Alternate interpretation for Exercises 3.35 thru 3.37 These exercises are based on the assumption that the converters are operating in DCM with a fixed duty cycle, which will not change as the load varies. This is not reflective of typical converter application where duty cycle is controlled to maintain a specified output voltage as the load varies. The exercises state that the load resistance is “... twice the critical value.” The definition given in the text for R crit is based on operating duty cycle, but the definition assumes that the converter is operating in CCM or at the boundary when computing R crit. If the converter is being controlled to maintain a specific output voltage, the duty cycle in DCM will be different than the CCM value, and an erroneous value for R crit will be computed. The question arises: what value of D should be used to calculate R crit ? Should one use the duty cycle of the converter operating in DCM, or use the duty cycle that the converter would be operating at in order to produce the specified output voltage in CCM? The latter seems more consistent with the text definition of R crit, but the tone of the text and the solutions provided use the former.

Alternate solution for Exercise 3.35 (Buck) Given: L = 25  H f s = 400 Khz V IN = 42 v D = 0.3 R load = 2R crit (DCM) Discussion: Assuming duty cycle is controlled to provide a constant output voltage, we may infer that the output current with twice the critical load resistance will be half of the critical output current. Duty cycle in DCM is given by: Thus, the regulated output voltage must be: Duty cycle in CCM is: Critical Resistance is: Critical Output Current is: Actual load resistance and output current:

Alternate solution for Exercise 3.36 (Boost) Given: L = 25  H f s = 400 Khz V IN = 12 v D = 0.4 R load = 2R crit (DCM) Discussion: Assuming duty cycle is controlled to provide a constant output voltage, we may infer that the output current with twice the critical load resistance will be half of the critical output current. Duty cycle in DCM is given by: Thus, the regulated output voltage must be: Duty cycle in CCM is: Critical Resistance is: Critical Output Current is: Actual load resistance and output current:

Alternate solution for Exercise 3.37 (Buck/Boost) Given: L = 25  H f s = 400 Khz V IN = 12 v D = 0.6 R load = 2R crit (DCM) Discussion: Assuming duty cycle is controlled to provide a constant output voltage, we may infer that the output current with twice the critical load resistance will be half of the critical output current. Duty cycle in DCM is given by: Thus, the regulated output voltage must be: Duty cycle in CCM is: Critical Resistance is: Critical Output Current is: Actual load resistance and output current:

Summary for Discontinuous Conduction Mode ParameterBuckBoostBuck/Boost ILIL IOIO I IN I IN - I O I L (max) D DCM 11 22

ParameterModeBuckBoostBuck/Boost V O RangeC/D0 < V O /V IN < 1V O /V IN > 1V O /V IN < 0 I IN C/DI O (V O /V IN ) = P O /V IN = (V O /R O )(V O /V IN ) I L (max) CI L + I L,cr D I L (min)CI L - I L,cr C/D Boundary I IN,cr C/D(V O /V IN ) 2 (V IN - V O )/2Lf(V IN /V O )(V O - V IN )/2Lf(V O /(V O - V IN )) 2 (V IN /2Lf) I L,cr C/DI O,cr I IN,cr I IN,cr - I O,cr I O,cr C/D(V O /V IN )(V IN - V O )/2Lf(V IN /V O ) 2 (V O - V IN )/2Lf(V IN /(V O -V IN )) 2 (V O /2Lf) Duty Cycle CV O /V IN (V O -V IN )/V O V O / (V O -V IN ) DV O /V IN (V O -V IN )/V O V O / (V O -V IN ) 11 C(V IN –V O )/V IN V IN /V O V IN / (V IN -V O ) D (V IN - V O )/V IN V IN /V O V IN / (V IN -V O ) 22 D Reverse diode for V IN < 0 DCM if I O /I O,cr < 1 Switching Converter Parameters