Buck converter design problem

Discussion in 'Homework Help' started by Jess_88, Oct 4, 2012.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
    Hey guys,

    I'm getting very confused with formulas for this problem.

    The single-phase ac grid rated at 240V, 50 Hz is feeding a diode full-bridge rectifier with a capacitor filter at its output. The capacitor then feeds power to a buck dc-dc converter. The ac supply voltage can vary in the range of -10% to +5% of its rated value. The dc-dc converter feeds power to a variable resistive load that draws power in the range of 100W to 5kW at 250V. The switching frequency of the converter has been selected as 10 kHz. Your task is to design the buck dc-dc converter so that it regulates the load voltage at 250 V with a maximum allowable peak-peak ripple factor of 2%, against any changes in ac supply voltage and load resistance. You may assume the rectifier output filter capacitor is large so that the voltage across it is ripple-free.

    A picture I made to mack it easier to see

    Calculate the:
    i) Required range of duty ratio of the switching device
    ii) Voltage and average current ratings of the switching device
    iii) Voltage and average current ratings of the diode in the dc-dc converter
    iv) minimum capacitance at the output of the dc-dc converter
    v) minimum inductance at the output of the dc-dc converter needed to keep inductor current continuous under all operating conditions

    What I have so far

    this is where i'm unsure about everything


    Id = Isw
    Vd = Vo/D = 250/0.7=357V

    for this I understand I need to calculate the critical inductance first, then work out the capacitance from there somehow

    is this the correct way to calculate it?
    Lcrit = [Vo * T * (1 - Dmin)] / (2 * Io)
    = [250 * (1/10k) * (1-0.7)] / (2 * 20)
  2. bountyhunter

    Well-Known Member

    Sep 7, 2009
    Your first problem is your description keeps saying "buck" converter but you are trying to go from 240V to 250V according to the drawing.

    I believe you have a fundamental error: if you take 240 VAC and run it trough a bridge and cap filter, you will get more like 360 VDC not 240 as you show. So, assuming you have 360 VDC to create a 250 VDC rail from, that is a buck design.
    Last edited: Oct 4, 2012
  3. RamaD

    Active Member

    Dec 4, 2009
    You have only mentioned wrongly as boost in the diagram, otherwise the calcs are ok.
    (i) & (ii) appear ok.
    (iii) Id = Isw?
    (iv) It is not Io, but is Iomin, where peak to peak ripple current will be 2*Iomin. Iomin can be calculated from minimum load given as 100W.