# Mobility of electrons and holes

According to experimental results it has been found that in Silicon holes are one-third as mobile as electrons. But if doping is considerably low such that phonon scattering is dominant over impurity scattering, then we know that mobility is then inversely proportional to thermal velocity(v_th) of carriers and phonon density. Then we know that since a hole has more effective mass(m_p*) than an electron, then according to that holes should have more mobility than electrons in that case. Then why does electrons have more mobility than holes in case of Silicon?

• What's the question? – GiorgioP Dec 30 '18 at 11:09

First, it is not generally true that "holes are one-third as mobile as electrons", but this is the case for specific semiconductors like silicon. There are other semiconductor materials like GaAs with other ratios (Mobility electrons ≤8500 cm2 V-1s-1 ; Mobility holes ≤400 cm2 V-1s-1). There are also semiconductors with hole mobility higher than electron mobility e.g. PbS or PbTe. In general, the mobility has to be calculated using the Matthiessen rule: $$1/\mu = \sum _i 1/\mu_i$$, where $$\mu_i$$ are the mobilities due to the scattering effect $$i$$. For your last statement I would like to look at the formulas: $$\mu = q\lambda/m^* v$$ and $$v \propto \sqrt{T/m^*}$$, hence $$\mu = q\lambda/\sqrt{m^*} T$$. This means, that in the end the mobility still goes with $$1/m^*$$ and therefore, electrons can still have higher mobility if their effective mass is lower.
• Well, as I've said, you can treat phonons and impurities independently using the Matthiessen rule. Assuming that you can neglect impurities (your post: "But if doping is considerably low such that phonon scattering is dominant over impurity scattering, then we know that mobility is then inversely proportional to thermal velocity(v_th) of carriers and phonon density" ), you can just focus on the effect of phonons. Either you use the velocity and mean free path $\lambda$ or the free traveling time $\tau = \lambda/v$. I used the formula with $\lambda$. – David N. Dec 30 '18 at 16:37
• I assume that the scattering cross section for holes and electrons depends on the band structure and therefore is different for each semiconductor, yielding to different $\lambda$ for holes and electrons for every semiconductor. – David N. Dec 30 '18 at 16:38