Electrical Engineering - Buck Converter, Prevent Saturation

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SUMMARY

The discussion focuses on calculating the inductance for a buck converter supplied by a 12VDC source, outputting 5VDC at 4A with a PWM frequency of 100 kHz. The continuous conduction mode inductance (L_CCM) was correctly calculated as approximately 3.646 µH. However, the calculation for the inductance to prevent saturation, which should exceed L_CCM, was incorrectly derived as approximately 2.431 µH. The correct approach requires that the inductance for saturation must be equal to or greater than L_CCM to ensure stable operation.

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YoshiMoshi
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Homework Statement



I need help with part b-

A buck regulator is supplied from a 12VDC source and outputs 5VDC at 4 A with a PWM switching frequency is 100 kHz. Answer the following then:
a- Determine the continuous conduction mode inductance (L_CCM).
b- Compute the inductance for the converter in with the requirement that the inductor current not exceed 10A to prevent saturation.

Homework Equations

The Attempt at a Solution



a.
V_s = 12 V, V_o = 5 V, I_o = 4 A, f = 100*10^3 Hz

D = V_o/V_s = 5/12
R = V_o/I_o = 5/4 ohm
L_CCM = ( (1 - D)R ) / ( 2f ) = ( (1 - 5/12)( 5/4 ) ) / ( 2*100*10^3 ) ~ 3.646*10^(-6) H

b.
R = V_o / I_o = 5/10 = .5 ohm
lambda = I_o / ( I_max - I_o ) = 4/(10 - 4) = 2/3
lambda = L / L_CCM, L = lambda * L_CCM = ( 2/3 )( 3.646*10^(-6) ) ~ 2.431*10^(-6) H

So this was my answers to the assignment. Apparently my answer to question a. is correct, and my answer to question b is wrong. I'm not sure what I'm doing wrong. My textbook provides me the lambda formula that I used in question b.

Since the value I found for L is less than L_CCM, and L has to be greater than L_CCM (not so sure), my answer to question b should be the same value as L_CCM upon realizing that the value I found for L is less than L_CCM?

Thanks for any help, much appreciated.
 
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