Application

            Let the circuit as indicated, for analysis reason, we may split the circuit into two partitions that is ac & dc

DC source with the value of V_DD

We replace the ac & replace diode along with constant drop model

For dc analysis the Circuit is as

In case of DC we let only DC current I_D is passing and no effect of the ideal diode due to this ideal diode is forward biasing and this results in short circuit so

                        From KVL eq

                                                V_DD -I_DR - V_Do - I_D r_d = 0

                                                V_DD = I_DR+ VDo + I_D r_d ---------------- (i)

                         In case of DC just DC current I_D is passing

For ac analysis we have to remove DC sources that are the part of the main circuit and also remove the DC source that appear in the earlier effective circuit and we also eliminate the ideal circuit therefore  our circuit became  as

                        For analysis of ac

                                                                        v_s = i_d (R + r_d) ------------- (ii)

                                     by joining (i) & (ii)

                        Total analysis is

                                                V_DD + v_s =I_DR + V_Do + I_D r_d + i_d (R + r_d)

                                                               = I_D(R + r_d) + i_d (R + r_d) +V_Do

                                                V_DD + vs = (R + r_d)( I_D + id) + V _D0  ------------ (iii) 

However

                                                             I_D + i_d = i_D

             So

                                                            V_DD+ vs = (R + r_d)( I_D + i_d) + V _D0 

                        Separating the dc & signal quantities on both of the sides of equation (iii)

                                                             V_DD = I_DR + V_D0

                                    That is shown by the circuit in the diagram below

            And for the signal

                                             v_s = i_d (R + r_d)

that is shown by the circuit in the diagram below

            but, if we carefully analysis the ac equation circuit, this is nothing more than a voltage divider.

            So the diode signal voltage is

                                                                        V_d = V_s r_d/(r_d +R)