I have a question: Do the voltage and the current for a cetrain load have to come from the same source?
Or could I, example given, use two independent sources like photopholtaic and wind or whatever, then go out of phase with each one by -90 resp. +90 degrees with a cap and a coil, then ac- sync this so in none of them is actally both (v+a) present, then simply but these together , maybe trough some diodes, to get a complete, working source of V+A ?
Quote from: dieter on February 24, 2014, 02:05:51 PM
I have a question: Do the voltage and the current for a cetrain load have to come from the same source?
Or could I, example given, use two independent sources like photopholtaic and wind or whatever, then go out of phase with each one by -90 resp. +90 degrees with a cap and a coil, then ac- sync this so in none of them is actally both (v+a) present, then simply but these together , maybe trough some diodes, to get a complete, working source of V+A ?
Huh? Voltage is the pressure that pushes current through a load. It is impossible to separate voltage and current. Even the highest impedance voltmeter there is (electrostatic voltmeter) needs some tiny current put into it by the voltage it is measuring. And even the slightest possible current needs some tiny voltage difference between the source and sink in order to flow.
Now.... you can certainly do weird things like putting sources in series. For example, my old RM503 oscilloscope has a CRT with a 12 volt filament, that draws a few hundred mA when operating. BUT.... this filament is raised to 3000 volts over the system's ground level, because it plays a role in the emission of the electron beam in the CRT. This is done by feeding the filament with a winding from the main power transformer, biasing it from the scope's HV section and limiting the voltage _difference_ across the filament so that only the small current flows, not the current that would happen (very briefly!) if the filament was hooked directly across the 3000 volt source. The scope didn't work when I first got it, it kept blowing fuses and I finally figured out it was because of insulation breakdown within the main (expensive and complicated) power transformer: the 3000 V biased winding was shorting to another winding within the transformer somehow. How to fix this without replacing the transformer (impossible to do anyway). So I got a good quality 12 volt transformer from Radio Shack, powered its 120 v primary from the mains, and let the secondary be elevated to the 3000 volt level by the scope's voltage multiplier, and I drove the filament from that. It works 100 percent fine now. So... in this sense yes, the 3000 volt elevation is provided from the HV section of the scope, and the actual current through the filament is provided by the 12 volt winding on the RadioShack transformer. Since the RS transformer's secondary is not inside the main transformer the original short is eliminated, and the insulation in the new transformer is good enough to keep the 3000 volt bias elevation of the secondary from leaking through to the primary and causing the same problem as before. The filament current is running on the voltage difference between 3012 volts and 3000 volts and this current comes from a different source than the 3000 volt elevation.
There--- are you completely confused even more now? Don't feel too badly, it took me quite some time to track down this problem and understand it to the point where I could repair the scope. It's a classic antique though and has some unique features, so I really was motivated to fix it, and I learned a lot while doing it.
So. that's a yes?
actually, isn't voltage and current sperated when they are out of phase? As in a coil the voltage is established earlier than the current and in a cap later, there is a moment in the ac cycle where both, the coil and the cap contain only v or a (?)
What if we tap this very moment only, by switching? Both is there, v + a, although maybe not as a perfect sinus wave.
Quote from: dieter on February 24, 2014, 03:25:17 PM
So. that's a yes?
actually, isn't voltage and current sperated when they are out of phase? As in a coil the voltage is established earlier than the current and in a cap later, there is a moment in the ac cycle where both, the coil and the cap contain only v or a (?)
What if we tap this very moment only, by switching? Both is there, v + a, although maybe not as a perfect sinus wave.
Well, not exactly. Yes, you can certainly "have" a voltage without current, as in a disconnected battery.... but any _measurement_ you make of the voltage will require a tiny current at least to flow through or into your instrument. This also means that any _use_ of the power requires current to flow. The phase shift in AC power feeding a reactive load means that the overall power is split into Real or True power (that powers or is dissipated in a load) and Reactive Power which isn't dissipated in the load, and Apparent Power which is the total power. The distribution of the input into these parts is dependent on the phase angle.
http://www.allaboutcircuits.com/vol_2/chpt_11/2.html
That is of cource the pre historic view on electricity and most of the old lad's think of it that way.
But in the real world, there is no voltage nor is there current flow and these elements only show up when the time component is taken into the equation.
What it means is that if you were to freeze time, all voltage would dissapear, and all current would stop flowing.
If you look at for example the pre historic equation, Power(in Watts) = Voltage(in Volts) x Current (in Amps) you can already see that there is no mention of the time component, and that therefore this equation is pre historic and incomplete.
Because from this equation, you can not tell if the total sum is per second, minute,hour,day or whatever and if you need to build a perfect model in a perfect world where time does not exist, you will run into problems with this.
In the classical sense it would be Joule per second, but when you are unable to measure the time component, you will run into problems with this.
I think it is interesting to see that the only place where this is really important, is in the area most people on this board are intrested in, and for the rest of the world this would not make any difference as long as the light comes on when they flip the switch...
But only so few understand this and only so few know what it is really about it all depends on your point of view so, from which point you are looking and measuring, to understand this you need to develop a complete shift in perspective because the 'normal' view is too limited to be able to explain and understand it.
Quote from: TinselKoala on February 24, 2014, 02:40:31 PM
It is impossible to separate voltage and current.
Only if you think it is.
It's kind of funny that you go into a treatise about time because all serious electronics analysis is based on time. In fact, often the analysis is based on analyzing what happens during an infinitely thin slice of time.
So a huge chunk of science is built around understanding what happens during an infinitely thin slice of time or with an infinitely thin wedge of a circle or with an infinitely thin slice of solid and so on. It dates back to the Renaissance. For example, how does current flow through a conductor? You cut up the conductor into infinitely small cubes to find out.
MileHigh
Quote from: Turbo on February 25, 2014, 01:00:04 AM
That is of cource the pre historic view on electricity and most of the old lad's think of it that way.
But in the real world, there is no voltage nor is there current flow and these elements only show up when the time component is taken into the equation.
I guess you haven't heard of static electricity, and you probably think that batteries don't have any voltage in them until they are connected to something. Your refrigerator light probably stays on even with the door closed, too.
Quote
What it means is that if you were to freeze time, all voltage would dissapear, and all current would stop flowing.
Sure, and if pigs had wings and a takeoff clearance from the tower, they could fly. BUT THEY DON'T, and YOU CANNOT FREEZE TIME.
Quote
If you look at for example the pre historic equation, Power(in Watts) = Voltage(in Volts) x Current (in Amps) you can already see that there is no mention of the time component, and that therefore this equation is pre historic and incomplete.
Oh come on, stop right there. What is the definition of a Watt? Answer the question. What is the definition of an Ampere? Answer the question. Now... does time enter into the "prehistoric equation" or not?
Quote
Because from this equation, you can not tell if the total sum is per second, minute,hour,day or whatever and if you need to build a perfect model in a perfect world where time does not exist, you will run into problems with this.
In the classical sense it would be Joule per second, but when you are unable to measure the time component, you will run into problems with this.
What is an oscilloscope? What is a frequency counter? How does a receiver like your cellphone know to tune into a GigaHertz frequency? How does GPS work? Just who is unable to measure a time component, I wonder. I can measure time down to the single nanosecond with my "garage sale" equipment. I'm sorry that you cannot, and I'm sorry that I won't be able to teach you how to do it, since you already know you can't.
Quote
I think it is interesting to see that the only place where this is really important, is in the area most people on this board are intrested in, and for the rest of the world this would not make any difference as long as the light comes on when they flip the switch...
But only so few understand this and only so few know what it is really about it all depends on your point of view so, from which point you are looking and measuring, to understand this you need to develop a complete shift in perspective because the 'normal' view is too limited to be able to explain and understand it.
Only if you think it is.
I'll bet money that you would have been unable to repair my RM503 scope, with the depth of your understanding.
See you are talking only from the physical layer.
There is a problem with your understanding.
You just don't get what i am trying to say becuse of your limited mindeset.
I am sorry for you.
Maybe you read it again so you understand you are in a place where time itself does not exist instead of dragging oscilloscopes and frequency counters to this place....
Becuse they are useless if there is no time then they will read 0 and i think you are a fool for not understanding that.
I was clear about what i ment.
Quote from: Turbo on February 25, 2014, 05:25:34 AM
See you are talking only from the physical layer.
There is a problem with your understanding.
You just don't get what i am trying to say becuse of your limited mindeset.
I am sorry for you.
Maybe you read it again so you understand you are in a place where time itself does not exist instead of dragging oscilloscopes and frequency counters to this place....
Becuse they are useless if there is no time then they will read 0 and i think you are a fool for not understanding that.
I was clear about what i ment.
Improved concepts need to account for behaviors explained by the concepts that they would replace. Before anyone can consider that you may have a superior idea you need to reconcile the fact that you assert incorrectly that: Power, the
time derivative of energy does not involve time, or that current, the
time derivative of charge crossing a point does not involve time. If your ideas depend on these false assertions, then your ideas fail to explain observations covered by existing models and therefore fail.
You have a limited mindset also you can not grasp the concept because you are looking at a 4 dimensional picture in 3 dimensional space and as a result you can never solve that equation.
let me make one thing clear.
ELECTRIC ENERGY FLOWS FROM A PLACE WHERE NO TIME EXISTS TO A PLACE WHERE TIME DOES EXIST.
YOU are in the place where time flows, with all your meters scopes frequency conter and all the other bullshit equipment that KNOWS how LONG A SECOND IS.
But taker all your measuring devices into that OTHER dimension and they will FAIL to work because A SECOND ITSELF does not EXIST there so the MEASURING DEVICE CAN NOT MEASURE ANYTHING.
I know it's beyond your scope of imagination, yet
Let me put it this way, you need to get thinking OUTSIDE the physical layer you are currently in...if you existed in the other realm, things would heve been easier for you.
If time moves at lightspeed and electricity moves at lightspeed and they move in the same direction then YOU MIGHT figure out some time.
So yea
I am not going to discuss this any further you all hang on to you pre historic models and go fuel up you car.
It seems you have to die first in order to gain this understanding.
Quote from: Turbo on February 25, 2014, 05:59:33 AM
You have a limited mindset also you can not grasp the concept because you are looking at a 4 dimensional picture in 3 dimensional space and as a result you can never solve that equation.
let me make one thing clear.
ELECTRIC ENERGY FLOWS FROM A PLACE WHERE NO TIME EXISTS TO A PLACE WHERE TIME DOES EXIST.
YOU are in the place where time flows, with all your meters scopes frequency conter and all the other bullshit equipment that KNOWS how LONG A SECOND IS.
But taker all your measuring devices into that OTHER dimension and they will FAIL to work because A SECOND ITSELF does not EXIST there so the MEASURING DEVICE CAN NOT MEASURE ANYTHING.
I know it's beyond your scope of imagination, yet
Let me put it this way, you need to get thinking OUTSIDE the physical layer you are currently in.
If time moves at lightspeed and electricity moves at lightspeed and they move in the same direction then YOU MIGHT figure out some time.
So yea
I am not going to discuss this any further you all hang on to you pre historic models and go fuel up you car.
It seems you have to die first in order to gain this understanding.
You make claims that existing observations contradict. For example E/M waves do not move at a constant speed. You are free to howl at the moon as you will.
I howl:
You did not read what i wrote you just replied to it.
You do not know what a EM wave really is.
If there is not time how can it vary in speed?
You do not know what electricity is.
Fin.
QuoteSo yea
I am not going to discuss this any further
Please keep your promise.
Quote from: Turbo on February 25, 2014, 06:06:57 AM
I howl:
You did not read what i wrote you just replied to it.
You do not know what a EM wave really is.
If there is not time how can it vary in speed?
You do not know what electricity is.
Fin.
Set up an experiment and show that your ideas make more accurate predictions than conventional theory. Show that your ideas make predictions that are at least as accurate as conventional theory for at least common circumstances.
Look at this PM I just got from newbie "Turbo":
Quote
I do not know your name, but i think you should stop posting to this board because you transfer your faulty and limited old fashioned ways of thinking onto the newcomers which causes problems.
You have two options, one quit with it, and two, learn the real deal.
Do you even read what someone posts before jumping up and start quoting what is said to then press through your pre historic view on electric energy transfer?
I am dead serious you do not even understand 0.0000000001% of this whole thing.
You think you know but let me tell you,
You don't know shit.
That gets a ROFL for sure! Come on, Turbo, tell us another one.
(I think it is absolutely HILARIOUS that someone can sit at a COMPUTER and tell electrical engineering professionals that the concepts they used to design and build COMPUTERS are wrong... but he can't show _anything_ that has ever been made to work using his own concepts.)
Hey... I just realized something. In a place where there is no Space, you don't need to put gasoline in your car, because everything is really REALLY close... and you can just walk to wherever you are going. And if there is no Time.... you'll get there at the same instant you left. This will really revolutionize travel, and put the Big Oil Powers That Be completely out of business.
I wonder if this is Rose's sibling. the irrationality, the arrogance and the unnecessary rudeness certainly match. In the meantime it's diverted Dieter's thread.
There is nothing wrong with making assertions as long as they don't contradict what is known to be true already. If one makes a claim or assertion that does contradict something we think we know, that's fine too.... as long as _evidence_ is presented to support the claim. Real evidence, checkable, repeatable evidence. Once the evidence for a claim is presented, that evidence can be examined for validity and relevance, by reasoned discussion, experimental research and demonstrations, and references to checkable outside sources.
If one presents no evidence, but instead relies on ad-hominem abuse and more contradictory claims.... then they are fair targets for harsh criticism.
(Did you know that I have a herd of winged pigs in my backyard? They are invisible, silent and they don't smell badly, so I'm not going to show you an image of them, but I assure you that they are there (if they haven't flown away.) Now... prove me wrong, or you must be an idiot no-nothing.)
See what I mean?
Quote from: Tseak on February 25, 2014, 09:55:33 AM
I wonder if this is Rose's sibling. the irrationality, the arrogance and the unnecessary rudeness certainly match. In the meantime it's diverted Dieter's thread.
I'd love to see a discussion between Atommix and Turbo. That would be a real hoot.
Quote from: Turbo on February 25, 2014, 05:25:34 AM
(snip)
Maybe you read it again so you understand you are in a place where time itself does not exist instead of dragging oscilloscopes and frequency counters to this place....
Becuse they are useless if there is no time then they will read 0 and i think you are a fool for not understanding that.
I was clear about what i ment.
By golly.... I do believe you are right!
@Dieter
Do you mean, for instance can one connect a (high amperage availability), 12 volt lead acid battery in series
with a (high voltage source) Vandegraff generator and get giant power ?
Current flows in loops (eventually).
Makes me wonder, has any body ever measured the resistance across their Vandegraff (not running) ?
I have, it's an open circuit. Sometimes the belt can become slightly conductive after long running, from carbonizing the materials, or from moisture adsorption. Then, if the test voltage is high enough, like from a "hi-pot" tester, you may be able to measure some finite conductance. But since most VDGs don't make physical contact from the pickup combs to the belt, you need the HV test even to see this small conductance. An ordinary DMM will read infinite resistance, or zero conductance.
QuoteDo you mean, for instance can one connect a (high amperage availability), 12 volt lead acid battery in series
with a (high voltage source) Vandegraff generator and get giant power ?
That's a good question. I don't know if the VDG, running, will allow the complete circuit from the 12 volt battery. I'd do the experiment instantly, but I don't have my VDGs available at the moment. Giant power? No, the most you could get would be the power from the VDG plus the power from the 12 volt battery. You will get (V
vdg x A
vdg) + (V
batt x A
batt) = P
total. I really don't think you will get (V
total x A
total) = P
total.
@dieter
QuoteI have a question: Do the voltage and the current for a cetrain load have to come from the same source?
Or could I, example given, use two independent sources like photopholtaic and wind or whatever, then go out of phase with each one by -90 resp. +90 degrees with a cap and a coil, then ac- sync this so in none of them is actally both (v+a) present, then simply but these together , maybe trough some diodes, to get a complete, working source of V+A ?
I have wondered much the same. What would happen if you were to pulse a transformer primary with high current DC and smack that in the rear with an immediate pulse of HV AC, in effect giving it a single duration pulse?
Can't combine AC & DC together? Try telling that to this guy.
http://www.energeticforum.com/renewable-energy/14116-plasma-rocket-engine.html#post241432
That whole thread is an interesting read.
QuoteCan't combine AC & DC together?
Just who said that, please? But first... how do you define AC, and how DC?
Since AC involves current reversals, and DC does not..... well, I think you can see where this is going.
I was just asking because several projects mentioned it, eg. current from the ground an voltage from the sky, wasn't that tariel guy doing something like that? And it was mentioned in that 4 mb FE collection pdf from this other russian guy.
If we compare electricity to a river, voltage be the speed of flow, curtent the size, it would be absurd to say it flows down that hill overthere but it has never been there.
Dieter:
QuoteIf we compare electricity to a river, voltage be the speed of flow, curtent the size, it would be absurd to say it flows down that hill overthere but it has never been there.
Sorry, but you are not even close. The difference in height of the river from some starting point A to some ending point B would be like the voltage. Voltage = height. The amount of water flow in cubic meters per second would be be like the current. The physical river itself would be like a large resistor.
MileHigh
That's right. Like charges repel, and this is the _source_ of what we call voltage. Pack a bunch of like charges together and they will repel with more force the more charges you pack in there. This is voltage. Provide a path for the charges to relieve this pressure, and they will flow along the path until the charge pressure is equal everywhere. This flow of charge is current. Charge is carried by free electrons, which carry the unit negative charge, and also by "holes" which are places where an electron "should" be in a molecular lattice, as in semiconductors. "Holes" are positively charged, electrons negatively charged. Because of the naming convention assigned originally by Benjamin Franklin, the electrons are called negative. This is somewhat unfortunate, because what we call "conventional" current is taken to flow from positive voltage regions to more negative regions... but the actual flow of electron charges is in the other direction, from the packed "negative" voltage region full of electrons and depleted of holes, over to the more positive voltage region, which is deficient in electrons and has a surfeit of "holes". This was not known in Franklin's time, though, and by the time the electron was actually characterized by several researchers like JJ Thomson and Robert Millikan in the years around the turn of the 20th century, we were stuck with Franklin's "negative" terminology.
"I am dead serious you do not even understand 0.0000000001% of this whole thing."
(Quoted from Turbo's PM)
Wow!
Check out the sig. digits here. I wonder how this was calculated? 10 places...not 9...not 6...not 3...but 10.
If it were me, I would have went 11 places out, just to be sure I got my point across. 12, of course, would have been overkill, but 11 would have done the job well.
Bill
Yeah, precision is important. Well, I guess it's time... or maybe NO time... to go out and feed my herd of invisible pigs. There are at least ten thousand of them in the back yard. I'm not sure how many, really. Have you ever tried to count invisible pigs? It's not easy. But it beats making Ainslie debunk videos, there's no percentage in those.
http://www.youtube.com/watch?v=jg2_yE5dEQg
Quote from: MileHigh on February 25, 2014, 09:36:08 PM
Dieter:
Sorry, but you are not even close. The difference in height of the river from some starting point A to some ending point B would be like the voltage. Voltage = height. The amount of water flow in cubic meters per second would be be like the current. ...
MileHigh
MileHigh, old sceptic I was starting to miss you. Not even close? Well, the speed of flow is determined by the height diffrence you mentioned, so you agree with me.
As for the cubic meter per second, that my friend, would be Watt and not Ampere, as you said so eloquently. :P
Quote from: dieter on February 26, 2014, 05:34:35 PM
MileHigh, old sceptic I was starting to miss you. Not even close? Well, the speed of flow is determined by the height diffrence you mentioned, so you agree with me.
As for the cubic meter per second, that my friend, would be Watt and not Ampere, as you said so eloquently. :P
No: the fluid flow rate is equivalent to electric current. The applied or lost power is the product of force and flow rate. The force is the result of the height change, the mass and the acceleration due to gravity.
Quote from: dieter on February 24, 2014, 02:05:51 PM
I have a question: Do the voltage and the current for a cetrain load have to come from the same source?
Or could I, example given, use two independent sources like photopholtaic and wind or whatever, then go out of phase with each one by -90 resp. +90 degrees with a cap and a coil, then ac- sync this so in none of them is actally both (v+a) present, then simply but these together , maybe trough some diodes, to get a complete, working source of V+A ?
It works like this as I see it. If the two supplies are electrically isolated until the outputs are merged, then each circuit will take it's own return current only.
ie, The HV circuit supply voltage causes the HV circuit's current and so the current caused by the HV will return to the HV supply.
The LV supply circuit's voltage causes the low voltage circuits current and that current will return to the LV supply.
Now if the HV and LV output are simply superimposed and have common grounds the HV will see the LV as a load. Like connecting the output of a flyback transformer to a battery. LV is load to HV.
If the reverse current is blocked to prevent that then the load determines what gets loaded the most.
If the load is high resistance the LV supply will do no work.
It's really quite simple and comes down to what pushes what, and current loops in separate supply's. It's not rocket surgery.
The voltage causes the current to flow in a (loop). Each supply will have it's own loop.
If we talk AC and we apply 10 000 volts positive half cycle to 150 volts negative half cycle, there will be a potential difference of 10150 volts, if the two circuit grounds are common the HV supply will see the LV as a load.
Cheers
EDIT: I talk in 'conventional' positive to negative current, not "electron current'. Flow of 'charge' (or whatever), not electron movement. Just to say how I think with respect to "current" flow.
It should be noted and logic that if the two were superimposed on the same conductors then the HV and the larger current will likely be measured, but they won't perform work together, the HV and it's associated current will do most of the work if the load is high enough resistance.
It basically conforms to work done at input and energy output. V x A only applies to the A caused by the V that caused it, and means power not energy anyway..
Spock and Kirk (Logic), Make up your own minds who is Spock and who is Kirk. ;D
http://www.youtube.com/watch?v=Kc6Vu-NLu0M
..
If we add a Flea to an Elephant will we end up with a Flea the size of an elephant ( would be a nasty animal), or an elephant the size of a flea, or will we just end up with, a Flea and an Elephant ?
..
This is kinda interesting and I think discussing it and thinking about it would help the less experienced to better visualize voltage and current.
IF we could post some drawings of situations where the mixing of HV-LC and LV-HC may be achieved then the circuits could be evaluated. And the effects seen in a picture.
The drawings would need both HV and LV supplies included with any groundings and the load/loads, supply - load isolation if any, ect. ect..
Pencil paper and eraser are cheap and work like a simulator to a degree, we just need to add the visualization based on what we know happens, though it can get complicated, it's often simple when seen on paper. I draw the supply itself then the circuit from positive to negative as much as is possible.
..
Quote from: Farmhand on February 26, 2014, 07:16:16 PM
This is kinda interesting and I think discussing it and thinking about it would help the less experienced to better visualize voltage and current.
IF we could post some drawings of situations where the mixing of HV-LC and LV-HC may be achieved then the circuits could be evaluated. And the effects seen in a picture.
The drawings would need both HV and LV supplies included with any groundings and the load/loads, supply load isolation ect. ect..
..
The relationships are not complicated. Assign whatever voltage you like to each of the two sources. Assign the corresponding internal resistance values. The problem that you will run into is where the current limit of the high voltage supply causes the value of the associated resistor to be variable, and the resulting external voltage across the combination of Vx and Rvx to collapse.
But Mark those supplies are in series. I'm not sure that is what is meant by the original poster.
..
Quote from: Farmhand on February 26, 2014, 07:39:16 PM
But Mark those supplies are in series. I'm not sure that is what is meant by the original poster.
..
If you want the voltage of two supplies then you have to put them in series, and the loop current will be limited by the supply with the lower current limit. A 25kV 1mA supply in series with a 2V 150A welder will still only deliver 1mA. If you want the current of two supplies then you connect them in parallel, and you will be limited by the lower voltage of the two.
Quote from: MarkE on February 26, 2014, 07:45:11 PM
If you want the voltage of two supplies then you have to put them in series, and the loop current will be limited by the supply with the lower current limit. A 25kV 1mA supply in series with a 2V 150A welder will still only deliver 1mA. If you want the current of two supplies then you connect them in parallel, and you will be limited by the lower voltage of the two.
Umm you don't need to tell me that. ;) But is that what the original poster wants or is talking about ?
I think he is talking more about a HV in parallel with a LV maybe the HV as HF as well, I'm not sure. I'm hoping the original poster will post a drawing of what he is visualizing.
This below is what I think they mean when they talk of mixing HV with LV. Don't forget, it is no me making the claim here. First I am trying to understand what the original poster actually means.
Maybe he will say if my drawing represents his thoughts. It could be considered without the smoothing capacitor or with.
In my opinion it's a pointless exercise. Unless there is some reason for it that I cannot see. :)
..
Here's an example, If we begin with the LV supply turned off and the Load resistor drawn near the ? on the drawing is 10000 Ohms then the capacitor will charge to about 1000 volts and about 0.1 Amps will flow in the load resistor. about 100 Watts. Then if we turn on the LV supply nothing will happen, because the capacitor is already charged to 1000 volts. The LV supply will draw idle power. That's what I say.
Now if we do it the other way and have the LV supply powering a 100 ohm load and dissipating about 1 Watt. Then we turn on the HV supply we could have a problem with fire or supply capabilities, or just use more power (10 kilowatts) and drive the load harder with the HV supply. ;D
Cheers
Quote from: Farmhand on February 26, 2014, 08:22:06 PM
Umm you don't need to tell me that. ;) But is that what the original poster wants or is talking about ?
Those are the only two options available: Series or parallel.
Quote
I think he is talking more about a HV in parallel with a LV maybe the HV as HF as well, I'm not sure. I'm hoping the original poster will post a drawing of what he is visualizing.
This below is what I think they mean when they talk of mixing HV with LV. Don't forget, it is no me making the claim here. First I am trying to understand what the original poster actually means.
Maybe he will say if my drawing represents his thoughts. It could be considered without the smoothing capacitor or with.
In my opinion it's a pointless exercise. Unless there is some reason for it that I cannot see. :)
..
I think that the person proposing the scheme does not understand that all practical voltage sources have internal impedances that limit their current.
Mark ,Even to a novice it should make no sense to connect a thin wire from say a flyback secondary in series with the thick wires from a step down transformer secondary.
It is true the impedance of the supplies will see the HV supply restricted in output current and at some point the voltage will drop. But for the sake of the argument we can consider both transformers capable of outputting say 1000 watts each or we could consider them ideal and thus they would handle any load, just to keep it simple.
We can see that in the first case of the LV off to start with then turning the LV on while HV is already on and the load resistance is high makes little difference due to the counter emf of the already charged to 1000 volts capacitor would prevent any power output from the LV supply.
I think if we want people to understand more we need to give examples and explanations in layman's terms, eg, in my example the counter emf situation shows up another myth which is that counter emf (the evil twin or the evil witch as UFO calls it) consumes power, when in fact it only restricts the power, counter emf does not consume power and any emf countering an applied emf is counter emf, be it induced or already present.
If we just say no it won't work and give technical explanations that most people do not understand then next to nothing is achieved. People don't just trust the words of a "naysayer" such as we, like they trust the words of the guru scammers. We must work to get the point across.
I am always amazed when a trained guy comes along and lumps me in with the claimants of OU or suggests I am making claims, when all I am doing is trying to explain how I see things in layman's terms.
I suggest if you want a strictly engineering interaction then go talk to an engineer. No offence but you cannot expect to talk engineer speak to laymen or complete novices and have them just accept it. If it bothers you not that the same fallacies keep arising then keep not explaining it so people can understand. This mixing of HV-LC and LV-HC has been claimed before by others and i challenged them to show it but was personally abused.
You should be aware Mark that I am on the side of (proof of claims and better understanding of basic electronics) as a means to reduce the effect of the Guru scammer type people and the eternal promise with no evidence - keep a thread going for ever type people, as well as helping others to be more discerning experimenters able to spot a fake or a mistaken claim.
Cheers
The system in the RM503 scope is something like that shown below. I haven't put in the capacitors, etc. and I've shown it as a positive elevation instead of the negative "elevation" used in the real scope. The filament is elevated to -3000 v wrt the scope's chassis ground; this supplies the electron current needed to make the electron beam. The filament gets its AC heating current from the 12 v transformer, which can supply its full rated output to a load, and the filament gets the DC beam current from the HV transformer (really a voltage multiplier system) through the HV rectifier tube and this can only supply a small current.
Needless to say, the insulation in the 12v transformer has to be good.
The loops described by Farmhand are here; they share one "leg" in common. If you could be an electron in that common leg... you'd probably get pretty confused. But it works, anyhow.
What I was talking about was: in AC a cap as well as a coil will bring the current out of phase. There is a moment in the AC cycle where there is full current at the cap, but zero voltage, and there's a moment where there's full voltage at the coil, but zero current. If you would use a mechanical or electronic switch to tap that very moment only and then "mix" the currentless voltage with the voltageless current, would anything happen?
But I do assume, even if the current is out of phase, energy is consumed nonetheless.
Quote from: dieter on February 27, 2014, 08:02:39 AM
What I was talking about was: in AC a cap as well as a coil will bring the current out of phase. There is a moment in the AC cycle where there is full current at the cap, but zero voltage, and there's a moment where there's full voltage at the coil, but zero current. If you would use a mechanical or electronic switch to tap that very moment only and then "mix" the currentless voltage with the voltageless current, would anything happen?
Uh.... no. Or maybe yes. You see, when you set up impossible initial conditions, it's impossible to predict outcomes. Sort of like dividing by zero.
Quote
But I do assume, even if the current is out of phase, energy is consumed nonetheless.
Not necessarily. As we covered before, AC power fed to reactive circuit components will exhibit phase shift between voltage and current. This results in the "power" being partitioned into "real" power and "reactive" power. The reactive power isn't dissipated.
From Wiki:
Quote
In the diagram, P is the real power, Q is the reactive power (in this case positive), S is the complex power and the length of S is the apparent power. Reactive power does not do any work, so it is represented as the imaginary axis of the vector diagram. Real power does do work, so it is the real axis.
The unit for all forms of power is the watt (http://en.wikipedia.org/wiki/Watt) (symbol: W), but this unit is generally reserved for real power. Apparent power is conventionally expressed in volt-amperes (http://en.wikipedia.org/wiki/Volt-ampere) (VA) since it is the product of rms (http://en.wikipedia.org/wiki/Root_mean_square) voltage (http://en.wikipedia.org/wiki/Voltage) and rms current (http://en.wikipedia.org/wiki/Electric_current). The unit for reactive power is expressed as var, which stands for volt-ampere reactive (http://en.wikipedia.org/wiki/Volt-ampere_reactive). Since reactive power transfers no net energy to the load, it is sometimes called "wattless" power. It does, however, serve an important function in electrical grids (http://en.wikipedia.org/wiki/Electrical_grid) and its lack has been cited as a significant factor in the Northeast Blackout of 2003 (http://en.wikipedia.org/wiki/Northeast_Blackout_of_2003).[2]
Understanding the relationship among these three quantities lies at the heart of understanding power engineering. The mathematical relationship among them can be represented by vectors or expressed using complex numbers, S = P + jQ (where j is the imaginary unit (http://en.wikipedia.org/wiki/Imaginary_unit)).
http://en.wikipedia.org/wiki/AC_power
So if voltage is pressure and current is flow tell me how there can be a phase shift between the two,
Quote from: Dave45 on February 27, 2014, 07:04:26 PM
So if voltage is pressure and current is flow tell me how there can be a phase shift between the two,
That's easy, set up a channel with an accumulator attached. If you increase the pressure in the channel, more water will store in the accumulator, and the flow downstream will not increase in lock step with the pressure increase. Likewise, when you decrease the upstream pressure, the accumulator will release water and the flow downstream will not decrease in lock step with the pressure decrease. This is why municipal water systems have water tower accumulators.
Ok sounds plausible for water, tell me if alternating current is just current alternating through a conductor then why does this circuit build a negative ion cloud on one end and a positive ion cloud on the other.
If negative current were the only energy moving in the coil wouldnt we have a negative ion cloud only on one end of the circuit.
Quote from: Dave45 on February 27, 2014, 09:06:52 PM
Ok sounds plausible for water, tell me if alternating current is just current alternating through a conductor then why does this circuit build a negative ion cloud on one end and a positive ion cloud on the other.
If negative current were the only energy moving in the coil wouldnt we have a negative ion cloud only on one end of the circuit.
Capacitance
Quote from: MarkE on February 27, 2014, 09:15:39 PM
Capacitance
:) comon Mark you can do better than that.
Quote from: Dave45 on February 27, 2014, 09:06:52 PM
Ok sounds plausible for water, tell me if alternating current is just current alternating through a conductor then why does this circuit build a negative ion cloud on one end and a positive ion cloud on the other.
If negative current were the only energy moving in the coil wouldnt we have a negative ion cloud only on one end of the circuit.
Mark Is right, Dave we kinda need to think in a way that isn't so (positive-negative), not so (either-or) kind of thing. If one plate becomes more negative then the other plate becomes less negative in relation to it, or the same thing if one plate becomes more positive then the other plate becomes less positive and a potential difference occurs between them. Simple, the potential difference is between the two plates and they both share the charge differential.
Cheers
P.S. Dave maybe Mark is thinking, Come on Dave you can do better than that. I think you are getting ahead of yourself a bit, you're talking ion clouds, but you don't understand charge displacement. Maybe reading to too much Utikin, he is there to fill peoples heads with fallacies and make claims for Tesla he didn't make. Better to not read him, better to get a good basic understanding of electricity ect. However we all make our own choices. In that way we are free.
..
And that happens because the circuit pumps electrons into the capacity at the negative end. This could be a large shape or just a point, with the other "plate" being the Earth itself or the other end of the circuit if, again, a shape or point is provided. Voltage is charge pressure, which comes from packing like charges together into some _capacity_. If you have sharp points or edges, the _gradient_ of the electric field becomes large enough to break down the air molecules into ions - at the "negative" end, electrons are emitted, latch onto air molecules and give them a negative charge. Opposite on the other end, the depletion or holes "suck" electrons out of the air molecules leaving them positively charged. The diodes in the circuit prevent a discharge path that way. Just as with any capacitor, as you pack more charge into the region, it becomes harder and harder to put more charge in because the charge already there repels the charge you are trying to stuff in.
Really, if you want to understand current and voltage you need to think about the movement of and forces between like unit charges. Electrons and holes. Water doesn't exactly repel itself and there's no "water holes" like there are unit positively charged "holes" in conductors and semiconductors, so the hydraulic analogy breaks down if you try to take it too far.
Voltage is "relative charge pressure", which Farmhand is describing, aka potential difference. If you stick both your voltmeter leads to the same charged surface you won't measure any voltage, even though the charge pressure might be very high compared to the "ground" ... the Earth is essentially an infinite sink and source for electrons and so it is the usual "zero reference" to which higher voltages... charge pressures... are compared.
But charge itself isn't relative, it is conserved. You can pack a bunch of charge onto a capacitance, and it won't discharge if there is no path to a lower potential (charge pressure). But it still takes more and more work to pack more charge into the capacitance to raise its voltage. Eventually you get so much charge into or onto a capacitance that the very air breaks down into ions which are very conductive, and the charge leaks away. The voltage rating of capacitors is essentially the dielectric strength that can withstand the potential difference between the plates.
Lets keep it simple
If current moves from neg to pos then current is moving through a diode in this direction
But the bemf from a pulsed coil moves through a diode in the reverse direction.
We can see this using a freewheel diode across a coil, or using this simple experiment.
So tell me what polarity is the energy moving through the diode from pos to neg.
Now I see the diode reroutes the energy from the bemf to the pos in a freewheel diode but the energy is moving through the diode in the wrong direction to be a neg polarity.
Dave:
I will take a stab at this one. For starters, yes, "we" got it backwards and current flow is normally flowing electrons. However, that is analogous to driving on the right in North America and driving on the left in the UK. Does it really matter in the final analysis? The resounding answer is NO, it does not matter. Literally any problem solved with the standard electric current convention we use will WORK FINE, so there is no need whatsoever to keep harping on the fact that "(electron) current runs backwards through diodes" and all that stuff. We simply accept the convention for what it is and stick to it in order to avoid mass confusion. EVERY SINGLE PROBLEM WILL WORK using the standard electric current convention. After you solve the problem, then if you really want to you can then say that in reality the electron current is opposite to the conventional current. This whole thing is a useless tempest in a teapot.
For your diagram, I will assume that the left side of the voltmeter is the negative sense input, and the right side of the voltmeter is the positive sense input. Believe it or not, it is a huge mistake for you to not put the polarity designations on your voltmeter inputs.
The diode in your diagram is not rerouting energy. That is not proper terminology. The diode either conducts or it doesn't conduct, and there is a measurable voltage across the two terminals of the diode. When the diode conducts we will assume there is a voltage drop of 0.6 volts. That means relative to your diagram the right side of the diode will be at <some voltage relative to the battery ground> and the left side of the diode will be at <some voltage relative to battery ground> minus 0.6 volts. Note that when the switch is closed the left side of the diode will be at the battery ground potential.
From your diagram, let's assume the switch is closed and some current is flowing. When the switch opens, the voltmeter will measure a large increase in potential while the current continues to flow through the air which has been broken down and is ionized and becomes a plasma. When the coil has discharged all of its energy the current flow stops and the voltmeter will measure the battery voltage. (there is a slight complication relative to the diode that I am ignoring here for the sake of simplicity)
That's it, the problem is solved. If you really want to split hairs you can do the whole electron current business but there is no point whatsoever in going there.
MileHigh
Quote from: Dave45 on February 28, 2014, 06:35:59 AM
Lets keep it simple
If current moves from neg to pos then current is moving through a diode in this direction
Let's keep it even more simple and recall that "conventional current" is _defined_ as moving from Positive to Negative in circuits. Again, this is a _naming convention_ that we are stuck with due to the time delay between Benjamin Franklin's work with electricity, and the actual discovery and characterization of the electron, the actual carrier of the charge that does move in the circuit. All electrical engineering analyses like Kirchoff's circuit rules, Ohm's Law, the designation of Cathode and Anode in circuit components, etc. use the CONVENTION, even though we know that the actual electron charges are moving in the other direction. The CRT, for example, illustrates this quite well. Even though the Cathode is strongly _negatively charged_ , it _emits electrons_ in a beam, rather than sucking them up like some kind of vacuum.
http://www.mi.mun.ca/users/cchaulk/eltk1100/ivse/ivse.htm
http://en.wikipedia.org/wiki/Electric_current
http://web.engr.oregonstate.edu/~traylor/ece112/lectures/elect_flow_vs_conv_I.pdf
http://www.allaboutcircuits.com/vol_1/chpt_1/7.html
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html#c3
I don't make this stuff up!
I know how current flows in a diode normally, what I am try to get across is that the bemf from a negatively pulsed coil has a pos polarity, and we know this by the direction it moves through the diode.
You say Im splitting hairs well when looking for free energy your darn right, Im looking in every nook and cranny.
Quote from: Dave45 on February 28, 2014, 05:04:09 PM
I know how current flows in a diode normally, what I am try to get across is that the bemf from a negatively pulsed coil has a pos polarity, and we know this by the direction it moves through the diode.
You say Im splitting hairs well when looking for free energy your darn right, Im looking in every nook and cranny.
Your IRF510 MOSFET is upside down. Connected like that with the source to the coil and the drain to the low side of the circuit current flows continuously through the MOSFET body diode.
What we know is that in the presence of a changing impedance inductor voltage changes so as to maintain the current direction and magnitude. Current was flowing clockwise through the coil. Increasing the impedance through the switch causes inductor voltage to rise such that the current continues. The diode eventually offers such a continuation path. Before it does, energy gets stored in the local parasitic capacitance, and energy gets dissipated in the MOSFET as it transitions from a low impedance to a high impedance, and in the diode as it transitions from a high impedance to a low impedance.
Lets add a diode for clarity,
If the bemf from the coil had a neg polarity the freewheel diode would be reversed, but it isnt this indicates the polarity of the bemf is of a pos polarity.
Quote from: MarkE on February 28, 2014, 05:15:35 PM
Your IRF510 MOSFET is upside down. Connected like that with the source to the coil and the drain to the low side of the circuit current flows continuously through the MOSFET body diode.
irrelevant to the discussion, this came from some circuit schematic I had on hand
Quote from: Dave45 on February 28, 2014, 05:20:58 PM
Lets add a diode for clarity,
If the bemf from the coil had a neg polarity the freewheel diode would be reversed, but it isnt this indicates the polarity of the bemf is of a pos polarity.
That configuration is actually used in a number of half-bridge power circuits to prevent conduction through the MOSFET body diode.
Why is this important, because we can use the pos energy to pulse a coil, now the diode works bidirectional it passes the pos bemf from the negatively pulsed coil into another coil, pulsing this coil with pos energy its bemf will be neg and will pass through the same diode, now your coils can ring back and forth.
Quote from: Dave45 on February 28, 2014, 05:23:27 PM
irrelevant to the discussion, this came from some circuit schematic I had on hand
Dave, if you don't draw things correctly then it is going to throw the conversation off.
Quote from: Dave45 on February 28, 2014, 05:55:11 PM
Why is this important, because we can use the pos energy to pulse a coil, now the diode works bidirectional it passes the pos bemf from the negatively pulsed coil into another coil, pulsing this coil with pos energy its bemf will be neg and will pass through the same diode, now your coils can ring back and forth.
Even if you fix the MOSFET orientation you are in trouble. That is a good way to ruin a perfectly good MOSFET.
Quote from: MarkE on February 28, 2014, 05:59:49 PM
Even if you fix the MOSFET orientation you are in trouble. That is a good way to ruin a perfectly good MOSFET.
Ok fix the mosfet orientation - the mosfet is not what we are discussing here. This is how a discussion turns into miles of senseless posts.
We are talking about the bemf of a negatively pulsed coil and its polarity not the mosfet orientation, so its wrong, I told you I took it from a schematic that was handy.
Quote from: Dave45 on February 28, 2014, 06:13:15 PM
Ok fix the mosfet orientation - the mosfet is not what we are discussing here. This is how a discussion turns into miles of senseless posts.
We are talking about the bemf of a negatively pulsed coil and its polarity not the mosfet orientation, so its wrong, I told you I took it from a schematic that was handy.
Dave, all coils have real resistance. All diodes have resistance and forward voltage drop. Therefore, anything that you do to connect a coil through a diode dissipates energy that you put into the coil to build up its magnetic field.
Forty years ago when transistors were expensive so called bi-level circuits were popular for things like solenoids and stepper motor drives. A drive voltage would be applied to a coil for a certain period of time to build up the current, and then a lower sustaining voltage would be applied that would ideally match the I*R drop of the coil in order to hold the current. The power drawn from that second, lower voltage supply is the ongoing loss in the magnetics. Since the mid 1980's we have had wonderful switching drivers that deliver the same effect by periodically connecting the coil to a power supply and then allowing the current to recirculate either through the coil or back through the power supply. These devices are carefully designed to minimize losses. And there are always losses. The bottom line is that any combination of coils and switches you can think of, it is almost certain that someone else has already used in some form or another. Resistance, eddy current losses, and hysteresis losses aren't going away. They eat at the energy stored in any non-superconducting magnetics ever built.
Dave:
My request is that if you are going to put up a diagram, then make it complete. Assuming that it is being driven by an external voltage source, what are the polarities of that voltage source? You put arrows up for current, are you talking about electron current or conventional current? You say "positive this" or "negative that" with reference to something in your simplified schematic, but you put no polarity indications on the components you are talking about. It would take you five more minutes to mark up your schematic segment, so why not do it? Why should the readers of the thread have to do mental gymnastics trying to figure out what you really mean?
Just take the example of a "positively pulsed coil." You can just as easily call that a "negatively pulsed coil." If you don't define which of the two terminals of the coil is your voltage reference, then your statements about the polarity of the pulsing of the coil are meaningless.
QuoteWhy is this important, because we can use the pos energy to pulse a coil, now the diode works bidirectional it passes the pos bemf from the negatively pulsed coil into another coil, pulsing this coil with pos energy its bemf will be neg and will pass through the same diode, now your coils can ring back and forth.
Yeah, sure, that's really clear. I can do mental gymnastics trying to figure out what you are saying, or you can do a proper diagram or sequence of diagrams to effectively communicate what you are trying to get across. About the only thing that I can say is that relative to schematic snippets I have seen, there will be no coils ringing back and forth. Why don't you draw out a timing diagram along with a properly done schematic diagram to explain what you are trying to get across.
MileHigh
Quote from: MarkE on February 28, 2014, 05:57:02 PM
Dave, if you don't draw things correctly then it is going to throw the conversation off.
Well this is all a bit odd, I have evidence to show that when a coil is snubbed by a freewheel diode the current in the coil stops immediately as compared to when the coil is allowed to discharge into a higher voltage.
Dave you are trying to use electron movement "current" for the coil charging then applying conventional current to the coil collapse. The coil is charged with positive potential and discharges current in the same direction of current flow when the coil is switched off, very simple the discharge of a coil is not Back emf is is forward emf. Understand that and you'll get somewhere.
Here these scope shots show the voltage and current through the coils when the coil is allowed to discharge through a higher voltage load and also what happens when the coil is snubbed by a free wheeling diode. As we can see the snubbed coil's current ceases immediately with no extra current ( the energy is wasted) likely burned off in the diode and coil, where the coil discharging to a higher voltage takes time to discharge.
Not talking in absolutes but to me the theory that the current recirculates in the snubbed coil falls flat on it's face in reality.
The top two shots are from a snubbed coil with a free wheeling diode (current stops immediately on switch off), and the bottom two the coil discharges into a load at double the supply voltage and current continues after switch off. Yellow are current traces across a CSR and blue are drain voltage traces.
The bottom of the "on" times looks angled because the scope grounds were connected to the positive rail at a discharge capacitor, I can assure you the switching was sharp as anyone would want. I grounded the scope there for ease of connection so I could measure the current and voltage together and keep the grounds in the same place.
..
Farmhand:
When a coil is snubbed by a freewheel diode then by definition the current decay will take much longer as compared to when a coil discharges into a high voltage EMF source or it discharges into a high resistance which causes the coil to generate a high voltage.
Unfortunately you are in the same territory as Dave. You put up a scope screen grab and you aren't saying what trace is what and what case is what. You have no related schematic showing where the scope probes are connected to the schematic, signal and ground connections.
So it's back to doing mental gymnastics to try to figure out what you are saying. All that I can state is that your comments about the discharge time for the coil are wrong and I can perhaps make some inferences about what your traces mean. But I am not going to spend five, ten, or 15 minutes trying to figure out the pieces to the puzzle.
MileHigh
P.S.: I can see how you were still editing your posting when I commented and added more information. It's an improvement but I will remain a stick in the mud and would prefer an accompanying schematic snipped with probe positions and all that jazz so that I can try to understand your points without the mental gymnastics. This is just an editorial comment, I am not literally asking you to do this.
Quote from: MileHigh on February 28, 2014, 07:48:54 PM
Farmhand:
When a coil is snubbed by a freewheel diode then by definition the current decay will take much longer as compared to when a coil discharges into a high voltage EMF source or it discharges into a high resistance which causes the coil to generate a high voltage.
Unfortunately you are in the same territory as Dave. You put up a scope screen grab and you aren't saying what trace is what and what case is what. You have no related schematic showing where the scope probes are connected to the schematic, signal and ground connections.
So it's back to doing mental gymnastics to try to figure out what you are saying. All that I can state is that your comments about the discharge time for the coil are wrong and I can perhaps make some inferences about what your traces mean. But I am not going to spend five, ten, or 15 minutes trying to figure out the pieces to the puzzle.
MileHigh
I do have a related schematic and you have seen it before, but you must have forgotten. The top shots show a coil pulsed and the discharge clamped by a freewheeling diode, the bottom shots show a coil pulsed and the discharge directed to a higher voltage load. I can find the schematic or draw another. There should be no need.
I don't see any discharge of current continuing after the snubbed coil is switched off do you ?
There is no puzzle, have Conrad do the experiment for you. It's very simple.
The top shots show a coil pulsed with a diode clamping the discharge to the supply rail, where is the current after switch off.
P.S. you reacted too quickly and didn't give me time to edit the post. re read it now. Slow down champ.
..
Farmhand:
You can't expect people to look at a scope screen capture from you and then pull the correct schematic from a "mental memory file of previous Farmhand postings" and then visualize that in their mind while looking at your scope capture. That doesn't make any sense. When you make presentations you have to come out of your own personal bubble and pretend that you are an outside observer with no preloaded preconceptions.
Anyway, my points have been made. When presented with incomplete data like this I normally simply tune out.
MileHigh
Farmhand:
I don't know why your tests are giving you contrarian results. However, the coils doesn't lie. The higher the resistance the coil discharges into, the higher the output voltage and the faster it discharges. An ideal coil discharging into a zero ohm resistance will never discharge and have an output voltage of zero, and the current will flow forever. When a coil discharges across a diode, that's akin to a very low resistance, hence a larger discharge time.
MileHigh
Actually I "may" be wrong in where I placed the current sense resistors, so you are probably right, I will need to do that experiment again to get a better result. No problem.
Point taken. Will provide revised shots as soon as I can with a setup made just for the experiment. And the schematic.
Cheers.
So in the mean time can it be established that the discharge from a coil is forward emf ? Not Back emf ? That would be a big step.
Cheers
This is how coils behave when connected to switching circuits.
Yes please. If we use the term "current" can we agree to stick to the established convention of the direction of "travel"? And if we are talking about charges, then let's call them charges and accept that they go the other way around the circuit. But if we are analyzing circuit behaviour from the standpoint of electromagnetism, power dissipation, and all that, we should stick with the established convention, even though we know it is "wrong". Drive on the left side of the road if you are in Britain or Japan, please, and if you are on Earth.... current (consider "conventional" as always implied) goes from positive to negative.
Quote from: Dave45 on February 28, 2014, 06:13:15 PM
Ok fix the mosfet orientation - the mosfet is not what we are discussing here. This is how a discussion turns into miles of senseless posts.
We are talking about the bemf of a negatively pulsed coil and its polarity not the mosfet orientation, so its wrong, I told you I took it from a schematic that was handy.
Indeed. If we get the _basics_ wrong, it is difficult to make any progress at all. Posting wrong schematics and then trying to discuss them as if they were right, or that the wrong parts don't matter, makes it very difficult for us poor hidebound folks that build and experiment. And it can really_really_ lead people to stray very far from reality if it's not nipped in the bud, so to speak.
Would you like to see some good examples of this? Just go over to
http://www.energy-shiftingparadigms.com/index.php and read the "papers" that are posted there.
Quote from: TinselKoala on February 28, 2014, 08:34:54 PM
Yes please. If we use the term "current" can we agree to stick to the established convention of the direction of "travel"? And if we are talking about charges, then let's call them charges and accept that they go the other way around the circuit. But if we are analyzing circuit behaviour from the standpoint of electromagnetism, power dissipation, and all that, we should stick with the established convention, even though we know it is "wrong". Drive on the left side of the road if you are in Britain or Japan, please, and if you are on Earth.... current (consider "conventional" as always implied) goes from positive to negative.
Dave, which convention is used does not end up mattering mathematically. Negative current convention dictates that current flows from the negative pole of a power source back to the positive pole. Positive current convention follows the opposite path. Negative going CCW or positive going CW yields the same results.
Very funny Dave. If I were to use a cartoon to describe your viewpoint, the head would be stuck up a different hole.
Circuit Diagram posted by Dave45 here: (http://www.overunity.com/14343/silly-question-about-voltage-and-current/msg390144/#msg390144)
(http://www.overunity.com/14343/silly-question-about-voltage-and-current/dlattach/attach/134173/image//)
Open the above link in a New Tab in order to see the
image.
There really isn't anything wrong with the depiction
of an N-Channel MOSFET functioning as a Low Side
Switch. The arrow depicting the direction of switched
inductor current flow, when considering the polarities
of the semiconductor components, must refer to
"electron flow." The associated power supply polarity
is therefore self evident.
The red-letter label starboard of the Spike Suppression
Diode which parallels the inductor is incorrect, however.
The Inductor Discharge Kickback will forward bias the
suppressor diode with a resultant current flow consistent
with the Time Constant of that portion of the circuit.
The direction of current flow through the inductor in
both cases (charging and discharging) will be the same.
OK so I was mistaken in my first test and admit it, I'll explain where I went wrong later.
Anyway With the circuit posted below, I see that there is indeed a current through the coil snubbed by the diode to the supply, and it seems to cause a continuous current in the coil I used, which heated it right up. ;D However when the coil was discharged to a higher voltage the current went to zero. Which is desirable to me.
I don't like coil snubbing, I was never sure why, then I thought I knew why, but now I know I know why I don't like it. ;D
First picture is the circuit the dashed portion is the alternative/second arrangement.
Second shot is with the diode snubbing the coil to the supply.
Third shot is coil discharging through a higher voltage.
..
P.S. I better recheck things. :)
..
Seems correct Yellow trace is inverted on the scope to make it correct I think.
With the diode snubbed to the supply rail it uses 500 mA (maybe more) and heats up the coil but with the coil discharged to the battery it uses only 200 mA (maybe 300 mA) and doesn't heat up the coil and the current goes to zero.
I did the original test to try to show why I think snubbing is bad and this test shows even more reason fo me so I am very glad I did it.
..
I know if I were switching a coil to turn it on and off which waveform I would prefer to see.
Look at all that current in the snubbed to supply shot, maybe I should do a Rosemary and claim OU on that. Circulating continuous current. hehehehe.
It's a heater ! ;)
..
P.S. The previous test I tried this on had a charging circuit which allowed me to make a bad probe placement which simply did not read the current through the diode back to the supply rail. Operator error. Thanks to Milehigh I know see what I think is a real result and it still back up my dislike of the practice of coil snubbing to the supply. So I am happy in two ways. I realized a mistaken result and also keep my dislike of coil snubbing for a good reason.
My apologies MilHigh, and Thanks. :)
..
Now, please FOCUS, this is very simple question : if electrons movement is the reason for electric current then why it require CLOSED CIRCUIT ? Take a charged electrolytic capacitor with HUGE energy stored (like those big ones used for power factor correction) , connect negative terminal to some place having less electrons like maybe other capacitor positive terminal), then measure how much electrons moved to that empty place by simple voltmeter. I think you would find nothing changed, capacitor still charged ,energy still there, electrons won't move easily...
Waiting for your comments
Quote from: forest on March 01, 2014, 04:46:33 AM
Now, please FOCUS, this is very simple question : if electrons movement is the reason for electric current then why it require CLOSED CIRCUIT ? Take a charged electrolytic capacitor with HUGE energy stored (like those big ones used for power factor correction) , connect negative terminal to some place having less electrons like maybe other capacitor positive terminal), then measure how much electrons moved to that empty place by simple voltmeter. I think you would find nothing changed, capacitor still charged ,energy still there, electrons won't move easily...
Waiting for your comments
If you do not apply a potential across something, why do you expect charge to move?
Quote from: TinselKoala on February 28, 2014, 09:32:29 PM
Very funny Dave. If I were to use a cartoon to describe your viewpoint, the head would be stuck up a different hole.
Im sorry the sarcasm was rude, just take a little time study electron flow, the direction it moves through a diode, the bemf and the direction it moves through a diode.
You will realize the bemf has an opposite polarity, this imply's that a neg pulsed coil will have a pos return and a pos pulsed coil will have a neg return.
Study ionization, the plarity of the electrodes, the ion clouds formed around the electrodes, how that relates to a pos and neg pulsed coil.
Alternating current is more that it imply's, a coil pulsed with alternating current receives a pos and neg pulse.
Check out static neutralizers and you will see how your neutralizing any extra energy you may be drawing from the ambient by powering a transformer with both legs.
Split the pos and neg so they cant neutralize each other.
Quote from: Dave45 on March 01, 2014, 06:14:10 AM
Im sorry the sarcasm was rude, just take a little time study electron flow, the direction it moves through a diode, the bemf and the direction it moves through a diode.
You will realize the bemf has an opposite polarity, this imply's that a neg pulsed coil will have a pos return and a pos pulsed coil will have a neg return.
Study ionization, the plarity of the electrodes, the ion clouds formed around the electrodes, how that relates to a pos and neg pulsed coil.
Alternating current is more that it imply's, a coil pulsed with alternating current receives a pos and neg pulse.
Check out static neutralizers and you will see how your neutralizing any extra energy you may be drawing from the ambient by powering a transformer with both legs.
Split the pos and neg so they cant neutralize each other.
Dave why don't you study up on things like solenoid drivers? Low-side only, high-side only, and "X" drivers have been in the market for decades. Variants of single sided drives using bi-voltage, single sided diode clamps, zener clamps, and simultaneous two switch switching are old hat. If you are listening to music on any kind of computer, or portable device, the Class D amplifier that is driving your speakers or headset is switching the output up to 1 million times per second. Class D amplfiers do not pull energy out of the ambient environment. Engineers carefully design the PCB layout and EMI filtering so that they do not emit excessive RF into the environment.
Quote from: forest on March 01, 2014, 04:46:33 AM
Now, please FOCUS, this is very simple question : if electrons movement is the reason for electric current then why it require CLOSED CIRCUIT ? Take a charged electrolytic capacitor with HUGE energy stored (like those big ones used for power factor correction) , connect negative terminal to some place having less electrons like maybe other capacitor positive terminal), then measure how much electrons moved to that empty place by simple voltmeter. I think you would find nothing changed, capacitor still charged ,energy still there, electrons won't move easily...
Waiting for your comments
I think you would find that the _other_ terminal of the capacitor has the opposite charge, and opposites attract, across the cap's dielectric. There is no current pathway in your hypothetical question for the charges to equalize _across the capacitor's dielectric_. Why should the electrons, or rather the charge they carry, travel away from someplace that they are strongly attracted to by the presence of all that opposite charge on the other side of the dielectric? You've offered them an extremely high-resistance non-pathway in the other direction along a very strong electric field gradient .... an open circuit as you have set up. There is no pathway for the equal and opposite charges in the capacitor to neutralize.
Quote from: Dave45 on March 01, 2014, 06:14:10 AM
Im sorry the sarcasm was rude, just take a little time study electron flow, the direction it moves through a diode, the bemf and the direction it moves through a diode.
You will realize the bemf has an opposite polarity, this imply's that a neg pulsed coil will have a pos return and a pos pulsed coil will have a neg return.
Study ionization, the plarity of the electrodes, the ion clouds formed around the electrodes, how that relates to a pos and neg pulsed coil.
Alternating current is more that it imply's, a coil pulsed with alternating current receives a pos and neg pulse.
Check out static neutralizers and you will see how your neutralizing any extra energy you may be drawing from the ambient by powering a transformer with both legs.
Split the pos and neg so they cant neutralize each other.
Dave take a good look at the scope shot and the circuit that produced it ( a few posts back), now notice that all the voltage is positive and all the current is positive,
If you look at the grounding point you will see that when the switch turns on the scope probe for voltage drops to zero, however the current rises positive and then declines positive. The applied voltage is DC and on switch on the scope probe drops to zero volts then on switch off the voltage rises positive as the current declines positive.
There is no negative to it. Unless a wanted or unwanted oscillation causes it.
People say the voltage of the coil reverses but going by my scope shot I do not see how, the voltage is scoped across the coil and the voltage is always nil or positive.
If someone else would like to point out where the voltage across the coil reverses I would be glad to try to understand it.
Cheers
What happens is the coil changes from being a "load" to a "supply" and so the actual voltage "polarity" does not change nor does the direction of the current the only thing that changes is the end of the coil we think of as positive with relation to "load" "supply" situations, this is due to voltage level not positive negative as such. In other words when the coil is taking current the dot to designate the positive end of the coil is at the top, then when the coil is supplying current the dot to designate the coil polarity is moved to the bottom. It's a change in the mind only, to rationalize the thing we are seeing and so like minded people can communicate about it. There is much disinfo about DC pulsing of coils. And much is not explained by some of the other trained guys or too many people ignore them.
I don't mind to argue with them, because I feel anyone can be mistaken, I don't mind being wrong if I learn something, but most of the time they are correct, it's just that most of us do not understand how they say it. Or misunderstand what they say. And the misunderstanding goes both ways of course.
The discharge of a coil is forward emf (or just plain emf) as opposed to counter or back emf. This is seen by the current remaining to flow the same way, the voltage reversal across the coil happens when the coil discharge causes a higher voltage than the supply, but it does not go negative.
Cheers
Farmhand, your coil circuit appears to be "overdamped" or even critically damped, in that there is no trace of an oscillatory ringdown when the supply is cut off. Other circuit parameters could be adjusted so that a ringdown at the resonant frequency of the circuit could appear, maybe. This may or may not have a good or bad effect on the performance depending on what the goal is.
When an L/R or L/C circuit is switched must faster than the reactive network time constant, trapezoidal waveforms such as seen in farmhand's oscilloscope captures result.
The coil was just one I had at hand to make the demonstration, the wave forms look much the same as the diagrams that you posted Mark. It's not a coil I would choose to use as a boost converter or something, it was just there so I used it. Apart from the absence of a ring down on the voltage trace it all looks normal to me, I only used 100 Hz to pulse it and it's 35 mH, to get past resonance It would require about 65 uF of capacitance. It's resistance is a bit much as well.
I make inductors for boost converters and such from small Iron powder cylinder cores wound with one or two strands of 1 mm Wire to get whatever inductance I want considering the frequency i want to use, my boost converters are high efficiency, and they do ring a bit.
eg. I designed one with a pair of 65 uH inductors to use at 40 kHz, I found this online app helpful. http://www.daycounter.com/Calculators/Inductor-Current-Power-Calculator.phtml It's a real handy piece of kit.
Cheers
Ah, so there simply isn't enough time to see the ringdown if it is there. Thanks.
Quote from: Farmhand on March 01, 2014, 11:49:09 PM
The coil was just one I had at hand to make the demonstration, the wave forms look much the same as the diagrams that you posted Mark. It's not a coil I would choose to use as a boost converter or something, it was just there so I used it. Apart from the absence of a ring down on the voltage trace it all looks normal to me, I only used 100 Hz to pulse it and it's 35 mH, to get past resonance It would require about 65 uF of capacitance. It's resistance is a bit much as well.
I make inductors for boost converters and such from small Iron powder cylinder cores wound with one or two strands of 1 mm Wire to get whatever inductance I want considering the frequency i want to use, my boost converters are high efficiency, and they do ring a bit.
eg. I designed one with a pair of 65 uH inductors to use at 40 kHz, I found this online app helpful. http://www.daycounter.com/Calculators/Inductor-Current-Power-Calculator.phtml (http://www.daycounter.com/Calculators/Inductor-Current-Power-Calculator.phtml) It's a real handy piece of kit.
Cheers
Farmhand:
Thanks for the calculator link...I have added it to some others that I use.
Bill
Speaking of calculators, I found one that allowed to enter some simple data to calculate the required capacitance to compensate a certain angle of inductive phase lag, but forgot to bookmark it, anyone?
BTW. what are you talking about, a collapsing magnet field does not cause a back emf?? Every cheap voltmeter will proof it does. It takes some time until the current has built up the magnet field. If you disconnect the power source, it will flow back like a rubberband. At least that is what I observed, I have no idea if it contradicts any theories. The voltage is ahead of the current in a coil when you turn it on, that "investment" of energy is the same amount as in the back emf. So no extra energy from the back emf.
I wanna say people should not trust their oscilloscopes too much, but then again, I know that I know nothing.
If we study classical physics we will probably never have free energy, think of a way and try it. V8Karlos came here in 2012 and gave some great suggestions a direction but not sure if anyone really tried his experiments.
He was using bifilar and caduceus coils like capacitors and getting some anomalous results, I have been studying his circuits, would suggest others do the same.
DC pulsed into a coil is negative energy, but the bemf from that coil will be pos.
AC pulsed into a coil is negative on one leg and pos on the other.
Ionizer's show us AC has a pos and neg leg, they also show us how to pull apart the ether so the pos and neg can be collected and static neutralizers show that if you do this with only one coil the effect is neutralized.
What does this mean and how can u make it work to your advantage.
Is voltage and current just pressure and flow, are negative and pos energy both flowing in a circuit
Will a diode pass both electrons and holes (positron)
Quote from: dieter on March 04, 2014, 04:58:24 PM
Speaking of calculators, I found one that allowed to enter some simple data to calculate the required capacitance to compensate a certain angle of inductive phase lag, but forgot to bookmark it, anyone?
BTW. what are you talking about, a collapsing magnet field does not cause a back emf?? Every cheap voltmeter will proof it does. It takes some time until the current has built up the magnet field. If you disconnect the power source, it will flow back like a rubberband. At least that is what I observed, I have no idea if it contradicts any theories. The voltage is ahead of the current in a coil when you turn it on, that "investment" of energy is the same amount as in the back emf. So no extra energy from the back emf.
I wanna say people should not trust their oscilloscopes too much, but then again, I know that I know nothing.
It depends on what "you're" definition of Back emf is, if you are defining it as returned emf then yes ok, the emf is returned for reuse if the circuit allows it to, but if you're definition of Back emf is the same thing as counter emf then no,
the counter emf happens at the same time as the applied emf and the
coil discharge happens after, the current is in the same direction and is therefore (forward emf or just emf).
Cheers
QuoteDC pulsed into a coil is negative energy,
Uhhhh..... no.
The diode drawing that mentions "negative energy" is also a facepalmer.
Sorry, dude, but you aren't going to be making much progress if you think like that. What is the current flowing through a reverse-biased Zener diode, I wonder, Orange Energy with the pips separated out?
Dave, others:
QuoteDC pulsed into a coil is negative energy, but the bemf from that coil will be pos.
AC pulsed into a coil is negative on one leg and pos on the other.
There is no such thing as "negative energy" and "positive energy" like you are thinking. Those comments are ridiculous.
It's the same old cliche. If you actually understood how a coil works, you would be making some real progress. But then people get defensive and all pouty when they hear that, or they refuse to even acknowledge the issue.
Some of you build circuits with coils all the time without knowing how they work. Dave says that he has his "own understanding" but it's clearly the wrong understanding.
As a bare minimum, you should undertake to understand how coils and capacitors actually work, and not how you believe or fantasize how they work. It's like you are a bunch of people in a wood shop and you don't understand how chisels and files work.
It's really that bad. What makes it worse is that there are probably 250+ perfectly good clips on YouTube and 1000 web pages were you could learn how these components work.
The rant is over and nothing will change!
MileHigh
Yes energy is energy. If we think of it like this - If positive energy does or is work then is negative energy undoing work ? Or is negative energy doing negative work ;D. Makes no sense.
..
Here is the best that you can do with ideal switches.
We live in a polar world the earth is negative the ionosphere is positive when lightning strikes leaders form from both, two forces seeking neutralization.
The air we breath the atmosphere we live in is neutral, pos and neg joined to collect from the environment we have to seperate the charges.
You laugh when I mentioned pos and neg charges and yet its everywhere, the ionization circuit is self explanatory.
Quote from: Dave45 on March 05, 2014, 07:18:37 AM
We live in a polar world the earth is negative the ionosphere is positive when lightning strikes leaders form from both, two forces seeking neutralization.
The air we breath the atmosphere we live in is neutral, pos and neg joined to collect from the environment we have to seperate the charges.
You laugh when I mentioned pos and neg charges and yet its everywhere, the ionization circuit is self explanatory.
Dave, it is self-explanatory when you add the parasitic capacitor at either end that exists whether you choose to draw it or not. So what do you want to do with this other than build a mosquito zapper? Are you out to try and harvest lightning strikes? Some of the challenges there are that the power levels are enormous and the form factors are very low. IE it's a struggle to keep stuff from getting vaporized, and what does not get vaporized still loses tremendous energy to heat.
QuoteYou laugh when I mentioned pos and neg charges
No, Dave... we "laughed" when you talked about positive and negative _energy_. Charge is not energy, although it may be close, and charge is a conserved quantity. Positive and negative Charge is a very different concept than positive and negative Energy.
One of the big problems that we encounter here is this laxity of meaning. Check the definitions of your terms, please. You can certainly make up your own language (splitting the positive? Whaat?) but if you do then nobody but you will know what you are talking about. If that many.
You say electrons are negative and that current flows from neg to pos then turn around and study circuits right the opposite, how can you ever understand whats really happening.
You said we laughed, do you really think that bothers me, I guess that makes me the minority, I like it, I really never fit in anyway and I like to follow my own path.
Ya'll keep patting each other on the back and agreeing and keep paying for something that is free.
Quote from: Dave45 on March 05, 2014, 09:16:49 AM
You say electrons are negative and that current flows from neg to pos then turn around and study circuits right the opposite, how can you ever understand whats really happening.
Why don't you ask the people who designed your computer? They do exactly that: they know which way electrons go, and they use the conventional circuit analysis that is "right the opposite". In fact, every single electronic device that you have ever seen that actually works, is designed by people that you think are stupid and wrong.
Quote
You said we laughed, do you really think that bothers me, I guess that makes me the minority, I like it, I really never fit in anyway and I like to follow my own path.
No, Dave, YOU said WE laughed at you for your positive and negative charges, and I corrected you because what I was laughing at, if I laughed, was your negative ENERGY. Charge is not energy, electron flow is not energy, conventional current is not energy, and especially when you pulse a DC coil you are NOT GIVING IT NEGATIVE ENERGY. Please try to keep your story straight.
Quote
Ya'll keep patting each other on the back and agreeing and keep paying for something that is free.
Whatever, dude. You just go ahead, keep on driving on the left when everybody else is driving on the right. I'm staying the hell out of your way.
Quote from: Dave45 on March 05, 2014, 09:16:49 AM
You say electrons are negative and that current flows from neg to pos then turn around and study circuits right the opposite
You say well we know its wrong but it works, have you ever asked yourself why it works, could it be that there is a flow from pos to neg as well as a flow from neg to pos.
What would happen if two charges a pos and a neg charge were separated, what happens when combined.
Quote from: TinKoala
...they know which way electrons go,
Aye, they do. In a vacuum or near vacuum as inside
a vacuum tube there is no question. :)
Within a conductor the movement is somewhat
different as it functions as a queue; each individual
initiating electron moves a very short distance but causes
the entire queue to move very rapidly the same short
distance. 8)
Now, that appears to be some sort of incredible power;
a single electron is able to move billions or even trillions
of others in the queue! :o
Quote from: Dave45 on March 05, 2014, 09:35:28 AM
You say well we know its wrong but it works, have you ever asked yourself why it works, could it be that there is a flow from pos to neg as well as a flow from neg to pos.
What would happen if two charges a pos and a neg charge were separated, what happens when combined.
I have asked myself that, and I have also "asked" much more clever folks than I, by studying and working problems in electrical engineering. You may want to locate the book called "Circuits, Devices and Systems" by Ralph Smith, to get an idea about just what such study entails. Look, it is available on the internet:
http://pdf7835.chuobooks.com/circuits-devices-and-systems-edition-PDF-3098440.pdf (http://pdf7835.chuobooks.com/circuits-devices-and-systems-edition-PDF-3098440.pdf)
Separating positive and negative charges and holding them apart is what a Capacitor does. When you combine them by shorting the capacitor or connecting it to a circuit, you get an arc, or usable current in your circuit elements, until the charges have completely neutralized each other or you put more charge in.
Next question?
(Little NE2 neons will tell you the polarity of charge. Only the negatively charged electrode glows in those. If you see both electrodes glowing you know you have AC in there at some frequency. If not... the electrode that IS glowing is the one that is negatively charged. It is emitting electrons that ionize the neon and when the neon atom "de-ionizes" it emits a photon of energy equal to the ionization energy.)
http://www.youtube.com/watch?v=T5I_BM4E00E
Quote from: SeaMonkey on March 05, 2014, 02:49:37 PM
Aye, they do. In a vacuum or near vacuum as inside
a vacuum tube there is no question. :)
Within a conductor the movement is somewhat
different as it functions as a queue; each individual
initiating electron moves a very short distance but causes
the entire queue to move very rapidly the same short
distance. 8)
Now, that appears to be some sort of incredible power;
a single electron is able to move billions or even trillions
of others in the queue! :o
Yep, the vacuum tube, especially those like CRTs that are designed to produce an electron beam that has effects outside the tube, shows the truth. Filaments and cathodes are connected to the Negative, the Plate is connected to the Positive and a potential of 400 V or more (the "B+" supply) is used to pull those electrons out of the cathode and boost them towards the plate across the electric field gradient. They are formed into a nice tight beam by magnets and coils and fieldshaping electrodes... the magnets bend the beam because it is a current-carrying thing just like a plasma wire, sort of. But the "conventional current" used to analyze the circuit is still taken to be "flowing" from the plate Anode to the Cathode, just like the tube were a diode. Which it is, with other things thrown in there to make it a triode or pentode.
Now let me make a very bold statement. There is nothing wrong with making up your own theories about stuff like electricity! But if you do so, you should still understand enough about the hidebound "conventional" theory, which is Quantum Electrodynamics in its various forms, to show how your theory is different, more interesting, that it makes different testable predictions than the existing theory, and so on. Having experimental evidence of a successful prediction that goes against conventional theory's prediction.... now that will ring some bells and that's why we are all here, I think.
Quote from: Dave45 on March 05, 2014, 09:16:49 AM
You say electrons are negative and that current flows from neg to pos then turn around and study circuits right the opposite, how can you ever understand whats really happening.
You said we laughed, do you really think that bothers me, I guess that makes me the minority, I like it, I really never fit in anyway and I like to follow my own path.
Ya'll keep patting each other on the back and agreeing and keep paying for something that is free.
Dave, the sign of charge is arbitrary for purposes of circuit analysis. It is only important to maintain the same convention throughout the analysis. The vast majority of the world chooses to follow positive current convention for the simple reason that it reduces the number of sign reversals when analyzing a circuit. Tracking electrons drifting from the negative pole of an energy source say CCW around a circuit is mathematically equivalent to following 'electron holes', IE positve charge CW around the same circuit. This has been explained to you and illustrated for you. If the concept still doesn't make sense to you, then please pick up a book on circuit analysis.