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bootstrap circuit.. how it works...

Discussion in 'Electronic Basics' started by noviceguy, Dec 5, 2005.

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  1. noviceguy

    noviceguy Guest

    Hi all...

    I am very new to circuit design.. and is currently looking at some
    application notes available on the net.. about DC-DC flyback
    converters... now i'm looking at Renesas's HA16384 PWMIC.. inside it i
    saw a sample application circuit..
    however, there is a particular circuit that i am having a hard time to
    maybe its my overall understanding that is lacking..
    i search the internet but.. i'm still confused about it... :(
    i am really hoping that someone could advice me on the issue...

    my problem lies in the boot-strap circuit to provide initial power to
    the PWM IC...
    i'm trying to understand why they are using a 220kohm resistor and a
    10uF capacitor (rating 50V). The resistor is connected directly to the
    rectified high voltage input of the transformer.
    how does this circuit function?
    it is also connected to the Auxilary winding of the transformer.. but
    initially, there would not be any power from this as the PWM is still
    not operating right?
    so the PWM will get its power from the resistor-capacitor path.. is
    this correct?
    but this path there will be very high voltage..
    does the resistor and capacitor combination reduces it to the PWM IC
    safe operating voltages?the IC max operating voltage is 30V.
    i'm really confused....i've search the internet... and the say this is
    a simple boot-strap circuit... but i still dont get it...
    how they come out with the 220kohm and 10uF values...
    what kind of calculation involved?

    also.. after the Auxilary winding starts operating and supply Vcc to
    the PWM IC, what happen to the 220kohm resistor connection?

    i try to represent the circuit below..
    141V (from example)
    rectified voltage -------------|--------------------- To TF
    R = 220kohm (1/4W)
    other cct -----------------|------Diode------TF Auxilary Winding
    PWM IC Vcc------------------|
    C = 10uF (50V)

    i hope my question make sense..

    thanks in advance for any advices...

    Best Regards
  2. First of all, I doubt very much that the 220k 1/4 watt resistor is
    rated for continuous rectified line voltage. I would use a 1/2 watt,
    just for the voltage rating.

    The idea is that the control chip remains off until the capacitor
    charges up to some predefined operating voltage via the 220k resistor.
    Then the control circuit is switched on and it has a millisecond or
    so to operate the supply before the capacitor is discharged to the
    point that the control circuit is switched back to a dormant state.
    If the diode from the auxiliary winding contributes power before that
    happens, it takes over as the power supply to the capacitor and the
    start up current from the 220k is incidental.

    The trick is to have a clean shut down of all loads on the capacitor
    (to well below the current supplied by the 220k resistor at low line
    voltage) except for a very low current voltage measurement that wakes
    everything up, cleanly. And there must be a considerable hysteresis
    in the voltage sensing, so that once switched on, the capacitor
    voltage can decay quite a bit before everything is switched off to
    wait for another start charge cycle.

    20 years ago, I designed a circuit made of transistors, zener diodes
    and resistors that performed this voltage sensing and power switching
    function for the supply in a commercial printer. I wish I had a
    schematic of it to show you.
  3. Tim Williams

    Tim Williams Guest

    1. It takes some time for the voltage to rise (T = RC).
    2. - What is the current draw when the voltage reaches, say, 10V? 30V?
    What does this current do to the voltage across the 220k resistor?
    Just in case, there's probably a protection circuit. The UC3842 has an
    internal zener diode. You could run it forever from the +320VDC rail
    through a resistor, but that would be wasteful.
    R = V/I
    It's ignored. The aux. winding supplies more current to the chip, which it
    requires once it comes online and starts moving things. Otherwise, this
    current draw would pretty well consume the voltage supplied by the 220k and
    it would not function very well.

    The aux. winding also often supplies a figure of how much energy is going
    through the transformer, important for maintaining regulation and isolation.

  4. noviceguy

    noviceguy Guest

    Thanks very much for the info John and Tim :)

    i think i got a rough idea on how this circuit works..

    you guys insight really help a lot...

    again..thank you very much

    Best regards
  5. Jasen Betts

    Jasen Betts Guest

    220K is quite a large resistor, with a supply of 141V only 0.64mA will flow
    through it as long as the PWM IC is using atleast that much the capacitor
    won't overcharge.

    What I expect happens is that the PWM chip lies dormant until the capacitor
    charges to about 12V (for example) then it turns on and with a pulse or two
    to the driver transistors the whole powersupply wakes up, and from then on
    most of the power to the PWM chip will be through the diode, but there'll
    still be a little coming through the resistor.

    becacuse the eaxilary winding has a lower voltage than the 50V capacitor's
    limit it won't overflow the capacitor, and because the PWM chip is using
    more than the 0.64mA the resistor can supply it won't either.

  6. noviceguy

    noviceguy Guest

    Thanks Jasen...

    also.. i read in the HA16384 specs, got a characteristic that states
    Stanby current of value 230uA (max)
    will this current value come into consideration in selecting the
    resistor?in addition to the capacitor charging time..

    anyway.. i think in a week time or so.. i'll give this circuit a try..
    well if i can get the transformer :)
    still sourcing for it...
    not so easy where i'm at unfortunately...

    anyway.. thanks a lot for all you guys advices and insights..
    really appreciate it... :)

    best regards...
  7. Jasen Betts

    Jasen Betts Guest

    Yes, the standby current (or some portion of it when the capacitor is
    depleted) will be flowing into that that chip while the capacitor is
    charging up to the chips starting voltage.

    230uA is well below thr 600-ish uA that will flow through the resistor,
    leaving 370 or so to charge the capacitor.
    something from inside an old PC powersupply?
    when they fail it's almost always not the transformer.
    I appreciate your respoonse too... but don't rely on everything I post
    I'm learning too.

  8. noviceguy

    noviceguy Guest

    no worries Jasen...

    hmm.. PC power supply... duh!! why did i think of that....
    thanks for the info... i always missed out on obvious things.. lol!

    actually.. i'm trying to build a battery charger kind of thing..
    so looking at some application notes on related stuff...

    afterward.. maybe i'll try inverter circuit too :)

    hmmm.. power supply stuff is quite fun...
    who would have thought that.. lol!

    well fun.. but dangerous as well..

    anyway.. thanks again... for all the advices John, Tim and Jasen.

    best regards
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