Buck Converters in DCM in Steady State

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SUMMARY

The discussion focuses on the analysis of buck converters operating in Discontinuous Conduction Mode (DCM) under steady-state conditions. Key equations derived include the relationship between input and output voltages, specifically Vo=Vin/2 (sqrt(M^2+4M) -M), where M=D^2 R/k. The confusion arises from the transition to a quadratic equation for solving Vout, which was not clearly communicated by the instructor. Understanding these equations is crucial for effectively analyzing buck converters in practical applications.

PREREQUISITES
  • Understanding of buck converter topology
  • Familiarity with Discontinuous Conduction Mode (DCM)
  • Knowledge of quadratic equations and their applications in circuit analysis
  • Basic principles of electrical power (P=V^2/R)
NEXT STEPS
  • Study the derivation of equations for buck converters in DCM
  • Learn about the implications of using quadratic equations in circuit analysis
  • Explore the differences between buck and buck-boost converters
  • Investigate practical applications of buck converters in power electronics
USEFUL FOR

Electrical engineers, power electronics specialists, and students studying circuit design who are looking to deepen their understanding of buck converters and their operational characteristics in DCM.

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Homework Statement
This problem comes out of a EET course called Power Electronics. See equation 3.48a. The text doesn’t show the derivation and I could not figure it out even after asking the instructor. This is for learning how DC-DC buck converters operate in a discontinuous mode. A buck converter uses transistors and passive components to step DC voltages down.
Relevant Equations
Power in = Power out
Let K=2Lfs and Pin=Pout,

Have,

(Vin-Vo)/k D^2 Vin =Vo ^2/R

(I am fine up to this part. I am equating input power to output power)

V^2=D^2 + D^2 R/2 Vo Vin - R/k D ^2 Vin =0

(This is where the instructor takes over. Can’t figure out how he got to V^2. Where did that come from? Thought I was just dealing with Vin and Vout.)

Let M=D^2 R/k

Vo=Vin/2 (sqrt(M^2+4M ) -M)
 

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Is the problem a buck boost converter feeding a resistive load R?

I'm thinking of ##P=\frac{V^2}{R}##
 
anorlunda, After much headache, I realized the book is using a quadratic equation to solve for Vout. The instructor didn’t explain very well. If he would have said word quadratic, a light bulb would went off.
 

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