Q:    Design the circuit in the diagram. So that V₀ = 3V while I_L =0,& V₀  changes by 40mV per 1mA of load current. Determine the value of R .(suppose four diodes are equal) relative to a diode having 0.7 drop at I mA current. Suppose n =1

SOLUTION:

V₀ = 3V, while I_L =0, so each of diode should exhibit a drop of 0.75V. If I_L =1mA, then V_o modified by 40mV and a change because of each diode is 10mV.

                                    Therefore  

                                                            r_d = 10mV/1mA=10 Ohms

                                                However

                                                 r_d = nV_T/I_D

                                                            10 = 1 x 25m/I_D

                                                            I_D = 2.5mA

                                    Therefore

                                                            15 - 3 - I_DR_D = 0

                                                                       R = (15 - 3)/I_D

                                                                                 = (15 - 3)/2.5m= 4.8k Ohms.