# Intrinsic semiconductor having less conductivity than extrinsic conductor

The question is quite simple.Why intrinsic semiconductor has less conductivity than extrinsic semiconductor? I want to know the exact doping concentration per atoms in extrinsic semiconductor relative to room temperature excited intrinsic ions.

• I don't have actual data, but it is only obvious why extrinsic semiconductor has greater conductivity: they have charge carriers in addition to intrinsically generated charge carriers (holes and electrons). – samjoe Mar 12 '17 at 6:02
• @samjoe Thanks for your reply.I know about the number of charge carriers being more in extrinsic than in intrinsic.Its just that i want to know how many dopant atoms are injected per Si or Ge atoms.And how that differentiates from the number of electrons being free in intrinsic semiconductor,both at room temperature – debo.stackoverflow Mar 12 '17 at 6:10
• Intrinsic Si populations are order 10$^{12}$ per cc. Doping levels are 10$^{15}$ to 10$^{20}$ per cc. See the difference? – Jon Custer Mar 12 '17 at 17:37

Intrinsic semiconductors have a dissociated population (a bunch of holes and electrons that separate due to temperature, and can contribute to conduction until they recombine). Because a high population of holes and electrons would cause a very FAST rate of recombination (faster than thermal generation occurs) , and a very low population of holes or electrons would cause very SLOW recombination (slower than thermal generation of pairs), it should be no surprise that at equilibrium, the fractional population of holes $n_p$ and electrons $n_e$, is related by an equation $$constant = n_p \times n_e = n_i^2$$ where the $n_i^2$ symbolizes the at-thermal-equilibrium numbers of holes and also of electrons.
For intrinsic silicon, $$n_p = n_e = n_i$$
Doping generates a large number of (for instance) electrons, pushing $n_e$ up, and by the equilibrium equation, forces $n_p$ down. But, conduction of electricity depends on the SUM of the holes and electrons. If one has undoped material conduction is $$K \times (n_e + n_p) = K \times 2 n_i$$ but for $n_e = 100 \times n_i$ doped material, that conduction goes up to $$K \times (n_e + n_p) = K \times 100.01 \times n_i$$