how to reduce voltage burden and measure microamps?

Discussion in 'The Projects Forum' started by brumac57, May 13, 2008.

  1. brumac57

    Thread Starter Member

    Apr 10, 2008
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    0
    Hi
    I'm trying to measure the performance of a charge pump 3V -> 5V (and other circuits) in a battery powered microprocessor device.

    Every multimeter I have tried imposes too much voltage burden when in series with the battery.

    I'm trying to measure from around 0.1uA to about 5-10 mA (depending on the state of microprocessor - sleep states, etc)

    Google tells me that I need a feedback opamp design meter.

    I've spotted a Keithley 480 Picoammeter on Ebay, but it only goes up to 2mA

    Does anyone have a suggestion about a suitable meter that I won't need a mortgage for?

    Maybe I could build a suitable measuring circuit??
    Can anyone point me to one?
    How would I calibrate it?

    Thanks
    Bruce
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    How much is too much of a voltage burden? A 1ohm resistor drops 10mv at 10ma, .1ohms drops 1mv.

    I know there is something out there, it is tickling the back of my brain, but I can't pull it up right now. Home brewed is extremely doable, and cheap besides, this I am sure. As to calibration, that would depend somewhat on your resistors, but baring that, you could still used your existing current meter to transfer the calibration, it measures current OK, just has too much resistance.
     
  3. brumac57

    Thread Starter Member

    Apr 10, 2008
    18
    0
    the burden is around 1mV per uA for the better meter I have.

    It seems to drop the voltage from the batteries to a level below that at which the charge pump will start.

    That's during the start up phase when the pump pulls more. I've tried getting it started with a direct connection then swapping the meter in line but it won't work.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    Now I remember, it is simple, to the extreme. Assuming the power supplies are totally independent of the following circuit (think qty 2 9VDC batteries) then the voltage out of the op amp is R*I. In other words, a 1Kohm resistor will output 1 Volt for 1ma input with no apparent resistance.

    The following limitations for this circuit apply,

    1st: The op amp HAS to be able to source the current. If you need to measure 100ma, the op amp will be outputing 100ma (most can't).

    2nd: The op amp has to be able to keep up with changes in current. Op amps are slow, so this could be a problem. This is an active circuit, the 0 ohms virtual resistance depends on the op amp having the instantaneous voltage needed to match the current.

    [​IMG]
     
    Last edited: May 13, 2008
  5. brumac57

    Thread Starter Member

    Apr 10, 2008
    18
    0
    Thanks for that, but it's a bit abstract for me.

    I'll keep studying it...

    I was hoping for a pointer to a circuit diagram or an instrument?

    Cheers
     
  6. Caveman

    Active Member

    Apr 15, 2008
    471
    0
    Here is a method.

    Put a potentiometer or active load on the output to adjust the current draw of the 5V side.

    Then power the 3V side with a variable linear regulator like the LM317. Place a resistor in series with the Vin so you can measure the voltage across it to determine the current. The current through that resistor will divide between your circuit and R2, so you will need to measure the current through R2 as well and subtract it from the current in your series resistor at Vin.
     
  7. brumac57

    Thread Starter Member

    Apr 10, 2008
    18
    0
    Ok

    Thanks a lot, I'll give it a go!

    Cheers
     
  8. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,535
    The diagram I drew is a current to voltage converter. Where it says current is the leads just as you would use for the DVM to measure current. If you measure between in input you will find a virtual resistance of 0 ohms, because the op amp is making it so.

    The output of the op amp is a voltage equivalent of the current going through the op amp. It is instrumentation, of the simplest sort. Instead of a shunt resistor you have the op amp and it's feedback resistance.

    I suspect you're going to have trouble seeing .1uA through the regulators approx 20ma draw. .0200001 vs .0200000 isn't viewable through most DVMs, but you should be able to see 100uA OK.
     
  9. brumac57

    Thread Starter Member

    Apr 10, 2008
    18
    0
    OK thanks for your help
    I'll cogitate on all of this!

    Cheers
     
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