Physicists, please help a humble electronics engineer understand his electrons better!
What I was taught in my recent electronics degree-> A voltage drop is indicated over a component (e.g resistor) the higher potential end is labelled (+) and the lower (-) The potential energy difference causes charged particles to 'fall' through the electric field created by the potential difference, conventional current falls from + to – and electrons from – to +
This is where I am confused; one end of a resistor doesn't have more charge that the other i.e one end isn't more charge-positive than the other like a battery or capacitor, the same charge carriers are at different energetic states after experiencing the resistor and electrical energy is different to charge (gravitational energy and mass analogy), so is this +/- labeling used differently to mean both charge and potential difference just to confuse me?
And if so, does this mean that electrons are falling uphill to the higher energy state? or is this where the gravitational energy analogy breaks down, i.e is high and low energy purely relative in electrical domain like +/- charge are relative to eachother and it;s just a matter of defining 'ground', so high potential for a positive charge is low potential for a negative charge?
Best Answer
The labeling has perfect physical meaning.
Consider we have fixed positive charge which creates electric field $\vec E$. Electron in this field will be attracted to positive charge with force $\vec F = q\vec E$, where $q$ is electric charge (with sign!). From the latter we can see that force and field have opposite directions for electron. In particular, it follows that $\vec E$ points from positive to negative charge.
From the point of view of energy gain, the electron has lowest energy close to positive charge. The energy of the former is related to the electric field potential $\phi$ (integral of force) as $\epsilon = q\phi$ and is the lower the higher $\phi$ is. Coulomb force drags electron towards positive charge, i.e. in direction opposite to electric field $\vec E$. This is the physical illustration of the relation $\vec E = -\mathrm{grad}\phi$ (gradient of some scalar field is the vector pointing towards the direction of increasing field).
The density of current passing through a conductor cross-section can be written as $\vec j = qn\vec v$, where $q$ is charge, $n$ - density of charge carriers, $v$ - speed. The latter is linked to applied electric field as $\vec v = \mu \vec E$. Note that electrons move in the direction opposite to electric field, as was shown above. Here $\mu$ is carrier mobility and is defined as $\mu = q\tau/m^*$, where $\tau$ is an average scattering time and $m^*$ is "effective" mass of the carrier. Using the above one can finally arrive to the differential form of the Ohm's law: $\vec j = (q^2n\tau/m^*)\vec E$, where the quantity inside the brackets is electric conductivity or inverse resistivity.
This allows to answer the current "sign" question.
Voltage drop between points 1 and 2 is the difference of electric field potential in this points, $V_{12} = \phi_1 - \phi_2$. It is positive for $\phi_1>\phi_2$, hence it is opposite in sign to electric field. So here is the final conclusion: current flows from highest potential to lowest.
As you can see, you do not need any conventions, and all can be figured out from mechanics. Hope this clarifies the confusion about signs.
Concerning charges, non-zero electric field means charge imbalance. It also means, there should be current as a reaction to remove this charge imbalance. When you are building energy diagram in this case, keep in mind that for positive charges the higher the field potential, the higher the energy, so that their energy diagram is inverted compared to that of electrons.
UPDATE: The energy diagram difference for positive and negative charges follow from the energy definition $\epsilon = q\phi$. While an electron has lower energy near positive charge $Q$, a hole (positive charge) will have larger energy near positive charge $Q$, whereas potential $\phi$ of the field created by $Q$ stays the same.
This is explicitly illustrated in p-n diode (see picture). p-doped region has negatively charged ions, and n-doped region has positively charged ions. Here the energy well for electrons looks as usual: electrons are accumulated in the energy well, since they indeed have lower energy near positively charged ions in the n-region and larger energy in p-doped region near negatively charged ions.
On the other hand, holes are accumulated near the the negatively charged ions in the p-region, where they have lower energy, and not in the n-region near positively charged ions, where the have higher energy. If you look at their energy diagram from electronic point of view, it looks counterintuitive: holes sit at higher energy. However, if you reverse energy axis, it will make perfect sense: holes sit in their energy well.
Image legend: Vertical axis - energy, horizontal axis - distance. White dots - negative ions, black dots - positive ions. Shaded areas show population with carriers. Image is derived from here.