A query regarding Thermal Voltage V = KT
Yes. A volt is also a Joule per Coulomb. A fact I never forget for a moment. Think of it this way:

If you have two plates separated by some gap in a vacuum in outer space, and place 100 volts across them, and then imagine placing one Coulomb of charge at the negative plate, that Coulomb of charge will be accelerated towards the positive plate. Eventually, the charge will strike the positive plate and the impact energy will be just 100 Joules. (And so it must also be that their velocity at impact is the same, so that the kinetic equivalent \$\frac{1}{2} m v^2\$ works out, correctly.)

You can move the plates further apart and the acceleration will be less, but the distance will also be just enough further apart to allow that Coulomb of charge to have exactly the same impact energy. (Because it will have exactly the same velocity in the end and therefore the same kinetic energy.) It doesn't matter how far apart, or how close the plates are, the results are always the same: 100 Joules of kinetic (impact) energy converted from the prior 100 Joules of potential energy. The only difference is the time it takes for that conversion.

It's consistent. 1 volt will impart 1 Joule on 1 Coulomb in this situation.



Since you are into units and dimensional analysis (and I think that is a very good thing), you might also now look into magnetics and the concept of Webers (or Joule-seconds per Coulomb or Volt-seconds.) A Joule-second is a unit of angular momentum. Electrons have quantized 'spin,' and they also have orbitals around an atom and there is something there to consider about angular momentum.

In fact, electron spin and orbital motion (or, at least, under the assumptions of models for these) both contribute to magnetic dipole moment. In magnetic materials, these tiny moments do not add up to zero.

In all these situations, they have angular momentum and the magnetic dipole moment will be proportional to the angular momentum, by some proportionality factor.

So, supposing this relationship and using \$\mu\$ for the magnetic dipole moment and \$L\$ as the angular momentum, we have (using \$x\$ for now as the factor):

$$\mu = x\cdot L$$

For an electron in a circular orbit (Bohr's early idea for atoms) and using \$\vec{p}\$ as the momentum vector of the electron and \$\vec{r}\$ as the vector from the center of the atom:

$$\mid \vec{L}\mid\: =\: \mid \vec{r} \times \vec{p}\:\mid \: =$$

But as this is circular and using \$R\$ as the scalar for the orbital radius,

$$L =  R\: p\:\operatorname{sin}\left(90^\circ\right) = R\: m_e\: v_e$$

For a current loop, and you can think of an electron in a circular orbit as being just such a current loop, we know that \$\mu=I\left(\pi R^2\right)\$. Also, the time it takes to make an orbit is \$T=\frac{2\pi R}{v_e}\$ and of course the electric charge of an electron is \$q_e\$, so \$I=\frac{q_e v_e}{2 \pi R}\$ (charge per unit time.)

Therefore the magnetic dipole moment of a single, circular orbiting electron is:

$$\mu=I\left(\pi R^2\right)=\frac{q_e v_e}{2 \pi R}\left(\pi R^2\right)=\frac{1}{2} q_e R v_e$$

(Assuming that \$v_e \ll c \$.) It now follows:

$$\mu = \frac{m_e}{m_e} \frac{1}{2} q_e R v_e= \frac{1}{2} \frac{q_e}{m_e} R \:m_e\: v_e= \frac{1}{2} \frac{q_e}{m_e} L$$

Clearly then, the missing factor must be \$x=\frac{1}{2} \frac{q_e}{m_e}\$!!

Angular momentum is quantized. (The revolution in physics from the early 1900's.) Therefore, \$L=N \hbar\$ (where \$\hbar=\frac{h}{2\pi}\$.) \$N\$ is any integer from 0, up. Assuming \$N=1\$, \$L=\hbar\$. So it follows that \$\mu=\frac{1}{2}\frac{q_e}{m_e}\hbar\$. For a single electron, this works out to \$\mu_e\approx 1\times 10^{-23}\:\textrm{A}\cdot\textrm{m}^2\$.

Suppose you have a bar magnet weighing \$69.5\:\textrm{g}\$ and that you go through a process of measuring its dipole magnetic moment using a compass and a yardstick and measuring the deflection relative to the local magnetic field of the Earth. You find that this is \$3.5\:\textrm{A}\cdot\textrm{m}^2\$.

Now suppose that you assume that almost all of the atoms in this magnet are iron atoms. The mass of one mole of iron is about \$56\:\textrm{g}\$. So the number of atoms here, using Avagadro's number, would be \$n=\frac{69.5\:\textrm{g}}{56\:\textrm{g}}\cdot 6.02\times 10^{23} \approx 7.5\times 10^{23}\$. So now you'd compute:

$$\mu = n\: \mu_{atom} = 7.5\times 10^{23}\cdot 1\times 10^{-23}\:\textrm{A}\cdot\textrm{m}^2 = 7.5\: \textrm{A}\cdot\textrm{m}^2$$

This is actually pretty close to the bar magnet measurement. The interesting thing here is that there are also some small contributions to the spin of an electron (ignored here) and also assumptions that all of the atoms contribute exactly one such electron moment. Also, we assumed circular orbits. Bohr's model was modified by Sommerfeld to first include elliptical orbits to explain certain hyperfine transitions and later modified to include motion in three dimensions by Sommerfeld, still later. And then it was again modified when Uhlenbeck and Gaudsmit added the concept of spin. (Pauli then came up with his exclusion principle based on all the quantum numbers by then.) But it turns out in practice that you can really do some remarkable predictions about such behavior using simplified model approximations.

Note that I had to introduce quantum theory and the simple Bohr atomic model to get here. It's things like this that helped force physicists into developing and accepting quantum theory.



I guess I just want to call your attention to how much can be done with dimensional analysis and thinking about the units in different ways (not the one obvious way, but by changing things out and combining units in different ways that you'd otherwise expect, using a LOT of imagination.) It takes you places you might not otherwise go and you might discover new ways of looking at the world around you!

Now look at the Ohm, which is a Joule-second per Coulomb^2. Where can your imagination go with that?

I recommend you read a book called Matter &amp; Interactions, 3rd edition, if interested further.

Note: 1 Joule = \$6.24150913 \times 10^{18}\$ eV. You are using \$\frac{eV}{kelvin}\$ for your Boltzmann's constant rather than \$\frac{volt}{kelvin}\$. So the value of q, the charge on an electron, should be used consistently. If you use \$\frac{eV}{kelvin}\$ for the units of Boltzmann's constant, then you must q=1 for the unit of charge (eV is in very small units associated with the smallest unit of charge, 1 electron.) If you use \$\frac{volt}{kelvin}\$ for the units of Boltzmann's constant, then you must use q in units of Coulombs, instead. Just keep the dimensions in mind, is all.