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voltage reference

Discussion in 'Electronic Design' started by Jamie Morken, Nov 24, 2003.

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  1. Jamie Morken

    Jamie Morken Guest

    Hi,

    I am using a precision 5Volt reference chip (max6133) and in the datasheet
    it says that the quiescent supply current is virtually independant of the
    supply voltage because the load current is drawn from the input only when
    required so supply current isn't wasted. I know that a standard voltage
    regulator dissipates (Vin-Vout)*current, so how can this vref chip dissipate
    the same power independant of the supply voltage, and why don't they make
    voltage regulators this way since it is more efficient!? :)

    cheers,
    Jamie Morken
     
  2. Ban

    Ban Guest

    Jamie Morken wrote:
    || Hi,
    ||
    || I am using a precision 5Volt reference chip (max6133) and in the
    || datasheet it says that the quiescent supply current is virtually
    || independant of the supply voltage because the load current is drawn
    || from the input only when required so supply current isn't wasted. I
    || know that a standard voltage regulator dissipates
    || (Vin-Vout)*current, so how can this vref chip dissipate the same
    || power independant of the supply voltage, and why don't they make
    || voltage regulators this way since it is more efficient!? :)
    ||
    || cheers,
    || Jamie Morken

    Jamie, try to think logically. If the supply current is constant it means
    exactly P=Vin*Is, so the power is proportional to Vin.
    Only with a switching regulator and some energy storing component(coil) you
    can achieve a constant power. Otherwise the energy is wasted in some
    pass-transistors or in case of 2-terminal references in dropping resistors.
     
  3. That's for an ideal regulator. In reality, there's an additional term
    in there related to current the regulator uses for its own purposes-
    that flows from input to ground, or current that flows from input to
    output and must be wasted in a divider network etc.

    In general, that current is a function of both the input voltage and
    the output current. With 78xx regulators, it's a fairly constant
    current of 4 or 5 mA. So, if the output current is 1mA, the regulator
    draws 5 or 6 mA. With others (especially early bipolar LDO regulators)
    it can do strange things, especially at high output currents or close
    to dropout.
    It's only the waste part they are talking about.

    Best regards,
    Spehro Pefhany
     
  4. Spehro Pefhany wrote...
    Right. As Spehro says Jamie, Pd = (Vin-Vout) * I is a best-
    case theoretical dissipation for a linear regulator, assuming
    that no additional power is required for the circuitry.

    To be fully accurate for LM317-style 3-terminal regulators
    we must add a Vin * Iadj internal-dissipation term for the
    reference (Iadj = 50 to 100uA), and a Vout * 1.22 / R1 term
    for the reference resistors. Notice that these terms don't
    depend on output current. These terms may be small, but this
    doesn't mean the LM317 is performing magic. With a '317 we
    have two serious constraints: a typical 2V minimum Vin-Vout
    overhead (the dropout voltage), and a minimum load current
    (5mA max) required to operate the circuitry. (A safety rule
    is to make the second reference-resistor term large enough to
    meet the minimum load current spec.)

    Comparing the '317 to other types of regulator ICs, they have
    different Pd extra terms and different constraints. Common
    780x-style regulators have the Vin * Iq quiescent-current
    term, where Iq = 8mA max, and is unaffected by load current.
    780x types also require a 2V minimum Vin-Vout overhead (the
    dropout voltage), but they have no minimum load current.

    The 2V dropout-voltage issue for '317 and '7800 regulators
    forces a considerable additional loss of power wasted in the
    heat sink. We can solve this issue with a low-dropout (LDO)
    regulator style. These also have the Vin * Iq quiescent-
    current term, plus a Vin * Iout / beta term if they use
    a bipolar pass element (beta is the PNP transistor's gain).
    Types using p-channel MOSFETs don't have an Iout/beta term.
    LDO regulators have a greatly reduced dropout requirement,
    some as low as 50 or 100mV. They generally don't have any
    minimum load current requirement. Sounds pretty good, but
    they do require large output capacitors, without which they
    may have inferior load transient response.

    Jamie, let's consider the issue of dropout voltage, which is
    related to the minimum input - output voltage in a regulator
    circuit. Just because you choose an LDO IC with a low 50mV
    spec. doesn't mean you can design a power supply with a small
    input-output voltage difference, no-way-Bob! Thinking of AC
    line power supplies, you have to provide large overheads for
    transformer load sag, diode resistance and storage-capacitor
    voltage ripple, all of which increase with load current, plus
    an allowance for brownout operation at one extreme and excess
    line voltage at the other extreme. Considerations like these
    can swamp the one or two-volt savings offered by an LDO, but
    hey, you take what you can get! Certainly an LDO regulator
    can be a real life saver in battery-operated systems.

    Thanks,
    - Win

    whill_at_picovolt-dot-com
     
  5. Will, you have such a clear way of explaining, my compliments.
    I wanted to ask Spehro about that Iq (Iq????) and ended up
    playing all kinds of online IQ tests I found...
     
  6. Win, you have such a clear way of explaining, my compliments.
    I wanted to ask Spehro about that Iq (Iq????) and ended up
    playing all kinds of online IQ tests I found...
     
  7. Robert Baer

    Robert Baer Guest

    Do *not* mess with Maxim; they do not seem to want to make what they
    advertise.
    Now if you have tens of thousands of dollars to toss at one part type,
    and are willing to wait 6 months to a year or so, then.....
    Many distributors refuse to carry Maxim at all, due to this problem.
     
  8. Rich Grise

    Rich Grise Guest

    The Vref chip doesn't "dissipate the same power independent of the
    supply voltage," it says "The quiescent supply current is virtually
    independent of the suplly voltage." The (Vin-Vout)*current term is
    just the load current that they're talking about, and you're not
    supposed to put much of a load on a reference chip - just a few
    milliamps, so the power dissipation is practically negligible.

    (say, a +5 reference with a +12 supply, supplying 15 mA to your
    comparator or whatever, that'd be (12 * Iq) + (7 * 0.015) watts.
    Not a whole lot. :)

    Cheers!
    Rich
     
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