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Resistive BJT

Discussion in 'Electronic Design' started by Jeff Johnson, Feb 7, 2011.

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  1. Jeff Johnson

    Jeff Johnson Guest

    Just for fun, suppose you were trying to make a BJT act as a resistive
    element. For the simple case of a voltage divider replace the lower
    resistor, R2, with a BJT. (Vcc---R1--R2---Gnd)

    Is it possible to get the BJT to behave independently of R1? That is, the
    BJT behaves as a fixed resistor and mimics a true voltage divider?

    Best I can do is show that if the input voltage of the BJT is inversly
    proportional to the square of R1 then the BJT will have an approximate
    constant effective resistance over whatever reasonable range(say R1 from 10k
    to 1M). It does need emitter degeneration to work though. I could't come up
    with a simple circuit to program the voltage though.

    Obviously one can probably program the input of the BJT in such a way to
    make the BJT act as a resistor in a voltage divider but the goal is to do it
    simple as possible.

    Anyone actually see a simple way to do this? Since this is more of a
    "homework" problem than real world problem one doesn't have to worry about
    thermal stability, drift, etc. Although lets not assume beta is fixed to
    make it a little more complicated.
  2. Jeff Johnson

    Jeff Johnson Guest

    Alright I guess not to many are interested. In any case by using eber-molls
    we can find the expression to make the bjt behave as a pure resistance,

    Vbe = Vt*ln(Vcc/Is/(R + Rq) + 1)

    So if one can program Vbe to follow the above expression then the bjt will
    have an effective and constant resistance of Rq independent of R. That is,
    the bjt will behave as a pure resistor.

    The question is can one actually do this in practice to make it work over a
    practical range(say from 1k to 1M)?

    It seems like one can use matched transistors to cancel out Vt and another
    bjt possibly setup the log relationship. To get R we'll obviously need some
    type of "sensing" mechanism(we'll assume R is not fixed). Even if it is
    possible to create such a circuit is it possible to do it in a practical

    One can follow a similar analysis with fets that work over a much broader
    range(much more practical):

    I = K*(Vgs - Vt)^2

    so, for a voltage divider,

    Rq = Vcc/I - R = Vcc/K/(Vgs - Vt)^2 - R

    which gives

    Vgs = Vt + sqrt(Vcc/K/(R + RQ))

    Which may or may not be easier to implement.

    Any ideas how one can implement this in practice? Obviously one can use
    analog or digital circuit blocks to implement the mathematical expression
    and feed the gate/base of the transistors. The real question is, is anyone
    smart enough to see any tricks to do it as simply as possible without any
    real complexity(no uP's, op amps, etc...). e.g., just use a few other
    transistors and resistors to compute the expressions to an relatively
    accurate degree. Can assume using matched transistors though. The point is
    not to implement such a circuit as that can easily be done but to implement
    it eloquently. I guess that means trying to do it with as few parts as
    necesssary at the cheapest cost with a practical degree of approximation and
    minimum complexity. Any bozo can use a dsp to compute those expressions but
    how many can do it with one or two transistors and a few resistors?
  3. Jeff Johnson

    Jeff Johnson Guest

    The question is, can this be done with just a bjt or two and a few other
    discrete and cheap components? Essentially one is linearizing the current.
    Making vbe depend on the log of something makes the current depend linearly
    on it. Can one setup such a circuit? A diode depends on the log of the
    current so if we had a current to voltage(e.g. a resistor) setup we should
    be able to do it. This, I suppose, would be similar to the LM13700's
    linearizing diodes.
  4. Jeff Johnson

    Jeff Johnson Guest

    I didn't just say fewest parts only. Read what I said again. Sometimes a
    sledge hammer is not the best thing for a 1 penny nail
  5. Jeff Johnson

    Jeff Johnson Guest

    Yes, there are many ways to make VCR's... that was pointed out but is not
    the point. I asked a question but so far have been given anything but an
    answer. NO ONE IS DEBATING THAT VCR's can be done! But there is, first off,
    no simple VCR that works over a wide set of circumstances(fets suck as VCR's
    BTW! Sure in some applications one can use them to good effect but 99% of
    the time in the real world they simply won't work as VCR's). Why? Because
    they are only in the omic region for low VDS voltages and drain currents. In
    the real world most resistors do not work at such low voltages and currents.

    For example, I can prove that a simple circuit such as

    If the diode(or subst a bjt) and bjt are matched.

    It's quite simple to prove that the bjt acts as the difference between R1
    and R2.

    Let RQ be the effective resistance of the bjt,

    RQ = Vc/Ic = (Vcc - Ic*R2)/Ic = Vcc/Ic - R2

    But Ic = Isq*(exp(Vbe/Vtq) - 1), But because of the "linearizing diode",
    Id = Isd*(exp(Vbe/Vtd) - 1)

    in which case

    Ic = Isq*((Id/Isd + 1)^(Vtd/Vtq) - 1)

    If we assume a matched pair then Isd = Isq and Vtd = Vtq so that

    Ic = Id

    That is, we have a simple current mirror,

    Since Id = I - Ib = (Vcc - Vbe)/R1 - Ib

    RQ = Vcc/(Vcc - Vbe - R1*Ib)*R1 - R2

    If Vcc >> Vbe + R1*Ib then

    Rq = R1 - R2

    So the effective resistance of the bjt is the difference between the two
    resistances. So if we can somehow add R2 to R1 then Rq would be independent
    of R2. So possibly using another current mirror or somehow modifying the
    existing one may allow one to effectively add a copy of R2 in series with R1
    making the BJT have a constant resistance of R1.

    More than likely there is some type of circuit configuration that will work
    well, at least in the ideal case, that is not too complex and is easier to
    implement than an OTA, uP, etc.

    The question is, who is smart enough to find it?
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