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Hi Rodolphe, 
You can always start a new thread and call it whatever you want!
Gary Hammond,
You can always start a new thread and call it whatever you want!
Gary Hammond,
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Hi Gary,
That would be a consideration if i cannot find a thread where it would fit in. But if it fits into another thread I rather include it in there: more threads with overlap in topics make this forum more chaotic in my opinion and a hassle to gather all information about a certain topic (especially for a newcomer).
I thought it had enough overlap with the initial posts on your thread "Gary's "Complete Advanced" SSG Build" so i posted it there:
Regards,
Rodolphe
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Dear Gary, RS,
While doing benchmarks for charge/discharge cycles in CG mode which I’m currently discussing with Gary on another thread, I started reading again through the theoretical explanation of operation of the CG mode in the 3rd handbook. While trying to understand that, I realized I had to take a step back and made sure that I understood the radiant mode correctly first. So opened up my 1ste handbook again, and I realized I have quite a couple of questions regarding the step where the radiant spike takes place. I hope you guys can help me explain the following.
My questions revolve around 3 references (Ref) of the beginners handbook which I struggle to unite with each other, these are:
(I inserted all these references as a PDF at the bottom of this post, so you can use that instead of skipping back and forth trough handbook):
Ref 1. Beginners Handbook, image on page 24, with explanation in first Alinea on page 25:
“…a high voltage spike…travels from the main coil winding back to the positive terminal of the 9 volt battery along 1 wire.”
Ref 2. Beginners Handbook, bottom image on page 72, with explanation in last Alinea on page 72:
“…and the Main Coil discharges into the second battery.”
Ref 3. Beginners Handbook, page 49, section Which Way does electricity Flow?
“…all of the circuit explanations in this book use the electron current model… currents are flowing from negative to positive in the circuits”
The 9V circuit from Ref 1
1.1 Ref 1 shows the arrows in the direction of the positive terminal of the 9V battery. Combining this with Ref 3, that would indicate that the battery is being drained instead of charged (because the arrows indicate a negative electron flow and if they move to the positive terminal, it drains).
1.2 However, if it is so that the 9V battery is being charged through the positive terminal, by the high voltage peak/magnetic field collapse, than this would mean that the electrons are moved ‘backwards’ (hence the term “negative energy”?), but then Ref 3 does not apply to the direction the arrows are indicated in Ref 1.
The 2x 12V circuit from Ref 2
From the forum here I understood, that not only the output battery is charged, but the input battery receives some charge too from magnetic field collapse. Combining that with my statement 1.2 (assuming it is true):
2.1 This would mean that the high voltage spike travels in 2 directions (in order to charge both batteries through the positive terminals): in both directions away from the main coil.
2.2 To stay in line with the arrow indications of Ref 1, and assuming my other statements are true, then the arrows in Ref 2 should look like I indicated them in Image 1 in the attachment.
2.3 If 2.2 is correct, that brings me to the question what happens in the area that I indicated with a “?”, because I don’t know which direction the arrow is pointing there anymore .
2.3.1 I was thinking maybe it is not going in any direction there, since the charging can take place “along 1 wire”, see Ref 1. And so to say “nothing” happens in this area, because the spike is completely
Apologies upfront for the cross-referencing, I really struggled to formulate my questions as clear as possible, but did it to my best efforts, hope you guys understand. If in doubt, please ask.
Looking forward to your input.
Many thanks in advance,
Rodolphe
P.S. on the last pages of the attachment I added some scope readings too.
2020-06-16 - Attachment.pdf
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Hi Rodolphe,
Yes, that's what is happening. The battery is being drained to run the circuit. It is, however, being hit by the high voltage radiant spikes every time the transistor turns off resulting in a partial recharge.The 9V circuit from Ref 1
1.1 Ref 1 shows the arrows in the direction of the positive terminal of the 9V battery. Combining this with Ref 3, that would indicate that the battery is being drained instead of charged (because the arrows indicate a negative electron flow and if they move to the positive terminal, it drains).
In your ref 1, you quoted the first paragraph from page 25 of the beginners manual. But the next paragraph is very important to understanding what is going on. I'll quote it here.It completely reverses the flow of the heavier ions in the battery and slows the accumulated "discharge rate" of the battery by over 95%! This allows the battery to run the "toy demonstration" at least 20 times longer than the battery normally would.Yes, ref 3 does apply to the direction of the arrows in ref 1. The arrows all show electron current flow. That is negative to positive. This is the run current. The radiant causes the heavier ions to momentarily reverse direction.1.2 ............... but then Ref 3 does not apply to the direction the arrows are indicated in Ref 1.
The high voltage radiant spike travels in all directions thru the circuit and can be detected with a neon bulb. It is mostly absorbed by the charge battery in a two battery system, but some of it gets back to the primary as well. You can also see this on the scope with a sniffer coil on each of the battery leads. It appears as a dampened high frequency oscillation. The one on the primary leads is inverted from the one on the secondary leads.The 2x 12V circuit from Ref 2
From the forum here I understood, that not only the output battery is charged, but the input battery receives some charge too from magnetic field collapse. Combining that with my statement 1.2 (assuming it is true):
2.1 This would mean that the high voltage spike travels in 2 directions (in order to charge both batteries through the positive terminals): in both directions away from the main coil.
No. Image 1 is incorrect. Ref 2 is correct. The arrows should all be going clockwise in your Image 1.2.2 To stay in line with the arrow indications of Ref 1, and assuming my other statements are true, then the arrows in Ref 2 should look like I indicated them in Image 1 in the attachment.
No, In the ? area the current is flowing left to right which is the clockwise direction of the electron flow during coil discharge. The radiant spikes are in addition to normal electron flow and are not all in the same direction.2.3 If 2.2 is correct, that brings me to the question what happens in the area that I indicated with a “?”, because I don’t know which direction the arrow is pointing there anymore .
2.3.1 I was thinking maybe it is not going in any direction there, since the charging can take place “along 1 wire”, see Ref 1. And so to say “nothing” happens in this area, because the spike is completely
For what it's worth, I tried replicating the Shawnee Baughman toy motor and couldn't get it to run more than a few minutes before it would blow the transistor. After three or four attempts, I gave up and made it into a two battery system with better transistors, diodes, and a neon across the transistor. It works great this way and charges very well for such a small device.
Gary Hammond,
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- Hi Rodolphe,
Because the Primary battery is connected to the top of the coil, and the bottom of the coil is connected through the output diode to the pos of the charge battery, the resultant ringing coming from the coil turn off will also back feed the primary with the opposite polarity then the charge battery is receiving, but not as high a spike/ring as the charge battery receives. Thus as Gary pointed out that the Ions in the primary battery ether slow down, or are stopped from the discharge mode and thus the primary battery does not discharge as fast...
in a back popping system, if you hit the primary battery with a big enough discharge to stop and reverse the ion flow in between the SG/SSG transistor discharge pulses from the primary battery, and put the primary battery into charge mode for a bit, it will gain in voltage and not run down.... the Watson Machine did this...... - you Have to backpop the primary battery in between the SG/SSG or what ever else motor/energizer discharge pulses and isolate the Primary battery from the drive circuit with a DPDT relay, or a quad mosfet pulser circuit acting as a DPDT circuit with a big cap powering / capturing the motor/energizer I/O during the backpop duration for this to work good, as the motor/energizer at high RPM's may need to fire a few times during the backpop duration...
Last edited by RS_; 06-17-2020, 08:49 PM. Reason: fixed the neg / pos output battery mistake as noted belowComment
- Hi Rodolphe,
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Hi Gary, RS,
Many thanks for your feedback!
I hope you guys were not under the illusion that I would understand it all just at once and have no more questions, because in that case I’m afraid I have to disappoint severely haha
Starting with your feedback Gary,
The 9V circuit from Ref 1
3. Please do correct me when I’m wrong again, but reading your replies to 1.1 and 1.2, I have the impression that we’re talking about slightly different things:
I do understand that in the 9V circuit the battery is drained during the “run” cycle, and that the current flow in the run cycle is flowing from the negative pole to the positive pole of the 9V battery as indicated on page 21 (I updated the attachment and added it as Ref 4). And I understand too the part that you’re referring to, explaining that because of the high voltage-/radiant spike, the battery runs longer than normal because of the discharge process being slowed down.
I’m talking however specifically about what is happening during the high voltage spike, which by my understanding is visualized and explained by Ref 1. (So not the Run cycle from Ref 4).
“When the transistor is turned off and even before the current starts to flow in the trigger winding to dissipate the energy from the collapsing magnetic field a high voltage spike, consisting of a longitudinal wave of pure potential travels from the main coil winding back to the positive terminal of the 9V battery along one wire”
So in my understanding the accompanying image shows this: the high voltage spike going from the top of the coil to the positive pole of the battery along 1 wire. And then the arrow cannot indicate the electron/current flow because as mentioned before; then the spike would be draining the battery during the high voltage spike moment, and it is charging or at least slowing the draining process at that moment (and besides I assume for a current flow there needs to be a loop instead of “one wire”).
If the above is correct, than I would argue that the arrows in Ref 1 only show the path for the radiant spike to the battery, but not the electron flow. Electron flow during the run cycle are show correctly in Ref 4.
The 2x 12V circuit from Ref 2
I intend to connect my sniffer coil to the SG tonight or tomorrow and have a look at my scope at all input and output wires.
If the top of the coil is connected to the positive pole of the input battery, and the bottom of the coil connected (via a diode) to the positive pole of the output battery (as in Ref 2):
4. How does the radiant spike knows to/can travel in both directions at once?
Maybe if I compare the radiant spike to a magnet instead of current/electricity I can understand it somewhat: If I put a strong magnet in the center of an iron strip, the magnet can attract iron at both ends of the strip. Exactly this is what I intended to visualize in the attachment with Image 1. Only after the radiant spike took place, the current starts to flow as shown in Ref 2.
5. How does the radiant spike know to be absorbed mainly by the output battery instead of divided equally over both input and output battery (assuming it runs in both directions at once as stated in 4).
6. Since the positive pole of the input battery and the negative of the output battery are connected, how does the radiant spike know to stop at the positive pole of the input battery, and not “shoot through” to the negative pole of the output battery, causing an opposite-/draining effect there. (assuming that if a radiant spike connected to the positive terminal of a battery causes a “charging” effect, then I would assume the opposite to be true too: a radiant spike connected to a negative terminal causing a depleting effect).
Regarding the feedback from RS:
I assume you meant here:Because the Primary battery is connected to the top of the coil, and the bottom of the coil is connected through the output diode to the neg of the charge battery
“Because the Primary battery is connected to the top of the coil, and the bottom of the coil is connected through the output diode to the POS of the charge battery”?
I think I have the same questions here as I wrote to Gary under “The 2x 12V circuit from Ref 2” but reading your story, I have the following additional questions:
In your story I get the impression that only because of the “resultant ringing” the input battery also receive a spike.
7. Why does the primary not receive the radiant spike regardless off the output battery / resultant ringing: The connection from the top of the coil to the input battery in the 2x 12V circuit is identical to the 9V circuit. Or does impedance play a role here? That the spike “discharges” to the location with the lowest impedance primarily? But if that was true, that would only be valid as long as the output battery is lower in voltage than the input battery. (While writing this I’m guessing a battery with a lower voltage level has a lower impedance). Or is the impedance also influenced by the fact that the input battery is part of the time discharging (causing the machine to run), in other words; does the internal “ion-/current flow” inside the battery because of the discharging cause extra impedance, overruling the lower voltage the input battery has at a certain stage compared to the output battery?
8. I’m not really familiar with ringing, just did a quick read on Wikipedia. What I understand of it here in connection with a coil is that it is an oscillation that damps out, meaning the current switching back and forth for a while. What I don’t understand here is how the radiant spike stands in relation to the direction of the current (because of the ringing) because of the following:
8.1. As stated in Ref 1, the radiant spike (in theory) takes place before the current starts to flow, giving me the impression that it is not dependent on the direction of the current (maybe related to the impedance as questioned in 6?). Or do I draw the wrong conclusion here, meaning that although the radiant spike does take place (in theory) before the current starts to flow, its direction is dependent on the voltage polarity, like the current is. (that would make my analogy with the magnet in 4. Invalid, and my impedance story in 7 too I guess), but would raise my question from 7 even more.
8.2. How can ringing take place here at all?: the diode to the secondary battery only permits current in 1 direction.
The whole back popping story I’m unfamiliar with so far, although I understand a hint of it from what you wrote, but not hoping to turn this post into a small booklet called “Rodolphe’s never ending questions regarding the radiant spike”, let me stop here (for now) .
Guys, again, many thanks for ploughing through with me here.…
Best regards,
Rodolphe
2020-06-17 attachment V2.pdfComment
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Hi Rodolphe,
You're getting into some theory that I don't fully understand. In the quote from Peter and in your explanation of fig 1, I get the idea that the "high voltage spike" is traveling thru a wire in a certain direction. Voltage doesn't travel. Only current travels. And when current travels there is an accompanying voltage drop, specifically V=I2R . But when pure voltage appears as a longitudinal wave of pure potential traveling thru the either, it is my understanding that it is everywhere at once ...... or nearly so. It forms dipoles between the source of the disturbance in the either and any nearby conducting materials. It is this dipole that then supplies the voltage to drive some current thru a conductor in a return direction. This current stops as soon as the dipole is depleted/neutralized. This is the radiant event. This is similar to the visible corona given off from very high voltage sources. Similar, but not the same.I’m talking however specifically about what is happening during the high voltage spike, which by my understanding is visualized and explained by Ref 1. (So not the Run cycle from Ref 4).
“When the transistor is turned off and even before the current starts to flow in the trigger winding to dissipate the energy from the collapsing magnetic field a high voltage spike, consisting of a longitudinal wave of pure potential travels from the main coil winding back to the positive terminal of the 9V battery along one wire”
So in my understanding the accompanying image shows this: the high voltage spike going from the top of the coil to the positive pole of the battery along 1 wire. And then the arrow cannot indicate the electron/current flow because as mentioned before; then the spike would be draining the battery during the high voltage spike moment, and it is charging or at least slowing the draining process at that moment (and besides I assume for a current flow there needs to be a loop instead of “one wire”).
If the above is correct, than I would argue that the arrows in Ref 1 only show the path for the radiant spike to the battery, but not the electron flow. Electron flow during the run cycle are show correctly in Ref 4.
This may not be correct, but that is how I visualize it. So the radiant dipole exists everywhere at once in the either until some circuit and/or circuits form to dissipate it. When this happens, work is done and electrical charges are transported in a battery, wire, resistor, etc.
4. How does the radiant spike knows to/can travel in both directions at once?
Maybe if I compare the radiant spike to a magnet instead of current/electricity I can understand it somewhat: If I put a strong magnet in the center of an iron strip, the magnet can attract iron at both ends of the strip. Exactly this is what I intended to visualize in the attachment with Image 1. Only after the radiant spike took place, the current starts to flow as shown in Ref 2.
See my answer above. The radiant spike is pure voltage and doesn't travel. There can be several different return paths for current to dissipate the dipole.
5. How does the radiant spike know to be absorbed mainly by the output battery instead of divided equally over both input and output battery (assuming it runs in both directions at once as stated in 4).
The radiant spike doesn't travel. But, the path of least resistance to kill the dipole will determine where current flows. Sort of like lightening in a thunderstorm.
Your question assumes that the radiant spike travels thru a conductor, which is not how I understand the radiant event. Current always flows the path of least resistance to kill any dipole that exists.6. Since the positive pole of the input battery and the negative of the output battery are connected, how does the radiant spike know to stop at the positive pole of the input battery, and not “shoot through” to the negative pole of the output battery, causing an opposite-/draining effect there. (assuming that if a radiant spike connected to the positive terminal of a battery causes a “charging” effect, then I would assume the opposite to be true too: a radiant spike connected to a negative terminal causing a depleting effect).
I'll let RS_ answer the rest of your questions.
Gary Hammond,
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its about the polarity of the coil's applied voltage and current flow / magnetic field, so that when the mag field collapse's the radiant voltage event (almost no current) goes through the output diode into the charge battery, and the 2nd neg polarity ring goes into the primary battery w/ more current at a much lower voltage.
"Pure voltage appears as a longitudinal wave of pure potential traveling thru the either, it is my understanding that it is everywhere at once......"
"The radiant spike is pure voltage and doesn't travel. There can be several different return paths for current to dissipate the dipole."
according to Tesla / Bearden, and most ele text books, Voltage is instantaneous, Current lags the voltage, and that lag time is dependent on the inductance of the coil, and the resistance of all of the conductors / components in the circuit closed loop path's.
The instantaneous voltage warps all of the ether around the wires and the entire circuit into a spiral that runs parallel to the wires, and is called the Heavyside flow in the ether, creating a dipole in the ether. Voltage does Not Travel through the ether, or down a wire, it instantaneously warps the ether into a spiral Dipole around the circuit wires and components. A Dipole is kinda like a high pressure system vs a low pressure system in our atmosphere, high pressure flows toward the low pressure and the pressure eventually equalizes, killing the dipole. Most High and low pressure systems are spirals in their air flow's.
It take's a while (nanosec's) for that spiral warped ether to cause electron current flow thru the closed loop circuit. What the wires and the circuit captures of the Heavyside flow in the ether is called Polyting (sp) electron current flow in the wires / circuit, and is a small fraction of the Heavyside flow in the rest of the spiral warped ether around the wires / circuit. When the current flow finely happens, it kills (equalizes) the spiral dipole in the ether......
The Tesla idea here is, that if you can can turn on the voltage (either warp) long enough to transfer that voltage potential to something (Cap / Battery) before current flows, and turn it off really fast, it transfers that Voltage Charge Potential to the target before current flow happens, Then very little to no energy in current flow is expended as E/R during current flow like in a normal circuit. The coil collapse radiant high voltage spike does this under the correct conditions as in JB's various circuits. and you end up with a higher Voltage Charge Potential in the target. You just have to expend a bit of E/R energy to trigger the Voltage Charge Potential to be created, and transferred to the target.
Hope this helps
It's late, will try to get back to this tomorrow.....
Last edited by RS_; 06-17-2020, 10:48 PM.Comment
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Thanks guys for your extensive answers!
I'm going to ponder on it a bit, read the theory in the first manual again, and then report back here, hoping it all starts to make more sense...
Many thanks again!
Best regards,
RodolpheComment
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It would help you to read certain chapters in Bearden's book Energy From The Vacuum in understanding what i posted, and learning a lot beyond the info in my post.Comment
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Hi RS,
I have the book on the shelf, but it is too thick to pick it up right now, that would mean I have to stop the experiments with the SG for a long time to plough through that one. In case there are a couple of chapters which you recommend specifically, I could have a look if I can manage to go through them....
I did read a little while ago the Bearden/Bedini book "Free Energy generator" and Tom Bearden's paper "TOWARD A NEW ELECTROMAGNETICS PART 4: VECTORS AND MECHANISMS CLARIFIED". I actually want to read at least the book again, but not quite sure if the book gave more specific information about whats' happening during the radiant spike as the handbook.
Anyhow, will read through your and Gary's messages above again and the explanation in the hanbook too and see if it will get any clearer. Will report back here on of these days (probably with more questions hahaha).
Best regards,
Rodolphe
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Hi Gary, RS,
I’ve been re-reading a lot of stuff the last week (beginners handbook, first chapters of the Bearden/Bedini book Free Energy Generation (FEG) ,earlier responses of Gary on this thread to my questions, the answers you gave me here last week). Things are getting slightly clearer I have to say. Hope can help me again to clear things up even further.
The signal on the scope over a coil and the output battery
From post #197 to #202 I talked with Gary about the scope signal. In regards with what we discussed here above, I have the following question (see attachment for scope signal/scope hookup):
The “charging” of the output battery consists of 2 steps:
Step 1 -> Moment “X” where the radiant spike takes place This is the part where the there is (almost) no current involved. 300+V spiked, where the ions in the battery get “hammered”.
Step 2.1 -> Section “C”, where the remaining coil energy discharges through ‘regular’ flow of electricity/current.
(Step 2.2 -> Section “C” A radiant spike also gets feedback to the primary battery during this period, will come back to this step later in my story)
(Step 4 ->Then after section “C”, when the voltage drops below the level that can still charge the output battery, the remaining coil collapse energy is dissipated in the trigger circuit. This is energy that is really lost/not recovered.)
1.1 Do I understand/formulate the above correctly?
1.2 What determines the voltage level at which section “C” takes place/starts (+/- 18V in my scope reading)?
1.3 What determines the peak of the radiant spike at “X” (+/- -50V in my scope reading)? I understood that the peaks can be 300V+, so is other words: why is only the 300V-50V section absorbed, and not more, like from 300V-14V? Is this because the “pressure difference” below 50V in comparison to the battery voltage is not high enough anymore to be absorbed?
The signal on the scope over the input battery and the radiant spike to the input battery
(see also attachment with scope reading of the spike over input battery)
By having a closer look at the circuit of the 12V 2-battery system and reading both of your comments from last week, I do understand better why the output battery receives the majority of the radiant spike.
2. What I still struggle with, is understanding why the input battery receives some of the spike too. Since the minus terminal of the input battery is not connected as long as the transistor I closed, which is during the radiant spike event. From your comments I understood that this has a relation with the “coil ringing” and I realized that based on the very limited information I found on the internet about this, I actually do not have a clear idea what this coil ringing is and what effect that it has in such a manner that the primary battery is able to receive some of the radiant spike too. Does this ringing quickly/briefly switches the transistor open again?
Could you explain this a bit more please?
@ RS, I would like to go a bit deeper into the last alinea of you post #263 in combination to what I’ve read in the book FEG, but first I like to have a clearer idea of the above mentioned, will come back to this at a later stage/post.
Many thanks in advance to both of you,
best regards,
Rodolphe
2020-06-27 Attachment.pdf
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Hi Rodolphe,
That is correct, or at least how I understand what's happening.The “charging” of the output battery consists of 2 steps:
Step 1 -> Moment “X” where the radiant spike takes place This is the part where the there is (almost) no current involved. 300+V spiked, where the ions in the battery get “hammered”.
Step 2.1 -> Section “C”, where the remaining coil energy discharges through ‘regular’ flow of electricity/current.
The radiant spike also hits the primary battery at moment X in your PDF scope trace. If you expand this event out on your scope, you'll see that the spike is a very fast, very brief ringing that takes place at the charge battery voltage. The positive peak of the spike hammers the charge battery, and the negative peak of the spike hammers the primary (run) battery. Here is an expanded scope shot of mine that shows this event on the blue channel 1 and the concurrent event picked up by my sniffer coil on the yellow channel 2. The sniffer coil is picking up it's signal from the charge battery positive lead and shows the battery ringing at 853.66 khz. This is where the magic happens.(Step 2.2 -> Section “C” A radiant spike also gets feedback to the primary battery during this period, will come back to this step later in my story)
And if you instead place the sniffer coil on the primary battery positive lead you'll see the same ringing but smaller and inverted compared to the charge battery.
One thing to note is that I have the channel 1 leads reversed from yours. That makes it inverted from what you show in your PDF.
Here's another scope shot taken at the same time that shows several successive discharges.
This is not quite correct. If you place channel 2 across the trigger winding (no sniffer coil used here) you will see the exact same wave form as channel 1, but inverted. Since the trigger circuit is so much higher impedance than the power circuit, it dissipates very little of the coil collapse energy. ...................... After section C, section A repeats. Section A is the sinusoidal wave being generated by the magnets going past the coil during that time interval after the coil discharge has taken place and before the transistor is turned back on.(Step 4 ->Then after section “C”, when the voltage drops below the level that can still charge the output battery, the remaining coil collapse energy is dissipated in the trigger circuit. This is energy that is really lost/not recovered.)
I think this would be the charge battery voltage plus whatever voltage drop exists in the charge battery circuit at the time. That voltage drop would consist of the .6 volt forward voltage drop across the diode and any I2R in the wiring.1.2 What determines the voltage level at which section “C” takes place/starts (+/- 18V in my scope reading)?
The total impedance of the discharge circuit determines how high the voltage peak will rise. The lower the impedance, the lower the peak voltage. Whatever is not absorbed by the battery is absorbed by the rest of the discharge circuit. So a large healthy battery and big wiring will keep the peak relatively low. Remember that this is a very fast ringing "spike" that the lagging current can't keep up with.1.3 What determines the peak of the radiant spike at “X” (+/- -50V in my scope reading)? I understood that the peaks can be 300V+, so is other words: why is only the 300V-50V section absorbed, and not more, like from 300V-14V? Is this because the “pressure difference” below 50V in comparison to the battery voltage is not high enough anymore to be absorbed?
Not sure I can explain this, as I don't fully understand it either. You are thinking in terms of current flow in a closed loop circuit through a wire. But it's my understanding that the voltage spikes hitting both batteries is the nearly instantaneous transverse wave outside the wire known as the Heavyside flow. It causes pointing flow and lamalar (sp) currents inside the conductors. I don't think the transistor is switched on but instead allows heavyside flow around it. Maybe RS_ has a better explanation?2. What I still struggle with, is understanding why the input battery receives some of the spike too. Since the minus terminal of the input battery is not connected as long as the transistor I closed, which is during the radiant spike event. From your comments I understood that this has a relation with the “coil ringing” and I realized that based on the very limited information I found on the internet about this, I actually do not have a clear idea what this coil ringing is and what effect that it has in such a manner that the primary battery is able to receive some of the radiant spike too. Does this ringing quickly/briefly switches the transistor open again?
Could you explain this a bit more please?
Gary Hammond,
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Hey Gary and RS,
Greatly appreciate the detail that both of you have layed out with respect to Rodolphe's questions. The mental stimulation here is excellent!!
Yaro
Yaro
"The Universe is under no obligation to make sense to you." -Neil Degrasse TysonComment
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Hi Rodolphe, Gary, Yaro,
I am working on a Reply.....
I have had a major sinus infection with pounding headache's off and on for weeks...... Got 2nd round of different antibiotics and a new nasal spray from the doc yesterday....
Make's it hard to concentrate on heavy duty topics like this subject.
it might be a day or more before i get this wrote upComment
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