power electronics

Question

A dc-dc buck converter has the following design parameters: 𝑉𝑖𝑛 = 120 V, 𝑅 = 25 Ω, 𝑃𝑜𝑢𝑡 = 150 W, 𝑓𝑠𝑤 = 200 kHz, and L = 0.4 mH. What should the load value be changed to in order for the converter to operate at the boundary of continuous conduction mode (CCM) and discontinuous conduction mode (DCM)?

Answer 1

To begin, the duty cycle and output voltage in order to determine the critical resistance between DCM and CCM.

Step 1 - Develop the V_l equation (eg. the voltage across the inductor) to find the duty cycle (D)

1. (Vin - V_out)D - V_outD'
2. VinD - V_outD -V_out+V_outD
3. Combining like terms and putting the equation in terms of D:
4. D = V_out/V_in

Step 2 - Find the output voltage using Ohm's Law

1. V_out = (P_outR)^0.5 = (150*25)^0.5 = 61.2 V

Step 3 - Calculate D and D'

1. Using the equation from step 1: D = V_out/V_in = 61.2/120 = 0.51 and D' = (1-D) = (1-0.51) = 0.49

Step 4 - Calculate the critical resistance between DCM and CCM

1. For this step I didn't provide how the equation for the critical resistance for the buck's DCM and CCM considering it involves a lot of algebra. With that said, the equation for the critical resistance is:
2. R_crit = (2Lf)/D' = (2*0.4*10^-3*200*10³)/0.49 = 326.5Ω

Answer: R_crit = 326.5Ω