Urgent HELP- Designing a buck converter!

Discussion in 'General Electronics Chat' started by diya khan, Nov 6, 2011.

  1. diya khan

    Thread Starter New Member

    Nov 29, 2010
    18
    0
    Hey,

    im required to design a buck converter that will step down the voltage to supply a piece of equipment.I also require to show the results of simulation of my circuit.

    My circuit should meet the following specifications:

    Vin=24V
    Vout=18V
    Max ripple compared to DC input=10%
    load resistance=5 Ω
    Switching frequency must lie between 50kHz to 200kHz
    Assume ideal devices have been used!

    ok..so to start of with, I no the circuit im required to design (not sure about the which switching device to choose yet).i have decided to use Pspice Student Version to simulate my circuit.
    I have calculated:

    Iout=Vout/Rload=18/5=3.6A
    ΔI=0.1*3.6=0.36A
    D=Vout/Vin=0.75

    Now for continuous conduction mode:

    we know that L=\frac{(Vin-Vout)*(D/Fs)}{Ir} whr Ir=ΔI


    Now the issue is that i really cannot figure out how to choose Fs!!Once i choose Fs correctly, i can find the value for my filter inductor.
    its mentioned in the ques that i can choose between 50kHz to 200kHz, but is there any appropriate logical way to selecting the Switching frequency?
    I would also like to know whether its better to use a p/n type MOSFET or IGBT for my circuit and why!

    thx
     
  2. crutschow

    Expert

    Mar 14, 2008
    13,003
    3,232
    A higher switching frequency allows the use of smaller inductors, but the switching losses so it's a tradeoff between those two parameters.

    IGBTs have a relatively slow switching speed so they are normally only suitable for lower frequency applications. MOSFETs are much faster so are typically used in switching power supplies, since they would have lower switching losses.
     
  3. diya khan

    Thread Starter New Member

    Nov 29, 2010
    18
    0
    ok. but could you give me a range, like as you said, MOSFETs are much faster which means that they have high switching frequency.So is 50kHz high enough??
    and what difference does it make whether i use a p-type or an n-type MOSFET?
     
  4. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    YES, but it requires understanding gained by actually building buck converters, not simulating them. The capacitors have losses and internal resonances, the inductor core has losses that increase with frequency, the FET and diode both have switching losses. You basically push the switching frequency up as far as you can before the switching losses get too high (and you define that, but probably less than 10% of load watt power). Understand, there are ways to reduce losses by "upgrading" to better caps and inductors but they cost more so it's all a matter of engineering judgement. No free lunch.

    In the real world, pushing a 3A buck converter above 100 kHz usually starts getting too pricey for what consumers want to pay for things, but it's possible to do. I would stay between 50 kHz and 100 kHz and live with a slightly larger inductor.
     
  5. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    Power P-FETs are usually more expensive for a given voltage/On resistance value because of larger die size.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Along with P-ch being more expensive, if all other parameters are roughly equal (Rdson, Vdss, logic/standard, Id rating) the gate charge of a P-ch will be roughly 2.5 times as much as that for an N-ch MOSFET. Also, the selection of P-ch is far more limited than for N-ch.
     
  7. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    The bigger die size of the P-FET means a larger gate capacitance which the switching driver circuit has to force to charge as fast as possible, so P-FETs take "stronger" drivers that can deliver higher peak currents to maintain the same fast switching speed.

    However, the P-FET buck circuit is simpler than the N because the N-FET buck requires some kind of way to generate a higher driver voltage to pull the N-FETs gate up to to turn it on fully. There are always trade offs. Most of the time they go with the N-FET to save money and use a cheap charge pump to boost up to get a gate driver rail.
     
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