Resistive BJT

If the bjt has its emitter grounded and you use the collector as the top of R2, and tune the base current, it will be ohmic in the saturation region, and more base current makes less collector ohms. But there will be a DC offset that varies with base current, in the

10s of millivolts maybe, and you'll see a lot of distortion if the signal swing at the collector is more than 10s of millivolts.

So for most applications, it won't work very well.

Fets work a little better; they at least don't have the dc offset and allow a little more signal range before they go nonlinear.

John

For small-signals, sure... but performance is rather better if you use a FET instead:

"Simple" is kinda relative... especially given that few applications require that the actual resistance of the BJT/FET is prescribed; usually they're part of some feedback loop where they get adjusted to "whatever resistance is needed" to get some particular output level.

Closed-loop feedback doesn't care about beta either, assuming you design for the worst case.

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 way?

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?

...assuming your input signal doesn't grow large enough to move the transistor out of its active range... ...and that some discrepancies due to, e.g., Early voltage are OK...

These days "fewest parts" is typically *not* "cheapest cost." (Particularly if you're counting, e.g., an op-amp as 1,000 transistor "parts" or a microcontroller as a million transistor "parts"...)

I'd likely spring for the op-amp feedback loop, servoing a pair of matched transistors as you suggested. It's still easy to understand and reasonably elegant...

---Joel

The DC bias point of a single transistor is awkward (REAL resistors work with any input polarity) so consider a differential pair instead. Four more tweaks will get you to the OTA (operational transconductance amplifier) which has, in a mass-produced chip, a roughly proportional output current to the input voltage difference, and takes a program current that scales the resistance over a few decades range.

There are weaknesses to this approach, of course. The linear resistance region is under 1V (maybe WAY under), and there's only one current-delivering pin (real resistors have two).

Look up CA3080 or LM13700 for application notes.

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.

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

Although it's always a bit risky to do so, in this case I would assume that if there were a 1 penny nail solution, it would have presented itself by now.

1d nail? One tap with a sledge hammer will do ;-) ...Jim Thompson

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The LM13700 has two transconductors for a dollar, it does qualify as cheap. As for discrete, an OTA uses matched pairs of transistors, so there's no benefit. Ten transistors per OTA, two OTAs per chip, means you'd have to beat 0.10 USD per matched pair (with zero assembly costs).

A transconductor without the level shifting (OTA output compliance is nearly rail/rail, as is the input common mode range) is just an emitter-coupled pair, but has limited range and requires DC shifting in many applications. Music synthesizers use that circuit, with the emitter current source modulating the impedance.

It's possible to bias a JFET with near-zero gate voltage, and for very small signals it acts like a voltage-controlled resistor. Always a resistor to ground, of course.

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?