It is quite easy to understand. An open wire takes a current to be loaded capacitively. But the current does currently not know that the wire is open at the end. But the current in an inductive is not allowed to jump we know from physics. So what should the current do if it reaches the open end? It increases the voltage so that the current is still running. It increases that much that the sparc is visible. The higher voltage at the end then acts like a generator and returns direction sender. If it reaches the end, it sees the transistor source-drain resistor to be loaded with the energy of that reflected wave. Is this complicated?
Lmao, that's awesome brother, this is the same type of thing I always wonder about. Hahahaha. Wonder what the amp looks like under the FLIR ir as you do that
Do you make your copper water cool blocks/ plates or do you have a supplier for them, as I am looking for a supplier for a larger water cool block to do 2 kw on vhf that will be doing 60 seconds TX /RX for very long periods, BTW nice work man, its very hard to find items like these here on this Rock in the Caribbean that I call home .
Yes I make them. If you think these ones would be suitable, or close to it, shoot me an email. Heat wise I have no doubt they could handle it. Mudducksharky@gmail.com
great show ! this is why you should not run pallet to max out put. RG 58 not rated much power. RG 8X or 8M not much better . If you kept going the coax would start on fife ! . 73's
