1.34V (at 45A).

Obviously your transformer is not suitable, however you might be able to modify it to get lower voltage at higher current. Some transformers have a gap between the windings and the core, which you may be able to thread a few turns of thick wire (at least 14AWG) through. If you can get enough turns to produce 1.34V then it should provide the required current.

If your transformer has no room for extra wire, but has separate bobbins for primary and secondary windings, you could cut off the secondary winding and replace it with your own thicker wire.

Here's an example:-

How to: High current Microwave oven Transformer

To fine tune the power you cauld either tap into several positions on the output winding, or add a variable length of cable to drop the voltage. You only need about 0.012Ω to drop 0.4V, which would cut power to the knife by half.