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Hi Warren, I will connect a Dc engine and see how the diode plug behaves with increasing load, the energy would be reactive again and the recorded power sink again theoretically. Greetings Sven
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-------- Ursprüngliche Nachricht --------
Von: "Warren Keillor [email protected] [EVGRAY]" <[email protected]>
Datum:11.02.2018 20:23 (GMT+01:00)
An: [email protected]
Betreff: Re: Aw: Re: Re: AW: Re: AW: Re: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.
Sven, Ole
Again, it makes me think of harvesting the energy using permanent magnet motors.. They seem to love pulsed DC input. The ac generator they drive must run at either1800 or 3600 rpm, depending on the type.
Pulleys and belts?
Cheers Warren
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On Sun, Feb 11, 2018 at 2:06 PM, [email protected] [EVGRAY]
<[email protected]> wrote:
Hi Sven,
"...sometimes I have the feeling that the diode plug is just a machine to generate free energy in the next stage ie the pushpull trafo by saturating it with sharp short pulses."
I thought the big ferroresonant transformer "transverter" had that purpose. Else get rid of it to make the device simpler. The ferroresonant transformer has a magnetic path that is supposed to go into saturation. This path acts as a very fast magnetic gate. The output push-pull transformer is then for impedance matching and putting the half phases together to be used in a single load.
"I have the feeling that the circuit is not yet complete as the diodes separate the inductor and capacitor, which are not necessary, but why are they drawn?"
Separation is only in the reverse conducting or blocking direction. The other half part has the diode in the opposite direction. Each half part is semi-resonant with one part for each direction of current flow.
"So the adjustment between inductor and capacitor must fit?!? Just how should that look on the scope that fits?"
They will oscillate at the resonant frequency and the curve is a sinusoidal wave or actually only a half oscillation for each half part of the diode plug. When the capacitance (or inductance) is decreased the oscillation becomes faster which is seen as a shorter distance in the repetition rate. I.e. the frequency is increased.
"But with smaller capacitors and the same trigger voltage less energy is taken, the frequency of 50 Hz remains, but I fear most of the energy is in the inductor Variac. When I tune to the best possible values, the lamps shine as brightly as with large capacitors."
But are the half sines (parabolic curves) not higher and of shorter duration even though the repetition rate is the same? The semi-resonant frequency should go up as well as the peak voltage if the discharge frequency stays at 50Hz and the load receives the same power.
"I do not have any ideas anymore. I wanted to connect the identical 230V / 24V transformers to each diode plug and shoot the pulses into the battery."
Transformers are for AC. If used at pulsed DC they may not demagnetize in the opposite direction and thus start to saturate. At least if used for pulsed DC (half phase) they must have an air gap to be able to proper demagnetize. Else the remanence of the magnetic core makes it a permanent magnet. Ed Leedskalnin calls the effect a perpetual motion hanger when the circuit is permanent magnetized.
Regards
Ole
---In [email protected], <s.friedrich@...> wrote :
Hello Ole,
SCR can be up to 200mA. If the voltage across the capacitor is 400V at the point of triggering then a gate resistor at 1.1kOhm makes the gate current around 364mA. With 400V across and 364mA through the gate resistor some 146W of power is dissipated in the resistor during the brief moment of triggering.
The trigger is adjustable via pots, I would like to cover a large voltage range so even switch below 200 volts or earlier trigger. But I also do not think that they are responsible for losses.
Otherwise, the circuit makes no sense. "That's right, but you should just give it a certain part of it.
The part given back is the reactive part of the power. The part being consumed is the active (or real) power. As long as only the free energy into active power becomes the active power then is free energy.
Well written, but how do you get it that way, sometimes I have the feeling that the diode plug is just a machine to generate free energy in the next stage ie the pushpull trafo by saturating it with sharp short pulses. Somehow I do not know how to continue. I have the feeling that the circuit is not yet complete as the diodes separate the inductor and capacitor, which are not necessary, but why are they drawn?
"I meant only with the shift of the power factor, the additional active power is taken from the reactive circuit, if one finds the point."
Ohmic resistances are the ones with the phase factor of one. Any reactive impedance or capacitance has the phase factor different from one. At one and the other is negative.
So the adjustment between inductor and capacitor must fit?!? Just how should that look on the scope that fits?
I always try to get a clean picture and adjust the distortions and tips are filtered out.
"I mean, the transformer pulls into the primary windings, and the load does not pull back into the primary windings." load on the secondary of the push-pull transformer stops the switching in the diode plug. "
Yes there could be a problem here. At no load the power is pure reactive being reflected back to source.
"I want to work with smaller discharge capacitors so the voltage on the diode plug and the transverter are pretty much the same."
It's not the capacitors that determine the voltage when to be discharged. It's the triggering circuit parameters of the voltage divider that fires the SCR. Smaller capacitance only makes the capacitors charge / discharge faster and use less cycles for given voltage build up.
But with smaller capacitors and the same trigger voltage less energy is taken, the frequency of 50 Hz remains, but I fear most of the energy is in the inductor Variac. When I tune to the best possible values, the lamps shine as brightly as with large capacitors.
I do not have any ideas anymore. I wanted to connect the identical 230V / 24V transformers to each diode plug and shoot the pulses into the battery. Unfortunately, the impedance of the transformers is so high that the diode plug does not start even with the battery as a load. I could try it again with freewheeling diodes and incandescent resistance.
Regards
Sven
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