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I am trying to resolve what appears to be contradiction:

  • When used in the active region, a BJT is modelled as a base-current-controlled current source. Given that it is acting as a current source, a BJT in the active region is thus theoretically an open circuit with very high impedance, and thus is not a base-current-controlled resistor across the collector/emitter.
  • An optocoupler (which combines a diode + phototransistor) can be used as a diode-current-controlled resistance. For example, there are products like the https://store.synthrotek.com/VTL5C3-Vactrol-Optocoupler_p_792.html. The implication is that the phototransistor is a photon-generated-current-controlled resistance across the collector/emitter when the incident light places it in the active region.

These two statements above make two different claims about transistors. In one case, the collector/emitter path is not a current controlled resistor. In the other case, it is a current controlled resistor. Can anyone help explain why this is? Am I missing something?

Edit: the optocoupler I link above uses a photoresistor, not a phototransistor.

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    \$\begingroup\$ The optocoupler you linked does not use a phototransistor. This is a photoresistive type (which is very uncommon these days because photoresistors are slow and use large amounts of cadmium, which is prohibited by RoHS and other directives). \$\endgroup\$
    – Hearth
    Sep 8, 2023 at 17:26
  • \$\begingroup\$ The offset voltage and/or current involved with using a BJT as a variable resistor is a complicating factor that makes it awkward. Sometimes, you'll see a JFET used as a voltage-controlled variable resistor when biased at \$ V_{ds}=0V \$ \$\endgroup\$
    – glen_geek
    Sep 8, 2023 at 18:19
  • \$\begingroup\$ Thanks for pointing out my mistake on the type of optocoupler. I have updated my post as such. \$\endgroup\$
    – cgiustini
    Sep 9, 2023 at 13:37

4 Answers 4

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Figuratively speaking, the output collector-emitter part of a BJT behaves as a constant-current non-linear resistor. It increases its static resistance Rce = Vce/Ic when the voltage Vce across it increases and v.v. Thus the collector current Ic stays almost constant (Ic = Vce/Rce = const since both the numerator and the denominator change in the same way).

The current set by this "resistor" can be controlled both by voltage (non-linear) and current (linear).

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Only a current source driven by completely decorrelated signal (e.g. a constant) has infinite impedance.

Given that it is acting as a current source, a BJT in the active region is thus theoretically an open circuit with very high impedance,

Theoretically, only a constant current source is an open circuit.

A variable (dynamic) current source can emulate whatever impedance you want, as long as you control it according to Ohm's law, i.e. current proportional to voltage across terminals. The constant of proportionality is the conductance (reciprocal of resistance).

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  • \$\begingroup\$ Exactly! We can name this conductance (resistance) static. Only to add that this dynamic current source changes its current when the load varies its resistance (simultaneously with it); then the load "sees" this emulated resistance. An emitter follower is a typical example where a constant current source (the collector) with varying driving voltage emulates a constant voltage source (the emitter). Thus a varying emitter load "looking" at the emitter "sees" a constant voltage source. \$\endgroup\$ Sep 9, 2023 at 5:13
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These two statements above make two different claims about transistors.

The optocoupler in your link does not contain a transistor, it's a LED placed in front of a LDR (light dependent resistor). Useful component when you need a variable resistor. Advantages: isolated drive voltage, low capacitance between drive side and output. Drawbacks: non linear resistance vs LED current, inaccurate (requires calibration), and a bit of distortion.

In a LED-phototransistor opto, if the phototransistor is not saturated it will behave as a controlled current source, like a normal BJT with light instead of base current. These optos are characterized by CTR (current transfer ratio) which is the ratio between LED current and output current.

It's a useful component to make a constant current source with an isolated drive voltage, but not accurate (current transfer ratio depends on temperature and it has wide dispersion). But if you can calibrate it or put it in a feedback loop, it works nicely.

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Yes, but maybe not very usefully?

Others have already covered the linear range: either a simple "no", or you consider it a function of both (Ib or Vbe or opto If) and Vce, nonlinear such that I(Vce) = Vce / Rce(Ib, Vce) (at fixed Ib, or Vbe or whatever) ends up flat anyway. Which isn't a very useful perspective; maybe just as a hand-wave for introductory purposes, but the truth quickly outs that it's more of a constant current, and we can decompose it into static bias plus dynamic (or incremental) resistance components, and the two are not related (i.e., dIc/dVce != Ic/Vce).

But going with that incremental measurement, we can confine ourselves to the base junction alone (on 5-terminal BJT optos like 4N35) and a small signal (typically 10s of mV).

Using a circuit like so:

schematic

simulate this circuit – Schematic created using CircuitLab

gives the following series of waveforms:

Variable incremental resistance waveform series
Source: my website, https://www.seventransistorlabs.com/Images/OptoBaseDynResistor_100k_18mA_to_180uA.gif

(Hm, could've stood to enable Hi-Res mode there, oh well; it's just in squigglevision now.) Ch4 reads 1x, Ch3 10x (hence, at the lowest point, Ch3 being 5.2mV is identical to Ch4's 52mV). "Ch3 Ampl" measurement goes down into the noise floor pretty well by the peak there, but suffice it to say it is being attenuated.

The reason this works is, the base junction in the 4N35 and similar parts is a photodiode -- it just so happens to also have a collector attached, so the photocurrent acts to induce collector current when so biased. In this case, the collector is open, and only the B-E junction is considered.

As Vbe goes up, Ib goes up; Ib isn't important here (it's driven internally by the photocurrent), but notice Ib(Vbe) is an exponential function, given by the Shockley equation, \$Ib = I_s e^{\frac{V_{be}}{n V_{th}}}\$. Taking the derivative of this and rearranging, gives us the incremental resistance dVbe/dIb, which we can express in terms of itself again to get \$r_b = \frac{V_{th}}{Ib}\$. The exact amounts, proportionality, etc., aren't really important (here, you need to know the LED and coupler efficiency anyway), just the fact that it varies inversely with current (and LED If will be close to proportional to Ib). As it happens, the value in this arrangement can be quite a bit less than 100k, hence making a voltage divider with R3, and attenuating the signal when bias is large.

Here, If varied from about 180µA to 18mA.

This was also done at 1kHz; at even 10kHz, phase shift is considerable, as Cbe is quite large, particularly in forward bias.

The signal level must be low, or else a significant portion of the exponential curve is explored by the signal amplitude; the result is a flattening of the top of the waveform -- considerable distortion.

Such biased-diode circuits have been used for audio and RF amplitude control and mixing for many years, but they are generally inferior to better ones like diode-compensated diff pairs (e.g. LM13700 and other OTAs, Gilbert cell mixers), or a few (MOS/J)FET based ones, or for that matter, these days, straight-up DSP (ADC-math-DAC). Or PWM when low bandwidth is required, or MDACs where the multiplicand shall be digital (which includes "digital pots" as well as R-2R and other DAC types).

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