Why are NPN Darlington transistors used to sink current?

Question

I notice that NPN Darlington transistors are commonly used to sink current. Wouldn't it make more sense to use PNP for the purpose? This would avoid shunting the load current through both junctions at once. Granted, we might want to share the current between two transistors; but in which case, please note that the second transistor is still carrying the full load (half via the C-E path, and half via the B-E path).

For that matter, why are transistors most commonly used for sinking current anyway; rather than driving it? I've never understood that.

In the above example, it seems more sensible to either (1) place the load below the transistor; (2) use a PNP Darlington; or even better (3) use a complementary PNP pair as shown here:

EDIT:

To clarify, one of the questions I'm asking is: Why can't we place this NPN transistor as-is above the load? Or, for that matter, place a PNP Darlington below the load? And also, why do Darlingtons even exist, when a complementary pair looks to be a cleaner solution?

Answer 1

Sinking load switches with an NPN darlington allow the control signal to be a GND referenced signal. If you use high side sourcing switches it is most typical that then the control signal needs translation down to a GND referenced signal domain.

These days when MCUs are controlling almost everything the GPIO pins on such devices are GND referenced signals. And so it should be obvious why many load switches use the syncing type components with a GND referenced input.

Answer 2

Regarding the use of NPN rather than PNP, Michael Karas answer is correct: you want ground-referenced control signals because the N-type transistors generally have better characteristics than the P-type equivalents.

Regarding other parts of your question: Darlingtons don't share the current between the two transistors 50-50. The one where the input signal arrives on the base carries maybe 1% of the current through it (assuming a beta of 100; most integrated circuit NPNs have betas much higher (~250) so the percentage is therefore even lower). The other transistor is therefore carrying 99%+ of the driven current.

This is a good thing, not a bad thing. Integrated Darlington pairs are configured in physical layout with a significant size differential, such that the main drive transistor has a much larger junction area than the first, allowing for much lower C-E on resistance for lower drive currents and much higher max current handling capability. This is without the need for pairing multiple transistors in parallel, which can cause uneven current splitting due to device differences, even on integrated circuits.

Lastly, NPN Darlingtons can be easily constructed on an integrated circuit effectively as a single meta-transistor; they share the same collector region but have different embedded base/emitter regions (with the size difference I mentioned earlier). Connecting the emitter of the smaller to the base of the larger is pretty trivial. I'm pretty sure this is what's done on the integrated multi-Darlington arrays e.g. ULN2k series (I don't have the details of access any more, but I did see some of this way back when doing my studies in this stuff).

Answer 3

In the Darlington configuration, the base current of the larger transistor helps to drive the load and is self-regulating. If one needs to drive a 10 amp load and wants to avoid assuming a beta greater than 40, one will need to be able to drive the base of the large transistor with 250mA. To get that 250mA, one would need to drive the base of the small transistor with 7mA. Using a Darlington configuration, if the load draws 10A, 9.75A will flow through the collector of the large transistor and 250mA will flow through the small transistor into the base of the large one. The 7mA driven into the base of the small transistor will be "wasted". If the load were to drop to 10mA, the base of the small transistor would still consume 7mA, which it would pass through the base of the large transistor, but little current would be required elsewhere.

In most other configurations, arranging for the large transistor to have 250mA available on its base when needed would imply that 250mA would be fed to the base of the large transistor even when it wasn't needed. In cases where the load is known to require 10A, that wouldn't be a problem, but in cases where the load might require anything from 10uA to 10A, wasting 250mA at times when the load requires 10mA may be undesirable.

Answer 4

You should be able to see for yourself from your own diagrams that the lower circuit needs access to the power rail, whereas the pure low side switch can be pre-packaged without needing that connection.