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

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

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  1. stevo

    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
     
  2. 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. MikeM

    MikeM Guest

    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
     
  4. John Fields

    John Fields Guest

     
  5. 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 Gt

    Roger Gt Guest

    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
    across the ground lead to read the current. This is generally
    less intrusive and changes the operating conditions the least.
     
  7. John Fields

    John Fields Guest

    ---
    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. 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 Borowski

    Al Borowski Guest

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

    Al
     
  10. 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
     
  11. John Fields

    John Fields Guest

     
  12. Fred Bloggs

    Fred Bloggs Guest

    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
    |
    +-------------------------------->(+)
     
  13. Rob

    Rob Guest

    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
     
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