Thank you! Itsu
It is appreciate .

Nice to see you have an Output Itsu.

If the Output Current is defined by: V / R = I

What would be needed to Increase the Current on the Output? 

According to my Calculations your Current is very small: 0.04779 through the 470 Ohm Resistor, and only 0.10209 through the 220 Ohm Resistor. Of course this is with the voltage of 22.46 Volts RMS.

Current is the Magnetic Field, so for a very small current like this, there is a very small Magnetic Field! B = μ₀NI/L

Where
B is the Magnetic Field B in Gauss.
μ₀ is the Permeability - μo = 1.26 × 10⁻⁷ T/m/At This is the Magnetic Constant. This could be replaced by your Relative Core Permeability μ_r: 24.3 in your case.
N is number of Turns in the solenoid,
I is Current in the coil,
L is the Length of the coil. Normally a Lowercase L (l)


So, your:

Input Magnetic Field is: 452.747 Gauss

Output Magnetic Field is: 73.6571 Gauss with the 470 Ohm Resistor

and

Output Magnetic Field is: 157.348 Gauss with the 220 Ohm Resistor


It is worth verifying there is a Magnetic Field Opposition also, just in case there is not. You have to get your Magnetic Fields way up, much higher than they are. I would wind thicker Wire Gague, 18-20 AWG or so. Drop the Impedance right down, getting the Fields way up, also drop your Load resistance some more.

Ideally, you want about the same or close to the same Ampere Turns. Now I have given away way too much.

If the Magnetic Fields are not in opposition, there can be no Gain!


   Chris Sykes
       hyiq.org

Changing the 220 Ohm base resistor into a 1KOhm potmeter (range from 150 Ohm to 1KOhm) does not improve the efficiency.
The frequency changes from 15KHz @ 150 Ohm to 63KHz @ 1KOhm and the input current (input voltage steady at 3.5V) changes from 800mA @ 150 Ohm to 130mA @ 1KOhm.

Measuring the secondaries inductance shows that the uncovered secondary still measures 990uH, and the covered (with the primary) secondary measures 948uH.
So we have an unbalans in the bucking coils inductances due to the primary.

I am thinking of removing the bifilar primary and install it on a free piece of the ferrite rod so the both secondaries will have balanced inductances.

Itsu

Sorry, Repost: Last post on the page...









Nice to see you have an Output Itsu.

If the Output Current is defined by: V / R = I

What would be needed to Increase the Current on the Output? 

According to my Calculations your Current is very small: 0.04779 through the 470 Ohm Resistor, and only 0.10209 through the 220 Ohm Resistor. Of course this is with the voltage of 22.46 Volts RMS.

Current is the Magnetic Field, so for a very small current like this, there is a very small Magnetic Field! B = μ₀NI/L

Where
B is the Magnetic Field B in Gauss.
μ₀ is the Permeability - μo = 1.26 × 10⁻⁷ T/m/At This is the Magnetic Constant. This could be replaced by your Relative Core Permeability μ_r: 24.3 in your case.
N is number of Turns in the solenoid,
I is Current in the coil,
L is the Length of the coil. Normally a Lowercase L (l)


So, your:

Input Magnetic Field is: 452.747 Gauss

Output Magnetic Field is: 73.6571 Gauss with the 470 Ohm Resistor

and

Output Magnetic Field is: 157.348 Gauss with the 220 Ohm Resistor


It is worth verifying there is a Magnetic Field Opposition also, just in case there is not. You have to get your Magnetic Fields way up, much higher than they are. I would wind thicker Wire Gague, 18-20 AWG or so. Drop the Impedance right down, getting the Fields way up, also drop your Load resistance some more.

Ideally, you want about the same or close to the same Ampere Turns. Now I have given away way too much.

If the Magnetic Fields are not in opposition, there can be no Gain!


   Chris Sykes
       hyiq.org

Why: Because it is the Magnetic Field (B), and the Time Rate of Change (t) of it, that Invokes Electromagnetic Induction, with Turns (N) is the Value of the EMF!!!

   Chris Sykes
       hyiq.org


P.S: A good resource for Calculating the Magnetic Field (B) is: http://www.calctool.org/CALC/phys/electromagnetism/solenoid