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Reactance measurment and current sources

Discussion in 'Electronic Design' started by john, Sep 22, 2007.

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  1. john

    john Guest

    Hi,

    I have build 32 constant current sources controlled by a DAC. Each
    source can provide upto +/- 1mA peak to peak with +/- 15volts
    compliance voltage. The sources had some DC current leakage problem
    which I solved using 6H inductors at the output of each source. The
    output becomes dependent on frequency and reactance. I could not go
    back and try to null the offset voltage thats why I had to use
    inductors. The load impedance is between 1 to 10kohm. I need to know
    that how much AC current is flowing through the load and how much
    current is flowing through the inductor practically at 10kHz. The
    inductor impedance at 10kHz is 370 kohm. What method should I use to
    measure the current? I looked into the oscilloscope current probes and
    they do not do that low current mesurement. I used the multimeter but
    it did not show me anything else except zero.. plus how can measure
    the reactance of the inductor at 10KHz?

    Please advice!

    John
     
  2. Guest

    hai u use shunt in which line u want measure the current
    across the shunt there will be voltage . the shunt rating example
    10:1 in this 10A current flow 1 volt is across it . by htis method r
    use
    hall device

    anand
     
  3. whit3rd

    whit3rd Guest

    Shunting excess current with an inductor WILL lower the 'leakage',
    but it also decalibrates your current sources. Either reprogram the
    DAC inputs so as to compensate the current offset, or find a way
    to make your load less sensitive to the 'problem'. I presume the
    DAC updates sample/hold amplifiers for each of the outputs?

    A multimeter takes tens to hundreds of milliseconds to complete a
    measurement,
    so it's unlikely to capture a current before the inductor conducts and
    the
    current disappears.
     
  4. Winfield

    Winfield Guest

    Thirty-two 6H inductors to ground? Arrgghh!!
    OK, trying to make do with a horrible configuration.

    "Ground" each of your inductors with a "virtual ground"
    made from the summing junction of an inverting opamp
    stage, instead of the real ground they're connected to
    now. A voltage will be developed across the opamp's
    feedback resistor, at the opamp output, that you can
    measure to determine the inductor current.
    Ahem! Have you accounted for the 6H inductor's winding
    capacitance? What if it's as high as 50pF? What then,
    grasshopper?

    This winding capacitance, BTW, may force you to add a
    little capacitance across the opamp's feedback resistor
    to stabilize the feedback loop.
     
  5. Robert Baer

    Robert Baer Guest

    ???? How in the he*** does one solve a "Current leakage" with an
    *inductor* ????
    Those six henry inductors must take up a *lot* of space!
     
  6. john

    john Guest

    Hi,

    Why do I need an opamp. Can I not just resistor in series with the
    inductor is enough?

    Regards
    John
     
  7. John, the opamp is to make a virtual ground, so the
    inductor will appear to be grounded, just as it is
    now, shorting out your current source at DC (I assume
    that's what you're doing now?). The resistor is in
    the feedback path of the opamp, carrying the current
    path, and the opamp output voltage lets you monitor
    the current without disturbing the inductor.

    You can learn about this stuff using Google. Or you
    could get a copy of our book, The Art of Electronics.

    :)

    Are you really using 6H inductors? Where did you get
    them? What's their internal resistance and ac loss?
    Did you know their ac resistance is much higher than
    their DC resistance? That's the whole poor Q thing.

    Also, did you know you can make simulated inductors
    from opamps? As for me, I like to make auto-zero
    servo loops when I have to kill a DC offset error.
     
  8. john

    john Guest

    Hi,
    Yes, I am using 6H inductors. I got them from soem comapny in
    california . I do not remember the name. The DC resistance is 2.5kohm.
    I do not know the AC loss. But the plan was to get high impedancw from
    them at 10kHZ.
    So, 2 x pi x f x L = 370Kohm, the intended load is between 1 to
    10kohm , so most of the current will go to the load not inductors.
    Now, I need to measure the current in both branches.

    Can you advice in detail that how to do it efficiently?

    Thanks
    John
     
  9. john

    john Guest

     
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