The final answer...

Home Help Search Login Register

Welcome,Guest. Please login or register.

OverUnity Research > Benches > giantkiller (Moderator: giantkiller) > The final answer...

2024-11-24, 15:45:58
News: Registration with the OUR forum is by admin approval.

Pages: 1 2 3 4 5 [6] 7 8 9 10 11 12 13 14 15 16 17 18

« previous next »

Author Topic: The final answer... (Read 135536 times)

Reiyuki	Re: The final answer... « Reply #125 on: 2018-06-20, 16:08:35 »
Group: Mad Scientist Full Member Posts: 185	Quote from: partzman on 2018-06-20, 15:44:07 This is true as a result of CoE and can be shown with bench experiments to exhibit gain. The problem is, the energy required to bring about the parametric inductance change equals the gain produced by the change. At least that has been my experience and I'm certainly open to any suggestions. The energy used to induce the parametric change can also be recycled (as in diode recovery or a tank circuit). I have had that same problem though (especially with a large core). It takes a lot of current or thousands of turns to fully saturate some cores. Saturating in air (or thin iron wire) is much easier. If solid-state isn't your thing, you can also consider using a rotary system to induce the change in L or C as in the bed1ni machines or the Rotary Electrostatic converter. Or something new and novel like a vibrator that compresses a series of spongy capacitor plates. « Last Edit: 2018-06-20, 23:26:45 by Reiyuki » --------------------------- When you say something is impossible, you have made it impossible

giantkiller	Re: The final answer... « Reply #126 on: 2018-06-21, 04:28:46 »
Group: Moderator Hero Member Posts: 1593 Frequency equals matter...	Tanks and resonance. Do not stray from Tesla. --------------------------- http://purco.qc.ca/ftp/

Centraflow	Re: The final answer... « Reply #127 on: 2018-06-21, 12:29:11 »
Group: Experimentalist Hero Member Posts: 2982	Quote from: Reiyuki on 2018-06-20, 16:08:35 The energy used to induce the parametric change can also be recycled (as in diode recovery or a tank circuit). I have had that same problem though (especially with a large core). It takes a lot of current or thousands of turns to fully saturate some cores. Saturating in air (or thin iron wire) is much easier. Yes I agree with you and still working on this when time permits. Meanwhile below is a simplified Janos Vajda paper Regards Mike ------------------------ Principle workings of a TPU.pdf (93.96 kB - downloaded 236 times.) --------------------------- "All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident." Arthur Schopenhauer, Philosopher, 1788-1860 As a general rule, the most successful person in life is the person that has the best information.

Centraflow	Re: The final answer... « Reply #128 on: 2018-06-21, 15:17:06 »
Group: Experimentalist Hero Member Posts: 2982	@ PM Quote:- This is true as a result of CoE and can be shown with bench experiments to exhibit gain. The problem is, the energy required to bring about the parametric inductance change equals the gain produced by the change. At least that has been my experience and I'm certainly open to any suggestions. The energy required is just a switch plus a LCR circuit. Depending how you connect it you can increase or decrease the charged inductance. The coils must be wound the same direction on the same core, or it will not work. If you use an iron core, 3 or more insulated turns, which is also your L2, and wind your L1 over the iron core, like a toroid, you should get some interesting results. Diagram to follow. Regards Mike --------------------------- "All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident." Arthur Schopenhauer, Philosopher, 1788-1860 As a general rule, the most successful person in life is the person that has the best information.

Centraflow	Re: The final answer... « Reply #129 on: 2018-06-21, 15:47:46 »
Group: Experimentalist Hero Member Posts: 2982	Diagram. Top increases inductance and the bottom one decreases inductance on discharge. Regards Mike ------------------------ The final answer... Parametric inductance change.png (16.17 kB, 1495x1060 - viewed 1273 times.) --------------------------- "All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident." Arthur Schopenhauer, Philosopher, 1788-1860 As a general rule, the most successful person in life is the person that has the best information.

partzman	Re: The final answer... « Reply #130 on: 2018-06-21, 17:47:44 »
Group: Experimentalist Hero Member Posts: 1808	Quote from: Centraflow on 2018-06-21, 15:47:46 Diagram. Top increases inductance and the bottom one decreases inductance on discharge. Regards Mike Mike, Thanks for the info and the schematic. I will do some bench experiments based on this and will post the results. I may have some more questions regarding the inductance values but will get started anyway. regards, Pm

Centraflow	Re: The final answer... « Reply #131 on: 2018-06-21, 18:16:38 »
Group: Experimentalist Hero Member Posts: 2982	Quote from: partzman on 2018-06-21, 17:47:44 Mike, Thanks for the info and the schematic. I will do some bench experiments based on this and will post the results. I may have some more questions regarding the inductance values but will get started anyway. regards, Pm Hi PM One thing, the cap across the switch has to be HV. Remember Itsu showed a 500v spike at the drain of his mosfet! a cap will do two things, one it will save the mosfet and two it adds to the tuning of the output. I have always said in the past, if you really look at this carefully it is a class E output before the diode and should be a sine wave (may have a flat top due to coil saturation). To bloody hot in my lab atm, 32ºc today. Regards Mike --------------------------- "All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident." Arthur Schopenhauer, Philosopher, 1788-1860 As a general rule, the most successful person in life is the person that has the best information.

giantkiller	Re: The final answer... « Reply #132 on: 2018-06-21, 19:26:37 »
Group: Moderator Hero Member Posts: 1593 Frequency equals matter...	The tpu pdf is based on thermal variance. Whats missing is kinetic transference. Transmission reflections do all the work without the need for amplfication by an implied or inserted circuit. Its all good but Tesla had no extra amplifiers to read the feedback. The tpu is a magnetic earth machine. --------------------------- http://purco.qc.ca/ftp/

Bob Smith	Re: The final answer... « Reply #133 on: 2018-06-22, 03:07:33 »
Jr. Member Posts: 71	Quote from: giantkiller on 2018-06-21, 19:26:37 The tpu pdf is based on thermal variance. Whats missing is kinetic transference. Transmission reflections do all the work without the need for amplfication by an implied or inserted circuit. Its all good but Tesla had no extra amplifiers to read the feedback. The tpu is a magnetic earth machine. GK I've been thinking back to Eric Dollard and Thomas Brown's Borderlands demo of the separation of magnetism and dielectricity that occurs when their L-C setup reaches resonance (tried to cue it up to where EPD addresses this separation phenomenon here): https://youtu.be/6BnCUBKgnnc?t=1385 So if magnetism at one end is pulsating at resonance, I'm surmising from what I've read in your posts and infer from others, that the magnetic field (and longtitudinal impulses) is pinging up against the magnetic field of the earth. I would presume that the earth's mag field is going to push back to re-establish equilibrium. This is where I'm at in trying to get my head around your statement "tpu is a magnetic earth machine". Kind of like TPU and Earth doing the Cha-Cha-Cha Regards, Bob

Centraflow	Re: The final answer... « Reply #134 on: 2018-06-22, 10:56:17 »
Group: Experimentalist Hero Member Posts: 2982	Quote from: giantkiller on 2018-06-21, 19:26:37 The tpu pdf is based on thermal variance. Whats missing is kinetic transference. Transmission reflections do all the work without the need for amplfication by an implied or inserted circuit. Its all good but Tesla had no extra amplifiers to read the feedback. The tpu is a magnetic earth machine. No the pdf is based on ALL waves, the electromagnetic is shown, it would also apply to water and ocean waves (how large waves are formed from interference of smaller ones), sound, light and thermal. SM knew about sound wave interference, that was exactly his forte, he then applied it to electromagnetic as he knew that all waves react the same in destructive or constructive. Once I had realised the only link between his real work and the TPU, things got clearer. Now it is realising the electronics of the system by substitution of the different magnetic loop configurations and pickup from the same so as to feedback a high current/low voltage to the input capacitor without affecting the performance of the circuit. IMO, loopback can't be from the output, it has to come isolated from within the circuit, it is a slow buildup and I think that will apply to "all" attempted loopback circuits and why they normally fail. Regards Mike --------------------------- "All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident." Arthur Schopenhauer, Philosopher, 1788-1860 As a general rule, the most successful person in life is the person that has the best information.

ion	Re: The final answer... « Reply #135 on: 2018-06-22, 16:01:13 »
Group: Elite Hero Member Posts: 3537 It's turtles all the way down	Quote IMO, loopback can't be from the output, it has to come isolated from within the circuit, it is a slow buildup and I think that will apply to "all" attempted loopback circuits and why they normally fail. Regards Mike It seems you are talking about local regeneration. How would you propose to implement this in your circuit? Regards --------------------------- "Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy

partzman	Re: The final answer... « Reply #136 on: 2018-06-22, 16:53:47 »
Group: Experimentalist Hero Member Posts: 1808	Mike, Here is an initial test result of your top schematic shown above which should show an increase in inductance on discharge. Two tests are shown taken with a 10vdc and 30vdc supply respectively. The probe IDs are CH1(yel) is gate drive to mosfet switch, CH2(blu) is drain voltage of switching mosfet, CH3(pnk) is output on pin1 side of Schottky diode, CH4(grn) is supply current, and Math(red) is input power consumption. Not shown is the dc voltage level on C? which is approximately equal to the max level of CH3. The circuit values used are C1 = 4.07nf, C2 = mosfet output capacitance, L1 = 47.9uH, and L2 = 27.2uH and the frequency is 342kHz which gave the largest peak output voltage on CH3. I'm not sure what I am supposed to be looking for although I do see a slight change in the rising slope of CH3 at turn off. Is this the indication of an inductance increase? Regards, Pm Added: The resonant frequency of the CH2 ringing is 3.22MHz. Considering the parallel inductance of L1 and L2 at 17.33uH, this would equate to an equivalent drain capacitance of 143 pf which is reasonable for the IRF636 mosfet I'm using. ------------------------ The final answer... Schema1 10vdc.png (35.82 kB, 800x480 - viewed 692 times.) The final answer... Schema1 30vdc.png (41.89 kB, 800x480 - viewed 646 times.)

Centraflow	Re: The final answer... « Reply #137 on: 2018-06-23, 09:16:45 »
Group: Experimentalist Hero Member Posts: 2982	Hi ION, PM ION I will get back to you on your question, I want to get PM hopefully in the same area that I am with the changing of inductance in charge and discharge using only a switch (mosfet). PM, you need to measure the individual inductance of each coil, then measure the inductance of the coils in series. I presume you have wound the L1 and L2 as I have said, L2 is 3 turns of insulated iron wire (+- 17cm dia: coil), L1 is a copper wire wound around the L2 like a toroid using L2 as the core. Don't make too many turns, about 20% of the circumference or less and make it so you can move it by sliding around the L2 (not too tight). Your idea of measuring the drain resonant frequency is good, but you need to put a cap across the source and drain and measure the capacitance. The combined series inductance of L1 and L2 is not a straight addition, it is a lot more as they are basically two coils in series on a magnetic core and not air coils (that is if you wound them like I have said). The other thing is to do circuit 1 and test, then change the connections and do to circuit 2, all at the same settings and see the difference. Back later, getting very hot here, reached 35ºc yesterday. Great work PM Regards Mike --------------------------- "All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident." Arthur Schopenhauer, Philosopher, 1788-1860 As a general rule, the most successful person in life is the person that has the best information.

partzman	Re: The final answer... « Reply #138 on: 2018-06-23, 14:16:04 »
Group: Experimentalist Hero Member Posts: 1808	Quote from: Centraflow on 2018-06-23, 09:16:45 Hi ION, PM ION I will get back to you on your question, I want to get PM hopefully in the same area that I am with the changing of inductance in charge and discharge using only a switch (mosfet). PM, you need to measure the individual inductance of each coil, then measure the inductance of the coils in series. I presume you have wound the L1 and L2 as I have said, L2 is 3 turns of insulated iron wire (+- 17cm dia: coil), L1 is a copper wire wound around the L2 like a toroid using L2 as the core. Don't make too many turns, about 20% of the circumference or less and make it so you can move it by sliding around the L2 (not too tight). Your idea of measuring the drain resonant frequency is good, but you need to put a cap across the source and drain and measure the capacitance. The combined series inductance of L1 and L2 is not a straight addition, it is a lot more as they are basically two coils in series on a magnetic core and not air coils (that is if you wound them like I have said). The other thing is to do circuit 1 and test, then change the connections and do to circuit 2, all at the same settings and see the difference. Back later, getting very hot here, reached 35ºc yesterday. Great work PM Regards Mike Mike, OK, for my previous post I used an existing coil assembly that I had from previous experiments with the specs- L2(iron) is 8T on 17.2cm diameter with L1(copper) at 282T wound over the top of L2. L1 = 47.93uH and L2 = 27.16uH. L1 and L2 in series, aid = 84.05uH and buck = 73.58uH. I see that I have L1 with too many turns so I will change that and run some more tests. Regards, Pm

Centraflow	Re: The final answer... « Reply #139 on: 2018-06-23, 15:35:17 »
Group: Experimentalist Hero Member Posts: 2982	Quote from: partzman on 2018-06-23, 14:16:04 Mike, OK, for my previous post I used an existing coil assembly that I had from previous experiments with the specs- L2(iron) is 8T on 17.2cm diameter with L1(copper) at 282T wound over the top of L2. L1 = 47.93uH and L2 = 27.16uH. L1 and L2 in series, aid = 84.05uH and buck = 73.58uH. I see that I have L1 with too many turns so I will change that and run some more tests. Regards, Pm OK, just curious but is this right what you said before Quote Added: The resonant frequency of the CH2 ringing is 3.22MHz. Considering the parallel inductance of L1 and L2 at 17.33uH, this would equate to an equivalent drain capacitance of 143 pf which is reasonable for the IRF636 mosfet I'm using. But L1/L2 in series measure 84.05uH (type of common core) not 17.33uH!! It is one reason you need to put a HV cap across the SD of the mosfet and measure the capacitance. This is quite informative:- http://www.crarc.ampr.org/uploads/Main/Archives/July%20Class-E%20Transmitters.pdf Regards Mike --------------------------- "All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident." Arthur Schopenhauer, Philosopher, 1788-1860 As a general rule, the most successful person in life is the person that has the best information.

Centraflow	Re: The final answer... « Reply #140 on: 2018-06-23, 15:57:27 »
Group: Experimentalist Hero Member Posts: 2982	My investigation into the TPU tells me 1. There is a magnetic core 2. The core heats up, so eddy currents, the heating problem SM had over time 3. The core is most probably iron 4. The core is also a coil or loop which emits magnetic waves 5. The core vibrates 6. Excess energy from the magnetic wave reaction around that loop/core (constructive) 7. The larger TPU's possibly have two ion coils and a second L1 etc which the two circuits are in series to increase power 8. Loop power comes from a second copper coil wound on the iron loop (high current but low voltage). Takes time to wind up the output Regards Mike --------------------------- "All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident." Arthur Schopenhauer, Philosopher, 1788-1860 As a general rule, the most successful person in life is the person that has the best information.

partzman	Re: The final answer... « Reply #141 on: 2018-06-23, 17:31:58 »
Group: Experimentalist Hero Member Posts: 1808	Quote from: Centraflow on 2018-06-23, 15:35:17 OK, just curious but is this right what you said before Quote Added: The resonant frequency of the CH2 ringing is 3.22MHz. Considering the parallel inductance of L1 and L2 at 17.33uH, this would equate to an equivalent drain capacitance of 143 pf which is reasonable for the IRF636 mosfet I'm using. But L1/L2 in series measure 84.05uH (type of common core) not 17.33uH!! It is one reason you need to put a HV cap across the SD of the mosfet and measure the capacitance. This is quite informative:- http://www.crarc.ampr.org/uploads/Main/Archives/July%20Class-E%20Transmitters.pdf Regards Mike Yes, this is correct. Considering the low reactance of C1 at 3.22MHz, this essentially means L1 and L2 are in parallel and therefore calculate to Lp or 17.33uH. Lp is in series with the drain capacitance of the mosfet and is grounded through the + terminal of the supply Any added capacitance from the drain to ground will resonate with Lp along with consideration of the reduced reactance of L1. Edit- Thanks for the Class E amp design article. I didn't have that one in my file. Regards, Pm

Centraflow	Re: The final answer... « Reply #142 on: 2018-06-24, 16:50:20 »
Group: Experimentalist Hero Member Posts: 2982	PM Please do continue with the second circuit, I would like to see what results you come up with! Regards Mike --------------------------- "All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident." Arthur Schopenhauer, Philosopher, 1788-1860 As a general rule, the most successful person in life is the person that has the best information.

partzman	Re: The final answer... « Reply #143 on: 2018-06-24, 20:54:16 »
Group: Experimentalist Hero Member Posts: 1808	Quote from: Centraflow on 2018-06-24, 16:50:20 PM Please do continue with the second circuit, I would like to see what results you come up with! Regards Mike Mike, Here is the 2nd circuit using the same components as in the 1st circuit except I've added a .022uf C2 in one of the tests. The frequencies are lower for these tests. The first scope pix is without C2 and the probe id are the same as before. The last pix is with the .022uf C2 in the circuit from drain to ground. These waveforms do resemble a class E amp without the output filter. Regards, Pm ------------------------ The final answer... Schema2 10vdc woC2.png (32.13 kB, 800x480 - viewed 614 times.) The final answer... Schema2 10vdc wC2.png (38.31 kB, 800x480 - viewed 588 times.)

partzman	Re: The final answer... « Reply #144 on: 2018-06-24, 21:17:54 »
Group: Experimentalist Hero Member Posts: 1808	Mike, This test is with a new coil arrangement where L2 =4.6uH with 3 turns in an 18cm diameter. L1 = 3.2uH with 62 turns (~20% of L2's circumference) wound over L2 and moved to the positions indicated in the scope pix. C1 = .022uf and C2 = 6800pf. With the power supply at 10v dc, the wiring of both L1 and L2 do begin to get warm. The first scope pix is with L1 positioned near the output of L2. The next pix is with L1 positioned in the middle of L2 and the last pix is with L1 near the junction of L1 and L2. Regards, Pm ------------------------ The final answer... Circuit1A 10vdc L1 at Output.png (36.36 kB, 800x480 - viewed 452 times.) The final answer... Circuit1A 10vdc L1 at Middle.png (36.57 kB, 800x480 - viewed 436 times.) The final answer... Circuit1A 10vdc L1 at Junction of L2.png (36.3 kB, 800x480 - viewed 428 times.)

giantkiller	Re: The final answer... « Reply #145 on: 2018-06-24, 22:51:50 »
Group: Moderator Hero Member Posts: 1593 Frequency equals matter...	Your circuit is singing. Make it bigger! --------------------------- http://purco.qc.ca/ftp/

partzman	Re: The final answer... « Reply #146 on: 2018-06-24, 23:07:19 »
Group: Experimentalist Hero Member Posts: 1808	Quote from: giantkiller on 2018-06-24, 22:51:50 Your circuit is singing. Make it bigger! GK, Could you elaborate on that a little bit more? PM

Centraflow	Re: The final answer... « Reply #147 on: 2018-06-25, 17:18:53 »
Group: Experimentalist Hero Member Posts: 2982	Hi PM Only seen this today, had visitors all weekend, some interesting results which I need some time to study. The last test shows that position does not seem to make much difference ATM. Are you going to do schematic 2 with the new coil setup? I know all this takes time and I appreciate what you are doing, but I think the results with your equipment are better than mine. One thing on the last tests you are showing I presume DC mean for CH3 PINK, but it does not seem to equate to the scope shot, or is there something I'm missing? Regards Mike --------------------------- "All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident." Arthur Schopenhauer, Philosopher, 1788-1860 As a general rule, the most successful person in life is the person that has the best information.

partzman	Re: The final answer... « Reply #148 on: 2018-06-25, 17:30:36 »
Group: Experimentalist Hero Member Posts: 1808	Quote from: Centraflow on 2018-06-25, 17:18:53 Hi PM Only seen this today, had visitors all weekend, some interesting results which I need some time to study. The last test shows that position does not seem to make much difference ATM. Are you going to do schematic 2 with the new coil setup? I know all this takes time and I appreciate what you are doing, but I think the results with your equipment are better than mine. One thing on the last tests you are showing I presume DC mean for CH3 PINK, but it does not seem to equate to the scope shot, or is there something I'm missing? Regards Mike Mike, Yes, the mean value shown on CH3 is the average or dc offset of the output waveform measured at pin1 of the Schottky diode. I will run the circuit 2 test with the new coil and post the results. Pm

partzman	Re: The final answer... « Reply #149 on: 2018-06-25, 17:47:32 »
Group: Experimentalist Hero Member Posts: 1808	Mike, Here is a test with Circuit 2 running with a 10v dc supply. Again notice the lower operating frequency verses Circuit 1. Probe ID's are same as before. Regards, Pm ------------------------ The final answer... Circuit2A 10vdc.png (37.82 kB, 800x480 - viewed 472 times.)

Pages: 1 2 3 4 5 [6] 7 8 9 10 11 12 13 14 15 16 17 18

« previous next »

Home Help Search Login Register

Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u

2024-11-24, 15:45:58

SMF 2.0.18 | SMF © 2021, Simple Machines
XHTML
RSS
WAP2