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Consider a bipolar junction transistor (either PNP or NPN). Which output would produce this circuit for various input (\$= V_{\mathrm{BE}}\$) voltages?
common base, zero voltage supply for the load circuit
It is common base, but without a voltage supply for the collector.

The resistor is assumed to have resistance on the order of \$V_{\mathrm{BE}}/I_{\mathrm E}\$ with \$V_{\mathrm{BE}}\$ typical for this base–emitter junction when forward biased.

Of course, Ī̲’m aware it’s no amplifier. The question arose from conditions considered in Why does the collector current direction remain the same in saturation and active region? and Transistor working with unusual biasing threads, but particular formulation of the questions hinders learning these specific things. Namely, Ī̲ seek arguments against the “two diodes model” as an universal answer for any question about BJT saturation.

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  • \$\begingroup\$ Without a voltage supply it won't do anything. Please draw a proper circuit diagram showing the actual voltage(s) applied. What do you mean by 'output'? \$\endgroup\$ – Bruce Abbott Aug 29 '16 at 18:55
  • \$\begingroup\$ @Bruce Abbott: «Vin» at the top left is the input voltage, (possibly) non-zero. There is no voltage source for the output part of the circuit. «output» is at the right (collector). \$\endgroup\$ – Incnis Mrsi Aug 29 '16 at 18:57
  • \$\begingroup\$ Where is the output, at the arrow? If you won't draw a proper circuit then at least put it on a line by itself (not confusingly embedded in the text). \$\endgroup\$ – Bruce Abbott Aug 29 '16 at 19:03
  • \$\begingroup\$ @Bruce Abbott: What is unclear in my circuit? One input (presumedly voltage-controlled, but it’s not important), one output. Why do you hesitate where the output for the common base is placed? \$\endgroup\$ – Incnis Mrsi Aug 29 '16 at 19:09
  • \$\begingroup\$ This resemble a circuit Bob Pease described on EDN March 18, 1996; (solution on EDN April 1, 1996). IIRC, one reverse biased junction would act as an LED (IR photons generated by zener breakdown) and the other as a photodiode, producing a current that will flow into the resistor to produce a voltage. (IIRC = I have a note with the schematics but not the actual magazines at hand) \$\endgroup\$ – Sredni Vashtar Aug 29 '16 at 21:58
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Here is your circuit redrawn more conventionally:-

schematic

simulate this circuit – Schematic created using CircuitLab

The Collector can pull current through R1 to the Emitter and Vin-, so the output voltage will (almost) equal the input voltage.

And now the same circuit, but with two diodes instead of a transistor:-

schematic

simulate this circuit

There is no way for the "collector" to pull down, so the output will remain at 0V.

Conclusion: the 'two diodes model' does not represent a BJT in saturation.

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  • \$\begingroup\$ Thanks. But why as much resistance as 100 kΩ? \$\endgroup\$ – Incnis Mrsi Aug 29 '16 at 19:56
  • \$\begingroup\$ I chose a typical resistor value that would normally be used for biasing the Base (producing the 'typical' Vbe you asked for). \$\endgroup\$ – Bruce Abbott Aug 29 '16 at 20:09
  • \$\begingroup\$ But the voltage drop on “a typical resistor that would be used for biasing the Base” isn’t normally V be. It would be (almost full) supply voltage, that is much higher. \$\endgroup\$ – Incnis Mrsi Aug 29 '16 at 20:15
  • \$\begingroup\$ I had another look at the circuit I mentioned above by Pease and if you add a 1k resistor on the emitter, reverse bias the BE junction with 12V, D1 emits photons that will reach the BC junction producing a tiiiiiny current. If you take R1 to be the 10 Mohm input resistor of a multimeter, then this is exactly the circuit proposed by Pease, and the collector gets pulled to -0.4 V. Source EDN issue of April first, 1996. (OMG, april's f...irst?) :-) \$\endgroup\$ – Sredni Vashtar Aug 29 '16 at 22:41
  • \$\begingroup\$ @IncnisMrsi 2N3904 'Typical' Vbe = 0.65V @ Ic = 1mA, / 300 (Hfe) = 3.3uA Ib. No supply voltage for Collector so Ic = 0 and Ie = Ib, so 'resistance on the order of Vbe/Ie' = 0.65V / 3.3uA = 195k. \$\endgroup\$ – Bruce Abbott Aug 29 '16 at 23:36
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Assuming it is a PNP transistor we have:

schematic

simulate this circuit – Schematic created using CircuitLab

We immediately see that the BE junction will be turned on if \$V_{in}>0.6\$, approximately. Since however we are assuming \$R\$ to be around \$V_{BE}/I_E\$, and since the base-collector junction has a lower forward-biased drop voltage than the BE junction, it means the BC junction will not be reversed biased, and hence the transistor is saturated, and \$V_o\$ will a little less than \$V_{BE}\$.

If it is a NPN transistor, we will reach the same conclusion if \$V_{in}<0\$.

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  • \$\begingroup\$ Ī̲’m a bit confused with “will be forward biased if *V*⁠⁠in > 0.6[V]”. In literature one can read that “unbiased” means “*V*⁠be = 0”. How should we call it while 0 < *V⁠*⁠be < 0.6 V? \$\endgroup\$ – Incnis Mrsi Aug 29 '16 at 21:01
  • \$\begingroup\$ @Incnis Mrsi: Sorry for the confusion. By "forward biased" I meant to say the junction is able to conduct a non-negligible amount of current. I think it's standard nomeclature for transistors (see for example this wikipedia article), but it is different when studying p-n junctions. In other words, the junction is "turned on". Otherwise, it is "turned off". \$\endgroup\$ – hcabral Aug 29 '16 at 21:19

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