Let me answer yesterday’s question first: “Is this circuit actually useful for anything?”

You bet: this is called a Current source.

The circuit will deliver a constant current by varying the voltage drop, even when the load varies. You can see this in the fairly flat curve on the Component Tester screenshot included yesterday: no matter what level positive voltage you apply to this thing, it’ll draw about 2 mA (just ignore the negative end of the scale).

Actually, I cheated a bit. The real two-transistor current source circuit looks like this:

By moving that 10 kΩ resistor away from the load, and tying it directly to “+” the circuit works even better. I’ve simulated it with an external power supply to drive that resistor separately, and get this CT screen:

Totally flat! – And that 2 mA current level is set by the 330 Ω resistor, by the way.

One use for this could be a constant-current LED driver (although its efficiency would be very low – you really need a switching circuit with an inductor to get good efficiency).

So how does this mind-bending circuit actually work?

The key point to note, is that the emitter-to-base junction is essentially a diode (which is probably why transistors are drawn the way they are!). And it has a fixed forward-drop voltage of about 0.65V. As long as the base is less than 0.65V above the emitter voltage, the transistor will be switched off. As soon as the base is raised higher, current will flow through that forward diode and the transistor will start to conduct.

This is also why you always need a current limiting resistor: the base voltage cannot rise above 0.65V, it’ll simply conduct more current. Until the current limits are exceeded and the transistor is destroyed, that is…

First, imagine that the leftmost transistor is absent: then the 10 kΩ will pull up the base of the rightmost transistor and cause it to fully conduct. The circuit now essentially acts as the load in series with the 330 Ω resistor. With a maximum load (a short-circuit), the whole supply voltage will end up across that 330 Ω resistor.

But…

With the leftmost transistor in place, something special happens: as soon as the voltage over the 330 Ω resistor rises above 0.65V, the leftmost transistor will start to conduct, pulling the base of the rightmost transistor down. It will continue to do so until the voltage over the 330 Ω resistor has dropped to 0.65V again. Because at some point the base of the rightmost transistor will be pulled so low that it no longer fully conducts – thus reducing the current through the 330 Ω, and thus lowering the voltage drop across it.

You’re seeing a neat little negative feedback loop in action. These two transistors are going to balance each other out to the point where the 330 Ω resistor ends up having a voltage drop of exactly 0.65V – regardless of what the load is doing!

To get 0.65V over 330 Ω, we need a 0.65/330 = 1.97 mA current.

And so that’s what this circuit will feed to the load. As you can see in that last scope capture, the regulation is extremely good between 0.65 and 9V.

By simply varying the 330 Ω value, we can set any desired fixed current level.

The reason I’m bringing this up, is that this circuit is in fact used in the OpenTherm gateway – see this schematic (look for the upside-down PNP version). With some extra circuitry to set the resistor to either 100 Ω or 28 Ω (100 Ω in parallel with 39 Ω). So the gateway is driving either 7 mA or 25 mA through the thermostat.

Welcome to the magical world of electronics – it’s full of clever little tricks like this!