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voltage reference

J

Jamie Morken

Jan 1, 1970
0
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
 
B

Ban

Jan 1, 1970
0
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.
 
S

Spehro Pefhany

Jan 1, 1970
0
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,

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.
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!? :)

It's only the waste part they are talking about.

Best regards,
Spehro Pefhany
 
W

Winfield Hill

Jan 1, 1970
0
Spehro Pefhany wrote...
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.

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
 
F

Frank Bemelman

Jan 1, 1970
0
Winfield Hill said:
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.

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...
 
F

Frank Bemelman

Jan 1, 1970
0
Winfield Hill said:
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.

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...
 
R

Robert Baer

Jan 1, 1970
0
Jamie said:
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

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.
 
R

Rich Grise

Jan 1, 1970
0
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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