Low Current HHO Production.

[Hydro-Cell]

30 percent KOH.
So, just to be sure, you are putting in 300 grams of KOH pellets for every liter (1000 grams) of water.
Right?

yes this is correct, be careful when adding the koh as the water will get hot. i use flakes to not pellets.
pics will follow as soon as i can get some sorted out

[mscoffman]

i have 1000watts of power going into this dipole. this consumes only 4.4 amps of power.
at this frequency and power the signal is capable of travelling round the world multiple times, it would be possible to send this frequency to australia on only 10 watts

@Hydro-cell

This experimental device sounds neat and I am going to remember this until I need hydrogen

Could you explain the 1000watts a little more clearly.  As 4.4amps @ 12Vdc is only
52.8 watts. If the power going into the "final" amplifier is only 53 watts then that's what
should be quoted as going out on the dipole even though a full wave dipole at 4km itself should
be very powerful. Is this; a) 1000watt final amplifier like an audio amplifier that is used?
or b) is this some calculation of ERP -  Effective "Radiated" Power?

VLF radio transmitters require more than ten watts because they compete with the
Lightning Radio noise floor.

I take it that the dipole is wound on the form like a electromagnet, feed as it's center. This
could be called a transducer rather then an antenna - so interference should not be much of an
issue. It's effect would be to produce EM waves that are heavily biased toward the Magnetic
component of the EM wave.

It's neat to see what you are accomplishing so far - all in one place (sort of like an advertisement).
please feel free to repeat what you have done with the 12Vdc stated and the 1KW figure
explained - if you do more in the future you can just repeat the part that stays the same.

:S:MarkSCoffman

[Hydro-Cell]

hi
my figures come out as follows......

using the ohms law circle i found that to find the amps used the formula is

power in watts / resistance in ohms ..squared

now i am using 1000 watts.
the resistance for the centre of a dipole is 50ohms

so thats 1000 / 50 = 20
now the square root of 20 is 4.4

so my current draw is 4.4amps


on the other hand the other to work out amps we can also simply divide power by volts which gives us 77amps.

so wich formula is correct???

since i am only reading low current i belive this to be correct at 4.4 amps.

[TinselKoala]

Ohm's law is V=IR
and power is Watts=I squared R
so I squared = watts/R
take the square root of both sides
I = (watts/R)^1/2
so I = (1000/50)^1/2
I = ~4.47 amps

This is a value for input power to the system--but using the 1000 watt figure is probably wrong.
What makes you think you are getting 1000 watts of power to the antenna?


This would make more sense if you measured the input voltage and current, then calculated the input power from that. Then, use an RF power meter to determine the effective power to the antenna.

[Hydro-Cell]

power / volts = amps