Q:    The circuit in the diagram utilize three identical diodes with n=1 & I_s = 10-14 A .determine the value of the current I to get an output voltage V₀ = 2V, if a current of 1mA is drawn away, what is the alteration in output voltage.

 Solution:

                        V₀ =2V, therefore the voltage drop across each of diode is 2/3V.

                        Therefore I ought to be

                                                 I = I_s e ^v/nVT

                                    Here,   V_T= 25 mV=0.025V

Putting the value in eq

                                                =  10^-14 e ^2/3x0.025  =3.8mA

If a current of 1mA is drawn from the terminals by means of a load, the current via the diodes decrease to

                                                            3.8 - 1 = 2.8

                        Therefore the voltage across each of diode will changes by

                                                            ?V = v₂ -v₁ = nV_Tln(i₂/i₁)

                                                                         ?V =nV_T ln(2.8/3.8) = -7.63mV

                                                The net decrease in value of V₀

                                                                         V₀ = 3 x 7.63= 22.9Mv