Q: A silicon junction diode having n =1 contain v = 0.7 volts at i =1mA . Determine the voltage drop at i =0.1mA &  i = 10mA.

Solution:

                                                v₂ -v₁ = nV_Tln(i₂/i₁)

                                                i₁ = 1mA

                                                       v₁ = 0.7Volts

                                                 & n =1   therefore v₂ = 0.7 +V_Tln(i₂)

                                    for

                                                   i₂ = 0.1mA

                                                     V_T= 25 mV=0.025 V

                                                      v₂ = 0.7 + 0.025 ln(0.1)

                                                                   v₂= 0.64 V

                                    Here

                                                      i₂ = 10mA

                                                     v₂ = 0.7 + 0.025ln(10) = 0.76V