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boost converter

Discussion in 'Electronic Design' started by [email protected], Nov 8, 2006.

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  1. Guest

    I was reading up on the theory behind the boost converter:

    http://en.wikipedia.org/wiki/Boost_converter

    and I thought, you know, there is a short circuit for a certain amount
    of time. Actually, for a fairly long time, since Figure 5 of the above
    link seems to show best performance for a duty cycle of 80% (please
    correct me if I'm wrong).

    Would having a resistor (even just one ohm) in the path of the
    switching transistor improve efficiency at all...?

    Thanks,

    Michael
     
  2. Joerg

    Joerg Guest

    Hello Michael,

    Figure 5 shows just a graph. Duty cycle versus performance depends on
    the output load (voltage and current), it cannot be generalized. 80%
    typically happens when the output voltage needs to be 4-5 times higher
    than the input.

    Nope, that would increase the losses because it would be like degrading
    the Rdson by 1ohm. When we design switchers there is always that hunt
    for the least in Rdson that a nickel or two can buy. No free lunches there.

    There really isn't a short. The inductor current begins to ramp up. At
    some point the switch opens and the inductor dumps its current into the
    load side capacitor. "Some point" means before the inductor core
    saturates. Else there will be an impressive and loud pyrotechnical show
    followed by a plume of smoke and wailing fire alarms. Same if it goes
    into continous current mode (CCM) and the inductor current "ratchets up"
    for some reason.
     
  3. Guest

    There's no short circuit. When the switch is on, the inductor limits
    the current, which builds up slowly in the inductor. Well, relatively
    slowly, anyhow. See Fig. 3 in your article.

    When the switch opens, the inductor "flies back" and dumps its energy
    into the output filter & load. Again, current ramps down, relatively
    slowly.

    Introducing resistance in the high current paths will cause I^2*R
    losses, decreasing efficiency. Resistance is not your friend.

    Best,
    James Arthur
     
  4. Guest


    Ok, thanks for the reply.

    Loud pyrotechnical show, smoke, wailing fire alarms... just *not* what
    I'd want in a boost converter...

    On that thought... put a fuse in the same path where I'd previously
    recommended a resistor?

    Michael
     
  5. This is a sci.electronics.basics question, so I am setting followups
    to there.

    The inductor is not a short circuit in the sense you imagine. There
    is an impedance to it, if not a lot of resistance. And you need to
    imagine in a dynamic (time derivative) sense.

    Let's say the inductor starts out with zero current when the switch is
    closed. The switch closure would attempt to have the current suddenly
    jump to some large value. But the inductor refuses to allow sudden
    changes. Instead: dI = V/L * dt, or dt = L/V * dI

    So for each fraction of time, only a small fraction of current change
    is permitted. Since it starts at zero in this case, it cannot go
    instantly to a large value. Assuming the voltage across the inductor
    remains constant, and it does in this ideal case you cite, a fixed
    change in current will require a fixed change in time. In other
    words, it cannot happen right away. To achieve this, energy is
    gathered up and stored in a magnetic field around the inductor.

    If you wait short enough time, the current will never manage to rise
    up too high. It will get to some point and then the switch opens.
    When that happens, the inductor will alter the voltage between its
    leads in any way required in order to maintain that current. Of
    course, the current will then have to flow through the diode and into
    the load. (Usually, there is a capacitor across the load at the other
    end of that diode in models like this to absorb the energy and supply
    current while the switch is closed.) This means that the energy
    stored in the magnetic field, as it now collapses, will be consumed.

    Again the law remains, dI = V/L * dt, or dt = L/V * dI. So it will
    also take time to decline to zero current. However, in this case, the
    V will not be the V of the battery, but instead the V that the
    inductor creates as the field collapses, needed to ensure a gradual
    decline in the current instead of a sudden decline.

    Does that make sense?

    Jon
     
  6. Joerg

    Joerg Guest

    Hello Michael,
    You could but by the time it blows the transistor is most likely already
    gone. A fuse is typically placed before the whole converter, in an
    attempt to prevent a major explosive event.
     
  7. john jardine

    john jardine Guest

    I've found inductors made from 5mm Copper tubing are ideal for use during
    much of the development process. (So trouble free, I've even pondered on
    specifying as the final component)
    john
     
  8. Guest

    Yes. Murphy says your $$ transistor will blow in microseconds,
    protecting your 10-cent fuse.

    Cheers,
    James Arthur
     
  9. Guest


    What did you wrap them on? (Iron? Just a PVC pipe?)
     
  10. Joerg

    Joerg Guest

    Hello James,
    A microsecond is actually a long time. A laser diode can go from fully
    functional to lalaland in one nanosecond. No kaboom, nothing, it's just
    dead.
     
  11. Joerg

    Joerg Guest

    Hello John,

    An air core of that size can trigger the Federales to show up at your
    lab in a jiffy ;-)
     
  12. john jardine

    john jardine Guest

    A bit of scrap plastic pipe. Then wind on turns to give the wanted
    inductance. A few dabs of hot glue to hold it together. Warm at 100Amps but
    no farting about with airgaps and saturation.
    john
     
  13. Fred Bartoli

    Fred Bartoli Guest

    Joerg a écrit :
    Don't think so. John's in UK :)
     
  14. Joerg

    Joerg Guest

    Hello Fred,
    Well, then it would be her Majesty's guards :)
     
  15. default

    default Guest

    The inductors are never run in a saturated condition, the voltage is
    applied for a relatively short time compared to what it takes to
    saturate the reactor.

    If you just closed the switch for a long time with respect to the size
    of the inductor - yeah, you'd be wasting power. Below saturation or
    well below saturation most of the energy is give up to the load when
    the switch opens.

    A resistor would decrease efficiency since all it can do is drop
    voltage and make heat. That energy is lost and not available to the
    load.
     
  16. Fred Bartoli

    Fred Bartoli Guest

    Joerg a écrit :
    Guard... as in shield?
     
  17. Joerg

    Joerg Guest

    Hello Fred,
    I guess they don't use iron clad body armor anymore these days ;-)
     
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