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Lenzless resonant transformer

Started by Jack Noskills, January 17, 2014, 04:58:15 AM

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Jack Noskills

Quote from: itsu on March 11, 2014, 03:56:03 PM
Thanks for your understanding.

I made a last measurement with the same setup as in the last video and setup the power into the bulb to exactly measure 1W (500mW * 2).
Then i measured the input voltage across the L3 with the blue probe and measured the current there, see the picture for the outcome.
We have 581mW * 2 = 1.162W into the core/coil.

The PA is an XT800.2  automotiv MOSFET amplifier class AB at max. 50KHz.

Regards Itsu


Seems to be close but no cigar.


Any idea how much power you can push through L3 ?
What is the watt ratting of light bulb in the output side ? It seems to be so bright that adding another should not be a problem. In my setup I put the 8 watt bulb first and then connected the 10 watt bulb on live circuit. Result was that 8 watt bulb dimmed a little when 10 watt bulb was lit. Nothing visible in input side 5 watt bulb.


As for the drilling, there needs to be something inside the toroid. When the blade comes through it will crack the epoxy unless there is wood or plastic taking the pressure from the drill. It should be interesting project, the coils on the holes behave funny. Magnetic loops are formed but not where one might expect. They always oppose the source even if there is 'free' path around the core. Flux seem to want balance in the core. This is why Gunderson uses magnets I quess': they form 'magnetic DC' bias in the core and then magnetic loops are formed elsewhere.




verpies

Quote from: Jack Noskills on March 12, 2014, 02:09:19 AM
Result was that 8 watt bulb dimmed a little when 10 watt bulb was lit. Nothing visible in input side 5 watt bulb.
Comparing the brightness of the output bulb to the brightness of the input bulb is tantamount to comparing Watts to Amps.
Meaningless mostly...

Quote from: Jack Noskills on March 12, 2014, 02:09:19 AM
Flux seem to want balance in the core. This is why Gunderson uses magnets I quess': they form 'magnetic DC' bias in the core and then magnetic loops are formed elsewhere.
As I understand the patent of the 2nd device, those magnets are merely sources of the vertical MMF. 
The author writes that the flux of these magnets (green) can take two paths through the core (see Fig.3 in the patent) and the primary coil (coils) are a source of the controlling horizontal MMF and flux (blue) that switches the green flux between two alternate paths of equal reluctance.

The blue horizontal flux closes circumferentially through the core, while the green flux passes vertically/diagonally through the core but closes through air, which is bad, unless a low reluctance path is provided for it by an external permeable spool or pot core.

itsu

Quote from: Jack Noskills on March 12, 2014, 02:09:19 AM
Any idea how much power you can push through L3 ?
What is the watt ratting of light bulb in the output side ?


No, no idea what the max. value will be.
The watt ratings of these bulbs are 0.6W at 6V, so pushing them a lot with 1W.

Quote
As for the drilling, there needs to be something inside the toroid. When the blade comes through it will crack the epoxy unless there is wood or plastic taking the pressure from the drill.

Good info,  thanks.

Regards Itsu

Jack Noskills

Quote from: verpies on March 12, 2014, 06:47:13 AM
Comparing the brightness of the output bulb to the brightness of the input bulb is tantamount to comparing Watts to Amps.
Meaningless mostly...


Here current and voltage are in the same phase in L3 because of resonance condition in the output. Does it make comparison any more meaningful in this case ?


Putting a diode bridge without smoothing cap in the input side should also make comparison easier ?

verpies

Quote from: Jack Noskills on March 12, 2014, 09:32:02 AM
Here current and voltage are in the same phase in L3 because of resonance condition in the output. Does it make comparison any more meaningful in this case ?
No, because the I-V phase relationship is not the problem in this case.
The problem with the power indicator on the input side is the uneven power sharing between the indicator and the DUT (a 2-device problem according to MPTT ) ...even if the DUT is a purely resistive load ( I & V in-phase).

On the output side this problem does not exist because the power indicator is the output DUT (a 1 resistive device problem).