Last night I found a good explanation on the web, this morning I can't find it!

Gist of it was Microwave Oven Transformers are designed to power a pulsed cavity magnetron with the minimum sized core and copper. The transformer is intended to deliver high-voltage / high amperage pulses and magnetic shunts are used to stop the transformer passing excess current and burning itself out. An inductive choke would do much the same job, but a few extra laminations inside the transformer are much cheaper.

Rewinding the secondary for other purposes the advice was to remove the shunts. In theory more turns should be added to the primary to compensate for removing the shunts, but in practice they work without. I suppose the worst that could happen is the primary burning out and junking the transformer,

Ages ago I built Tesla Coils and looked into MOTs because they're an attractive source of cheap beefy high-voltage power. Unfortunately, MOTs are exceptionally dangerous, perhaps the closest you can get in the home to a full-throttle industrial accident. Apart from the kilovolt shock hazard they deliver more than enough current to char flesh. Several amps! Those lucky enough to survive electrocution spend months in hospital recovering from severe burns.

Don't play with an unmodified MOT!

Dave

I took apart a modern microwave oven which used a HF + much smaller transformer to provide the EHT for the magnetron – the HF was derived by an electronic circuit fed from the rectified mains

The shunt is as has been said to provide a degree of protection – both to the magnetron and the transformer by providing current limiting. When current flows in the secondary, an opposing magnetic field is set up in the transformer core. This reduces the induction in the secondary, and therefore output voltage. This is overcome when there is only one magnetic path, and so all the primary flux cuts the secondary winding, and current is then only limited by the winding resistance ( and AC impedance which we will ignore for this discussion)

However, if a second flux path is provided – the 'shunt'- The secondary generates the opposing magnetic field, the strength of which depended on load current, but the Primary flux wishes to be maintained, primary flux is forced to pass more and more through the shunt, and less through the secondary leg, hence output voltage drops as does current, and a balance is achieved.

This is easily seen in the inexpensive buzz-box stick welders – these often use normal E-I core laminations, with the primary and high-amperage secondary windings on the center leg of the E laminations. All flux cuts the secondary all the time, current is max and limited only by wire and cable resistances. Typically the hi-current secondary open circuit voltage is 18 to 26 volts.

The lower current winding are in series with the high current primary and wound on one of the outer legs of the E laminations. The resulting voltage is higher – from 30 volts up to even 50 volts. This ensures a good arc is struck at the start. But then, after the arc starts, the current flow in the outer leg secondary generates an opposing flux which forces a portion of the primary flux to flow in the opposite leg of the E core, and not in the secondary leg, so dropping the secondary voltage an current.

( Contrary to intuition, the HIGHER the secondary voltage in the outer leg windings, the LOWER the secondary current – this is because the high voltage strives to force a high weld current, which in turn generates a higher opposing flux, forcing the voltage to drop and current with it.)

Here I converted a buzz-box core to a 5000 ampere spot welder – the core is simply a square core , similar in operation to a toroidal core, and as a buzz-box welder it had a shunt across the center which I removed. The secondary winding are over the primary. The white upper section with the black stripes are pieces of wood held in place with tie-wraps to stop the core laminations buzzing ( its a buzz box..) at high currents!

The yellow cables are 20mm diameter, two in parallel and in the central part of the core they occupy the space where the shunt sat.

The core is not ideal – it has lots of leakage – desirable in cheap stick welders to have some sort of welding 'control' in lieu of proper current regulation ( variable saturable inductors, etc) and so does not provide best performance in a spot welder application. You want the best magnetic coupling ( secondary under the primary, tight wound) , good silicon steel laminations, properly insulated ( lacquer/varnish) , very low resistance in the weld current path ( only copper, no brass, steel, aluminium, etc).

I have 3.8v at no load on the secondary – at 5000 amps flowing the voltage drops to 2 volts typically, at the weld, and is 3.2volts at the ens of the yellow secondary wires – 1 v is lost in the rest of the cables, the weld arms, connections, etc – thats only 0.2 milliohms!!! Resistance rules in a spot welder!

All very long winded,

The bottom line is if you leave that shunt in place, you will a Spot Welder Not Make…

And Removing it will raise the primary magnetisation current a fraction but can be ignored.

Joe

Edited By Joseph Noci 1 on 26/01/2021 06:38:29

Yep, collect those uWave ovens if you can – those transformers are very useful for all sorts of things – excellent for high current power supplies ( for the likes of radio hams) , beefy audio amplifiers, small welders , etc – If you try to purchase such a core with primary winding it costs a fortune!

Regarding bandsaw blade welding – The current required for blade welding varies hugely, and generally underestimated – the thin blade, sort of 0.25 mm thick wood blades, the shiny bright silver types, up to 10mm or so width, weld easily with around 300 / 400 amps – within the capability of an MOT style spot welder. The metal shop type blades – 0.6 to 0.8mm thick x 12 to 16mm wide are a different story – A 0.8 x 16mm blade yields a surface contact area of 12 sq mm – that is a large area and needs lots of current to heat orange/white! Easily 2000 amps or more. Too little amps creates a tendency to extend the weld time, which just heats up a larger portion of the blade, annealing the nearby teeth to uselessness – the weld is also invariably suspect. The weld region must be very limited in width and that you achieve with a very high, short blast of current.

I have an 'IDEAL BS-1 blade welder – it works well for the woodshop blades, but is very marginal for the metal saw blades…So, I hook the big spot welder up to the IDEAL's terminal for the heavy blades!

Joe

EDIT :

Even more reason to collect more transformers – if you can get a few identical ones, you can parallel the secondaries and double/triple, etc, the current easily.

Joe

Edited By Joseph Noci 1 on 26/01/2021 13:58:27

I silver solder my 6 x 4 band saw blades. I chamfer the two ends using my belt sander, I made a simple jig to hold each ends, one at a time, getting both the chamfer to the same angle. Then using another little jig ( a piece of angle iron this time) to hold both ends together, I apply the flux, heat things up and solder. A easy quick job and, I mix and match bits to make another blade, now have more than eight. Rarely do I get a broken one, worn out yes. Wouldn't be without the bandsaw, there was an article some where on making a larger table for ripping lengths of wood John