A p -type semiconductor has an acceptor density of 10^18 cm^-3. If ED-Ev=0.2 eV and me^*=mh^*=m

step 1 This is the problem number 2. The problem stated that a P -type semiconductor has an accepted de

A p -type semiconductor has an acceptor density of 10^18...

Key Concepts

Boltzmann Statistics in Semiconductors

Boltzmann statistics describe the probability of occupation of energy states in a non-degenerate semiconductor. The carrier concentration in intrinsic semiconductors depends exponentially on the ratio of the bandgap energy (or energy barrier from a dopant level) to the thermal energy (kT). This exponential dependency explains why intrinsic carrier generation is negligible at room temperature when the energy barrier is significantly larger than kT, ensuring that dopant-induced carriers dominate conduction.

Carrier Mobility and Electrical Conductivity

Electrical conductivity in a semiconductor is determined by both the concentration of charge carriers and their mobility. Carrier mobility reflects how quickly a carrier can move through the semiconductor lattice under an applied electric field, and, when combined with the carrier concentration and the elementary charge, directly yields the conductivity. In doped semiconductors, where one type of carrier predominates, the mobility of that majority carrier is key to calculating the overall conduction properties of the material.

Intrinsic vs. Extrinsic Conduction

This concept distinguishes between the carrier generation mechanisms in a semiconductor. Intrinsic conduction originates from electron–hole pairs thermally generated across the bandgap, and its concentration is highly sensitive to temperature due to the large energy gap. Extrinsic conduction, on the other hand, is dominated by carriers introduced via dopants (donors or acceptors) and is largely independent of the thermal generation process at moderate temperatures, making it the prevailing conduction mechanism in doped materials under typical conditions.

Dopant Ionization and Energy Levels

Dopants introduce discrete energy levels within the bandgap of a semiconductor. In p-type materials, acceptor dopants create energy levels slightly above the valence band that can accept electrons, leaving behind holes that contribute to electrical conduction. The relative position of these dopant levels, along with the thermal energy available, determines the degree of ionization of the dopants, which in turn influences the carrier concentration in the material.

Transcript

00:01 This is the problem number 2.

00:04 The problem stated that a p -type semiconductor has an accepted density of 10 power 18 per centimeter cube if the e .v is 0 .2ev and effective mass of the electrons and the holes are equal to the rest mass of the electron.

00:23 And the first part so that the intrancing conduction in the crystal is negligible at room temperature.

00:31 So here the receptor density is given for p type semiconductor that is p or that is na is given that is 10 power 18 for a centimeter cube and e d minus ev is given which is 0 .2 ev and the effective mass of the electrons and the holes are equal and it is equal to the rest mass of the electron which is 9 .1 into 10 power minus 31 kg and this difference is nothing suppose this is the bottom of the conduction band and this is the top of the valence band and here there is a donor level ed and this difference is is given which is 0 .2 now we have to show that the conduction in this type of the semi -contenture is negligible at room temperature the first part of the question so that the intrinsic conduction in the crystal is negligible at room temperature for this we have to find out first the n i the intrinsic calculation the intrinsic density at room temperature at room temperature the value of the kt is 0 .0259 a and we will use here h bar is equal to 6 .58 into 10 power minus 16 e .v second and the effective mass already used now we have the formula for intrinsic density that is n i n c and the a b n b square root exponential minus a g by k t we know the value this k t and we have to find out this effective density of state that is nc and the nv nc is nothing it is two and this is m e effectively on the electron k t divided by 2 pi h bar square where h bar is the h upon 2 pi and h is the plank constant m power 3 by 2 and here if you substitute the value of the m star that means it is 2 m into k t by 2 pi x b bai h bar square and 3 by 2 then the value is that 0 .016 per centimeter cube and because here k t is constant 2 pi as bar constant that m e is equal to m h bar hence m h is equal to since m s star is equal to m h star is equal to m hence and c that is n b n n c r equal and the value is 0 .0016 per centimeter cube now let's find n i and i is nothing it is 0 .0016 exponential minus e g by 2 k t because the value of this e g is around 4 k t so if we substitute this value, the value of this, let's assume first if eg is equal to 1 .12ev, then this value is minus eg by 2kt is around 4 into 10 power minus 10...

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