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Very low DC current measurement

Discussion in 'Electronic Design' started by habib bouaziz-viallet, Mar 23, 2005.

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  1. Hi All,

    In order to estimate power consumption and Lithium battery life cycle. I
    need to measure supply DC current drained by an MSP430 micro which is as
    low as 100µA (TI specs)

    Any methods ?

    Many thanks, Habib
    betula.fr
     
  2. Paul Burke

    Paul Burke Guest

    Just stick a 1k resistor in series with the battery and measure volts
    across it. The 0.1V or so dropped shouldn't make much difference to the
    consumption of the MSP430.

    If it has brief phases of high power consumption, measure those
    separately with a different resistor.

    Paul Burke
     
  3. PeteS

    PeteS Guest

    There are a number of ways.

    One of the ways I have done this (over 6 decades of current,
    incidentally) is to put the load current path as the programming side
    of a current mirror.

    The other side of the mirror will have the same current (within the
    tolerance of the device - see the ADI MAT-0x series for excellent
    matched mirrors, or perhaps choose a Wilson mirror - you should really
    only need this if you have large variations of programming current or
    supply voltage- a Wilson mirror is designed to negate Early effect, a
    variation of bipolar transistor gain proportional to Vce. Use one if
    the effective Vce of both sides of the mirror will differ
    significantly).

    This output current may then be amplified, converted etc., to your
    heart's content for the measurement system.

    Cheers
    PeteS
     
  4. Le Wed, 23 Mar 2005 03:31:19 -0800, PeteS a écrit :
    Hi Pete,

    Considering your idea, i drew a little schematic to clarify it.
    http://cjoint.com/data/dxnAJRjWNb.htm
    Is that conform about what you think ?

    In that schematic the two Vce's differ, what is Wilson I-mirror ? so far,
    i have never hear about.

    Many thanks, Habib
    betula.fr
     
  5. John Larkin

    John Larkin Guest

    Why not just use a DVM? The better handheld Flukes resolve microamps.

    John
     
  6. Luhan Monat

    Luhan Monat Guest

    A 1k resistor in parallel with a shotkey diode makes a useful bench
    tool. Put in series with the power feed for the device you are testing,
    it allows for both 'full power' operation and for measuring microamps
    for low power measurements.
     
  7. That may depend on the allowable burden voltage, meaning the
    voltage loss.

    Rene
     
  8. Guest

    Transimpedance amp?
     
  9. John Larkin

    John Larkin Guest

    Analog microammeter?

    John
     
  10. Ban

    Ban Guest


    This circuit has *no* voltage loss and can be placed in either leg, as long
    as the compliance is met.
    ___
    +-|___|------+
    | R |
    I | |\ |
    o----->----+-|+\ | ___ ___
    | >--+----)-+-|___|-+-|___|-+
    | +-|-/ | | | R1 | R1 |
    | | |/ .-. | | | ===
    | | | | | | | GND
    | | | |R | | |
    | | '-' | | | |\
    Ue=0 | | | | +-|+\ Uout
    | +-------+ | | | >--+----o
    | | | | | +-|-/ |
    | | .-. | | | |/ | Uout=2*R*I
    | | | | | | | |
    | | | |R | | | |
    | | |\ '-' | | | |
    | +-|-\ | | | ___ | ___ |
    V | >--+----+-)-|___|-+-|___|-+
    o-----<----+-|+/ | R1 R1
    I | |/ |
    | ___ |
    +-|___|--------+
    R
    (created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)
     
  11. Rob Gaddi

    Rob Gaddi Guest

    But, unless there's something I'm not seeing, does require a seperate,
    higher voltage source than what is being monitored. As a bare minimum,
    if you're measuring the current from Vcc, the power to the opamps will
    have to be Vcc + Vopamp_clearance + Imax * R.
     
  12. Ban

    Ban Guest

    That is exactly what *compliance* means. If you put the circuit into the
    ground leg, there is almost no voltage across it and you could also power
    the circuit from the same supply as the DUT. Nice is the bipolar capability
    for small ac currents.
     
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