help about DCDC converter please.

Discussion in 'General Electronics Chat' started by nandax, Mar 23, 2009.

  1. nandax

    Thread Starter Member

    Nov 20, 2008
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    I am using step down converter
    TPS62056 from texas instrument
    (http://focus.ti.com/lit/ds/symlink/tps62056.pdf)
    but i doesn't give me enough information how to determine Cin ,COut, and inductor for the circuit.
    can somebody help me..??
    I'd like to have 3.3 Output voltage with Iout=100mA, with Vin=9 Volt
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    Page 17 recommends L values in the range 10-22uH. Table 2 seems to give a range of typical off-the-shelf inductors that would suit. Most likely the higher the value of L you use (within the stable range) then the lower the output current ripple.

    Page 18 discusses the choice of capacitors - the minimum Cout seems to be 22uF but the low ESR demand seems to be the critical factor in component choice - hence ceramics are recommended over tantalum. Also Cin seems to be a minimum of 10uF with higher values improving the input filtering performance. Again ESR values are critical.

    Circuit component values required to give you the 3.3V output can be determined from the other design info.

    How experienced are you with circuit construction?

    Where will you source your components?

    Are you using a prototype board specifically designed for this chip?

    I think you will need to be extremely careful with your layout - conductor paths and earthing, etc.

    Good luck:)
     
  3. nandax

    Thread Starter Member

    Nov 20, 2008
    12
    0
    thanks for the reply..
    actually this is my first time dealing with DCDC converter circuit,
    but what is the function of PG pin? I still dont get it,
    and what is recommended value for pull up resistor in LBO pin?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    See page 13.

    From the text I believe PG is a logical level output signal that may be used to monitor a particular condition - whether or not the supply output voltage is within ~1.5% of the target design value. Being a straight logic level signal, it will either be somewhere near ground / low state ("power not good" or EN pulled low or unit is shutting down) or in a high state ("Power Good" - PG). At start up, PG would be low. As the output voltage rises and stabilizes near the target value PG will switch high.

    PG could be used to indicate normal operation to other circuitry connected with the supply. Other conditions apply to the functioning of the PG output - the EN pin must be set high for instance (i.e. connected to VI). I guess that's logical - no enable equals no output!

    Being an open drain output, the PG terminal requires a pull-up resistor to allow the signal to switch correctly between the logical low and high states. PG can only sink 1mA, so it is probably not intended to do any "hard work" - e.g. directly lighting a LED.

    You don't have to use PG - it's an unnecessary but potentially useful output. It is recommended that if PG is not used the user just leaves it open or ties /wires it to ground.

    LBO has a different purpose but is also an open drain output requiring a pull-up resistor. Again, you don't have to use LBO unless you want to monitor the particular condition - to detect the low battery state. The text doesn't say what the maximum current sinking capacity of LBO is, but I would think it could well be 1mA as for the PG pin.

    It is recommended that no greater than 6V be applied to PG or LBO. They can be tied to Vo through their pull-up R's provided Vo<6 [7V absolute maximum]. Regarding the 1mA limit, if Vo is 5V (say) the pull up R on either pin should be greater than ~5kΩ.

    Never tie PG or LBO directly to any voltage other than ground, without the minimum pull-up R included in the series path.

    Hopefully that helps.
     
  5. nandax

    Thread Starter Member

    Nov 20, 2008
    12
    0
    I'd like to use LBO pin to detect low battery state,
    but what will be the voltage accross LBO pin?
    because it says it will be low if LBI voltage below 1.21 V.
    if I set the LBI low if the Vin below 6 with resistor divider, what will be the voltage accross LBO pin?
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    If your battery input is 9V suppose then for argument's sake, you want to detect when the battery voltage falls to 7.5V.

    You need to create a voltage divider (R5 & R6) with VI as the source and LBI as the divider sense point.

    So with 7.5 V across R5 & R6 the centre point should have fallen to 1.21V

    i.e. VI x R6/(R5+R6) = 1.21

    or 7.5 x R6/(R5+R6) =1.21

    or R6/(R5+R6)=0.161

    or R5=5.21 x R6 - e.g. if R6 = 75K and R5 = 390K

    If LBO is pulled up to Vout (via R pull-up) then the value of LBO high will depend on the current Vout value with VI having dropped to the target 7.5 V detection point.

    You could, as per the data sheet notes, tie LBO through its pull-up to a another fixed 6V supply. But it's annoying needing another supply - albeit very low current.

    If Vout is still around 3.3V with VI having fallen to 7.5V, then LBO will transit from around 3.3V to around 0V at the low battery detection point. So LBO high (= battery still good) would be 3.3V.

    :)
     
  7. nandax

    Thread Starter Member

    Nov 20, 2008
    12
    0
    thanks for the reply..
    now I have a problem to determine Cin, because in page 18 (datasheet) it says that the minimum value is 10uF,
    and if I calculate max Irms = 72,2 mA (Io=150mA, Vo= 3.3 V, Vi= 9V)
    that means the capacitor ripple current rating should be above 72.2mA right?
    but how to calculate if the 10uF capacitor rating can handle that max Irms?
     
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