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

Discussion in 'Electronic Design' started by stevo, Apr 5, 2004.

1. ### stevoGuest

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

2. ### Walter HarleyGuest

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.

3. ### MikeMGuest

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

5. ### Spehro PefhanyGuest

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

6. ### Roger GtGuest

X-No-Archive: yes
: 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
less intrusive and changes the operating conditions the least.

7. ### John FieldsGuest

---
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.

8. ### Spehro PefhanyGuest

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

9. ### Al BorowskiGuest

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

Al

10. ### Spehro PefhanyGuest

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

12. ### Fred BloggsGuest

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
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
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
|
+-------------------------------->(+)

13. ### RobGuest

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