Silly question about voltage and current

[Dave45]

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.

[MileHigh]

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

[TinselKoala]

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!

[Dave45]

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. 

[MarkE]

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.