Q:     let a diode having n = 2 biased at 1mA. Determine the change in current as an outcome of changing the voltage by

                                    (i) - 20 mV (ii) - 10 mV (iii) - 5mV      

                                    (iv) +5mV      (v) +10mV  (vi) +20mV

In each of case perform calculations           

                                                 (i)By utilizing small signal model

                                                 (ii) By utilizing the exponential model.

Solution:

                        For little signal model we contain

                                                Δv = r_d Δi

                                                            Δi   =  Δv / r_d

                                    However

                                                r_d = n V_T / I

                                                                   =2 x 25m/1m   = 50ohms

                                                             Δi =  Δv / 50

                        From exponential model now

                                                  i  = I_s e ^v / nV_T

                        We also know that

                                                         I_D + Δi    = I_D e ^Δv /nV_T

                                                        I_D +  Δi    = I_D e ^Δv /nV_T

                                                                  Δi =  ID (e ^Δv/nVT - 1)

                         However

                                                             I_D = 1mA  therefore

                                                             Δi =  (e^Δv/nV_T -1)

                        (a)    For i = - 20 mA

                                                (1) Δi = -20/50 = - 0.4mA

                                                (2) Δi = e ^-20 m/50m ^-1= 0.33mA

                        (d)                    Δv = 5mA

                                                        (i)   Δi = 5/5o = 0.10mA

                                                (ii) Δi = e^5/50 - 1=0.11mA

                        (f)                    Δv = 20mA

                                                (i) Δi =20/50= .40mA

                                                  (ii) Δi = e ^20/50 - 1 = 0.49mA