Electron Reversing Device

[tinman]

@MileHigh
Quote:  In you example, the longer the pulse is on, the less of a resistance the coil appears to be.  In that sense Ohm's law applies but the results are all a function of time.

So if our input voltage and time per cycle is constant over that 1 ohm resistor,and our output voltage on one output resistor is higher for the same amount of time-then what dose ohm's law tell us?
Then on the other 1 ohm output resistor where the voltage is the same as the input,but the time that voltage is across that resistor is longer than the input pulse-what dose ohm's law say about that?

Quote:Have you measured the values of the three separate one-ohm resistors?  What is their tolerance band colour?

As stated in the video,video description and in this thread,i measured all 3 resistors to be 1 ohm-to the second decimal.
All three resistors were rotated,and the results are the same.
The tolerance band color is gold.

Quote:We need a schematic of the circuit also.  I understand how the circuit seems obvious to you.  But it's not obvious if you (meaning us readers) are not inside your personal "bubble."

In my post just above you last-In this video,i have removed the DMM's and input diode's-eliminating any part they may have been playing.
In there place,i have 3x 1 ohm resistors.
The schematic is posted a few time's in this thread.

Quote:Also note that a coil is not a resistor.  R = V/I for a resistor.  A coil does not have the property of resistance

??:-A coil is very much a resistor in the true meaning of the term resistor.
My pancake coil with 5.2 volt's across it draws 790ma
R=V/I  In this case my coil has a resistance of 6.58 ohms
My DMM says it's 6.6 ohms.
Every material has resistance unless it is super conductive.
As we know,the resistance will climb higher as the voltage climbs within the inductor (depending on pulse rate<time)due to the back EMF.
But as the current build's up within that inductor,it will reach a point where the voltage across the inductor and the supply voltage are the same.
The ringing at the start i see as being the supply voltage, inductor voltage and current trying to come to an equilibrium.Keeping in mind that at the frequency we are running at,that time period of ballance is so very very short.
We also have to take into account the skin effect with AC and pulsed DC(single direction altinating current)
As we know,the higher the frequency-the greater the skin effect,and the higher the resistance becomes within the inductor.

[MileHigh]

Tinman:

Thanks for the verification of the one-ohm resistors, I missed that.

So if our input voltage and time per cycle is constant over that 1 ohm resistor,and our output voltage on one output resistor is higher for the same amount of time-then what dose ohm's law tell us?
Then on the other 1 ohm output resistor where the voltage is the same as the input,but the time that voltage is across that resistor is longer than the input pulse-what dose ohm's law say about that?

It looks to me like your "current entrapment loop" is at play.  As the circuit runs the BPC sees a voltage waveform across it were positive voltage dominates over negative voltage.  That sets up a clockwise current loop through the BPC.   You could check the coil voltage with your scope.

When you look at the negative pulse only, you see it starts at a higher initial current level compared to the input and then slopes down.  The higher initial current and the sloping down comes from the BPC discharging its stored energy.  So the negative pulse shows two sub-currents, the sub-current due to the function generator pushing current through the left LED, and the decaying sub-current that is being driven by the discharging BPC.

By the same token when you look at the positive pulse only, you see three currents this time.  During the positive part of the positive pulse you see a sub-current that is going through the right LED and you also see a small sub-current that is going into the BPC to energize it.  So the total current that you see during the positive pulse can be divided into two sub-currents. There is a third observable current, and that is during the "off" part of the positive pulse.  That's the coil discharging it's stored energy again.  It's the same current I make reference for the negative pulse above as a sub-current, but this time you are seeing it through the right "M4" resistor instead of the "M3" resistor.

So where is the power coming from to keep the BPC "stoked?"  It comes during the positive output of the square wave generator.  When the signal is positive current flows through the right LED, and it flows through the BPC.  The two components are in parallel.  So the current is actually increasing slightly as the BPC charges up and the impedance of the overall circuit drops as time goes on.  I looked at the clip again and I think that you can see the current rising very slightly during the positive pulse.

Knowing that a half-cycle is 50 uS and it looks like we might be seeing about 4 time constants by looking at the waveform, that means the time constant is about 12.5 microseconds.  Ignoring the LED for a second, then we have L/R = 12.5 microseconds.  So that means that a guesstimate for the value of L is 107 micro-Henries.  (R is the two one-ohm resistors in series with the 6.6 ohm resistance of the BPC,  8.6 x 12.5 = 107.5.)

If you do the tests like I suggested you may be able to confirm this.

So in a nutshell, when you connect the coil, you clearly see increased power consumption by looking at the input current waveform.  The input current when the pulse is high increases slightly over time.  That's the "extra juice" that "powers" the coil (i.e.; energizes it during the positive pulse so it can discharge energy during the negative pulse) and keeps current circulating clockwise through the coil.  So when you look at each of the two current sensing resistors associated with the LEDs, in both measurements you see the same "extra" current.  You see the same "extra" current on the "M3" and "M4" resistor during the low output from the square wave generator.  Note when you look at the schematic that the current flows through "M3" and "M4" when the coil discharges because they are in series for the discharging current loop for the coil.

That's my story and I'm sticking to it unless somebody corrects me.

MileHigh

[MileHigh]

Tinman:

In my post just above you last-In this video,i have removed the DMM's and input diode's-eliminating any part they may have been playing.
In there place,i have 3x 1 ohm resistors.
The schematic is posted a few time's in this thread.

I am aware of that but the last time the schematic was posted was almost two weeks ago.  I used the schematic shown in post #51 but I am not 100% certain that that is the correct final-version schematic.   When you do a new circuit please just make a new schematic as a courtesy to your readers.  Also put designations on all you components so that in your clips you can say, "I am putting the scope across R1."  That would make life so much easier for everybody and the mental calisthenics required to follow what you are doing would be eliminated.

(http://www.overunity.com/Smileys/default/huh.gif)(http://www.overunity.com/Smileys/default/huh.gif)??:-A coil is very much a resistor in the true meaning of the term resistor.
My pancake coil with 5.2 volt's across it draws 790ma
R=V/I  In this case my coil has a resistance of 6.58 ohms
My DMM says it's 6.6 ohms.

Sorry I wasn't being clear enough.  In electronics talk it's commonly taken for granted that you are ignoring the resistance associated with a coil.  I was only talking about the inductance of the coil.  A real-world coil is modeled (simplified) as an ideal inductor in series with a resistor.  So when you first connect a voltage source to a coil, it looks like an open circuit because the inductance is an open circuit on first contact.  In other words, the inductance dominates on first contact and no current flows.

As we know,the resistance will climb higher as the voltage climbs within the inductor (depending on pulse rate<time)due to the back EMF.
But as the current build's up within that inductor,it will reach a point where the voltage across the inductor and the supply voltage are the same.

On first contact the effective resistance is infinity and then over time it drops to the electrical resistance of the coil itself.  Hence the effective "look alike" resistance of a coil drops over time.  What you say above is not clear and you would need to draw up a circuit to discuss it.  Even something that sounds simple is not so simple in this context.

The ringing at the start i see as being the supply voltage, inductor voltage and current trying to come to an equilibrium.

The ringing that you are seeing by definition is due to some inductance interacting with some capacitance and resistance in the circuit.  Exactly how it is being set up and why it happens and what the actual inductance component is and what the actual capacitance component is would require some more investigation.  The truth is that it's only a secondary issue and it's not significant so just as easy to ignore it for now.

If anybody is really serious about this, the basic analysis for what is going on in this circuit should be done on paper.  You use your scope and some graph paper.  You look at your schematic and you identify points where you want to measure voltage and current.  Then with your graph paper you sketch out the voltage waveform from the square wave generator.  All you need to do is sketch out is say three full cycles.  Then below that you sketch out the input current waveform.  Then below that you sketch out the current waveforms for the two LEDs.  Then you could add my suggestion for the voltage waveform across the BPC (that's also the voltage across the right LED.)  Then for sure you want to sketch out the voltage at the output side of the input resistor.  That's another mystery voltage point in your schematic because you didn't look at it.

That's how it's really done when you learn about electronics.  You can see that there are multiple voltage nodes and current loops in your schematic.  By drawing out the timing diagram for the circuit it should all start to come together.  You see the interrelationships between different waveforms and you put it all together and then it all starts to make perfect sense.

What's implicit in what I said is that I am not 100% confident that my take on your circuit is correct for every point and someone might correct me.  To be absolutely sure I would have to be sitting at the bench myself drawing out the timing diagrams.  Certainly something equivalent to using your scope and making up a timing diagram would be to simulate this circuit in pSpice.  Then you drop virtual probes onto the schematic and have it generate the timing diagrams.

The thing to keep in mind is to let cooler heads prevail.  You are playing with a basic circuit so there is no reason to assume that anything unusual is happening.  Also, just making a few measurements and drawing a preliminary conclusion is in almost all cases not going to cut it.  It's not as simple as "comparing input to output" and almost never is.  Plus you are just looking at the currents and not looking at the voltages.  The BPC coil in your circuit is converting voltage excitation into current flow and setting up an "extra" current loop in your circuit that got superimposed on top of your "regular" current output.  There is no challenge to Ohm's Law and there is no over unity.  It's just basic circuit analysis.

MileHigh

[MileHigh]

Poynt:

If you are still checking in, I finally decided to try pSpice Student for Tinman's circuit.  I didn't see the download on the Cadence website, so with some trepidation I downloaded pSpice Student 9.1 off of what appeared to be a legit engineering school web site.

I did the schematic without too much trouble.  However, I can't seem to get the simulation going.  I don't even see the simulation toolbar and I tried F11 and nothing happens.  Perhaps I need to import another library or something?

I am just using standard diodes for now.  I am so new I don't even know if there a models for LEDs readily available in the student version.

I am attaching my masterpiece.

Whoops I just noticed that my timing parameters for the square wave are wrong!

[poynt99]

It looks ok MH.

I would add a 200 Ohm in series with your FG to simulate tinman's FG pot.

Did you set up a simulation profile? i.e. how long the run is etc? What actually happens when you run the sim?

There are no decent models for LED's imo. I sometimes use 3 diodes in series instead. I did create my own model for an LED once on a circuit I was working on with ION.

You may have created a conventional schematic rather than a simulation schematic. That would explain why you don't see the simulation toolbar.