Question
Why do semiconductors obey Ohm's law for only low fields?
✓
Answer
The drift velocity of a charge carrier is proportional to electric field $E.$
$v=\frac{e E}{m} \tau$
i.e., $v \propto E$
But $v$ cannot be increased indefinitely by increasing $E.$
At high speeds, relaxation time $\tau$ begins to decrease due to the increase in collision frequency.
So drift velocity saturates at the thermal velocity $\left( v _{ th }=10^5 ms^{-1}\right)$ and becomes independent of the electric field at higher values of $E.$ At $300 K ,$
$v_{t h}=\left(\frac{3 k_B T}{m}\right)^{1 / 2}$
$=10^5 ms^{-1}$
$\tau=10^{-12} s$
An electric field of $10^6 Vm ^{-1}$ causes saturation of drift velocity.
Hence semiconductors obey Ohm's law for low electric fields $(E <10^6 Vm ^{-1}$ ) and above this field, $I$ become independent of $V.$
$v=\frac{e E}{m} \tau$
i.e., $v \propto E$
But $v$ cannot be increased indefinitely by increasing $E.$
At high speeds, relaxation time $\tau$ begins to decrease due to the increase in collision frequency.
So drift velocity saturates at the thermal velocity $\left( v _{ th }=10^5 ms^{-1}\right)$ and becomes independent of the electric field at higher values of $E.$ At $300 K ,$
$v_{t h}=\left(\frac{3 k_B T}{m}\right)^{1 / 2}$
$=10^5 ms^{-1}$
$\tau=10^{-12} s$
An electric field of $10^6 Vm ^{-1}$ causes saturation of drift velocity.
Hence semiconductors obey Ohm's law for low electric fields $(E <10^6 Vm ^{-1}$ ) and above this field, $I$ become independent of $V.$
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