Wanting to make some negative DC to watch conventional current flow

[saturated]

I wanted to get my hands on some negative DC and see (ok we know it will  ) if it would pass through a diode from the cathode side

The first "circuit" I came up with in the midst of drawing it I thought oh wait this is gonna be a normal circuit we are just looking at it from another perspective but then I realized yeah no not quite

You cannot view this attachment.

I kept trying to figure out how to wire this up to get negative DC and about the best I could do was having to utilize an external ground

You cannot view this attachment.

Which seemed to work
For the ground i connected the positive terminal of PACO to the Ground terminal of my scope
You cannot view this attachment.

I slowly dialed up and adjusted it to about 5C DC

So I had -5V DC at the anode side of the diode
You cannot view this attachment.

And then -4.4V DC at the anode

You cannot view this attachment.

So I think I did this right and seems to agree with what I was looking for

Then I remembered well there's nothing wrong with the first circuit it's just that the cathode will be at zero  so I hooked it up and get about 0.67V DC at the anode

[saturated]

So what's all the fuss about yeah wow the sky is blue  is this breaking news ?

The point being that earlier before this I only considered conventional current flow I just knew "the positive flowed" in the direction of the arrow

All of this kinda worked until I started trying to figure out alternating current and diodes 

What would be cool is to see some negative fluctuating DC pass through 

[saturated]

The next phase was a little easier

Wanting to see negative fluctuating DC travel through a diode

I connected a diode and resistor same ones in series and added a signal to the cathode of the diode

Then put my scope probe on the anode side

When I had the DC offset of the signal at zero or above no signal passed through

You cannot view this attachment.

(Sorry I have no idea why my phone camera only gets a fragment of the green lines)

Now if I turn the DC offset negative the signal can be seen at the anode side of the diode

You cannot view this attachment.

 

[g1]

What makes you say "this is not negative, it is ground" ?
From the PACO schematic, the (-) terminal appears to be floating, not grounded.  So you can connect the (+) output to ground and you have a negative supply at the (-) terminal.  This seems to be what you did anyway.

[g1]

When I had the DC offset of the signal at zero or above no signal passed through

Your signal wasn't big enough.  If it had been big enough, the peaks beyond the diodes forward voltage drop would have got through and you would have seen the negative alternations of the sinewave.  This would have been rectified but unfiltered DC.

When you added the DC offset, you 'biased' the diode to be on, so the complete sinewave was able to get through.

[saturated]

Yes sir thanks I will do that 

Ok I was gonna spare you the gory details but anyway

You cannot view this attachment.

You cannot view this attachment.

You cannot view this attachment.

I need to figure out why I have a complete green 🍏 line in real life but when I take a picture I only get partial fragments

[Kaz Kylheku]

What do you mean by "negative DC" though? Anti-conventional current, i.e. flow of negative charge?

Both currents are always present together, opposite and equal.

If there is a chemical cell battery in the circuit, inside a battery, both kinds of charged particles, anions and cations, are present and moving in opposite directions.

Imagine you have a circular ring of copper with current moving around in it (perhaps induced magnetically).  If, in our frame of reference, we spin the ring in the opposite direction of electron flow, we can get the electron current appear to stand stil. Then what is moving in circles is the nuclei of the copper atoms (together with any lower shell electrons that don't participate in current). That's a real flow of positive charges.

In other words, in a wire, the fact that protons are not moving while electrons are de facto means that the protons are moving in the opposite direction. If you could ride on top of an electron, you would be leaving protons behind in the rear view mirror, and that's essentially your conventional current flow.

It's like the wind in your face when you're on a bike. The air may be standing still on a windless day, but you have "conventional wind" in your face. 

What is an intepretation of conventional current in a vaccuum? Electrons are shooting from the cathode of a tube to the anode. Nothing is flowing backwards. But again, if you could ride on top of an electron, you would see that the more positive anode/plate is rushing toward you.

[saturated]

Yes sir thanks I appreciate the input to the discussion 

I kinda see your point

For instance I can pop the hood on my 1980 Bronco and measure 12.6V DC or whatever

Then I can switch the probes and go hey look now I've got minus 12.6V DC

You cannot view this attachment.

So perhaps it's all relative 

That's why we have to use reference to ground to make sense of it all

I guess

I mean c'mon man you a lot more about this stuff than me 

But obviously alternating current is real 

[J M Fahey]

There is not negative or positive by itself, voltages are always relative to some other value.
If you have a battery and ground positive end, the other end gives you negative voltage.
All normal voltage souces have both.

That said, *electrostatic* voltage sources, such as when you repeatedly wipe a piece of plastic with a piece of dry cloth, can be "unipolar" which can be checked by what they attract, such as dust, lint, etc, but they can not supply a current, light a lamp, run a motor, etc.

[Kaz Kylheku]

I have something useful in my background in that long before learning some electronics as a hobby, my senior year physics class in high school covered electricity.

We learned things like, oh, voltage being the electric potential of a charge: J/C (joules per coulomb).  And if we go in the most changing gradient of an electric field, the faster the voltage changes, the stronger the field is: strength is V/m (volts per meter). Electric field strength is also N/C (newtons per coulomb): how hard is the field exerting a force, per unit charge.

The two relate: if we start with V/m, and substitute the J/C units for voltage, we get J/Cm.  But energy J is force times distance: N・m.  Substitute that for J into J/Cm and we get Nm/Cm, where the m's cancel out to get N/C: newtons per coulomb field strength.

Space is permeated by electric fields. At any point in space, we can assign a voltage.  There is no absolute zero voltage: we must pick a point and call that zero, then express all other voltages relative to that: lower voltages negative, higher ones positive.

Say we have a charge sitting in a field. The electric force is acting on it. If we move the charge upfield against this force by some distance we do work on it (N・m), raising its potential energy. If we divide that potential energy by the amount of charge, we get the potential: voltage. We have raised the charge to a higher voltage. The greater the field strength (J/C) the harder we have to push, so the voltage change is higher. That's captured by the equivalent V/m units of field strength: the more volts we raise over a given distance, the stronger the field must be.

Also in that same high school course we covered how if we do a surface integral around some closed surface, of the electric field through that surface, if the result is nonzero, it means that the surface (e.g. sphere) contains a charge. That's what it means for there to be a charge somewhere: more electric field lines are coming out of a space than in, or vice versa. We can add them all up with a surface integral around the space. That was an advanced topic; all of us in that physics class were also in the calculus class, so it was okay to get into that sort of thing.

This stuff is good to know for being less mystified by certain aspects of circuits.