Confirming the Delayed Lenz Effect

[CRANKYpants]

HAPPY THANKSGIVING TO ALL MY FELLOW CANADIAN TURKEY GOBBLERS OUT THERE! 
CHEERS
T

[gotoluc]

Hi everyone,

I have an update on the Transformer Delayed Lenz Effect.

See Video for explanation: http://www.youtube.com/watch?v=ZZtIhOV00uU

The below Scope Shots are:
Ch 1 Yellow Probe is 10 Ohm Shunt Resistor, Ch 2 Lt. Blue is Coil Voltage, Ch 3 Purple is Secondary Coil 10 Ohm Load and Ch 4 is Resonator Coil 10 Ohm Load

First is Transformer @ 60Hz no Load
Second is Transformer @ 60Hz with Secondary on 10 Ohm Load
Third is Transformer @ 60Hz with Secondary on 10 Ohm Load and 29uf on Resonating Coil

Please post your comments and Power Calculations. Keep in mind the Shunt Resistor is 10 Ohms and not 1 Ohm.

Luc

[gyulasun]

Hi Luc,

When the secondary coils are unloaded the input power is approximately: P_in=V*I*cos82°   (I estimated from the scope shots the input voltage leads current by about 82°. How I got this 82°: the time difference between input voltage and current is about 3.8msec, this gives roughly 82°.)

So P_in=5.96*0.0508*cos82°=0.0421W

Loaded case, no capacitor, I noticed no change in the 82° phase shift :

input power  P_in=5.57*0.0548*cos82°=0.04248W

output power  P_out=0.577²/10=0.03329W

Loaded case with capacitor, I noticed the phase shift decreased to about 34.5° from 82°, this means it is decisive to watch on the scope only the voltage drop decrease across the shunt resistor, you have to consider any change in the phase shift too.

input power P_in=5.2*0.0457*cos34.5°=0.1958W

output power P_out=0.674²/10=0.04542W

circulating power in the LC (140mH 39uF) tank circuit, assuming resonance  P_LC=0.484²/10=0.02342W

Efficiency in the loaded case, no cap:
P_out/P_in=(0.03329W/0.04248W)*100=78.3%

Efficiency in the loaded case with cap:
(P_out + P_LC)/P_in=(0.04542W + 0.02342W)/0.1958W=0.3515*100=35.15%

So it seems this setup now behaves as a conventional transformer when there is no 39uF tuning cap connected. When the 39uF is connected to the till then 140mH idle coil, efficiency suffers, probably due to the change in the phase shift in the input voltage-current.

(I recall the 140mH coils have got a DC resistance between 120 to 180 Ohms?)

Gyula

[gotoluc]

Hi Gyula,

thank you for doing the power calculations.

As you can see I have no idea how to do AC Power Calculations yet but there is hope as I now understand that a higher voltage across the Shunt Resistor does not necessarily matter if the Phase (cos angle) stays the same. Do I have that correct?

I decided to look at Phase much closer on the Scope and found it did shift @60Hz.

In the First scope Shot below I expended to Scopes Voltage divisions so we could clearly see where they fall at the Zero point.
For loads I decided to separate the 2 Secondaries and used a 10 Ohm Load on each.
I found that it's only @248Hz that there is Zero Phase Shift when under Load. See Second Shot below. Frequencies above 248Hz the Phase slowly goes up and below 248Hz Phase slowly drops.

The First Shot below is the no load centered Reference. We have exactly 4 squares between each side of the rising and dropping voltage phase. Each square should be 22.5 degrees, so then one of the 5 divisions in each square should be 4.5 degrees. The Current is 3 divisions behind the Voltage (3 x 4.5 = 13.5) so 90 - 13.5 = 76.5 degrees.

So would you agree this is an accurate way to get Phase degree?

Third Shot is Complete view.

Can you re-calculate the power in vs power out with this more accurate data.

I still don't understand how you came up with the numbers above so I better let you do it. Hopefully in time I will be able to learn the equations

Thanks for your time Guyla... I'll get there one step at the time thanks to people like you who are willing to help.

Luc

[CRANKYpants]

The Current is 3 divisions behind the Voltage (3 x 4.5 = 13.5) so 90 - 13.5 = 76.5 degrees.
Luc

ACTUALLY LUC THE CURRENT LAGS THE VOLTAGE BY 17 DIVISIONS AND 17 x 4.5 = 76.5 DEGREES
I know it's anal but...

SO YOUR POWER IN = Vprimary x Iprimary x COS 76.5 (COS 76.5 = 0.233)
SO YOUR POWER IN = Vprimary x Iprimary x 0.233

Pout = Vload^2 / Rload

CHEERS
T