Common base circuit with zero supply voltage

Question

Consider a bipolar junction transistor (either PNP or NPN). Which output would produce this circuit for various input (\$= V_{\mathrm{BE}}\$) voltages?

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.

Answer 1

Here is your circuit redrawn more conventionally:-

^{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:-

^{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.

Answer 2

Assuming it is a PNP transistor we have:

^{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\$.