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current follower

Discussion in 'Electronic Design' started by john, Jun 8, 2007.

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  1. john

    john Guest


    Can anybody give some designing hints to make a current follower
    circuit using opamps or transistors. I do not know whether it is
    possible or not... I do know the voltage follower using opamps...

  2. A current follower, as I understand it, might better be called
    a current booster. You can make one from two resistors and a
    power opamp. Given there's already a current source, you just
    add a series resistor (which won't affect the load since it's
    being driven from a current source, after all) and an opamp
    follower with another resistor to the load. The ratio of the
    resistors determines the output current, Io = Ii (1 + R1/R2).
    There'll be an output current error from the opamp's offset,

    Like the Howland and other types of opamp current-source
    circuits, it becomes a voltage source at high frequencies, as
    if it was a current source with lots of parallel capacitance.
  3. john

    john Guest


    I designed the constant current source and it has some leakage current
    so I put an inductor acorss it. I got rid of the dc but its impedance
    changes with frequency and sometimes the impedance goes below the load
    impedance which is not desirable for my application. So, I thought
    that if I can make and add a current follower in series with the
    inductor. So for DC current the inductor will be shorted and the unity
    gain current follower will dump the charge into ground. and for AC
    whatever the inductor offers the impedance it will be in series with
    the very high impedance of the opamp . So, the total impedance will be
    higher than my load impedance.
  4. It depends a lot on the specs... maximum input resistance, output
    compliance, unipolar or bipolar, common ground or not, available power
    supply, etc.

    One simple bipolar method would be to put a load resistor and voltage
    follower ahead of a Howland current source (not that I'm a particular
    fan of that configuration).

    Best regards,
    Spehro Pefhany
  5. Joerg

    Joerg Guest

    Or it becomes an oscillator as one guy on the German NG just found out.
    So I pointed him to AoE because you guys were honest and called it a
    "Textbook current source ..." and at the end "... is not widely used".

    Anything that relies on tightly toleranced components gives me the goose
    pimples and it has the potential to give a CFO a heart attack.
  6. It might have exactly the same output current as input, but have more
    voltage compliance, isolation or other useful characteristics. Such
    things are common in process control.
    Yes, that seems to be important in this case. And it will only work up
    to a certain load resistance when you run out of voltage headroom.

    Best regards,
    Spehro Pefhany
  7. Rich Grise

    Rich Grise Guest

    I have actually seen it (Howland) used in a commercial product, but I
    don't like it.[/QUOTE]

    Guys, help me out here. I searched for almost 5 minutes on variations of
    "howland current source", and the closest I've gotten to a schematic so
    far looks like an LM317 in battery charger mode. ??:-/

    Anybody got a schematic of this Howland circuit, and maybe, a preferred
    circuit alongside it?

    Usually, when I need a current source, I actually make a current sink,
    with some constant Vb and some emitter resistor on an NPN. Admittedly,
    not too tight on the tolerance, but it got the job done. :)

  8. I have actually seen it (Howland) used in a commercial product, but I
    don't like it.

    Best regards,
    Spehro Pefhany
  9. If you go to Google books, and search with the term,
    "howland current source" you'll get the discussion in
    The Art of Electronics, right at the top of the list.

    Below that you'll see where it's discussed in a dozen
    other books plus some journals, etc. Happy reading!
  10. D from BC

    D from BC Guest

    Foggy memory.. :p
    Isn't the Howland with 2 transistors.. Ic1 = Ic2
    Something like that..
    The bases are tied together...
    D from BC
  11. Frank Miles

    Frank Miles Guest

    Do you need unipolar or bipolar?
    How accurate? Tempco?
    Input/output impedances?

  12. Jasen

    Jasen Guest

    Common base ampliifier?
    Current mirror ?

  13. Is this related to your other thread in sed,
    entitled "Inductive Circuits"? There you
    defined your cc-source as 0-600uA, 10 to 32KHz,
    (is it 10Hz or 10KHz?), and Rl = 20k to 400k.

    Several people remarked that 600uA into 400k
    requires a 240V compliance, which I'm sure you
    do not intend, but did not see a clarification
    from you.

    If the voltage compliance is within 12VPk then
    the LM13700 might be interesting. For higher
    voltages it will be an opamp plus transistors,
    several examples of which have been discussed
    here over the years.
  14. Winfield

    Winfield Guest

    With Tony's ASCII circuits shining, we might add.
  15. Fred Bartoli

    Fred Bartoli Guest

    Tony Williams a écrit :
    Also 400k at 32kHz is just 12pF, which isn't much.
    Surely not achievable with a Holland circuit and will require naturally
    constant current with low parasitic feedback output circuits, like a
    cascoded current source.

    The ticket is probably something along those lines:
    (hope it will not wrap)

    <-- 1mA
    1K ___
    /| | _ |
    /+|--GND /| >| / \ |
    .--+-< | ___ /+|--- GND |--(- +)--+
    | | \-|-+-|___|-+< | /| \_/ |
    | .-. \| | | \-|-. | |
    | | | | | ||| | | _ |
    | | | --- '-||--+ >| / \ |
    | '-' --- || | 47p |--(- +)--'
    | | | 1u .-. || /| \_/
    | '------' | | .--||----+ 5V
    | | | | || |
    | '-' /| | ___ | ___ ||
    | | /+|-+-|___|--+-|___|-||----> high Z
    | A <-+-< | 100K | || AC out
    | | \-|-. | 2uF
    | 10u 20K | \| | |/ film or ceramic
    | || ___ '------' .---|
    | IN >------||-|___|--. | |> || ___
    | || | | +--||-|___|-< from A
    | (5V/mA) ___ | |\ | | || 47K
    '---------|___|--------+-----|+\ | |/
    | | >-|---| C adj (pF range)
    10K | .--|-/ | |>
    | | |/ | |
    | '-------|-----+
    | | |
    .-. | .-.
    | | /+\ | |
    10K| | 5V( ) | |1K
    '-' \-/ '-'
    | | |
    '----------+-----+-< -Supply

    The cascode might not be required for the impedance you target, but it
    is so easily added that you don't want to delete it.

    The low frequency DC servo loop might be interesting to adjust over all
    the load impedance ranges if the low freq corner is 10Hz, and will lead
    to very low transient response.
    Alternatively, it might be interesting to make the first integrator a
    S/H circuit with somewhat higher GBW product, make the following opamp a
    simple inverter, and servo the output voltage only during the rest
    periods (if the working duration can be made not too long).

    The C adj capacitor is adjusted to compensate for (almost) all the
    output parasitics and will start rolling off at 340kHz (assuming a 10pF
  16. john

    john Guest


    I saw the following link,M1

    and found the howland current source. I am still confused about all
    the disscussions. My question is that if I use the current source
    ( that I built and cannot go back and change the design) in parallel
    with an inductor who is in series with the howland current source can
    solve the problem of saving the load from the DC leakage current and
    let the ac signal appear acorss the load completely. Please advice!

  17. Fred Bartoli

    Fred Bartoli Guest

    john a écrit : Your link doesn't work
    I can see this.
    I don't undestand what you mean. Once you place the inductor in // then
    you place it in series.
    I can't make sense of what you say. Maybe you could sketch something and
    post it somewhere.
    Maybe too you can tell us what you're trying to achieve with this
    current source you want.
    Maybe you can clear the questions you were asked instead of letting us
    guess and shoot in the dark.

    BTW, given the frequency bands and the impedance values you gave us,
    counting on inductors to do the job is totally hopeless.
  18. Tom Bruhns

    Tom Bruhns Guest

    This sounds like the same problem I posted a reply to recently. If
    you want to get rid of the DC into the load, just put a capacitor in
    series with the current source. Make it large enough to have a
    reactance low enough to not have unacceptable voltage drop at the
    lowest frequency of interest; but if driven by a true current source
    with infinite compliance, it will even pass DC. I think for your
    application, something in the area of 10uF should do well; I suggest
    polypropylene dielectric, though polyester/Mylar would probably do
    fine too. So anyway, you need some place for the DC to go: just add
    a resistor from the current source output to ground, or to a voltage
    such that the current source output DC voltage is what you want. You
    had earlier mentioned a low DC current, something in region of 100
    nanoamps as I recall. So you can use a very large resistance and have
    a small DC drop across the resistance. That's MUCH better than trying
    to get an inductor to do the job. Go look up your earlier thread, and
    look at my posting there. There are some additional caveats to

  19. Winfield

    Winfield Guest

    That's a good suggestion, Tom, I hope John sees it.
  20. john

    john Guest


    The capacitor will discharge through the load which is not desirable
    in my application. I put an inductor in parallel with the current
    source and it did take care of the DC leakage current but for low
    frequency AC signals, inducroe does not offer high impedance. So, I
    was hoping to find a circuitry that can short the DC and offer high
    impedance when current is AC.

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