How do I measure currents of ~500uA (need to measure power)

S

stevo

Guest
I need to measure the total power consumption of one of our circuit
boards. The board's current consumption varies over time (it is
battery powered and solar panel charged). I would like to set up an
experiment where the current and voltage are recorded over time.
Basically, I want to create a Power or power consumption monitor.

The problem is measuring the current. My multimeter (Fluke 177) does
a good job of this, but no data logging. My oscilloscope does not
have a current probe (and I doubt you can buy one that works in the
500uA to 1mA range). Can't use the data acquisition card either
(voltage only).

So, how do people accurately measure currents of around 1mA without
using a multimeter? If you use a small resistor and then measure the
voltage drop, this increases the power consumed (introduces some
error). Is there anyway to avoid this?

Thanks
Stephen
 
"stevo" <stephendeese@hotmail.com> wrote in message
news:25cad608.0404051143.5c6a7c80@posting.google.com...
So, how do people accurately measure currents of around 1mA without
using a multimeter? If you use a small resistor and then measure the
voltage drop, this increases the power consumed (introduces some
error). Is there anyway to avoid this?
Click to expand...
If you use a sufficiently small resistor, the amount of error introduced is
sufficiently trivial. 1 ohm at 0.5mA drops 0.5mV, and dissipates 0.25uW;
what proportion of your supply voltage/power is that? If it's too big, go
to 0.1 ohms.

This is precisely what the Fluke does. There's no magic there. If you read
its specs you can figure out what size resistor it's using. If you want to
do the same thing, but without using the Fluke, just use your own precision
resistor. If your voltage-only data acquisition card isn't sufficiently
sensitive, build a precision DC amplifier with an opamp.

In general it is not possible to make measurements without influencing the
quantity being measured.
 
stevo wrote:

I need to measure the total power consumption of one of our circuit
boards. The board's current consumption varies over time (it is
battery powered and solar panel charged). I would like to set up an
experiment where the current and voltage are recorded over time.
Basically, I want to create a Power or power consumption monitor.

The problem is measuring the current. My multimeter (Fluke 177) does
a good job of this, but no data logging. My oscilloscope does not
have a current probe (and I doubt you can buy one that works in the
500uA to 1mA range). Can't use the data acquisition card either
(voltage only).

So, how do people accurately measure currents of around 1mA without
using a multimeter? If you use a small resistor and then measure the
voltage drop, this increases the power consumed (introduces some
error). Is there anyway to avoid this?

Thanks
Stephen
Click to expand...
Assuming that you are willing to break a wire, and then insert a
low-value shunt resistance, how low do you want the resistance to be?
(i.e. how low is "good enough")?

(Hint, the Fluke also inserts a "shunt" in series with your measurement)

When you insert the shunt, which lead do you break? (High side or ground
side)?

MikeM
 
On Mon, 05 Apr 2004 16:24:55 -0500, the renowned John Fields
<jfields@austininstruments.com> wrote:

On 5 Apr 2004 12:43:01 -0700, stephendeese@hotmail.com (stevo) wrote:

I need to measure the total power consumption of one of our circuit
boards. The board's current consumption varies over time (it is
battery powered and solar panel charged). I would like to set up an
experiment where the current and voltage are recorded over time.
Basically, I want to create a Power or power consumption monitor.

The problem is measuring the current. My multimeter (Fluke 177) does
a good job of this, but no data logging. My oscilloscope does not
have a current probe (and I doubt you can buy one that works in the
500uA to 1mA range). Can't use the data acquisition card either
(voltage only).

So, how do people accurately measure currents of around 1mA without
using a multimeter? If you use a small resistor and then measure the
voltage drop, this increases the power consumed (introduces some
error). Is there anyway to avoid this?
---
If you use a small resistor it _decreases_ the power consumed and does
create an error, but if the error created is small enough, then it
won't matter much. It depends on how much error you can tolerate, as
others have pointed out.
Click to expand...
Hmm.. if the device has a switching power supply delivering a constant
voltage to a contant load, inserting some resistance in the source
will likely increase the total power consumption. If it's a linear
regulator then it should not affect the total power consumption much
(approximately constant current draw from a voltage source). If it's a
resistive load, then the total power consumption will go down.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
 
X-No-Archive: yes
"Spehro Pefhany" wrote
: John Fields wrote
: > (stevo) wrote
: >
: >>I need to measure the total power consumption of one of our
circuit
: >>boards. The board's current consumption varies over time (it
is
: >>battery powered and solar panel charged). I would like to set
up an
: >>experiment where the current and voltage are recorded over
time.
: >>Basically, I want to create a Power or power consumption
monitor.
: >>
: >>The problem is measuring the current. My multimeter (Fluke
177) does
: >>a good job of this, but no data logging. My oscilloscope does
not
: >>have a current probe (and I doubt you can buy one that works
in the
: >>500uA to 1mA range). Can't use the data acquisition card
either
: >>(voltage only).
: >>
: >>So, how do people accurately measure currents of around 1mA
without
: >>using a multimeter? If you use a small resistor and then
measure the
: >>voltage drop, this increases the power consumed (introduces
some
: >>error). Is there anyway to avoid this?
: >---
: >If you use a small resistor it _decreases_ the power consumed
and does
: >create an error, but if the error created is small enough, then
it
: >won't matter much. It depends on how much error you can
tolerate, as
: >others have pointed out.
:
: Hmm.. if the device has a switching power supply delivering a
constant
: voltage to a constant load, inserting some resistance in the
source
: will likely increase the total power consumption. If it's a
linear
: regulator then it should not affect the total power consumption
much
: (approximately constant current draw from a voltage source). If
it's a
: resistive load, then the total power consumption will go down.
:
: Best regards,
: Spehro Pefhany
: --

If the unit is connected with a wire from the power source, you
could measure the resistance of the line and measure the drop
across the ground lead to read the current. This is generally
less intrusive and changes the operating conditions the least.
 
On Mon, 05 Apr 2004 22:42:19 GMT, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

On Mon, 05 Apr 2004 16:24:55 -0500, the renowned John Fields
jfields@austininstruments.com> wrote:

On 5 Apr 2004 12:43:01 -0700, stephendeese@hotmail.com (stevo) wrote:

I need to measure the total power consumption of one of our circuit
boards. The board's current consumption varies over time (it is
battery powered and solar panel charged). I would like to set up an
experiment where the current and voltage are recorded over time.
Basically, I want to create a Power or power consumption monitor.

The problem is measuring the current. My multimeter (Fluke 177) does
a good job of this, but no data logging. My oscilloscope does not
have a current probe (and I doubt you can buy one that works in the
500uA to 1mA range). Can't use the data acquisition card either
(voltage only).

So, how do people accurately measure currents of around 1mA without
using a multimeter? If you use a small resistor and then measure the
voltage drop, this increases the power consumed (introduces some
error). Is there anyway to avoid this?
---
If you use a small resistor it _decreases_ the power consumed and does
create an error, but if the error created is small enough, then it
won't matter much. It depends on how much error you can tolerate, as
others have pointed out.

Hmm.. if the device has a switching power supply delivering a constant
voltage to a contant load, inserting some resistance in the source
will likely increase the total power consumption. If it's a linear
regulator then it should not affect the total power consumption much
(approximately constant current draw from a voltage source). If it's a
resistive load, then the total power consumption will go down.
Click to expand...
---
From his description of the circuit, Im ass-uming that he's driving
the load off of batteries and that he's talking about inserting a
resistor in series with the load and measuring the voltage drop across
that resistor in order to determine the current through it and the
[resistive] circuit it's in series with. Such being the case, with a
constant voltage supply across the sum of the load resistance and the
shunt, (strange name for a series resistor, no?) and since I = E/R, if
R goes up I _has_ to go down.

--
John Fields
 
On Mon, 05 Apr 2004 19:21:25 -0500, the renowned John Fields
<jfields@austininstruments.com> wrote:

From his description of the circuit, Im ass-uming that he's driving
the load off of batteries and that he's talking about inserting a
resistor in series with the load and measuring the voltage drop across
that resistor in order to determine the current through it and the
[resistive] circuit it's in series with. Such being the case, with a
constant voltage supply across the sum of the load resistance and the
shunt, (strange name for a series resistor, no?) and since I = E/R, if
R goes up I _has_ to go down.
Click to expand...
He says "circuit", nothing about it being resistive that I see.
Consider this (unlikely?) case:

1K
___
+-|___|-+
| | .-------------. +--+
+--+--o _/o+---| | 5V | | | 500uA
| / |85% efficient| ------ .-. v
9V--- Vin| SMPS |Vout | |
- | | | | 10K
| | | ------ '-'
+--------------| | 0V | |
'-------------' +--+


If the switch is closed, the output power is 2.5mW, and the input
power is 2.94mW, so the current from the battery is 0.327mA

If the switch is open, the output power remains 2.4mW, the input power
to the SMPS remains 2.94mW, but now there is a voltage drop across the
1K of 1V/mA, so the input voltage is 9 - i, and the current is
2.94mW/Vin. Solve for i (quadratic) and we get 0.339mA. Total power is
now 3.05mW. So current and total power both go up.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
 
So, how do people accurately measure currents of around 1mA without
using a multimeter? If you use a small resistor and then measure the
voltage drop, this increases the power consumed (introduces some
error). Is there anyway to avoid this?
Click to expand...
Could you use an active current to voltage convertor? (an Op-amp and a
few passives).

Al
 
On Tue, 06 Apr 2004 10:53:29 +1000, the renowned Al Borowski
<aj.borowski@erasethis.student.qut.edu.au> wrote:

So, how do people accurately measure currents of around 1mA without
using a multimeter? If you use a small resistor and then measure the
voltage drop, this increases the power consumed (introduces some
error). Is there anyway to avoid this?

Could you use an active current to voltage convertor? (an Op-amp and a
few passives).

Al
Click to expand...
You mean something like this? The inverting input is a virtual ground
(within a few millivolts of ground) provided the op-amp can supply the
current and provided the output has settled.



---------------+
| .----+------.
| | |
| | | Rx
| | | ___
| | Circuit | + -|___||
| | | | |
--- | | | +V |
- '-------+---' | |\| |
| +------+ -|-\ | Vo = - I * Rx
| | >-+-----o
+----------------------------|+/
|/|

-V


The problem I usually have in this general situation is not the small
drop (millivolts) from a sense resistor but the fact that few of the
things I design draw nice steady current, and usually you want to know
the average and/or the RMS current. Sometimes a passive RC filter
across the shunt can give a useful average reading in cases where the
current is drawn in short pulses or whatever.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
 
"stevo" <stephendeese@hotmail.com> wrote in message
news:25cad608.0404051143.5c6a7c80@posting.google.com...
I need to measure the total power consumption of one of our circuit
boards. The board's current consumption varies over time (it is
battery powered and solar panel charged). I would like to set up an
experiment where the current and voltage are recorded over time.
Basically, I want to create a Power or power consumption monitor.

The problem is measuring the current. My multimeter (Fluke 177) does
a good job of this, but no data logging. My oscilloscope does not
have a current probe (and I doubt you can buy one that works in the
500uA to 1mA range). Can't use the data acquisition card either
(voltage only).

So, how do people accurately measure currents of around 1mA without
using a multimeter? If you use a small resistor and then measure the
voltage drop, this increases the power consumed (introduces some
error). Is there anyway to avoid this?

Thanks
Stephen
Click to expand...
Can you beg, borrow or steal a hall effect current clamp. 10 loops of the
lead wire thru' the clamp will give you 5 -> 10 mA indicated current. What
about buying one of those Lemo (?) current transducers and trying the above.

rob
 
On Tue, 06 Apr 2004 00:46:32 GMT, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

On Mon, 05 Apr 2004 19:21:25 -0500, the renowned John Fields
jfields@austininstruments.com> wrote:

From his description of the circuit, Im ass-uming that he's driving
the load off of batteries and that he's talking about inserting a
resistor in series with the load and measuring the voltage drop across
that resistor in order to determine the current through it and the
[resistive] circuit it's in series with. Such being the case, with a
constant voltage supply across the sum of the load resistance and the
shunt, (strange name for a series resistor, no?) and since I = E/R, if
R goes up I _has_ to go down.

He says "circuit", nothing about it being resistive that I see.
Consider this (unlikely?) case:

1K
___
+-|___|-+
| | .-------------. +--+
+--+--o _/o+---| | 5V | | | 500uA
| / |85% efficient| ------ .-. v
9V--- Vin| SMPS |Vout | |
- | | | | 10K
| | | ------ '-'
+--------------| | 0V | |
'-------------' +--+


If the switch is closed, the output power is 2.5mW, and the input
power is 2.94mW, so the current from the battery is 0.327mA

If the switch is open, the output power remains 2.4mW, the input power
to the SMPS remains 2.94mW, but now there is a voltage drop across the
1K of 1V/mA, so the input voltage is 9 - i, and the current is
2.94mW/Vin. Solve for i (quadratic) and we get 0.339mA. Total power is
now 3.05mW. So current and total power both go up.
Click to expand...
---
Yes, I thought that was what you had in mind from your previous post,
and I also thought it an unlikely configuration. That's why, In lieu
of an update from the OP, I ass-umed a linear, resistive load on the
supply.

--
John Fields
 
I need to measure the total power consumption of one of our circuit
boards. The board's current consumption varies over time (it is
battery powered and solar panel charged). I would like to set up an
experiment where the current and voltage are recorded over time.
Basically, I want to create a Power or power consumption monitor.

The problem is measuring the current. My multimeter (Fluke 177) does
a good job of this, but no data logging. My oscilloscope does not
have a current probe (and I doubt you can buy one that works in the
500uA to 1mA range). Can't use the data acquisition card either
(voltage only).

So, how do people accurately measure currents of around 1mA without
using a multimeter? If you use a small resistor and then measure the
voltage drop, this increases the power consumed (introduces some
error). Is there anyway to avoid this?
Click to expand...
One way to measure a minuscule loading would be to use an auxiliary
source to power the load at battery/array voltage, and then mirror the
circuit-under-test loading to the battery, like so- this can be made as
accurate as required- OA1 buffers VBATT onto the load using AUX PWR, OA2
senses current supplied to the C.U.T. and level shifts a voltage onto
the voltage-to-current converter, OA3, loading the BATT/ARRAY by an
identical amount- the setup requires a single line break between BATT(+)
and C.U.T. (+)- the OAs run off AUX PWR- current mirror gain shown as
1, but the Rs's and Rc can all be scaled as required to suit the
specifics of data logger FS range, AUX PWR headroom and so forth:

Please view in a fixed-width font such as Courier.



(+)-------------------+
AUX PWR |
|
(-)--+ |
| |
--- |
com OA1 |
|\| I
+---|>|---+--------+-------------|+\ Rs --->
| | | | >--+---/\/\----+-------+ VBATT
| | | +-|-/ | | |
| | | | |/| | | |
| | | | | | | |
| | / +-------------------+ |
| | Rs | | | |
| | / | / | |
| | \ | R | |
| | | | / | |
| | | | \ | |
| | | | | | |
| | | | +-------+ | |
| | | | | | | |
| | +-----------+ | / | | +--------+
| | | | | R | | | |
| | d /| | | / | | | C.U.T. |
| |+ +-|| /+|--+ | \ /| | | | |
+------+ +---+ ||--< | | | /-|-+ | +--------+
| | | | B | +-|| \-|----------+--< | | |
|SOLAR | | A | s \| | \+|-----+ |
|ARRAY | | T | | | \| |
| | | | T | +---+ OA3 | OA2 |
+------+ +---+ | | | ---
| |- | / | com
| | | Rc |
| | | / |
| | | \ |
| | | | |
+---------+--------+---------------+------------>(-)
| | -
--- | Vm=I x Rc TO DATA
com | + LOGGER
|
+-------------------------------->(+)
 
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