Editing Langmuir probe (section)

===Exponential electron current===

As the voltage of the Debye sheath is reduced, the more energetic electrons are able to overcome the potential barrier of the electrostatic sheath. We can model the electrons at the sheath edge with a [[Maxwell–Boltzmann distribution]], i.e.,

<math>f(v_x)\,dv_x \propto e^{-\frac{1}{2}m_ev_x^2/k_BT_e}</math>,

except that the high energy tail moving away from the surface is missing, because only the lower energy electrons moving toward the surface are reflected. The higher energy electrons overcome the sheath potential and are absorbed. The mean velocity of the electrons which are able to overcome the voltage of the sheath is

<math>
\langle v_e \rangle = \frac
{\int_{v_{e0}}^\infty f(v_x)\,v_x\,dv_x}
{\int_{-\infty}^\infty f(v_x)\,dv_x}
</math>,

where the cut-off velocity for the upper integral is

<math>v_{e0} = \sqrt{2q_{e}\Delta V/m_e}</math>.

<math>\Delta V</math> is the [[voltage]] across the Debye sheath, that is, the potential at the sheath edge minus the potential of the surface. For a large voltage compared to the electron temperature, the result is

<math>
\langle v_e \rangle = 
\sqrt{\frac{k_BT_e}{2\pi m_e}}\,
e^{-q_{e}\Delta V/k_BT_e}
</math>.

With this expression, we can write the electron contribution to the current to the probe in terms of the ion saturation current as

<math>
j_e = 
j_i^{max}\sqrt{m_i/2\pi m_e}\,
e^{-q_{e}\Delta V/k_BT_e}
</math>,

valid as long as the electron current is not more than two or three times the ion current.