Confirming the Delayed Lenz Effect

[gotoluc]

Hi everyone,

I built a transformer with a Bifilar Coil as Secondary.

Thane has suggested he uses Bifilar Coils in his Generator so I was thinking good chances are that same effect could work in a Transformer.

I have tested it and it works the same
The big advantage is, now we can save Tons of Wire and have Low Coil Resistance which translates to more Current to the Load.
So put away your MOT's and make yourself a Transformer with a Bifilar Coil

See video Demo: http://www.youtube.com/watch?v=rUmVSf878aY

@Overunityguide, can you or someone do the math on the efficiency. Thank you

Luc

Data of Coils and Electrical:

Primary DC Resistance 79 Ohms
Primary Inductance 665mH
Secondary DC Resistance 8 Ohms
Secondary Inductance 418mH

Idle H-Bridge (not connected to primary)
Input Voltage 62.7vdc
H-Bridge Current wasted 30ua

H-Bridge Connected to Primary
Input Voltage 62.4vdc
Current to H-Bridge input 8.3ma (no load or with load)

Load on Secondary is 150 Ohms @ 4.10 Volts RMS

Scope Sots below are Orange is Current Probe across 1 Ohm Shunt, Light Blue is Voltage Probe across the Coil and Purple is Probe across the 150 Ohm Load on Secondary
First Scope Shot is No Load on Secondary
Second Scope Shot is 150 Ohm Load on Secondary

[nul-points]

Hi everyone,
[...]
or someone do the math on the efficiency. Thank you

Luc
[...]

hi Luc

assuming that the scope is presenting True (calculated) RMS values (which appears to be the case, looking at the 'square wave' values):

i/p power to primary would appear to be approx 2W

o/p power in 150 ohm load would appear to be approx 100mW

efficiency approx 5% - is this the calc you wanted?

HTH

[gotoluc]

hi Luc

assuming that the scope is presenting True (calculated) RMS values (which appears to be the case, looking at the 'square wave' values):

i/p power to primary would appear to be approx 2W

o/p power in 150 ohm load would appear to be approx 100mW

efficiency approx 5% - is this the calc you wanted?

HTH

Hi NP

No I don't think so

For Total Input I calculate 62.4vdc @ 8.3ma = 0.51792 Watts
This is the Total Input Power to the Transformer. However the 79 Ohm Primary is eating most of that in its Resistance. So to be fare we need to factor that out to find the true power transfer available to the Secondary. However, this is where I need help.

The output is 4.10 Volts RMS on a 150 Ohm load = 0.1121 Watts

Thanks anyways

Luc

[nul-points]

Hi NP

No I don't think so

For Total Input I calculate 62.4vdc @ 8.3ma = 0.51792 Watts
This is the Total Input Power to the Transformer. However the 79 Ohm Primary is eating most of that in its Resistance. So to be fare we need to factor that out to find the true power transfer available to the Secondary. However, this is where I need help.

The output is 4.10 Volts RMS on a 150 Ohm load = 0.1121 Watts

Thanks anyways

Luc

hi again Luc

you want the efficiency of the transformer, is that correct? 

if so, you don't need to include the power used by your H-bridge

your scope shows 32mV RMS across a 1 ohm shunt - this equates to a current of 32mA RMS

you don't state shunt location explicitly but it appears to be on the primary side of the transformer
(because the waveform is not in phase with the voltage across the 150 ohm output load resistor)

so my calc for i/p power = 61.5 * 0.032 = 1.97W (approx 2W)

my calc for your o/p power is 4.1 * 0.027 =  0.1107W (approx 100mW)

hence my calc of your transformer efficiency**  is approx (0.1 / 2) * 100 = 5%

HTH


[EDIT:  ** for this frequency and load]

[Overunityguide]

so my calc for i/p power = 61.5 * 0.032 = 1.97W (approx 2W)

@nul-points, sorry but I have to give you 'zero points' for your input calculation. (please don't take it personal)

I say this because you are forgetting one very important parameter which also should be added in your input power calculation. This one important parameter is the phase angle (power factor) between current and voltage at the primary side of the transformer... So when comparing the loaded state with the unloaded state of gotoluc's bifilar secondary's transformer. You can see that in the loaded state this phase angle is becoming higher.
(So read: more to 90 degrees). Further you can see that the amperage to the primary went up when connecting the load, but that in relation with this the phase angle (power factor) also becomes bigger.
Those two related parameters (primary amperage and phase angle) are why you don't see any change in the total input power in gotoluc's very need input power measurement setup.

my calc for your o/p power is 4.1 * 0.027 =  0.1107W (approx 100mW)

@nul-points, this is where I totally agree with you, so it looks like that there is for now 0.1107W coming out of the transformer for no noticeable input power change at all!

@Thane & Gotoluc... Great work and great cooperation guys. And if I may give one suggestion, maybe it is better to replace the input coil in the setup with a low Ohms / high current one, so that more usable power can be transferred to the secondary. And maybe try to run it at some higher frequency settings (1000Hz/1500Hz). But I am sure that you already thought about this...

So than again: Great work.

With Kind Regards, Overunityguide