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Help for negative resistance oscillator

Discussion in 'Electronic Design' started by Robert Baer, Apr 28, 2004.

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  1. Robert Baer

    Robert Baer Guest

    What i have: an inductor of unknown value, but most likely in the
    henry region, with resistance from 4K to 10K.
    It is a grounded inductor, and has so many turns that it cannot stand
    current thru the coil - as it would saturate.
    I would like to make an oscillator that uses the inductor as one of
    the frequency determining components, snd it seems a negative resistance
    oscillator would do the trick.
    However, i have no reference material that gives circuitry for
    solid-state devices that could do this - not even op amps.
    Suggestions and references, please.
  2. I think you may have been mislead about this "negative resistance"
    oscillator concept. Unless the oscillator is a *special* one that uses
    an actual negative *resistance* region of a device, e.g. a tunnel diode,
    it is still a conventional oscillator. The negative real part of an
    impedance concept, is no more than a mathematically re-interpretation of
    conventionally understood loop gain. For example, a Colpits oscillator
    can be analysed either directly by loop gain, or by noting that its
    input impedance at a node has a 1/s^2 term, i.e. a negative real part.

    A true negative resistance oscillator has an explanation based on an
    independent of frequency resistance. Very few oscillators have this

    It may be that such a negative resistance oscillator may have an
    advantage for high resistance inductors, but you need to supply a solid
    reason why, before departing from conventional approaches. What
    frequency do you want the oscillator at?

    Kevin Aylward
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
  3. Maybe you are thinking of the "negative resistors" frequently involved in
    chaotic oscillators? They are often referred to as "nonlinear resistors" or
    "Chua's diode", and generally have an I-V characteristic with a negative
    slope (usually with a 'breakpoint' in the middle). They are quite easily
    made with an op-amp and a few resistors. Have a look around here for
    starters: the resistors around the OP-27 give the desired effect:

    Googling on the latter two terms above should give other hits using similar
  4. John Jardine

    John Jardine Guest

    (Sorry, I can't see the original post)

    Have found the most reliable way to resonate high resistance big inductors
    is to use a ('balanced' or 'long tail') pair of transistors. The L-C tuned
    circuit is connected as a collector load and the oscillating voltage across
    the tuned circuit is fed back to the other transistor's base.
    Run the DC supply voltage in the 10-20V area and this allows many volts of
    oscillation headroom before the circuit starts clipping.
    (If you use PNP transistors then the oscillation will be centred on the 0V

    Static DC coil resistance isn't an issue as the tuned circuit is being fed
    purely by AC and no static voltage offset is developed. The coil resistance
    just appears as an oscillating 'loss component' causing the resonant
    frequency to drop a (calculable) amount.
    As a balanced pair is naturally "wideband" then the arrangement is also
    capable of resonating inductors out to many MHz.

    A neat side benefit is that the oscillation voltage can be very smoothly
    adjusted, just by altering the DC current in the commoned emitter
    This 'tail current' ma's equates directly to the coils "Q" value at the
    resonant frequency. Hence all the usual tuned circuit parameters can be
    extracted from the info generated by this simple setup.

    (I say this as just for fun, I'm playing with a similar design at the
    moment. It's quite amusing to see a big inductor resonating with 100uF at
    1Hz. :)

  5. Robert Baer

    Robert Baer Guest

    It would be nice if it could oscillate in the tens of KHz region, but
    the large inductance probably would not support that.
    The internal resistance is so high that i have not found a way to
    determine the "self-resonant" frequency (exciting with a zener in
    negative resistance mode was of no help).
  6. Robert Baer

    Robert Baer Guest

    I take it that the resistor-diode part acts as an amplitude limiter.
    Thanks for the idea; maybe i can get son=mething from it on an
    experimental basis.
    Since the final circuit canot have floating supplies, this exact
    scheme seems to be unuseable with a grounded inductor.
  7. Robert Baer

    Robert Baer Guest

    However, there are two problems (which i had originally mentioned):
    the inductor cannot support any DC current, as it will easily saturate,
    and it is grounded on one end.
  8. John Jardine

    John Jardine Guest

    A PNP pair -will- oscillate a 'one end grounded' inductor. And Yes; it
    generates a two terminal negative resistance.
    The negative resistance thingy applies to any oscillator. It's prime
    invocation is to cast fear and confusion upon people not of the
    brotherhood. (a la dB's and hexadecimal :)
    The inductor never sees a static-DC-current-offset hence no lopsided core
    magnetism and assymetric saturation. The circuit is -resonant-, the ma's of
    fluctuating collector current are just making up for the oscillating losses.
    The coil's DC resistance turns up as a balanced term in both the positive
    and negative half cycles.
  9. Fred Bartoli

    Fred Bartoli Guest

    So what ? Just use PNPs and active loads.

    | |
    | |
    | |
    |< >|
    -| |-
    |\ /|
    | |
    | GND
    | | |
    | .-. |
    --- | | || |/
    --- | | .---||---+---|
    | '-' | || | |>
    | | | |\ | |
    | `-----+-|-\ | |
    .------+ | >---+ .-.
    | .|. Ref---|+/ | |
    --- |C| |/ | |
    --- |C| '-'
    | |C| |
    | | |
    created by Andy´s ASCII-Circuit v1.22.310103 Beta

    I let you close the oscillator loop.

    BTW 10kHz with a 1H inductor is 250p. And a multi kOhm DC resistance
    inductor is probably much more than 1H, so don't count too much on the tens
    of kHz.

  10. Fred Bartoli

    Fred Bartoli Guest

    I must have missed something. Could you post a schematics please ?

  11. Do a google search for lambda diode and grid dip meter
    its a P chanel and a N channel fet kinda back to back
    and is used to replace tunnel diodes.

    or see

    your inductor sounds like it is large, I dont know if a lambda pair will
    oscillate that low but it may be worth a try.

    Steve Roberts
  12. It ought to be in the 10's of henry region with that resistance. IIRC an
    auto ignition coil is 20 henry's and several K ohms.
    Look into using a gyrator. Art of Electronics has one type, there are
    others. I think you might get some gyrators to go unstable (whether you
    want it or not, it seems).

    I've seen several articles for negative resistance oscillator design for
    microwave vco's, that might not be of much use.

    Although you might apply the concept of using a grounded-base
    oscillator. Make a series tuned circuit with your inductor and a
    resonant cap, and ground the base of a transistor. Then apply your
    positive-feedback from collector output-port to emitter input-port.



    DIY Piezo-Gyro, PCB Drill Bot & More Soon!

  13. John Jardine

    John Jardine Guest

    Maybe -I've- missed something :)
    It's pretty much the same as your ascii art but no series blocking cap. I've
    sent the circuit to A.B.S.E.

  14. Fred Bartoli

    Fred Bartoli Guest

    Ok, so at least I'm still good to something. But I still don't understand
    how you can say there's *no* DC current through the inductor.
    Do you intend to run it class C ?

    Oh, and I'm sorry but my ISP just dropped the abse group. Could you please
    send it directly to me ?

  15. John Jardine

    John Jardine Guest

    You're correct.
    Having just looked again at the diagram it's obvious that a DC current must
    flow throught the L. I'll keep my mouth shut from now on.
    (also said times before!).
    The error in my thinking occurred when the large inductors I was resonating
    were passing DC currents of a few uA and the meters weren't indicating it.
    I was also doing the sums from an equiv' circuit that had no bias.
    (if in error I blame the test gear :)
    I'll send the diagram on so you can see this.
  16. Fred Bartoli

    Fred Bartoli Guest

    So will I. My comment about class C was obviously equally wrong as I
    realised just after having it sent.

  17. John Jardine

    John Jardine Guest

    [I'll try again. Second time sent.]

    You're correct.
    Having looked at the diagram it's obvious that a DC current must
    flow throught the L. I'll keep my mouth shut.
    (also said times before!).
    The error in my thinking occurred when the large inductors I was resonating
    were passing DC currents of a few uA and the meters weren't indicating it.
    I was also doing the sums from an equiv' circuit that had no bias.
    (if in error I blame the test gear :)
    I'll send the diagram on so you can see this.
  18. [snip]

    I wonder if it could be done with a variation of the
    Baxandall sine wave oscillator. ie, Parallel-resonate
    the coil and drive it with a switched constant-current
    square-wave, with the polarity of the const-I being
    swapped at each zero-crossing of the voltage across the
    tank. Rough sketch below.

    -+---+-----+- +Vs
    | | R5
    | R1 |
    | | |/e
    | +---|pnp C2
    | | |\ +------||------+
    | R2 | | |
    0v--|\| | | C1 | RL L |
    |S>---+ +--||--+--/\/\--////--+--0v
    +-----+|/| | | |
    | | R3 | | +------+
    | | | |/ | | |
    | | +---|npn | +-|\ |
    | | | |\e | |O>--+
    | | R4 | +----|/ |
    | | | R6 |
    | -+---+-----+- -Vs |
    | |

    It's a pair of constant-current sources, alternately
    switched by comparator/gate (S), which is driven by
    the buffered sinewave across the resonant tank. C1
    is a large dc-blocker and C2 is the resonating capacitor.

    The Baxandall circuit requires a Q of roughly 5-10 for
    best operation. So the design frequency will initially
    be determined by Q = wL/R. With the rough numbers already
    given this suggest an Fosc in the 6KHZ to 10KHZ region.

    That then determines the value of C2, which will be around
    500pF. C2 includes the Cstray of the inductor.

    The impedance of the tank at Fres is Z = L/CR, so will
    be up in the 300k region. With 15v supplies, and a few volts
    across the tank, the required switched currents would then
    be of the order of +/- 20-40 uA or so.

    All numbers above winged on the fly, and to be confirmed.

    An interesting possibility could be to use a single OTA as
    the switching constant-current source. An OTA has a voltage
    adjustable current output and this could be used to
    stabilise the amplitude of the voltage across the tank.

    __ +Vs | |
    0v--+---| \| C1 | RL L |
    R1 |OTA>----||--+--/\/\--////--+--0v
    +--+|__/| |
    | | -Vs | +------+
    +----|->--+ | | |
    | +-------<---R2---|--+-|\ |
    | | |O>--+
    | C3 +----|/ |
    +--R3--+---||--+ |
    | | D |
    | /|--+--+---R4---|<|--+
    +--<O| |
    \|--+ +---R5---|>|--+
    | D |
    0v-+- |
    -+- -Vs

    The same +1 buffer looks at the voltage on the tank
    and now switches the polarity of the OTA's output
    current via R2 and R1. R1 and R2 should probably be
    sized for a 2V or so peak voltage across R1.

    Actually I'm not quite certain of the last sentence.
    An OTA has a linear (log) region of about 100mV or
    so. So there may be a need to check that there is
    enough loop gain for oscillation with R2:R1 ratios.

    The additional opamp is an erroramp/integrator that
    compares the half-wave rectified AC via R4 against
    a reference current from R5, and sets the required OTA
    output current via R3. R3 should be sized for about
    100uA through it when the integrator is at pk +Vout,
    (R3= 250k-ish with 15-0-15 supplies). R4 and R5 should
    be sized so that the ratio of R5/R4 is about 4.7/1
    (giving ACVpk= -Vs*2/3 approx).

    Ummm... haven't a clue whether the above will work or
    not. Suggest you SPICE it first.
  19. Robert Baer

    Robert Baer Guest

    I sort-of guess that i would be lucky to get a "self-resonance" in the
    KHz region...
  20. Robert Baer

    Robert Baer Guest

    OOOOh! *VERY* nice!
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