PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
OverUnity Research > Electrical / Electronic Devices > Solid-State Generators > Partnered Output Coils
2024-04-19, 05:23:02
News: If you have a suggestion or need for a new board title, please PM the Admins.
Please remember to keep topics and posts of the FE or casual nature. :)

Pages: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [15] 16 17 18 19 20 21 22 23 24 25 26
Author Topic: Partnered Output Coils  (Read 362862 times)
Topic Options img
img  EMJunkie
Re: Partnered Output Coils « Reply #350 on: 2015-02-04, 20:53:10 »
Group: Guest
Quote from: Itsu on 2015-02-04, 20:31:24
Output measurements / calculations done, see screenshot.

(input was looking the same, 6.7V on the Fluke, 235mA on the amp meter)

Had to isolate the FG from ground!!

Yellow is the voltage across the 10 Ohm resistor
green is the current sensor just above it  (current controller set to 500mA/div.)
Red is the math function Ch1 * Ch4 (yellow * green) but as the current value has to be taken x50, also this power value has to be taken x50, so 0.675W

Resistor gets to about 35 degrees C.


This is all without any tuning, 0.5mm gap between the cores, partner coils 1cm apart.

Video is taken and will be uploaded later.

Regards Itsu


Nice Work Itsu again!

I wonder, All the spikes are missing from your Scope Shots of the output. Do you know why these are missing?

Power in Watts = Voltage Squared / Resistance - Current = Voltage / Resistance

The 400+ Volt Spikes I am getting are directly across the 10 Ohm resistor. I don't see any of it in your scope shots? Can you explain why please?

You can see from my Schematic that the Resistor will see all of the E=½LI2 so All of the Primary, less the voltage drop across the Diode will be present across the Resistor. I see none of this in your Scope Shots?
   
img  Itsu
Re: Partnered Output Coils « Reply #351 on: 2015-02-04, 20:56:08 »

Group: Elite Experimentalist
Hero Member
*****

Posts: 4092

Finally, if i calculate the output from the measurement points in the diagram (voltage probe between crosspoint L2 and load resistor and PS return lead,
current probe just above load resistor), then the output calculates to be much higher, see screenshot.

Yellow is voltage between top of load resistor and PS return
green is current sensor above the load resistor
red is the math function Ch1 * Ch4 (yellow * green) but as the current value has to be taken x50, also this power value has to be taken x50, so 3.3W

But again,  i don't think this is the correct way to measure the output as it is NOT taken across the output resistor,  comments please.

Regards Itsu
   
img  Itsu
Re: Partnered Output Coils « Reply #352 on: 2015-02-04, 21:03:45 »

Group: Elite Experimentalist
Hero Member
*****

Posts: 4092
Quote from: EMJunkie on 2015-02-04, 20:53:10
Nice Work Itsu again!

I wonder, All the spikes are missing from your Scope Shots of the output. Do you know why these are missing?

Power in Watts = Voltage Squared / Resistance - Current = Voltage / Resistance

The 400+ Volt Spikes I am getting are directly across the 10 Ohm resistor. I don't see any of it in your scope shots? Can you explain why please?

You can see from my Schematic that the Resistor will see all of the E=½LI2 so All of the Primary, less the voltage drop across the Diode will be present across the Resistor. I see none of this in your Scope Shots?

Hi Chris,    no i don't know why these 700V spikes (in my case) are missing when directly measuring across the load resistor.
I guess because i am measuring floating across this resistor, without reference to ground, but thats just a guess.
I hope it will be explained

Regards Itsu
   
img  EMJunkie
Re: Partnered Output Coils « Reply #353 on: 2015-02-04, 21:09:11 »
Group: Guest
Quote from: Itsu on 2015-02-04, 20:56:08
Finally, if i calculate the output from the measurement points in the diagram (voltage probe between crosspoint L2 and load resistor and PS return lead,
current probe just above load resistor), then the output calculates to be much higher, see screenshot.

Yellow is voltage between top of load resistor and PS return
green is current sensor above the load resistor
red is the math function Ch1 * Ch4 (yellow * green) but as the current value has to be taken x50, also this power value has to be taken x50, so 3.3W

But again,  i don't think this is the correct way to measure the output as it is NOT taken across the output resistor,  comments please.

Regards Itsu


Hey Itsu,

Thanks, again excellent work!

Well, I will be interested here to see what the consensus is. I can see some Wave information is missing from the First set of Output Scope Shots! This is a given as there is no where for the Inductor to discharge to other than the Resistor and Partnered Output Coils!

Yes, Its a Delilah here!

We see on the Output 3.3 Watts above the 0 Voltage Terminal, on the Positive Terminal, 1.4 Watts above the 0 Voltage Terminal, but 3.3 still remains a mystery! (0.6 + 1.4 != 3.3!)

What's going on here?
« Last Edit: 2015-02-04, 22:49:57 by EMJunkie »
   
img  Jeg
Re: Partnered Output Coils « Reply #354 on: 2015-02-04, 21:13:54 »
Group: Guest
Excellent work Itsu!
Make me a favor please and change polarity of diode. Measure again the output power the same way as before. I see some difference between the two modes. Can you test it?

Oops sorry! I thought you had implement Meyer's circuit with feedback coil. Never mind. Nice job  ;)
   
img  partzman
Re: Partnered Output Coils « Reply #355 on: 2015-02-04, 21:27:01 »
Group: Experimentalist
Hero Member
*****

Posts: 1671
Hi All,

I am new to OUR and would like to thank Smudge for the invitation.  I hope I can be a positive contributor to this forum.

Itsu:

Nice clear waveforms you have provided on your latest build! These allow for a reasonably accurate output power measurement to be calculated.  I will use your CH4 output current measurement broken down into segments of energy calculations and then to final output power.

Looking at CH4 on your expanded "emj8.png" and based on 500ma/div as you stated, the current level at the rising edge of the gate drive voltage is approximately 275ma.

The output current level at the falling edge of the gate drive signal is approximately 450ma.  So, the total mean value of the current is ((.45-.275)/2) + .275 = 363ma. The period for this portion of the load current is 288us.  Therefore the energy developed in the 10 ohm load resistor for this period is (.363^2)*10*288e-6 = 380uJ.

We still have a smaller portion of output current that ramps down to zero from the rising edge of the gate drive that we need to include as well.  The starting peak of this ramp is 275ma and the duration is ~ 60us. The mean for this portion is .275/2 = 138ma.  The load energy for this portion is (.138^2)*10*60e-6 = 12uJ.

Therefore, the total output load energy is 380uJ + 12uJ = 392uJ. With the operating frequency of 1732Hz, the power output = (392e-6)*1732 = .68 watts.

The input power is 6.78*.24 = 1.62 watts.  Therefore the COP = .68/1.62 = .42.

I realize Itsu that you could have made these measurements so forgive me for preaching to the choir!  (NOTE: It took so long to write this that I see you have posted your own measurements!)

One more comment is that your replication and waveforms appear to be very close to EMJ's with one exception and that is the calibration of Chris's current probe.  Chris shows his calibration to be 1mv/10ma but if it is 1mv/1ma then the above calculations would be close to his waveforms as well.

I hope the above calcs are correct so feel free to point out errors, etc.

partzman     
   
img  Itsu
Re: Partnered Output Coils « Reply #356 on: 2015-02-04, 21:46:13 »

Group: Elite Experimentalist
Hero Member
*****

Posts: 4092

partzman, welcome.

well you sure know your math, thats no doubt a  positive contribution already  O0

My scope can do these calculations,  but many other scopes don't, so you get plenty of opportunities to practice your skills  ;)

Regards Itsu
   
img  EMJunkie
Re: Partnered Output Coils « Reply #357 on: 2015-02-04, 21:50:45 »
Group: Guest
Quote from: partzman on 2015-02-04, 21:27:01
Hi All,

I am new to OUR and would like to thank Smudge for the invitation.  I hope I can be a positive contributor to this forum.

Itsu:

Nice clear waveforms you have provided on your latest build! These allow for a reasonably accurate output power measurement to be calculated.  I will use your CH4 output current measurement broken down into segments of energy calculations and then to final output power.

Looking at CH4 on your expanded "emj8.png" and based on 500ma/div as you stated, the current level at the rising edge of the gate drive voltage is approximately 275ma.

The output current level at the falling edge of the gate drive signal is approximately 450ma.  So, the total mean value of the current is ((.45-.275)/2) + .275 = 363ma. The period for this portion of the load current is 288us.  Therefore the energy developed in the 10 ohm load resistor for this period is (.363^2)*10*288e-6 = 380uJ.

We still have a smaller portion of output current that ramps down to zero from the rising edge of the gate drive that we need to include as well.  The starting peak of this ramp is 275ma and the duration is ~ 60us. The mean for this portion is .275/2 = 138ma.  The load energy for this portion is (.138^2)*10*60e-6 = 12uJ.

Therefore, the total output load energy is 380uJ + 12uJ = 392uJ. With the operating frequency of 1732Hz, the power output = (392e-6)*1732 = .68 watts.

The input power is 6.78*.24 = 1.62 watts.  Therefore the COP = .68/1.62 = .42.

I realize Itsu that you could have made these measurements so forgive me for preaching to the choir!  (NOTE: It took so long to write this that I see you have posted your own measurements!)

One more comment is that your replication and waveforms appear to be very close to EMJ's with one exception and that is the calibration of Chris's current probe.  Chris shows his calibration to be 1mv/10ma but if it is 1mv/1ma then the above calculations would be close to his waveforms as well.

I hope the above calcs are correct so feel free to point out errors, etc.

partzman    

@Partzman,

Welcome!

My Current Probe was showing the 10x Cal on the scope. So you're right is saying that:

Quote

Chris shows his calibration to be 1mv/10ma but if it is 1mv/1ma then the above calculations would be close to his waveforms as well.


I calibrated to display output Amps inc Cal. I should have worded this better! Meter was set to: 1mv/10ma but calibrated to 1mv/1ma Apologies for the confusion.

I think we are missing something real important here. We are missing 1.3 Watts, please see: http://www.overunityresearch.com/index.php?topic=2760.msg45477#msg45477
   
img  ion
Re: Partnered Output Coils « Reply #358 on: 2015-02-04, 21:57:16 »
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
Itsu said:

Quote
Hi Chris,    no i don't know why these 700V spikes (in my case) are missing when directly measuring across the load resistor.

The reason the spikes are so high (on the primary) is that the inductor energy is not yet being coupled to the output load, due either to leakage inductance or diode turn on time or some of both. If it were being coupled to the output load resistor it would not rise so high. The inductor is uncoupled and freewheeling at this point.

Input power= power used to charge the inductor and that dissipated by the FET, and ohmic winding loss. (here we are not counting FET drive power)

Output power= power recovered from the inductor and coupled to the load resistor during the flyback phase minus diode loss and sum of all winding ohmic loss. (note: this is the usable power into the load, the other losses could be added in as heat production instead of subtracted out if it is desired to find true total power output)


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
img  EMJunkie
Re: Partnered Output Coils « Reply #359 on: 2015-02-04, 22:08:32 »
Group: Guest
Quote from: ION on 2015-02-04, 21:57:16
Itsu said:

The reason the spikes are so high (on the primary) is that the inductor energy is not yet being coupled to the output load, due either to leakage inductance or diode turn on time or some of both. If it were being coupled to the output load resistor it would not rise so high. The inductor is uncoupled and freewheeling at this point.

Input power= power used to charge the inductor and that dissipated by the FET, and ohmic winding loss. (here we are not counting FET drive power)

Output power= power recovered from the inductor and coupled to the load resistor during the flyback phase minus diode loss and sum of all winding ohmic loss.

Hey ION,

Yes, understand what youre saying. But the scope sees this voltage spike in the secondary Circuit! Just moving the Scope Ground cable shouldn't remove these Spikes. The Scope sees them there initially.

Does this make sense?

We are still missing something here, we are missing 1.3 Watts from the Input Power Source, to the Output, this cant currently be accounted for!

Quote from: EMJunkie on 2015-02-04, 21:09:11

We see on the Output 3.3 Watts above the 0 Voltage Terminal, on the Positive Terminal, 1.4 Watts above the 0 Voltage Terminal, but 3.3 still remains a mystery! (0.6 + 1.4 != 3.3!)


I see the removal of important Wave data on the output, and a loss of 1.3 Watts, there is issues that are not currently explained.
« Last Edit: 2015-02-04, 22:50:39 by EMJunkie »
   
img  Itsu
Re: Partnered Output Coils « Reply #360 on: 2015-02-04, 23:07:54 »

Group: Elite Experimentalist
Hero Member
*****

Posts: 4092

I have to agree with Chris, only move was the ground lead.


Anyway, here the Video of the above taken input / output measurements:

https://www.youtube.com/watch?v=4EbNr3czXdQ&feature=youtu.be


Regards Itsu
   
img  poynt99
Re: Partnered Output Coils « Reply #361 on: 2015-02-04, 23:15:49 »

Group: Administrator
Hero Member
*****

Posts: 3205
It's not as complicated as it may seem...
Itsu,

You yourself said that the output power measurement should be made with the current through the load, and the voltage across it, agreed?

Measuring the voltage between the denoted probe point and ground is not directly across the load, therefore it must not be correct.

Why would one expect the waveforms to be equal in two very different probe configurations? If the spikes are gone when measuring across the resistor directly, then they must not have any bearing on the load.
   
img  poynt99
Re: Partnered Output Coils « Reply #362 on: 2015-02-04, 23:30:27 »

Group: Administrator
Hero Member
*****

Posts: 3205
It's not as complicated as it may seem...
Itsu,

 O0 O0 for you!

Good job.
   
img  partzman
Re: Partnered Output Coils « Reply #363 on: 2015-02-04, 23:43:11 »
Group: Experimentalist
Hero Member
*****

Posts: 1671
All,

The voltage measurement for the 10 ohm load resistor must be made differentially to be accurate.  I have attached a sim that approximately represents the circuitry and have included some math calcs in the plot window.  Notice the difference in "eout" and "eout2".  Eout is taken differentially and is 2.022v average. Eout2 is taken at node V(n003) and is 8.898 volts average.  Eout2 is not an accurate representation of the load voltage due to the DCR and inductance of L2 in this example.

The sim traces do not match the circuits of Cris and Itsu exactly due to differences in circuit parasitics, K factors, coil inductances, etc,  plus the core in this example is purely linear.
Conceptually however they are close.

I hope the attachment isn't too large!!!

partzman
   
img  EMJunkie
Re: Partnered Output Coils « Reply #364 on: 2015-02-04, 23:52:33 »
Group: Guest
Quote from: poynt99 on 2015-02-04, 23:15:49
Itsu,

You yourself said that the output power measurement should be made with the current through the load, and the voltage across it, agreed?

Measuring the voltage between the denoted probe point and ground is not directly across the load, therefore it must not be correct.

Why would one expect the waveforms to be equal in two very different probe configurations? If the spikes are gone when measuring across the resistor directly, then they must not have any bearing on the load.

#Poynt99,

Maybe you can answer the Question at hand instead of passing it off. Well Done good job doesn't answer this question.

I ask again, slightly clearer:

Quote

How is it, that from the 0 Voltage Terminal, to the Output, we see 3.3 Watts, but from the 0 Voltage Terminal to the + Voltage Terminal we see 1.4 Watts - Where is the missing power gone? 0.68 has been accounted for!
   
img  Groundloop
Re: Partnered Output Coils « Reply #365 on: 2015-02-05, 00:02:29 »
Sr. Member
****

Posts: 336
@All,

I did dig out my old (slightly rusty) Metglas AMCC-320 core tonight. The two
partnered coils are in bucking parallel (both outer coils 300 turns 0,7mm wire).
Between the bucking coils is a 12 Volt 3,5 Watt LED bulb.

The white center coils (300 turns 0,5mm wire) is used as a JT oscillator running
from a 9 Volt battery with one switching transistor and a 1K8 resistor in series
with the base and trigger coil.

The input usage was approx. (slightly depleted battery) 7,9 V @ 0,05 Amp. = 0,375 Watt.
The LED lamp was at approx. 1/2 brightness.

Amazing much LED light for so little input! :-)

GL.
   
img  poynt99
Re: Partnered Output Coils « Reply #366 on: 2015-02-05, 00:16:19 »

Group: Administrator
Hero Member
*****

Posts: 3205
It's not as complicated as it may seem...
Quote
How is it, that from the 0 Voltage Terminal, to the Output, we see 3.3 Watts, but from the 0 Voltage Terminal to the + Voltage Terminal we see 1.4 Watts - Where is the missing power gone? 0.68 has been accounted for!

There could be a couple of reasons. One possible explanation is that when the output is measured from the denoted probe point and gnd, it may not only include the spike, but it may be in series with the power supply as well.

The actual cause may be interesting, but is irrelevant in terms of COP measurements.

Power measurements with a scope always involves a measure of the current through a device and the voltage directly across the same device. You do agree on that much I hope? As such, the measurement as depicted in your diagram can not provide a true representation of the actual output power being dissipated in the load resistor.
   
img  poynt99
Re: Partnered Output Coils « Reply #367 on: 2015-02-05, 00:25:07 »

Group: Administrator
Hero Member
*****

Posts: 3205
It's not as complicated as it may seem...
Thanks partzman (and good job yourself),

I'm sure with some tweaking of the sim, we could get the wave forms pretty much exact. I was tempted to do it myself, but Itsu has done a wonderful job with the actual circuit.
   
img  ion
Re: Partnered Output Coils « Reply #368 on: 2015-02-05, 00:36:57 »
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
I was working on the sim in LTSpice when I saw that partzman had already posted it. Good job partzman.

One thing that I believe is a point of interest is the drive voltage for the FET. From the prior waveforms that were posted by EMJ, I believe his drive voltage is too low leading to the FET quickly coming out of saturation and holding the inductor at some fixed level of current when the inductor begins to saturate. This may account for some missing energy.

Partzman, you have much more drive (15 v) in your sim so won't see the degenerative effect of Rds as the inductor saturates. Your FET will remain saturated.

Actually EMJ specs a deflection transistor, (post 335 NPN horizontal deflection transistor: D1555) not a FET, although his drawing (VIC) shows a FET and if the deflection transistor is driven from his SG it is possible there is not enough base current drive so the transistor comes out of saturation  when the core begins to saturate early as seen in his waveforms.

So the question to EMJ:
1) you have spec'd a deflection transistor earlier, but show a FET in your VIC drawing.
What device was actually used for the waveforms you posted?

2)Are you driving the transistor or FET from your SG? At what drive voltage?.
If it was a deflection transistor that was used for the posted waveforms, it would be helpful to know the drive current capability of the SG and if a base current limiting resistor was used.

All of this is needed for a proper simulation.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
img  EMJunkie
Re: Partnered Output Coils « Reply #369 on: 2015-02-05, 00:49:55 »
Group: Guest
Quote from: poynt99 on 2015-02-05, 00:16:19
There could be a couple of reasons. One possible explanation is that when the output is measured from the denoted probe point and gnd, it may not only include the spike, but it may be in series with the power supply as well.

The actual cause may be interesting, but is irrelevant in terms of COP measurements.

Power measurements with a scope always involves a measure of the current through a device and the voltage directly across the same device. The measurement as depicted in your diagram can not provide a true representation of the actual output power being dissipated in the load resistor. You do agree on that much I hope?

@Poynt99

Yeah, nice, a start to finding a solution.

Don't you think, this is and should be the most important thing to find the reason for before any OU can be achieved? I certainly do.

How can we know where to look if only know where half the Energy we dissipate is in fact going to? Or another way to look at this, we have an issue with accounting for energy between Point A and Point C! Where Point A is connected to B and then to C.

How can we just all of a sudden Loose half of it?

Take a pipe, three lengths, Water flows in one pipe, at the end, we see a "Y" Intersection, then we should be able to add the Water Volume out of both of the ends of the "Y" Intersection to get the total on the Input. 1/2 + 1/2 = 1. This is not what we are getting!
   
img  EMJunkie
Re: Partnered Output Coils « Reply #370 on: 2015-02-05, 00:50:43 »
Group: Guest
Quote from: Groundloop on 2015-02-05, 00:02:29
@All,

I did dig out my old (slightly rusty) Metglas AMCC-320 core tonight. The two
partnered coils are in bucking parallel (both outer coils 300 turns 0,7mm wire).
Between the bucking coils is a 12 Volt 3,5 Watt LED bulb.

The white center coils (300 turns 0,5mm wire) is used as a JT oscillator running
from a 9 Volt battery with one switching transistor and a 1K8 resistor in series
with the base and trigger coil.

The input usage was approx. (slightly depleted battery) 7,9 V @ 0,05 Amp. = 0,375 Watt.
The LED lamp was at approx. 1/2 brightness.

Amazing much LED light for so little input! :-)

GL.

Looking really Good Groundloop!

Any ideas on the current topic?
   
img  ion
Re: Partnered Output Coils « Reply #371 on: 2015-02-05, 01:01:53 »
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
from EMJ
Quote
We are still missing something here, we are missing 1.3 Watts from the Input Power Source, to the Output, this cant currently be accounted for!

EMJ, there are a few questions and comments in post #368, please consider them.

It is quite possible the loss is due to the FET coming out of saturation early because of high inductor current as the core saturates early. In this case the FET is acting as a current source, trying to limit current drive to the inductor L1, therefore having to dissipate it as heat for the remainder of that portion of the cycle.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
img  EMJunkie
Re: Partnered Output Coils « Reply #372 on: 2015-02-05, 01:05:06 »
Group: Guest
@ION,

Quote from: ION on 2015-02-05, 00:36:57
I was working on the sim in LTSpice when I saw that partzman had already posted it. Good job partzman.

One thing that I believe is a point of interest is the drive voltage for the FET. From the prior waveforms that were posted by EMJ, I believe his drive voltage is too low leading to the FET quickly coming out of saturation and holding the inductor at some fixed level of current when the inductor begins to saturate.

You have much more drive in your sim so won't see the degenerative effect of Rds as the inductor saturates.

Actually EMJ specs a deflection transistor, not a FET, although his drawing (VIC) shows a FET and if the deflection transistor is driven from his SG it is possible there is not enough base current drive so the transistor comes out of saturation  when the core begins to saturate early as seen in his waveforms.

So the question to EMJ:

1) you have spec'd a deflection transistor earlier, but show a FET in your VIC drawing.
What device was actually used for the waveforms you posted?


Apologies, The drawing I drew was a generic. No specific Parts were intended.

See: http://www.overunityresearch.com/index.php?topic=2760.msg45449#msg45449

All Part Numbers and Datasheets provided are correct as to the Scope Shots provided.

Quote from: ION on 2015-02-05, 00:36:57

2)Are you driving the transistor or FET from your SG? At what drive voltage?.
If it was a deflection transistor that was used for the posted waveforms, it would be helpful to know the drive current capability of the SG and if a base current limiting resistor was used.


Driven from Function Generator. I am sorry, I don't know exactly, guessing 5v roughly. It is something I did play around with a bit to get the wave form. No limiting Resistor, max output is 9V into 50 Ohms, so it could be somewhere around: 0.18amps - I don't know exactly I am sorry! I can re, measure it if you like. Parts are all still there.

Quote from: ION on 2015-02-05, 00:36:57

All of this is needed for a proper simulation.


Yes, understand.
   
img  EMJunkie
Re: Partnered Output Coils « Reply #373 on: 2015-02-05, 01:06:44 »
Group: Guest
Quote from: ION on 2015-02-05, 01:01:53
EMJ, there are a few questions and comments in post #368, please consider them.

Yes, yes, was prepping info.

Please let me know if you need more info.
   
img  Groundloop
Re: Partnered Output Coils « Reply #374 on: 2015-02-05, 01:09:44 »
Sr. Member
****

Posts: 336
Quote from: EMJunkie on 2015-02-05, 00:50:43
Looking really Good Groundloop!

Any ideas on the current topic?

EMJunkie,

I will leave accurate measurement to the good people of this forum.
So far I have just been playing with the circuit to see what will happen.

The main frequency of the circuit is approx. 40Hz. There are some
over harmonics up to approx. 10MHz.

I always try to loop my circuits to test for OU, so my next test will
be adding a diode bridge and a electrolytic capacitor to the output
of my circuit and loop it. We will see...............

GL.
   
Pages: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [15] 16 17 18 19 20 21 22 23 24 25 26
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-04-19, 05:23:02