In a semiconductor like GaAs, the saturation velocity for both holes and electrons is on the order of \$10^7\$ cm/s. However, under lower electric fields E, the drift velocity of electrons is much larger than that of holes, since \$v_d = \mu E\$, where the mobility \$\mu\$ of electrons is much greater than holes. Why then are the saturation velocities comparable?
Electronic – Why is the saturation velocity for holes and electrons comparable
pn-junctionsemiconductors
Related Topic
- Electronic – Why can current flow through the reverse biased base-collector junction (N-P junction) in a BJT with a forward biased base-emitter junction
- Electronic – How the Depletion Region of PN Junction changes under Bias
- Electronic – Extrinsic semiconductors: why can’t electrons/holes in the band gap conduct
- Electronic – Diffusion current causes drift current
Best Answer
In different semiconductors, different mechanisms are primary drivers of velocity saturation.
In materials without accessible higher bands (Si) this is inelastic scattering of charge carriers with the emission of optical phonons of energy \$E_{optic}\$. This mechanism predicts the saturation velocity \$\sqrt{E_{optic}/2m_{carrier}}\$, \$m_{carrier}\$ is an effective mass of electron/hole. In silicon, effective masses of electrons and holes are close to the free electron mass (\$1.08m_0\$ and \$0.81m_0\$, respectively), and the saturation velocity values are close for both quasiparticles.
In gallium arsenide, the quasi-electron effective mass (\$0.067m_0\$) is seven times less then the hole effective mass (\$0.47m_0\$). This low effective mass achieves for GaAs quasi-electrons a higher saturation velocity value (as compared to Si) at moderate electric field strengths. Notice that, because of much higher mass, the hole velocity at the same moderate electric field strengths is not saturated. This is the reason why we do not have a GaAs analog of Si-MOSFET devices.
As electric field strength grows higher, another mechanism of velocity saturation comes into play and not only brakes the velocity increase as phonon emission does, but even decreases the velocity values. Because of the band interaction, the quasiparticle's dispersion curve (energy vs. wave vector) is no more parabolic, in contrast to simple dispersion curves for free particles (\$ω=\hbar k^2/2m_0\$). As electric field strength grows higher, the impact of this non-parabolicity on the quasiparticle motion becomes more pronounced. At the electric fields high enough, the velocity asymptotically approximates its saturation value \$v_{sat} \sim Wa/2{\hbar}\$, where \$W\$ is the width of the energy band and \$a\$ is the semiconductor lattice constant. As you see, this value does not depend on the quasiparticle effective mass and is identical for both electrons and holes. Notice once more, this explanation is only acceptable for materials with accessible higher bands.
This answer in no way pursues the goal to be essential for your education. I only hope that it provides you with keywords for searches of textbooks and papers on the subject.