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Inductive Circuits

Discussion in 'Electronic Design' started by john, Jun 7, 2007.

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  1. john

    john Guest


    I designed a constant current source that can produce current in the
    range of 0 to 600uA with frequency range of 10 to 32kHz. The current
    source has a leakage current of about 240nA which I do not want to go
    to the load. My load impedance is between 20Kohm to 400kohm. Inorder,
    to protect the load from the leakage current I placed an inductor in
    parallel with the load. I did not know the exact inductor value, found
    some old inductor in the cabnet. The inductor did get rid of the
    leakage and let the ac waveforms appear across the load. I started
    calculating the inductor value that I needed using formula X = 2pi FL.

    I choose X = 10 Mohm for which L = 50 H at 32 Khz but X drops to 159
    Kohm when frequecy drops to 10 Hz. which is less than the load
    impedance so the current will choose the least resistive path. can
    anyone adivce me that how to compensate this drop of impedance ( X )?

    I can not trimmed the offset of the constant current source using
    potentiometer due to mechanical problems.

  2. If it's a constant DC 240 nA, maybe just a DC constant current source
    of that magnitude in parallel with the load.
  3. Why don't you put a capacitor in series with the load, instead? A few
    microfarads should do it, use an unpolarised type (i.e. not an
    50H is very large you know! You would be better off designing a
    current source without the leakage.
  4. Guest

    Erm, that's 240V .... Isn't the load impedance for a current source
    usually 0 ohms???
  5. Although 600uA at 20K is 12V. Perhaps the limits are not independent

    No, I don't think so. It is supposed to stay constant over some range
    of load resistances (the "compliance").
  6. That is like saying that the normal load impedance for a
    voltage source is infinity ohms.
  7. Guest

    Hm, I was thinking transducers...
  8. [snip the inductor solution. Probably not viable.
    It sounds like you need a low frequency, (<10Hz),
    dc voltage offset servo loop. An inverting opamp
    integrator with 10meg resistor looking at the o/p,
    balancing the cc source towards zero voltage offset.
    Use a low offset fet input opamp. Reducing to <1mV
    dc offset voltage would be equivalent to <50nA into
    the 20kohm load.
  9. john

    john Guest


    What I understood is that I have to add another opamp at the output of
    my current source or build a new one...

  10. john

    john Guest


    I am also thinking to add a switch in series with the inductor. When
    DC current flows into the inductor, the switch should get closed and
    when AC apperas across the inductor then switch should offer very high
    impedance ( like in mega ohms) so the reactance of the inductor will
    not be a issue anymore. Can anyone suggest what kind of circuitry I
    need to do it!

  11. Tom Bruhns

    Tom Bruhns Guest

    600uA into 400kohms is 240 volts. RMS or peak or what? In any event,
    it's quite a lot. Does your current source really have that much

    Keeping DC out of the load it almost trivial: just add a blocking
    capacitor. Its reactance at the lowest frequency of interest should
    be small compared with the load, so that you don't have too much drop
    across it, though since it's a constant current source, it doesn't
    really matter that it be especially low reactance... except that you
    also need a place for the leakage current to go. It's seriously
    impractical to use an inductor that will be megohms reactance at low
    frequency and not show some potentially bad resonances and the effects
    of shunt parasitic capacitance at higher frequency. Instead, just use
    a resistor. The DC drop in 22 megohms for 240nA current is only a
    little over 5 volts, and any current source that can deliver 240 volts
    should have no trouble with that. You can always return the resistor
    to a voltage that keeps the output of the current source centered near
    0, if you want. I'm assuming that the error caused by 22 megohms in
    parallel with your load is not significant in your application, but
    perhaps that's not correct; in that case, you can use a more
    complicated boot-strapped circuit to accomplish the task. If you use
    such a resistor (say 22Mohms) to absorb the DC current, be aware that
    stray capacitance from the output node of the current source to ground
    will represent a significant load: even just 10pF is only about
    500kohms at 32kHz. People who try to keep high impedances at high
    frequencies tend to use guards that are driven to the same voltage as
    the node they guard, so that the effective capacitance from the
    guarded node to ground or common is very much less than the
    capacitance from the guarded node to the guard. Another way of saying
    that is that the driven guard keeps capacitive currents from the
    guarded node to very low levels, compared with what they would be if
    the node were not guarded.

    Sounds to me like you're trying to learn to swim in water that's
    pretty deep. I wish you luck. For example, 50 H at 32kHz is
    (theoretically, but difficult to achieve in practice) about 10Mohms as
    you say, but at 10Hz, 50H isn't anything close to 150kohms.

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