Author Topic: RESONANCE EFFECTS FOR EVERYONE TO SHARE (Read 299348 times)

duff · « **Reply #510 on:** February 06, 2009, 07:57:48 AM »

Quote from: gotoluc link=topic=6225.msg155690#msg155690

Yes, I see what you mean .... this is why I got this USB scope... the software does the RMS surface calculations... I don't know how you can calculate that

Luc,

I could be wrong but I'll bet your scope does not know the difference between a square wave vs sine.

I'd think that the rms value is only valid for sine waves....

You might want to check your manual or with the mfg. on that one.

Also, thanks for those precise instruction on winding the inductor. Without the dimensions I'd still be banging my head trying to get it to work!!

-Duff

duff · « **Reply #511 on:** February 06, 2009, 10:06:55 AM »

Quote from: gyulasun on February 05, 2009, 10:45:43 AM

(In fact the generator should feed either a tap on L1 or another coupling coil as a matching means between 50 Ohms and the resonant impedance, this would be the first to achive for Luc also and the second goal is to match the lamp's impedance to the resonant circuit with L2 number of turns at a fixed coupling.)

@Gyula,

I am interested attempting to impedance match this circuit.

I've been a ham for 25yrs - matched lots of different type of antenna but this little inductor is small with fine wire and not easily tapped. Coupling to it with another inductor would mean more losses.

Thinking that the primary is resonant and reactances canceling, I tried incrementally inserting resistors in series with the SG.

Ohms Volts
49 3.001
43 3.014
20 3.095
10 3.221
5.6 3.250
3.9 3.269

The output voltage (across load) did exactly opposite of what I thought would happen. This circuit is very happy with the current sensing resistor (0.84 ohms), meaning that it must already look like 50 ohms to the SG (if my reasoning is correct).

I'm wondering if the load impedance is being reflected back into the primary.

Quote from: Montec on February 06, 2009, 12:03:15 AM

Hello Duff
This may or may not apply to your coils but a Tesla Coil (TC) is usually made up of two air coils that have the same resonant frequency. The voltage increase/decrease is not based on the turns ratio but is based on a ratio of the inductance's of the two coils. See http://www.scribd.com/doc/3876818/Denicolai-Tesla-Transformer-for-Experimentation-and-Research-96pp2001 for more info.

@Montec

Thanks for the link though I am familiar with that article.

If you found that interesting check out the following. I thought is was an excellent article and was posted earlier in this thread:

Tesla Coils and the Failure of Lumped-Element Circuit Theory
http://www.ttr.com/corum/index.htm

-Duff

gyulasun · « **Reply #512 on:** February 06, 2009, 01:08:34 PM »

Hi Duff,

Thanks for the reports on your findings. I was not aware of Luc's earlier activity on picking the best diode type for such circuits, though I mentioned using Schottky as first and 1N4148 and the like as second. I looked for the data sheet on SBL3040 and it justifies the good result: its forward voltage drop is only 0.2 - 0.24V for a current from 10mA to 100mA range, see Fig.3 on Page 2 in data sheet: http://www.vishay.com/docs/88732/88732.pdf

I am also puzzled by the half wave rectifier's performance here, maybe the oscilloscope waveform across your L2 would reveal something but I would expect a normal sinusoidal 305 kHz waveform with symmetrical zero crossings...

Re on the input impedance. Maybe you could place again resistances with the hot output of the SG in series (replacing the 0.84) but in increasing values towards several hundred Ohms maybe up to 1kOhm in 100 Ohm steps and see on the scope when the divided input voltage is halved across the series L1C1 (nodes 2 and 0)?
You surely know this method when the series inserted resistance equals the unknown input impedance (at resonance of course) then the input voltage from the SG gets halved across the unknown, (a normal voltage divison happens).

Quote

I'm wondering if the load impedance is being reflected back into the primary.

Well, energy must come from somewhere and it should be coming from L1, it must have a big electromagnetic near field around it due to its sizes. Have you tried to short circuit L2 and see the voltage amplitude across L1? Maybe overcoupling occurs between L1 and L2?

Regards, Gyula

EDIT: Once you managed to find the input impedance, you could use a Pi filter between the SG and this circuit?

gyulasun · « **Reply #513 on:** February 07, 2009, 12:15:36 AM »

Quote from: duff on February 06, 2009, 07:57:48 AM

Luc,

I could be wrong but I'll bet your scope does not know the difference between a square wave vs sine.

I'd think that the rms value is only valid for sine waves....

You might want to check your manual or with the mfg. on that one.

-Duff

Hi All,

Unfortunately, a PC scope software bug has misled member nul-points on pulsed capacitor discharge measurements and this indicates we have to be cautios and triple check measurements that show unusual results.

See his findings here: http://www.overunity.com/index.php?topic=4419.msg150394#msg150394

I do not mean Luc has also got a problem in his PC scope I mean only it is advisable to check some results with another type of measurement device (normal analog scope, etc).

Regards, Gyula

gyulasun · « **Reply #514 on:** February 07, 2009, 12:16:39 AM »

Double post

duff · « **Reply #515 on:** February 07, 2009, 09:25:46 AM »

Hi Gyula,

Quote from: gyulasun on February 06, 2009, 01:08:34 PM

I am also puzzled by the half wave rectifier's performance here, maybe the oscilloscope waveform across your L2 would reveal something but I would expect a normal sinusoidal 305 kHz waveform with symmetrical zero crossings...

Unfortunately my camera has lost the ability to communicate with the computer. I have yet to determine whether its the camera or the software.

I did look at the waveform across the secondary and its more like a square wave with rounded corners with jagged sides. Yesterday I looked at the output of the FWB and is seemed ok.

If I can get the camera talking again I'll upload the pics.

Quote from: gyulasun on February 06, 2009, 01:08:34 PM

You surely know this method when the series inserted resistance equals the unknown input impedance (at resonance of course) then the input voltage from the SG gets halved across the unknown, (a normal voltage divison happens).

Yes I know the method and I'm embarrassed to say it did not occur to me. <cowering in shame>

@Luc

Using the above principle I was able to determine the impedance at resonance. I started at 302KHz where the impedance was 741 ohms (741 +j0) and by adjusting the capacitance and varying the frequency the impedance decreased and the voltage increased. Finally I reached the point where the voltage started droping again. At the sweet point I had the following output:

V_out = 4.521VDC
F = 121.9 KHz
C1 = .0012uF
L1 = 1.38mH
R_L1 = 8.26Î©
R_L = 63.6Î©
Zin = 139Î© @ 0 degrees.

SG = square wave 23Vpp / 8.13Vrms

The above voltage was with the full-wave bridge in the circuit.
Note: I determined yesterday that a half-wave bridge, in this circuit, yields more volts out.

Impedance measurements were taken using a sine wave.

Gyula suggested a matching network and I look at that some today. That would provide a little more output . Also wondering how harmonic attenuation (due to the matching network) would influence the output???

-Duff

[Edit]
Change component names to correspond with schematic
Corrected value of L1
Now LC works in resonance formula

gyulasun · « **Reply #516 on:** February 07, 2009, 11:02:42 AM »

Hi Duff,

Would you mind updating your schematics with the new component values because at the moment I am puzzled a little by your new symbols like R_L C_series etc. ? (I mean the circuit you measured Z=139 Ohm real input impedance.)

I can assure you: you DO NOT have to feel yourself embarrased at all for anything , you do an excellent and outstanding job here!

Thanks, Gyula

EDIT:

Quote

I did look at the waveform across the secondary and its more like a square wave with rounded corners with jagged sides. Yesterday I looked at the output of the FWB and is seemed ok.

Would you recall that you used a square wave input in this case? It is odd you found waveform like that, though the loaded Q of L1C1 might justifies this? It would be also good you could check the 3dB bandwidth in your latest circuit when you have time for that.

duff · « **Reply #517 on:** February 07, 2009, 12:00:49 PM »

Quote from: gyulasun on February 07, 2009, 11:02:42 AM

Would you mind updating your schematics with the new component values because at the moment I am puzzled a little by your new symbols like R_L C_series etc. ? (I mean the circuit you measured Z=139 Ohm real input impedance.)

Sorry for the confusion. I renamed the components to be inline with the previous posts...

Schematic added above...

Quote

EDIT: Would you recall that you used a square wave input in this case? It is odd you found waveform like that, though the loaded Q of L1C1 might justifies this? It would be also good you could check the 3dB bandwidth in your latest circuit when you have time for that.

I'll get the 3db bandwidth for you - no problem

Yes - I used square wave input for voltages across RL
I used sine when taking impedance readings.

As always I value your input.!

-Duff

duff · « **Reply #518 on:** February 07, 2009, 01:07:20 PM »

Gyula,

Vpeak = 5.6V
3db voltage = 5.6 * .707 = 3.96V

Fr = 121.9 KHz
3db points: 92.7, 140.6 KHz

3db BW = 140.6 - 92.7 = 47.9KHz

Q = 121.9/47.9 = 2.545

I think I did that correctly - it's been a long time...

-Duff

gyulasun · « **Reply #519 on:** February 07, 2009, 04:54:49 PM »

Hi Duff,

Yes you did it correctly, and the loaded Q of 2.5 means something ruins too much the unloaded (high) Q of L1... and this low value also explains the waveform you described across L2.

(The unloaded Q of L1 in itself is (2pi*f*L1)/R_L1=1056.43/8.26 gives about Q=128.)

Because basically the SG output of 50 Ohm is loaded by a series resonant circuit that is further loaded by the transformed bulb or your transformed 66.6 Ohm resistance, the big mismatch seems to come from the 50 Ohm that is in series with the coil: if you consider the loaded Q like (2pi*f*L1)/(R_L1+R1+50) then you get Q=1056.43/59.1=17.8 and this is loaded further by the transformed load of 66.6 to your value of 2.5, agree?
So improvement can come by matching somehow the primary tank circuit to 50 Ohm. One possibility is to make some taps on L1 by unwinding about 1/6 or 1/5 part of it, make a few cm long tap and continue rewinding again what was unwound to receive back the 145 turns; another possibility is indeed use a matching network (can be a pi filter or L+C low pass filter etc).

However we should not forget what Luc was asking: verifying his anomaly finding. So I do not suggest any further alteration in his original circuit for now, it is up to you how to proceed. At this stage you are now, we know pretty much correct data about this circuit. I will be available further on.

Regards, Gyula

duff · « **Reply #520 on:** February 10, 2009, 04:39:25 PM »

I don't have anything of real significance to report. I have been refining what I have and trying a few variations.

I found that changing the wire size of the primary resulted in an increase in volts out and the resonant frequency decreased.

Changing the wire size from #28 to #24 resulted in the voltage output going from 4.521VDC to 4.705VDC across RL.

Wire: #24
Turns: 93
Height: 2.185", 55.5mm
Inductor OD: 3.15", 80mm
Inductance: 520uH
Resistance: 1.97Î©

F= 196.6KHz
C = .00118uF
V_Capacitor = 155 Vpp
V_out 4.705 V_DC
Z_in = 138.6Î©

I tried adding 4 taps to the bottom of this inductor with 7 turns between taps. What I found is that the voltage output across RL dropped as I moved up the taps, and the impedance increased meaning my inductor was too short.

Since I'm out of #24 wire I'll have to order some more which will take a few days. I have some #10 magwire but that would be a drastic change in inductor size.

Replacing RL (66.6Î©) with a 100ma lamp results in the lamp burning a full brightness. Something to note about using lamps is the resistance changes with current.

Resistance Votage Drop Current
Across Lamp
2.65Î© 0.0 0
9.158Î© 0.4579V 50ma
23.35Î© 2.335V 100ma
25.09Î© 2.760V 110ma

Quote from: gyulasun on February 07, 2009, 04:54:49 PM

Because basically the SG output of 50 Ohm is loaded by a series resonant circuit that is further loaded by the transformed bulb or your transformed 66.6 Ohm resistance, the big mismatch seems to come from the 50 Ohm that is in series with the coil: if you consider the loaded Q like (2pi*f*L1)/(R_L1+R1+50) then you get Q=1056.43/59.1=17.8 and this is loaded further by the transformed load of 66.6 to your value of 2.5, agree?

Yes, that sounds logical to me.

-Duff

gotoluc · « **Reply #521 on:** February 10, 2009, 04:40:58 PM »

Hi everyone,

I know things have been quiet for some days now but don't think the topic is dead. What we are doing at this time is re-evaluating the direction the research is going to take.

Once that has been done we will let you know by posting the new test results.

Thanks for all your support and interest.

Luc

minde4000 · « **Reply #522 on:** February 11, 2009, 01:30:48 AM »

We all waiting gotoluc. Tnx for your efforts and info.

sparks · « **Reply #523 on:** February 11, 2009, 03:26:30 AM »

@gotoluc

Please checkout some of Tesla's work. Below is a post I made couple of days ago. The second half is a quote from Tesla under questioning from some lawyer. If you learn to couple your load and allow the waves to go undamped you should be good to go.

The simple charging of a capacitor can run with gain. Tesla told us how a loong time ago. Charge her up slow. Let her go fast. Make sure things oscillate a bit from your hammer blow. Mother nature remembers. What SM was doing is no different. Charge up the cap. Dump her into a helical resonator. Tap the current so you don't screwup the primary oscillations. You can throw an arrow at a target or you can invest it into pulling a string on a bow. Which is the biggest bang for the buck/target.

Yes; I charged the condenser with 40,000 volts. When it was charged full, I discharged it suddenly, through a short circuit which gave me a very rapid rate of oscillation. Let us suppose that I had stored in the condenser 10 watts. Then, for such a wave there is a flux of energy of (4 x 104)2, and this is multiplied by the frequency of 100,000. You see, it may go into thousands or millions of horsepower.

gotoluc · « **Reply #524 on:** February 11, 2009, 04:02:20 AM »

Thanks for the input sparks

Pun intended

... I hope you don't mind. We will get to it soon.

Thanks for sharing

Luc

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Author Topic: RESONANCE EFFECTS FOR EVERYONE TO SHARE (Read 299348 times)

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