Discontinuous Conduction Mode with ConstantInput Voltage V_in

In applications such as speed control of DC motors, the input voltage (V_in) remains constant and the output voltage (V₀) is controlled by varying the duty ratio D. Since V₀ = DV_in, the average inductor current at the edge of continuous conduction mode is obtained from equation 43 as

                                                                                          (44)

In Figure 9 the plot of I_LB as a function of D , keeping all other parameters constant, is shown. The output current required for a continuous conduction mode is maximum at D = 0.5 and by substituting this value of duty ration in equation 44 the maximum current ( I_LB,max) is obtained as

                                                                                                   (45)
From equation 44 and equation 45 , the relation between I_LB and I_LB,max is obtained as

                                                                                      (46)

To understand the ratio of output voltage to input voltage (V₀ /V_in) in the discontinuous mode, it is assumed that initially the converter is operating at the edge of the continuous conduction (Figure 7), for given values of T,L,V_d and D. Keeping these parameters constant, if the load power is decreased (i.e., the load resistance is increased), then the average inductor current will decrease. As is shown in Figure 10 , this dictates a higher value of V₀ than before and results in a discontinuous inductor current.

 

Figure 10: Discontinuous operation is buck converter	Figure 11: Buck converter characteristics for constant input current

 

In the time interval the current in the inductor L_f is zero and the power to the load resistance is supplied by the filter capacitor alone. The inductor voltage e_L during this time interval is zero. The integral of the inductor voltage over one time period is zero and in this case is given by

                                                                     (47)

In the interval ( Figure 10 ) the current ripple in L_f is

                                                                                                   (48)

From Figure 10 is can be seen that

(since the current falls)                                                                                         (49)

                                                                                                                                 (50)

Substituting the values of and e_L from euqation 49 and equation 50 into equation 48 gives

                                                                                           (51)

                                                                                      (52)

Hence,                                                                                                             (53)