Reason to explain why $n-p$ region of zener diode is heavily doped Calculation of current through zener diode $n$ and $p$ regions of zener diode are heavily doped so that depletion region formed is very thin and electric field at the junction is extremely high even for a small reverse bias voltage. Current in the circuit is : $I=\;\frac{V}{R}=\;\frac{5}{250}=\;\frac{1}{50}$ $=0.02\,\mathrm{A}$ Current through resistor of $1\,\mathrm{k}\,\Omega$ is: $I=\;\frac{15}{1000}=0.015\,\mathrm{A}$ As zener diode and $1\,\mathrm{k}\,\Omega$ resistor are in parallel, current through the zener diode is : $I=0.02-0.015=0.005\,\mathrm{A}$