# Does diffusion current "consume" thermal energy?

My (possibly erroneous) understanding of solid state physic has led me to the belief that diffusion current may "consume" thermal energy.

Here is my (possibly erroneous) understanding. When an E field exists within a conductor, it causes free electrons to accelerate. These electrons then interact with the conductor's atoms, and exchange some of their energy. The algebraic average movement of the electrons constitutes the drift current in the conductor. An E field in a conductor thus causes drift current and consequently Joule heating.

Now let's consider an unbiased PN junction. Because the junction is unbiased, the drift and diffusion currents are equal in magnitude, but opposite in direction. Thus there is no net current. However, there is still an builtin E field. This E field presumable still causes electrons (or electrons and holes if you want to think about it that way) to accelerate, and gain energy. And this extra energy is still transferred to the atoms, causing Joule heating. But by the conservation of energy, there is no net heating. This leads me to believe that diffusion current must be "consuming" the very heat the drift current is creating?

Is my understanding flawed? Is my reasoning flawed? Or have I arrived at a valid conclusion? And if I have in fact arrived at a valid conclusion, can you explain the mechanism by which the thermal energy created by the E field acting upon the electrons is ultimately "consumed" by the diffusion current?

• Inside a macroscopic body the equilibrium of a charge is not when it is placed in a zero E-field but when it is in zero electrochemical field. In other words the work function of a mass $m$ carrying $z$ charge in an electric potential field $\phi$ and chemical potential field $\mu$ is $\delta w = zd\phi + md\mu$, so that its free energy change is $dF = -SdT+\delta w$. Isothermal equilibrium demands $\delta w = 0$ so that for fixed $r=\tfrac{z}{m}$ you must have $rd\phi+d\mu=0$. A shocking revelation to all EEs: a voltmeter measures electrochemical potential not electric potential. Commented May 14 at 15:26

There is no heating or cooling involved in the diffusion-drift currents in equilibrium. Essential is that at any electrical potential the electron ensemble is in thermal equilibrium with the crystal lattice of the semiconductor constantly exchanging (gaining and losing) energy and momentum with the crystal lattice vibrations (phonons) in microsopic collision processes so that on average the electrons are at the same temperature as the lattice. Electrons have a considerable thermal velocity. When an electron moves in the direction of the electric field, it loses kinetic energy which is rapidly resupplied by kinetic energy gain from collisions with phonons. When it moves against the electric field, it gains kinetic energy which is rapidly lost to the lattice by phonon collisions. These collisions with phonons are essential for both the drift and the diffusion current, which are connected by the Einstein relation between diffusion coefficient and mobility $$D= µkT/q$$.