Use for the Tesla Switch

[nvisser]

After a lengthy rework of my PCB, I was able to test this out. While in theory the cap should charge to the total of the series batts (it does). It also provides a "pulse" to the parallel batts. However, each time the cap charges, it consumes more current that it can restore. End result, Dead Batts.

Bit's

I guess a smaller cap like the two that JB uses in "The small solar charger" would charge up much faster so that you can discharge every 500msec into the battery.
Thanks for trying. I get so frustrated without the pic. Got no idea how long it will take to get here from the UK.
I`m busy putting a pc board and cable together for it.

[nvisser]

Like the scaler charger

Maybe the 2 caps should also be charged in parralel to 24v and then switch in series before discharging the 48v into the parralel 12v batteries. It will be a lot a switching!!

[JohnStone]

Hy - seekers of the truth
Searchting for the "truth" as well I just red the following explanation by T. Bearden regarding JB's lead acid negative resistor - again and again:
Bearden on Bedinis' Negative Resistance Effect - 04/09/00
I wonder if this might be the missing link.
He explains the "mechanics" of the negative resistor - Bedini type - for us:
Some quotations:
"First, to understand John's work one must be aware that there are several currents in a lead acid battery…heavy lead-type ion current in the battery, and the electron current in the battery…
Suppose we "hit" a battery's terminals with an instantaneous leading edge rise ...
During that "ion response lag" time, the electrons continue to furiously rush in and pile-up on the plates…
So there is obviously a hysteresis (time delay) in the response of the massive ion current…
In short, the delay can be manipulated to freely "regauge" the system…
But that increased potential at the pileup is actually a change to the ambient potential of the vacuum…
This is the way that John creates a negative resistor directly inside a lead acid storage battery….
Then we deliberately cut off the pulse sharply,….. produces a very interesting effect here also…
…system suddenly and remarkably increases the negative resistor...
There are several other schemes that can be used at this point.....
....."
Please study the link!
That might explain why every battery and load condition needs to be tuned to a specific switching, frequency and delay and why JB spend tons of time in order to search for a specific reaction of the batteries. TB explains that the battery itself forms the negative resistor as reaction of proper driving.
Additionally there seems to be some kind of resonant effect in the battery - not because of inductance and capacitance - but because of ion mass and potential and inrush current as well. That explains why the negative resistor can be initiated by SSG or discharging capacitors or tesla switch or additional intermediate pulses .... but well tuned!

What is your opinion regarding TB's explanation? Will it help to understand?

JohnStone

[JANGYD]

Thanks

If you are capturing the scope shots while dumping a single capacitor, I would not expect to see the upward movement. If you had two capacitors in parallel, (like the scalar charger) or two batteries in parallel and switching them into a series configuration, I'd expect to see the upward movement first, because you are putting the devices in series when you trigger the semiconductor. The upward movement is when the semiconductor device is turned "on" and potential is moving from collector to emitter through the device. (The positive of the lower cap/battery is more positive in potential than negative of the upper cap/battery.) The difference in potential between the capacitors (or batteries) is causing a move in potential across the device as the positive (lower) and negative (upper) are brought together, i.e. bringing the negative of the upper cap/battery above the lower cap/battery positive.

You do not have this (I think) with a single capacitor discharge, it is just straight down from the current level if I am correct on what you are doing.



Charges (I did not say current) always flow from a more positive to a more negative potential, so I'd expect it to flow "backward" through the device after the "connection" is made from the lower positive potential to the upper negative potential. Once the device is switched "on", you now have two objects (caps or batteries) in series and the only way for this to look to the device is that the upper negative is more positive than lower positive. So I guess in reality, it must "go negative" and allow charges to flow in the opposite direction or the device would just "shut itself off" and you'd lose the "series" connection on the batteries and/or capacitors.



I think I've finally "got it", at least I hope I have.

I only have two transistors on each side right now, one to do the series switching and one to go across on the top. I am looking over the series transistor.

I believe, based on JBs two transistor TS, (with more diodes) that that is the only device that goes "negative", because of the switching of a more positive collector to a more negative emitter. And, it is not doing it when there is "no" signal on the transistor. The device went negative even when there was a signal applied to the base which is different than what Naudin was saying, i.e. that there is not a signal attached to the base of the transistor. Naudin stuff was running in 2nd emitter breakdown, JBs is just running backward...."normally".

Leroy

P.S. I've got some more testing to do....

*EDIT* The going negative is the reason, you can actually see a potential of the two batteries in series when the transistor is switched on. Normally there would be a big drop over that transistor, which would mean that only 12 + 12 - 4V (tranny drop) = 20 volts is possible on the series side, but I've seen the entire thing plus a little on occasion...When it goes negative, you get the entire thing and maybe a little more depending on how the switching is done which should not "normally" be possible. Maybe the trick is that when it goes negative, to the max point and then starts to come back up, you need to shut the transistor off you need to take the signal to the base away.

Thanks Leroy,

You make me encouraged and save my money and time. Thanks again.
Unfortunately, I don't have time to experiment this week, New Year's Day in the lunar calendar.

I hope I get a working system in my house until this coming cold winter.

My purpose of this research is the last phrase on JB's post #597

Regards,

JANG

[ldissing]

...My purpose of this research is the last phrase on JB's post #597

597 or 598. I think you mean 598?

Leroy

[JANGYD]

597 or 598. I think you mean 598?
Leroy

598 is good and 597 is better

JANG

[John_K]

Tesla Switch

Hi Folks,

I (too) was going through JB's posts and reviewing some old stuff.

@Bit's (mainly, but anyone can chime in here),

JB said it was very important that we had a diode (1N4001) and monolithic (1uF) or electrolytic (2.2uF) on each of the optos (H11D1). Even the photos JB posted of his TS prototype showed them on his board. On your schematics Bit's you only show them on Q1 & Q2. Are they also on the Q9 & Q10 optos as well, or have you been able to get your D-TS working without them?

JB also posted very early on that to get the TS to work, it has to mimic the cap pulser circuit. Qutoing JB:

"The Tesla switch is something much different in the way charging is performed. The switch must be set to perform the same function as a capacitor discharge system allowing the differential across the load. So if the system is 12 volts each battery you must dump 24 volts across 12 volts very quick and cut off before things boil. 24 volts across three collector emitter drops + the diode leaves about 18 volts or less depending on junction loss, now the load must leave the differential to charge that 12 volt battery so you do not have much left. "

One thing that has been bugging me Bit's, is that in your very informative "Working D-TS-1" video, you state that increasing the pulse time actually raises the voltage being dumped to the 12V side. I'm not arguing, but could it be that our meters are just too slow to see the pulse fi the pulse time is shortened?

I've noticed if I have an analog volt meter placed across B1 & B3 (or B2 & B4), when it switches the needle flies up to about 28V, before bouncing back to about 24V. Then when the pulse is done, the needle bounces to about -5V. I'm not sure if this is a characteristic of analog meters, but a digital meter does not show this - even ones with a bar meter like in Bit's video.

I added diodes and caps on my Q9 & Q10, but am yet to see much difference. I'm contemplating adding the 5th and 6th transistors to make a more contemporary TS as well.

Another thing I though of - JB also states that the transistors should be fired (biassed) at the same time. I know Bit's has has success with firing Q1 before Q10, but is this not wasting energy with the voltage drop and associated heat being generated by the transistors and the load?

Just some things to think about, which I'll figure out on the bench - but I thought I'd share them with the group anyways.

Keep tinkerin'


John K.

[ausev]

Try caps on your relay circuit?

Hi Matt,

I am a complete newbie here, I want to power an electric car with the tesla switch circuit.

Congratulations on getting this solid state version going.

I have been studying this forum for a while now and you seem to have come up with the best solution so far.

I would like to replicate this and have purchased the parts I can see in your pictures but the wiring is hard to work out.

Could you please post a circuit diagram of your solution for me so I can give this a go?

Regards,

AusEv

Hi Matt,

I have been studying John Bedini's circuit diagrams and trying to work out your wiring from the pictures you posted but not making much headway there. Would really appreciate even just a pencil sketch of the wiring layout you are using with your solid state relays.

I got 6 x HFS15's for mine and 4 x 12V 6AH batteries for my replication project.

I was also wondering if you have considered putting caps accross the inputs of the relays to increase the on time of the switch to see how that affects the circuit like this:



You could try different caps to see if that improves the efficiency. If anything it would give you another variable to control within the circuit. I have a theory that the timing of the switching is what is making this circuit work. I will be trying this myself once I can replicate your circuit.

Also from the pictures I cannot see any diodes in your circuit yet they are all over John's circuit diagram. Am I just not seeing them there?

Regards,

AusEv

[Matthew Jones]

@AusEv

I have not had good luck charging batteries with the relays. It is efficient for running a load but it does not do the same thing a Bits setup which will charge batteries to a point. Mine cannot self power itself and overcome the load of driving the relays. It will power a load about 12 times longer if switch at 1hz.
The relays are also very temperamental. They cook at the drop of a hat. At 30 dollars a pop, that hurts.

But if you want the setup here it is. Its the wiring from the Free Energy Guide, modified.

Matt

[Matthew Jones]

Iddising wrote...*EDIT* The going negative is the reason, you can actually see a potential of the two batteries in series when the transistor is switched on. Normally there would be a big drop over that transistor, which would mean that only 12 + 12 - 4V (tranny drop) = 20 volts is possible on the series side, but I've seen the entire thing plus a little on occasion...When it goes negative, you get the entire thing and maybe a little more depending on how the switching is done which should not "normally" be possible. Maybe the trick is that when it goes negative, to the max point and then starts to come back up, you need to shut the transistor off you need to take the signal to the base away.

One funny thing I am seeing. Like you are. The entire voltage shows up immediately, I have checked this on my scope, then the resistance shows up and you get a voltage drop.
But the funny thing is if you wire multiple transistors (MJL's) you get only the same voltage drop.
In my case the 2 batteries in series equaled 25.74 volt.

Just switching into series and discharging into a cap allowed for 24.87 volts to hit the cap. 1 transistor firing. The scope showed the 25.74 spike.
But firing 4 transistors as serial switch gave me the exact same voltage. Whats that about??? Its probably somthing simple.

So I was reading and found this article about gunn diodes and there negative resistance.
If you read it you'll see the gunn diode has to be wound up.

Maybe that the trick. Wined up the MJL's at its natural resonates, (Like JLNaudins setup) Then use that transistor to fire the serial connection, while its at its peak.
It shouldn't be that hard to setup.

Also I can't find it in the thread cause it was a while ago but I remember Mr John saying to fire the The transistor he was attempting to use a coil and cap. To get it to fire correctly???. He also gave me that Naudin link and said we'll get the idea from it.

I don't know might be worth looking at.

Matt

[wings]

Hy - seekers of the truth
Searchting for the "truth" as well I just red the following explanation by T. Bearden regarding JB's lead acid negative resistor - again and again:
Bearden on Bedinis' Negative Resistance Effect - 04/09/00
I wonder if this might be the missing link.
He explains the "mechanics" of the negative resistor - Bedini type - for us:
Some quotations:
"First, to understand John's work one must be aware that there are several currents in a lead acid battery…heavy lead-type ion current in the battery, and the electron current in the battery…
Suppose we "hit" a battery's terminals with an instantaneous leading edge rise ...
During that "ion response lag" time, the electrons continue to furiously rush in and pile-up on the plates…
So there is obviously a hysteresis (time delay) in the response of the massive ion current…
In short, the delay can be manipulated to freely "regauge" the system…
But that increased potential at the pileup is actually a change to the ambient potential of the vacuum…
This is the way that John creates a negative resistor directly inside a lead acid storage battery….
Then we deliberately cut off the pulse sharply,….. produces a very interesting effect here also…
…system suddenly and remarkably increases the negative resistor...
There are several other schemes that can be used at this point.....
....."
Please study the link!
That might explain why every battery and load condition needs to be tuned to a specific switching, frequency and delay and why JB spend tons of time in order to search for a specific reaction of the batteries. TB explains that the battery itself forms the negative resistor as reaction of proper driving.
Additionally there seems to be some kind of resonant effect in the battery - not because of inductance and capacitance - but because of ion mass and potential and inrush current as well. That explains why the negative resistor can be initiated by SSG or discharging capacitors or tesla switch or additional intermediate pulses .... but well tuned!

What is your opinion regarding TB's explanation? Will it help to understand?

JohnStone

nice picthure here page 5-9:

http://www.free-energy-info.co.uk/Chapter5.pdf

[Bit's-n-Bytes]

Hi Folks,

I (too) was going through JB's posts and reviewing some old stuff.

@Bit's (mainly, but anyone can chime in here),

JB said it was very important that we had a diode (1N4001) and monolithic (1uF) or electrolytic (2.2uF) on each of the optos (H11D1). Even the photos JB posted of his TS prototype showed them on his board. On your schematics Bit's you only show them on Q1 & Q2. Are they also on the Q9 & Q10 optos as well, or have you been able to get your D-TS working without them?I did not use them on Q9, Q10, but it does not seem to make a difference, IMO

JB also posted very early on that to get the TS to work, it has to mimic the cap pulser circuit. Qutoing JB:

"The Tesla switch is something much different in the way charging is performed. The switch must be set to perform the same function as a capacitor discharge system allowing the differential across the load. So if the system is 12 volts each battery you must dump 24 volts across 12 volts very quick and cut off before things boil. 24 volts across three collector emitter drops + the diode leaves about 18 volts or less depending on junction loss, now the load must leave the differential to charge that 12 volt battery so you do not have much left. "

One thing that has been bugging me Bit's, is that in your very informative "Working D-TS-1" video, you state that increasing the pulse time actually raises the voltage being dumped to the 12V side. I'm not arguing, but could it be that our meters are just too slow to see the pulse fi the pulse time is shortened? I have experminted with many pulse times, loads, configurations, etc. The pulse times, when shortend worked in some cases (lighter load) where I found the best rate at 800ms for the 1157 bulbs. While this was the best rate, it only maintained the batts. I am still searching for the magic.

I've noticed if I have an analog volt meter placed across B1 & B3 (or B2 & B4), when it switches the needle flies up to about 28V, before bouncing back to about 24V. Then when the pulse is done, the needle bounces to about -5V. I'm not sure if this is a characteristic of analog meters, but a digital meter does not show this - even ones with a bar meter like in Bit's video.I belive your right on the meters. DMM are slow due to the sampling rate, but don't be fooled by the meter swing on the analog meter. Yes it is impressive but momentum from the galvometer carry's it past what the actual voltage is. Unless you have a spike or cap surge, when you series up the batts, the voltage will be V1 + V2 - loss.

I added diodes and caps on my Q9 & Q10, but am yet to see much difference. I'm contemplating adding the 5th and 6th transistors to make a more contemporary TS as well.

Another thing I though of - JB also states that the transistors should be fired (biassed) at the same time. I know Bit's has has success with firing Q1 before Q10, but is this not wasting energy with the voltage drop and associated heat being generated by the transistors and the load?

Just some things to think about, which I'll figure out on the bench - but I thought I'd share them with the group anyways.

Keep tinkerin'


John K.

Commented above in red.

Thanks.

Bit's

[Matthew Jones]

@Bits

In your circuit, I am trying to find the point at which we can find the potential after the load is done with it. The potential left over to hit the battery (IE 14.00 +- volt).
(Version 2.2.6)
I am checking after Q10 before D3 and D7 with the hot probe and grounding at batt 4 ground. When the direction of the circiut allows.

Is this correct?

I am not getting the correct voltage from that point. After D3 and D7 open I only get battery voltage. of the charging battery. It continues to go up while that side is being charged.

So how are we supposed to know its working?

Thanks
Matt

[Bit's-n-Bytes]

@Bits

In your circuit, I am trying to find the point at which we can find the potential after the load is done with it. The potential left over to hit the battery (IE 14.00 +- volt).
(Version 2.2.6)
I am checking after Q10 before D3 and D7 with the hot probe and grounding at batt 4 ground. When the direction of the circiut allows.

Is this correct?

I am not getting the correct voltage from that point. After D3 and D7 open I only get battery voltage. of the charging battery. It continues to go up while that side is being charged.

So how are we supposed to know its working?

Thanks
Matt

First;
What do have installed for the values of R7 and R8? I think that I had to change them to 50 Ohms to allow for the tranny to turn fully on.

Second;
What load are you using? The 1157's?

Also, what is your pulse rate?

Thanks

Bit's

[Matthew Jones]

First;
What do have installed for the values of R7 and R8? I think that I had to change them to 50 Ohms to allow for the tranny to turn fully on.

Second;
What load are you using? The 1157's?

Also, what is your pulse rate?

Thanks

Bit's

I am using 47 ohm 2 watts. They work good.

Yes 1157's.

I was trying to find what voltage was available after the load.

Is this going to turn into a math problem? LOL

Thanks
Matt