Use for the Tesla Switch

[JohnStone]

High Side Transistor Drivers

Hey Bits & nvisser,

still occupied understanding the schematic. My concern is the control of the power transistors.

Please have a look to the pspice simulations attached. The optos were replaced by bare wire in order to have a look to genuine currents and voltages.




If the simulation isn't wrong it is very obvious that the load and the voltage of the battery loaded will have an essential influence to the control of the power transistors. R7 = 1K should be definitely of lower value because it stops Q10 from switching to low voltage difference (CE). On the other hand a big load and a discharged battery J4/J6 can destroy the opto.

Independent of this the data sheet of the H11 opto (optos are current amplifiers/deminishers as transistors) shows that:
- 9mA of diode current can produce 5mA max of output current
- 60ma of diode current can produce 16mA of output current.
This might be a quite severe limitation for switching the transistors. They need more base current in order to switch high current like above 1A.

Additional you should take in account, that using a bulb as load introduces a highly variable load (cold curent approx. 10 times the hot current) having an major effect on driving the transistors (see simulations).

I remember reading somewhere, that the load should be a inductor or motor - is that true?

These effects mentioned above might overwrite and hide the expected true overunity effects.

Besides:
Designing a high side switch is a very demanding story if losses require to be low. Once again: JB knew why he introduced the transformers( Mueller Report). Mys guess is that these 8 Ohm / 1000 Ohm ratio refers to the AC impedance for audio frequencies. The DC impedandce will be much lower. My measurements at such transformers show about ca. 2 Ohm / 30 Ohm.

Please comment the simulations. I wonder if you measure similar voltages in your circuits.

[nvisser]

I add the diodes in the diagram to prevent cap discharging into the top battery through the loads when Q1 switch off.
With the 6 transistor switch this will not be needed

[nvisser]

Hey Bits & nvisser,

still occupied understanding the schematic. My concern is the control of the power transistors.

Please have a look to the pspice simulations attached. The optos were replaced by bare wire in order to have a look to genuine currents and voltages.
[ATTACH]4816[/ATTACH]
[ATTACH]4817[/ATTACH]
[ATTACH]4818[/ATTACH]
[ATTACH]4819[/ATTACH]
If the simulation isn't wrong it is very obvious that the load and the voltage of the battery loaded will have an essential influence to the control of the power transistors. R7 = 1K should be definitely of lower value because it stops Q10 from switching to low voltage difference (CE). On the other hand a big load and a discharged battery J4/J6 can destroy the opto.

Independent of this the data sheet of the H11 opto (optos are current amplifiers/deminishers as transistors) shows that:
- 9mA of diode current can produce 5mA max of output current
- 60ma of diode current can produce 16mA of output current.
This might be a quite severe limitation for switching the transistors. They need more base current in order to switch high current like above 1A.

Additional you should take in account, that using a bulb as load introduces a highly variable load (cold curent approx. 10 times the hot current) having an major effect on driving the transistors (see simulations).

I remember reading somewhere, that the load should be a inductor or motor - is that true?

These effects mentioned above might overwrite and hide the expected true overunity effects.

Besides:
Designing a high side switch is a very demanding story if losses require to be low. Once again: JB knew why he introduced the transformers( Mueller Report). Mys guess is that these 8 Ohm / 1000 Ohm ratio refers to the AC impedance for audio frequencies. The DC impedandce will be much lower. My measurements at such transformers show about ca. 2 Ohm / 30 Ohm.

Please comment the simulations. I wonder if you measure similar voltages in your circuits.

John Stone The attachments does not open.
The main circuit as you see it was mostly given to the thread by John Bedini and was already proved to work without problems

[Bit's-n-Bytes]

Hey Bits & nvisser,

still occupied understanding the schematic. My concern is the control of the power transistors.

Please have a look to the pspice simulations attached. The optos were replaced by bare wire in order to have a look to genuine currents and voltages.
[ATTACH]4816[/ATTACH]
[ATTACH]4817[/ATTACH]
[ATTACH]4818[/ATTACH]
[ATTACH]4819[/ATTACH]
If the simulation isn't wrong it is very obvious that the load and the voltage of the battery loaded will have an essential influence to the control of the power transistors. R7 = 1K should be definitely of lower value because it stops Q10 from switching to low voltage difference (CE). On the other hand a big load and a discharged battery J4/J6 can destroy the opto.

Independent of this the data sheet of the H11 opto (optos are current amplifiers/deminishers as transistors) shows that:
- 9mA of diode current can produce 5mA max of output current
- 60ma of diode current can produce 16mA of output current.
This might be a quite severe limitation for switching the transistors. They need more base current in order to switch high current like above 1A.

Additional you should take in account, that using a bulb as load introduces a highly variable load (cold curent approx. 10 times the hot current) having an major effect on driving the transistors (see simulations).

I remember reading somewhere, that the load should be a inductor or motor - is that true?

These effects mentioned above might overwrite and hide the expected true overunity effects.

Besides:
Designing a high side switch is a very demanding story if losses require to be low. Once again: JB knew why he introduced the transformers( Mueller Report). Mys guess is that these 8 Ohm / 1000 Ohm ratio refers to the AC impedance for audio frequencies. The DC impedandce will be much lower. My measurements at such transformers show about ca. 2 Ohm / 30 Ohm.

Please comment the simulations. I wonder if you measure similar voltages in your circuits.

Thanks JohnStone, I did in fact change R7 and R8 to 50 Ohms a while back. Sorry, I should have updated the group.

Bit's

[redcar1957]

'Magnacoastermotors'

'Magnacoastermotors' who is this Richard?
anybody know?

[nvisser]

Hey Bits & nvisser,
Independent of this the data sheet of the H11 opto (optos are current amplifiers/deminishers as transistors) shows that:
- 9mA of diode current can produce 5mA max of output current
- 60ma of diode current can produce 16mA of output current.
This might be a quite severe limitation for switching the transistors. They need more base current in order to switch high current like above 1A.
.

Do you think by putting another transistor in the darlington way should solve the current problem. Specially once we start to drive bigger loads from the switch.
I think 60mA is the absolute max for the diode part and not good for it.
The manufacturers of the 4n35 opto told me that 10mA is sufficient to switch a transistor on fully. They did not comment on the current passed by the driver transistor
The way I see a npn. With the load and more positive voltage on the collector and the more negative voltage (normally ground) on the emitter you just need 0.7V above the emitter voltage on the base to switch it fully one and pass whatever current the transistor can safely handle with heat sinks. I can be wrong

[JohnStone]

Sorry, I don'n know why the attachments do not open.
Try this Picasa-Webalben - Gross Family - TS simulation link.

@nvisser #2076
I agree that JB gave this schematic.
Question: Do you refer to the Mueller report? This was a low power application with well tuned transformers - an ingenious idea - but not for low frequency (i.e. seconds).
Did JB give a power-schematic as well?

I do not know if my simulation is kompletely right but it gives a deep insight in that what transitors do in this schematic. I think that if this switching is not perfect in our visible realm - the scalar energy realm will be very distant. This is what I undertood from JB.

@Bit,s #2077
In my simulation I tried 330 Ohm for R7 getting 16mA into the Base Q10 - this conforms to 50 or 60 mA diode current through the opto. This is true for 5 Ohm Load resistance.
At 100 Ohm load resistance the base current of Q10 is 2,4 mA only. This proves the enormous dependency on the load and battery voltage (J3 and J4).
This is the burden of all high side drivers living out of their switching voltage.

@nvisser #2080
Yes the Imax for the opto diode is 60 mA. at 25 °C. 40mA will give 15 mA at output - no relavant tradeoff. So I recommend to drive 40 mA (a problem for PIC!). If the opto dies - than because too high Ice in the output (empty battery and load well below 5 Ohm.

The answer from the manufacurer is not very clear. It depends ... on the real schematic.

Have a look to the transistor itself:
The MJ2119x is good up to 5 A when low loss is needed (Uce). But the amplification will reduce considerably starting with 3 A = amplification ca. 60.
15 mA will make max 1,5A Ic. For 3A you need considerably more base curent.
Now I ask the manufacturer: What do you guess ist a full switched transistor? - it depends .

But it goes worse: A high side switch can not make use of the full 24V. It is supposed to have a low Uce (12V battery and 12V load - intended). And that is the real problem here. The load will influence the base current in Q10 in a very unwanted way. At low load current the loss in the transistor is several Volts at 300 mA. Sometimes the base current of 15 mA can't be driven because of the low voltage difference. The Transistor control is highly floating. I hope you can open my pics in order to understand.

This effect will be even worse if the load is an inducor - no current possible in the very beginning. Sudden switch on is not possible.

I will try simulating PNPs (normal and darlington) being driven to ground An alternative will be to use (at least for research) extra voltage sources i.e. DC/DC converters. - I will come back whithin some days.

At the moment I have no real suggestion how to solve the problem. I detected it not long ago and started with simulation. This took some days to set it up at home (freeware - pspice student version, models for the components - I can post the schematic data for own further trials).

So far the mechanical switch seems to be superior and fast and the transformer principle as well (for lower power) because they do separate the switching from the intended effect indeed.

While my simulation corresponds to my knowledge and experience I still am a bit suspicious regarding my simulation as long I get no confirmation that your reality corresponds somehow.

Due to the time shift the Monday arrives soon here and I try to forward some more simulations ASAP if required. zzzzzzzzz

JohnStone

[ldissing]

Sorry, I don'n know why the attachments do not open.
Try this Picasa-Webalben - Gross Family - TS simulation link.

@nvisser #2076
I agree that JB gave this schematic.
Question: Do you refer to the Mueller report? This was a low power application with well tuned transformers - an ingenious idea - but not for low frequency (i.e. seconds).
Did JB give a power-schematic as well?

I do not know if my simulation is kompletely right but it gives a deep insight in that what transitors do in this schematic. I think that if this switching is not perfect in our visible realm - the scalar energy realm will be very distant. This is what I undertood from JB.

@Bit,s #2077
In my simulation I tried 330 Ohm for R7 getting 16mA into the Base Q10 - this conforms to 50 or 60 mA diode current through the opto. This is true for 5 Ohm Load resistance.
At 100 Ohm load resistance the base current of Q10 is 2,4 mA only. This proves the enormous dependency on the load and battery voltage (J3 and J4).
This is the burden of all high side drivers living out of their switching voltage.

@nvisser #2080
Yes the Imax for the opto diode is 60 mA. at 25 °C. 40mA will give 15 mA at output - no relavant tradeoff. So I recommend to drive 40 mA (a problem for PIC!). If the opto dies - than because too high Ice in the output (empty battery and load well below 5 Ohm.

The answer from the manufacurer is not very clear. It depends ... on the real schematic.

Have a look to the transistor itself:
The MJ2119x is good up to 5 A when low loss is needed (Uce). But the amplification will reduce considerably starting with 3 A = amplification ca. 60.
15 mA will make max 1,5A Ic. For 3A you need considerably more base curent.
Now I ask the manufacturer: What do you guess ist a full switched transistor? - it depends .

But it goes worse: A high side switch can not make use of the full 24V. It is supposed to have a low Uce (12V battery and 12V load - intended). And that is the real problem here. The load will influence the base current in Q10 in a very unwanted way. At low load current the loss in the transistor is several Volts at 300 mA. Sometimes the base current of 15 mA can't be driven because of the low voltage difference. The Transistor control is highly floating. I hope you can open my pics in order to understand.

This effect will be even worse if the load is an inducor - no current possible in the very beginning. Sudden switch on is not possible.

I will try simulating PNPs (normal and darlington) being driven to ground An alternative will be to use (at least for research) extra voltage sources i.e. DC/DC converters. - I will come back whithin some days.

At the moment I have no real suggestion how to solve the problem. I detected it not long ago and started with simulation. This took some days to set it up at home (freeware - pspice student version, models for the components - I can post the schematic data for own further trials).

So far the mechanical switch seems to be superior and fast and the transformer principle as well (for lower power) because they do separate the switching from the intended effect indeed.

While my simulation corresponds to my knowledge and experience I still am a bit suspicious regarding my simulation as long I get no confirmation that your reality corresponds somehow.

Due to the time shift the Monday arrives soon here and I try to forward some more simulations ASAP if required. zzzzzzzzz

JohnStone

@JohnStone,

There is a difference between driving the DIODE and driving the TRANSISTOR in that OTPO. The maximum current through the transistor is 100ma (H11D1), and is that dependent on the current through the diode?

Leroy

P.S. Good simulations though!

[ldissing]

@JohnStone,

You are correct on the transformer version. You can NOT have LONG "on" times, so how in the world did it work? The circuits shown and the loads shown do not really "fit" with that short of a time (the time to charge the cap and discharge the cap), and especially if you consider the loads used? You could not have had 1 second on times, although, you could have had delays, but the "on" times are very short, so how could you ever have had any real charging? Or at least sustained loads and constant potentials over the batteries? In order to maintain the load, the switching would have had to have been pretty fast, at least 500 times a second, which is a possiblity, but JB said, I believe, that it was 10 times a second. The "on" time for that would have still been about 1 or 2 milliseconds max. So you are talking about 1% or 2% duty cycle.

I'm at a loss to figure out what JB has been saying, I have the switching he is talking about, I have the negative going pulse, and it is bigger than the positive going pulse, but charging...I don't think so. He is a lot smarter than me though, and I won't say he is wrong, because I'm an idiot in comparison, but I wish he would come back and point us further in the right direction.

Leroy

***EDIT P.S. I'm working on the transformer version now ***

[JANGYD]

Potential Switch

@Leroy

Attached scopeshots are captured form inverted potential switch on FEG.

Thesewave forms looks like JB's drawing for the Negistor,
but I don't think these are right.

Regards,

JANG

[ldissing]

@Leroy

Attached scopeshots are captured form inverted potential switch on FEG.

Thesewave forms looks like JB's drawing for the Negistor,
but I don't think these are right.

Regards,

JANG

@JANG,

Thank you for the pictures.

It seems like you are missing the positive going spike in the waveforms. It may have been previous to the downward movement however. My scope probes (maybe the scope) are messed up, and the voltage measurements are not right (it will only cost me $500.00 to replace them...don't have the cash for that), but I can see the waveforms. I do have on the TS, an upward movement, followed by a downward movement to the negative side (below 0) which is at least 1 1/2 times (more like 2 times) the positive going movement, then back up a little, then it stays there for a short time, then back to the positive side. I'm pretty sure this is the correct waveform but all bets may be off due to the equipment. I may have to just bite the bullet and try to find the cash to buy the probes. The upward spike should happen a little before that downward movement, but the downward spike looks about right.

What does it mean? What is the correct way to utilize it? Does that downward movement mean that current is flowing backward through the transistor when no bias is present (opto triggered control and the opto is "off", so the base is floating at this moment)? I'm still trying to figure it all out, so maybe we can all work together to get this. Nobody else chimed in about how brilliant they are and how they all have it figured out, so I guess I'm in good company...in the dark!

In the dark,

Leroy

****EDIT P.S. Picture 1 and 3 look the most right to me. You said FEG book...what are you doing when you capture these pictures?****

[JANGYD]

@JANG,

Thank you for the pictures.

It seems like you are missing the positive going spike in the waveforms. It may have been previous to the downward movement however. My scope probes (maybe the scope) are messed up, and the voltage measurements are not right (it will only cost me $500.00 to replace them...don't have the cash for that), but I can see the waveforms. I do have on the TS, an upward movement, followed by a downward movement to the negative side (below 0) which is at least 1 1/2 times (more like 2 times) the positive going movement, then back up a little, then it stays there for a short time, then back to the positive side. I'm pretty sure this is the correct waveform but all bets may be off due to the equipment. I may have to just bite the bullet and try to find the cash to buy the probes. The upward spike should happen a little before that downward movement, but the downward spike looks about right.

You are correct, Leroy, that's one of reason why I'm not right.
There was no upward movement on my circuit.
I always think why?
Do you think(know) why upward movement shoot?
Is it possible to get that waveshot with only one switch?
Can you test it with only one transistor-making other 2TRs short on E-C.

What does it mean? What is the correct way to utilize it? Does that downward movement mean that current is flowing backward through the transistor when no bias is present (opto triggered control and the opto is "off", so the base is floating at this moment)? I'm still trying to figure it all out, so maybe we can all work together to get this. Nobody else chimed in about how brilliant they are and how they all have it figured out, so I guess I'm in good company...in the dark!

That's the second reason why I am not right.
There was some reverse current when the votalge went negative.
There should be no current when it goes negative.
But I don't know why the upward movement happen.
Can you show me the result if it happens with just one swithch working.


Regards,

JANG

[ldissing]

There was no upward movement on my circuit.
I always think why?

But I don't know why the upward movement happen.

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.

There was some reverse current when the votalge went negative.
There should be no current when it goes negative.

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.

Can you show me the result if it happens with just one swithch working.

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.

[Bit's-n-Bytes]

Hey Bits
I hope you got your switch functional again.
As I am still waiting for my picaxe chips I cannot test all the ideas I got
I am posting another mod on Bits diagram. I just add 2 large caps and then the switching is important.
It does not use the bipolar switch and is very easy to implement if you got a pic pulser
As Q1 switch on the cap will charge up. Q1 switch off and Q10 switch on and discharge the cap to the right hand parallel batteries.
Then Q2 switch on charge the 2nd cap.Q2 switches off and Q9 switch on and dump the cap.
That gives you no changes to the circuit. Only add the caps and of coarse the switching.
I did not specify pulse times as we have to see how long it takes to charge the cap up to 22-23V.
You still have to keep the loads to prevent the short in the bottom as there are no transistors. I don’t like the no transistors in the bottom idea as it looks to me as if we load the bottom batteries with the load or short it with no load and the bottom battery are actually driving the load when the batteries goes in series. Does anybody agrees with me on this one

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

[ldissing]

...
Can you show me the result if it happens with just one swithch working.
...

@JANG,

I have a hand drawn graph of the waveform, I'll try to get it out here for you to see soon.

The first one will be with the transformers. The transformers will not leave it in the negative state nearly as long as the opto/diode/cap version, because the "signal" can only last until the capacitor is charged in the transformer version. It decays as you would expect like in a capacitor discharge, only reverse.

The second one is with only one transistor, diodes in for the upper transistors. I've got a 12V 50W load, but I'm not trying to light it up, just see the waveform. As I increase the length of "on" time for the opto, the length on the negative side increases.

Leroy

*EDIT Graph uploaded. The lines going down could have been more vertical, as they were on the scope.