Selfrunning cold electricity circuit from Dr.Stiffler

[Freenrg4me]

*Removed* by RStiffler

[fritz]

If we stop looking at the trees and look at the forest for a moment, there is not enough current to saturate that huge chunk of core mass in the slightest. Ron states it won't work without the core and there is only one wire off the secondary to boot. Where is the 25ma of induction?

As I pointed out in my mail above - the 25mA are from the generator.
"Normally" this energy cannot be consumed with 10k input resistance
and 2Vpp. The circuit generates an "inner" negative resonance voltage
which makes the signal generator work against these voltage.
The inner voltage maybe somewhere at 50 V (peak).
This means that the power sucked from the generator is
(50+2)^2 / 10kOhm - which is in the some hundred milliwatt range.
This (getting more power from the generator) is achieved by using C1.
L2,L3 transforms the high voltage/low current/100MHz(fundamental) peak pulse into
a low voltage/ high current pulse.
Because of the high frequency/lots of harmonics (probably up to some 100MHz),
its no problem to couple the power using some parasitic capacitor of the breadboard.

The level and art of tricking around with rf circuits is of a high degree.
Discussing the most funny 2048 best rf tricks will keep us busy for years ...
(this was a warning)

Measuring the heat emitted by the 50 Ohm output resistor of the generator
would be the immediate proof (whats going on here)

rgds,

wolfgang

[edork]

fritz, Hi
How you tell what go into circuit by measure heat on resistor across that is across source and not inline?

I can not do, how?

[fritz]

OK,

2Vpp square equals 1Vp equals 1Vrms is the generator
voltage without load.
The generator source resistance is 50 Ohms.
The maximum power you can get out is if terminated with
another 50 Ohms.
Means we have a generator with 1Vrms on 100 Ohms gives
10mA output current if terminated)
20mA output current if shorted

You see - even a shorted generator provides less current
as needed for a bright led.

The output resistor of the generator dissipates 20mA^2 * 50 Ohms
= 20mW. (if shorted)

The LED needs with 25mA@2.73V 68.2kelvin 5mW
Means we would need 68.25mA current from a 1Volt
supply to have equal power.
This 68.25mA would induce a power dissipation of 68mA^2 *50 Ohms
= 232mW on the output resistor.
To have 68.25mA@50 Ohms, you need 3.4125 V on this resistor -
means that the circuit must generate a voltage of -2.41V to
achieve this.

A 1/2 watts resistor can have up to 50 kelvin temperature rise
on nominal load - 500mW.

This means - if happens what I suppose - the generator output resistor
gets measurable hot. - hotter than if generator is shorted.

The 10kOhm resistor should definitly get hot, too.

rgds,

Wolfgang

[edork]

fritx, Hi again.

What circuit you look at? 10K in parallel with input cap?

He show 1ohm in series with input and only 0,0004 mv acreoss it?