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OTA revisited

Discussion in 'Electronic Design' started by D. Graveson, Jan 18, 2006.

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  1. D. Graveson

    D. Graveson Guest

    Remember operationnal transconductance amplifiers?
    So my question: Is it suitable to use a one megohm resistor at the output
    of an OTA (before the darlington buffer) ?

    LM13700 documentation says "output voltage swing can be set at any level by
    selecting RL" but fails to mention any recommended upper limit.

  2. joe

    joe Guest

    Hi Danny

    I know another OTA very well, the CA3080, made by fairchild for
    I guess that national chip works quite similar:
    The bias current fed into the control input is mirrored to the
    unbuffered output.
    If you put 1uA into the control-pin, the maximum output current is
    to +/- 1uA. So the maximum voltage swing would be +/-1V when loaded
    with a resistor of 1M.

    By changing bias current and load resistance you can set the output
    according to to your needs, obvously there is no absolute limit of the

  3. D. G.

    D. G. Guest

    Tanks Joe!
    It answers my question. Not many people out there knowing OTAs.

    I would have another question or two about linearizing diodes which CA3080
    don't have.
    Are you familiar with this too?

  4. joe

    joe Guest

    hi Danny

    Yes, I know about these linearization technique which was part of RCA
    CA3280 dual OTA
    which I designed into some audio gain-control-circuitry in the early
    80s of the last century.
    At that time, there existed some fine application notes from RCA and it
    might be a good idea to search them, perhaps you are lucky and some
    people have scanned them in.
    Nowadays I use CA3080 without linearization to achieve a perfectly
    distorting guitar PreAmp. Anyway, you might put your question and I can
    try to answer them.

  5. joe wrote...
    Intersil has marked the CA3280 and CA3280A as "inactive" although
    they still have web pages up, with datasheets and app notes, see

    But they don't seem to have AN6818 up any longer. Sadly, I don't
    seem to have saved that one in time.

    Gerber has 22 CA3280 in stock, at $2.58 each. There's also lots
    of aftermarket supply.
  6. D. G.

    D. G. Guest

    I'm just concerned about the input current (Is) calculation to the amplifier
    input as well as the peak-to-peak voltage at the same input .
    Correct me if I'm wrong:

    With current (Id) injected into the diodes, datasheet says the amplifier now
    behaves as a current amplifier because the transfert function is:
    Iout = Is * 2 * Iabc / Id where Is is the input current into OTA

    Datasheet says Is should not exceed Id/2. I use Thevenin to find out Is and,
    once simplified, the final equation I use is:
    Is = Vin / Rin where Rin is the limiting resistor betwwen the OTA
    input and the voltage source.

    I pick Vin peak value to find Is peak value. I assume it is the peak value
    that is of concern. For example, if Vin=1Vp-p and Rin=10K, then Is peak =
    0.5V / 10K or 50uA.
    There is also a very small value resistor called Rp shunting the input to
    ground and it does not seem to do anything in that calculation. The value of
    Is seems to be related to Rin only. According to Thevenin...

    Even though they describe the OTA as an current amplifier when diodes are
    used, it seems to me that we still have to limit the input voltage. No
    (apparent) word about this in datasheet. I assume this because of a small
    graph showing harmonic distorsion with and without diodes, related to
    peak-to-peak input voltage.

    My final assumption is that Rin must meet two conditions:
    1) limit Is below Id/2
    2) limit peak-to_peak input voltage lower than 100mVpp (with the help of Rp
    to do a voltage division)

    Hope my explanation is coherent and correct...! :)

  7. joe

    joe Guest

    Hello Danny

    I think your maths is ok, but I will try to give you some brief
    Did you ever see the equivalent 'circuit diagram' of the OTA?
    It is not too complicated and shows the following basic stages:

    Both inputs are the base-electrodes of a differential
    npn transistor pair. The sum of their emitter currents
    Ie1+Ie2 is fed by a common npn current mirror set
    by Iabc. With a current amplication of 1A/1A,


    -> Emitter-, collector- and base currents of the
    differential amp are all biased by Iabc.

    The collector currents Ic1 and Ic2 are fed into two complementary
    current mirrors with a current amplication of 1A/1A.
    Their collectors are tied together and form the current driven output.

    All in all, the OTA works similar to an ideal differential amp, where
    the output current equals the diffence of the collector currents of
    the input stage, which is an exponential function of the differential
    voltage between base1 - base2 of the input stage, multiplied
    by the sum Ie1 + Ie2 = Iabc. Without doing extended maths,
    it evident that

    1) with Vbase1=Vbase2:
    Iout=Ic2-Ic1=0 -> perfectly balanced

    2) with Vbase1=Vbase2 +/- (>100mV)
    Ic1 = Iabc/0
    Ic2 = 0/Iabc
    Iout= +/-Iabc -> 100% overdriven

    As with any differential amp, rising the base1-base2 differential
    voltage beyond 100mV does not make any difference, as the whole
    Iabc is fed through one single input Transistor. Keep in mind that at
    that stage output CURRENT of the OTA is controlled in the linear
    region by an input VOLTAGE in the range of few 10 mVs.

    On the other side, if you use the linearizing diodes parallell to the
    base-inputs, you shunt the inputs by the low impedance of a diode,
    which decreases linearly as diode bias current increases.
    Now the OTA is converted to an input current controlled current source!

    Combined with the current limiting series resistors, it turns back to
    voltage driven current source again.

    Without linearization diodes, I recommend a differential voltage
    limitation done by to antiparallell 1n4148 or similar. This is a good
    practise that does not affect the transfer function anyway.

    Using linearization diodes, there is no need for an input voltage
    limitation provided that the input is driven by some opa or similar
    with a restricted voltage drive capability:
    Match the current limiting input resistor to the diode biasing current
    and everything should run fine. So I agree to your first condition,
    the second one being adjusted by me.

  8. D. G.

    D. G. Guest

    This gives me something to digest.
    But things are clearing up. The next step is to get my protoboard, grap my
    scope and my audio generator and... to experiment.

    What I find quite admirable about this kind of device is the mere simplicity
    of the circuitry. A differential pair and four current mirrors. That's about

    Thanks for shading some light!
  9. D. G. wrote...
    That's assuming perfection of the circuits. In the case of the
    CA3280, this may be close to the truth, but in the previous
    incarnation, the CA3080, the important current-biasing mirror,
    not being a Wilson-mirror type, had huge Early-effect errors.
    "That's about it," did not apply, unless one was very careful.
  10. D. G.

    D. G. Guest

    I should've said the topology of the OTA is quite simple and identical. All
    the models I know (I don't know about recent models... if any) used the same
    topology. Indeed, deep inside the chip, there must be a little bit more like
    internal biases, references and compensations tricks; that is caracteristic
    to integrated circuits.

  11. Ben Bradley

    Ben Bradley Guest

    Walt Jung described these in his "Audio Ic Op-Amp Applications" 2nd
    and 3rd editions (and probably 1st too). This book (all editions) has
    been out of print and used prices have been rather high ($50 and up)
    for several years.
    Jung described both the 3280 with its internal linearizing diodes,
    and a 3080 circuit using external diodes on the input to do the same
    thing the 3280 does internally.

    Also, a 'standard' way of connecting the output is to go directly
    to the negative input of an opamp, with the positive input grounded
    and a feedback resistor, scaled for gain, from the output to the
    negative input. I forget if the output had any extra distortion caused
    by varying voltage as opposed to going into a zero-impedance summing
    input, but the op-amp does gives direct conversion of current output
    to a low-impedance voltage output, which is what's wanted in most
    There's a practical upper limit, with the limiting factor probably
    being stray capacitance and the current output's ability to drive it.
    Higher resitance will result in a lower max frequency response.
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