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'[EE]: Multimeter question (Fluke)'
2004\07\20@101830 by 8859-9?B?1m1lciBZYWxo/Q==?=

I have Fluke 189 Digital multimeter and I am currently measuring the current consumption.  When I put the dial in the uA range, the unit shows the consumption as ~4000 uA, when the dial is mA range, it shows the consumption as 2.1 mA.  Which one is true?  This problem is not specific to fluke digital multimeter, I have seen this behaviour on many multimeters.  Which reading should I be trusting?
Any help is appriciated.
Omer

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2004\07\20@103909 by Mike Harrison

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On Tue, 20 Jul 2004 17:18:58 +0300, you wrote:

>I have Fluke 189 Digital multimeter and I am currently measuring the current consumption.  When I put the dial in the uA range, the unit shows the consumption as ~4000 uA, when the dial is mA range, it shows the consumption as 2.1 mA.  Which one is true?  This problem is not specific to fluke digital multimeter, I have seen this behaviour on many multimeters.  Which reading should I be trusting?

The answer should be 'both'.
The difference is probably due to the DMM using different shunts on the 2 ranges - the shunts will
have different resistances, and so will affect the actual current passing. If you hook a second
meter in-line you will probably see the current change as you swap range on the other meter.
In your example, I would expect that the circuit you are measuring uses something like a switchmode
regulator, as you would normally expect current to drop with the higher-resistance uA range.

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2004\07\20@105605 by David VanHorn

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At 05:18 PM 7/20/2004 +0300, =?iso-8859-9?B?1m1lciBZYWxo/Q==?= wrote:

>I have Fluke 189 Digital multimeter and I am currently measuring the current consumption.  When I put the dial in the uA range, the unit shows the consumption as ~4000 uA, when the dial is mA range, it shows the consumption as 2.1 mA.  Which one is true?  This problem is not specific to fluke digital multimeter, I have seen this behaviour on many multimeters.  Which reading should I be trusting?

Are you sure the current is constant?
If the processor is running, there is likely some AC component, which may be an issue. Also the shunt resistance may be playing a part.

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2004\07\20@105811 by Omer YALHI

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>In your example, I would expect that the circuit you are measuring uses
something like a switchmode
>regulator, as you would normally expect current to drop with the
higher-resistance uA range.

The circuit I am testing is running on 3.6V lithium batteries and has no
regulator.

>The difference is probably due to the DMM using different shunts on the 2
ranges - the shunts will
>have different resistances, and so will affect the actual current passing.
If you hook a second
>meter in-line you will probably see the current change as you swap range on
the other meter.

So how can one accurately measure the current consumption of their circuit?
The difference I am seing right now is too much to work with, 2.1 mA against
4.0 mA.

Thanks, Omer

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2004\07\20@110153 by Omer YALHI

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>Are you sure the current is constant?
>If the processor is running, there is likely some AC component, which may
be an issue. Also the shunt resistance may be >playing a part.

I have powered up everthing and measuring the unit, the current consumption
should be stable (it is stable anyway, both in uA range and mA range).
There is just that difference, and do not know which value is correct(er)!

:-)

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2004\07\20@110812 by Mike Harrison

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On Tue, 20 Jul 2004 09:56:20 -0500, you wrote:

>At 05:18 PM 7/20/2004 +0300, =?iso-8859-9?B?1m1lciBZYWxo/Q==?= wrote:
>
>>I have Fluke 189 Digital multimeter and I am currently measuring the current consumption.  When I put the dial in the uA range, the unit shows the consumption as ~4000 uA, when the dial is mA range, it shows the consumption as 2.1 mA.  Which one is true?  This problem is not specific to fluke digital multimeter, I have seen this behaviour on many multimeters.  Which reading should I be trusting?
>
>Are you sure the current is constant?
>If the processor is running, there is likely some AC component, which may be an issue. Also the shunt resistance may be playing a part.

I just measured  the shunt R on my fluke 87 III - 100 ohms on uA, 1 ohm on mA.
Unless you are using a current regulator, adding 99 ohms is certainly going to change your current,
although you woul usually expect it to decrease.

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2004\07\20@111432 by Mike Harrison

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On Tue, 20 Jul 2004 17:57:30 +0300, you wrote:

>>In your example, I would expect that the circuit you are measuring uses
>something like a switchmode
>>regulator, as you would normally expect current to drop with the
>higher-resistance uA range.
>
>The circuit I am testing is running on 3.6V lithium batteries and has no
>regulator.
>
>>The difference is probably due to the DMM using different shunts on the 2
>ranges - the shunts will
>>have different resistances, and so will affect the actual current passing.
>If you hook a second
>>meter in-line you will probably see the current change as you swap range on
>the other meter.
>
>So how can one accurately measure the current consumption of their circuit?
>The difference I am seing right now is too much to work with, 2.1 mA against
>4.0 mA.

Both answers are correct (assuming your DMM is not faulty)
One reading is the current draw with 1 ohm in series, the other is the reading with 100 ohms in
circuit (or whatever shunt values your DMM uses). In general, the higher current range will have
less resistance, and therefore affect your circuit less, so you should use the mA range wherever
possible.


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2004\07\20@112511 by Nigel Orr

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> I have powered up everthing and measuring the unit, the
> current consumption
> should be stable (it is stable anyway, both in uA range and mA range).
> There is just that difference, and do not know which value is
> correct(er)!

You could use your own shunt and measure the voltage across it as another
check.  For example, if your circuit can cope with 0.1V drop on the power
supply at 4mA, add a 27R shunt and measure the voltage across it.
Calculate the current as (V/27) and measure the supply voltage.

Then try it with a 10R shunt, and calculate the current as V/10 and measure
the supply voltage.  You should be able to plot a few points, draw a graph,
and work out what the expected supply current is with no shunt.

Then you can decide which meter to believe :-)

Sometimes when measuring current, the circuit will not appreciate the extra
cable length to the meter and back, so a shunt soldered in line is the
easiest way to measure current.

Nigel
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2004\07\20@115835 by Omer YALHI

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Interesting.  It does not make sense to pay $450 or so on a meter and do all
the calculations.  Shouldn't the meter at least do the necessary math to
show the current consumption as same in mA and uA (taking into account of
different resistor values)?

The units spec shows that it has the
DC uA accuracy:
 up to 500 uA is 0.01 uA
 up to 5000 uA is 0.1 uA

DC mA accuracy
 up to 50 mA is 0.001 mA
 up to 400 mA is 0.01 mA

From what I understand, these values and accuracies in the spec do not mean
a thing!  Why put it then?

Thanks and Regards,

Omer

{Original Message removed}

2004\07\20@120702 by Mike Harrison

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On Tue, 20 Jul 2004 18:59:13 +0300, you wrote:

>Interesting.  It does not make sense to pay $450 or so on a meter and do all
>the calculations.  Shouldn't the meter at least do the necessary math to
>show the current consumption as same in mA and uA (taking into account of
>different resistor values)?
>
>The units spec shows that it has the
>DC uA accuracy:
>  up to 500 uA is 0.01 uA
>  up to 5000 uA is 0.1 uA
>
>DC mA accuracy
>  up to 50 mA is 0.001 mA
>  up to 400 mA is 0.01 mA
>
>From what I understand, these values and accuracies in the spec do not mean
>a thing!  Why put it then?

The meter IS giving you an accurate figure for the current flowing in your circuit - the current
CHANGES when you change range. With ANY meter you must understand and take into account the effect
that the meter has on the circuit.
I'm sure that somewhere in the spec it states the resistance that the various current ranges add,
and this must be taken into account.

You must understand the limitations of any measuring instrument in order to get a meaningful
reading.
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2004\07\20@120913 by Alan B. Pearce

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>Shouldn't the meter at least do the necessary math to
>show the current consumption as same in mA and uA
>(taking into account of different resistor values)?

The meter is, but your circuit under test does not know that it should draw
the same current with a 100x value resistor in the supply lead.

Remember also that when measuring on the mA range, you are going to be
trying to measure a current that will be of the magnitude of the inaccuracy
of the meter on that range, hence you see the current go down when instinct
tells you that if anything, it should go up. Your accuracy for the meter on
that range is probably +/-1mA +/-1 digit.

When measuring on the microamp range you are measuring at close to full
scale of the meter, so the inaccuracy will disappear into the weeds, and be
such a small part of the reading it does not matter.

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2004\07\20@121251 by Nigel Orr

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pic microcontroller discussion list <> wrote:
> Interesting.  It does not make sense to pay $450 or so on a
> meter and do all the calculations.

I would expect that the meter meets all it's specs, especially as it is a
Fluke.

>  Shouldn't the meter at least do the
> necessary math to
> show the current consumption as same in mA and uA (taking
> into account of
> different resistor values)?

It can't.  There is no way for it to know what your circuit would draw
using a different shunt.  It's up to you to realise the limitations of any
measuring device when you use it.  I would expect that another ammeter in
series would agree with the two separate readings, so the meter is reading
the current correctly.  The unfortunate thing is that it is also affecting
the reading significantly.  There's no way to measure something without
affecting it, it's just that in your case, the effect is too big to be
satisfactory.

It would only take a few minutes to get a more accurate estimate, if it's
important to the product.

> From what I understand, these values and accuracies in the
> spec do not mean a thing!  Why put it then?

They mean that you can reliably get the same reading in the same circuit.
You need more than that, so you need a different way to read it.

Nigel
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2004\07\20@122743 by Olin Lathrop

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> the current CHANGES when you change range.

Except that his is changing backwards, so something else is going on.  The
lower current ranges will have high shunt resistances.  Normally a high
shunt resistance will cause less current to be drawn, however he is seeing
more.  This may be due to a nearly dead battery in the meter, which is a
suspicion particularly when the resistance ranges don't overlap right.

Some circuits do exhibit negative resistance to their power source, so a
larger shunt can result in higher current.  This is usually do to a
switching power supply, but at 2 and 4 mA I doubt that's what's going on.


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(978) 742-9014, http://www.embedinc.com

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2004\07\20@150848 by Robert Rolf

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His circuit may also be oscillating when it sees the
higher impedance power supply (uA shunt), and so draws
more power.

I had a Maxim 1676 switch mode upconverter that had this
similar behaviour (Fluke 87). Turned out that the extra
drop in the supply lead from the higher shut resistor
made the convertor turn on longer, which
made the drop worse and so it drew a lot more current to
get the output back up where it wanted to be.

The net power was the same, but because the starting voltage
was lower, it pulled more current. That lost power went
into the shunt.

Robert

Olin Lathrop wrote:
{Quote hidden}

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2004\07\20@191808 by Jinx

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> show the current consumption as same in mA and uA

Omer, I've recently had the same thing happen to me. A circuit was
apparently using 1100uA on one range and 760uA on another. Like
Olin hinted at, check your battery. Mine came right

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2004\07\20@192847 by Bob Axtell

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I once worked for Fluke, and what you are experiencing is the "range
error" problem. Its a bit like the paradox in particle physics called
"Schroedeger's (sp?) Cat"; the "fact that the measurement is being made"
affects the outcome.

Each range has a different value resistor used to develop the
differential voltage needed to measure the circuit. The tested circuit
will act differently when each different resistor is inserted. The
effects are rarely noticeable at large voltages 25V+ but at 1-2V the
effect is quite noticeable.

There is very little that can be done, except buy a very expensive meter.

--Bob

Jinx wrote:

{Quote hidden}

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2004\07\20@194129 by Bob Blick

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> I have Fluke 189 Digital multimeter and I am currently measuring the
> current consumption.  When I put the dial in the uA range, the unit shows
> the consumption as ~4000 uA, when the dial is mA range, it shows the
> consumption as 2.1 mA.  Which one is true?  This problem is not specific
> to fluke digital multimeter, I have seen this behaviour on many
> multimeters.  Which reading should I be trusting?

Probably neither. I've noticed the Fluke meters have a considerably
strange impedance on the current ranges that influences anything but low
frequency current readings.

The really cheap multimeters seem to work better for current(and the fuses
in them are cheaper, too!)

Cheerful regards,

Bob

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2004\07\21@022530 by Omer YALHI

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I would like to thank everyone for their input.  The meter is brand new so
the battery is not dead.  I will stabilize the software and the circuit so
that it draws constant current (I did, but will check it again), also I will
try measuring the voltage across different shunts, as suggested to me, and
calculate the current consumption.  Draw a graph and estimate the current at
0 shunt, this seems to be the best approach.

Regards,

Omer

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2004\07\21@032912 by Bob Axtell

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Here is the solution Omar:

Pick a resistance value that you can live with, such as 1 ohm 1%. Place
that in series with your low-power circuit that you are trying to measure.

Now, to make your measurement of current by reading differential
voltage, you have removed the range uncertainty problem. To measure the
current, measure the voltage drop across the 1 ohm resistor. To solve
for current:

If the voltage across the 1 ohm resistor is 0.10V, then I=E/R or I =
0.1/1 or the current through the resistor is 100mA. See?

But be aware that even the 1 ohm resistor affects the circuit you are
measuring. But you now KNOW how it is interacting

--Bob

Omer YALHI wrote:

{Quote hidden}

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2004\07\21@041342 by Omer YALHI

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Thank you Bob Axtell, I will do that and do it with several different
resistor values.

--Omer.

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2004\07\21@171027 by John N. Power

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> From:         Omer YALHI[SMTP:spam_OUToyalhiTakeThisOuTspamTEKSAN.COM.TR]
> Sent:         Tuesday, July 20, 2004 10:57 AM
> To:   .....PICLISTKILLspamspam@spam@MITVMA.MIT.EDU
     > Subject:        Re: [EE]: Multimeter question (Fluke)


> So how can one accurately measure the current consumption of their circuit?
> The difference I am seing right now is too much to work with, 2.1 mA against
> 4.0 mA.

> Thanks, Omer

The emphasis on low power consumption of PIC circuits has caused me to
think about ways to measure sub-microamp supply currents. I have been
considering building a meter to measure the idling and operating currents of
battery powered microcontroller circuits, and I believe that the same idea
may be useful for your problem.

The basic idea is to make a virtual ground opamp circuit and use it to
measure the current under test. Connect the non-inverting input of an opamp
to ground. Connect a feedback resistor Rf between the output of the
amp and the inverting input.Connect the positive side of your circuit to
the positive side of the battery. Ground the negative side of the battery.
Connect the low (normally grounded) side of the load (your circuit) to
the inverting input of the amp.

The return side of the load will be kept at ground by the amplifier. The
load current will be supplied through Rf,. which puts the amplifier
output at   - Rf * load_current. Measuring the amplifier output voltage
and dividing by Rf gives the current you want to measure. The voltage
across your circuit will be kept at the battery voltage. You have the
option of putting a capacitor across Rf to smooth the reading. A
capacitor across the load (normally this would be the supply bypass
cap in your circuit) does not affect the reading, since the total
current of the load and bypass cap is the real current you need to
measure anyway.

Switching in various values of Rf gives you range selection. If the load
current exceeds the capacity of the opamp to supply it, use a PNP
transistor to buffer the amp output. Connect the base to to the amp
output, the emitter to the "output" end of Rf, and the collector to the
negative supply voltage for the opamp. A pullup resistor from the
emitter to the positive supply with stabilize things. Output voltage
is now measured from the emitter end of Rf; this may require a
buffer amp (voltage follower). Make sure that you have enough
negative supply to keep the reading linear.

John Power

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2004\07\21@181510 by Mike Harrison

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On Wed, 21 Jul 2004 17:10:17 -0400, you wrote:

>> From:         Omer YALHI[SMTP:oyalhispamKILLspamTEKSAN.COM.TR]
>> Sent:         Tuesday, July 20, 2004 10:57 AM
>> To:   .....PICLISTKILLspamspam.....MITVMA.MIT.EDU
>      > Subject:        Re: [EE]: Multimeter question (Fluke)
>
>
>> So how can one accurately measure the current consumption of their circuit?
>> The difference I am seing right now is too much to work with, 2.1 mA against
>> 4.0 mA.
>
>> Thanks, Omer
>
>The emphasis on low power consumption of PIC circuits has caused me to
>think about ways to measure sub-microamp supply currents. I have been
>considering building a meter to measure the idling and operating currents of
>battery powered microcontroller circuits, and I believe that the same idea
>may be useful for your problem.

I've been thinking about this problem for several years, and everyone I've spoken to in this
business agrees that it's a problem, in particular the issue of measuring low currents, but without
having voltage-drop problems when the product wakes up and does something that sucks a much greater
current.
I've now started serious work on designing a 'bench PSU for microcontrollers', which I intend to
manufacture when I've got it all sorted - hopefully before the end of the year.
The main feature is that it will have a 7 digit current display (999.9999mA), able to display
currents of a few microamps, up to 1 amp with fast re-ranging and no shunt changing. It won't give
huge (7-digit) _accuracy_, you don't need that, but it does give a large _range_, and will  display
3 significant digits anywhere between 99.9uA to 999mA, with at least 1% accuracy
It will also do true avaraging of current in the above range over a selectable time period, to give
a good indication of true battery consumption, and fast min/max. Current reading and ranging will be
at something like 20 microsecond intervals.

I already have all the analogue measurement stuff basically working - Watch this space - I'll post
more info as it gets nearer completion.



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2004\07\21@223421 by Ken Pergola

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For those that have not seen it, Keithley has a nice handbook entitled:
"Low Level Measurements: Precision DC, Current, Voltage and Resistance
Measurements". It's a keeper. Perhaps it might help out with regard to this
thread.

I have the paperback version of this. I'm not sure if Keithley still offers
this for free in paperback, but I believe you can get it in PDF format on
their CD they offer.

In any event, Keithley has a lot of interesting articles covering a wide
range of topics:

www.keithley.com/main.jsp?action=keithleysearch&searchType=document&k
eywords=article&clickPath=What%27s%20New%5EArticles&role=Public%20Document

Enjoy.

(Beware of URL wrap -- some re-assembly may be required)


Best regards,

Ken Pergola

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2004\07\22@031604 by Omer YALHI

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Hi Mike,

This sounds like a product I would like to invest in, if it works and does
the true averaging of current consumption.

>3 significant digits anywhere between 99.9uA to 999mA, with at least 1%
accuracy

99.9uA seems a little high though, most of my circuits are in the range
2-3uA when sleeping.

-Omer

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2004\07\22@040116 by Mike Harrison

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On Thu, 22 Jul 2004 10:15:35 +0300, you wrote:

>Hi Mike,
>
>This sounds like a product I would like to invest in, if it works and does
>the true averaging of current consumption.
>
>>3 significant digits anywhere between 99.9uA to 999mA, with at least 1%
>accuracy
>
>99.9uA seems a little high though, most of my circuits are in the range
>2-3uA when sleeping.
>

Yes, but you don't often need much accuracy at the bottom end - how often does it actually matter
whether you are drawing 2uA or 2.2uA? You will still be able to see changes of 0.1uA, which you need
when doing things like looking for floating pins, but absolute accuracy down this low is rarely
needed, and gets really tricky (i.e. more costly) to achieve. The spec at this end is still to be determined (other than the displayed resolution will be 0.1uA),
as I don't yet know what the noise performance will be like.








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2004\07\22@050453 by Omer YALHI

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>Yes, but you don't often need much accuracy at the bottom end - how often
does it actually matter
>whether you are drawing 2uA or 2.2uA? You will still be able to see changes
of 0.1uA, which you need
>when doing things like looking for floating pins, but absolute accuracy
down this low is rarely
>needed, and gets really tricky (i.e. more costly) to achieve.
>The spec at this end is still to be determined (other than the displayed
resolution will be 0.1uA),
>as I don't yet know what the noise performance will be like.

Oh, ok, now I understand.  I thought the scale started from 99.9uA, what you
were actually saying was the accuracy is 1 decimal to the right.  1 decimal
actually is more than enough, as you say, "what difference does it make
whether you are drawing 2uA or 2.2uA?", not much really.

What is the lowest current it will read, to the 1% accuracy?  Nano amps?

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2004\07\22@053435 by Mike Harrison

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On Thu, 22 Jul 2004 12:04:46 +0300, you wrote:

>>Yes, but you don't often need much accuracy at the bottom end - how often
>does it actually matter
>>whether you are drawing 2uA or 2.2uA? You will still be able to see changes
>of 0.1uA, which you need
>>when doing things like looking for floating pins, but absolute accuracy
>down this low is rarely
>>needed, and gets really tricky (i.e. more costly) to achieve.
>>The spec at this end is still to be determined (other than the displayed
>resolution will be 0.1uA),
>>as I don't yet know what the noise performance will be like.
>
>Oh, ok, now I understand.  I thought the scale started from 99.9uA, what you
>were actually saying was the accuracy is 1 decimal to the right.  1 decimal
>actually is more than enough, as you say, "what difference does it make
>whether you are drawing 2uA or 2.2uA?", not much really.
>
>What is the lowest current it will read, to the 1% accuracy?  Nano amps?

At present, I'm planning on the lowest range having a FSD of around 400uA, with 0.1uA resolution. Remember it's aimed at battery applications, so unless you're working with the extreme life-end of
the smallest coin-cell you don't need any more sensitivity.
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