the Darlington pair
/ DAR-ling-tun /
The Darlington pair is a simple way to make a transistor with an absurdly large current gain: wire two transistors so the first one's output feeds the second one's input, and their betas MULTIPLY. If each transistor has a beta of 100, the pair behaves like a single transistor with a beta of 100 times 100 = 10,000. So an almost negligible base current can control a hefty load current — handy when your control signal is very weak or your load is very hungry.
The connection is neat: the emitter of the first transistor drives straight into the base of the second, and their collectors are tied together. The tiny base current of the first transistor is amplified by its beta, and that already-amplified current becomes the base current of the second transistor, which amplifies it again. The combined gain is beta1 times beta2, plus a bit. The price is paid in voltage: because the input now passes through TWO base-emitter junctions in series, the Darlington needs about 1.2 to 1.4 V from base to emitter to turn on, double the usual 0.7 V. And when used as a switch it cannot saturate as fully — its 'on' collector-emitter voltage is a relatively high 0.7 V or more (the first transistor cannot saturate), so it dissipates more heat than a single transistor carrying the same current.
Where you meet it: in power output stages, motor and relay drivers, and touch sensors where the input current must be vanishingly small. It is sold as a single three-legged package (like the TIP120) so it drops in exactly where an ordinary transistor would. The honest caveats: the doubled turn-on voltage and the poor saturation make it lossy and warm; it is slower than a single transistor because the first transistor cannot quickly pull charge out of the second's base to turn it off; and stacking two unreliable betas gives an even more unpredictable (though always large) total gain — yet another reason to design so the exact value does not matter.
Drive a 5 A motor from a microcontroller pin that can only source 5 mA. A single transistor with beta 50 needs 100 mA of base current — far too much for the pin. A Darlington with effective beta 5000 needs only 5 A / 5000 = 1 mA, comfortably within the pin's budget. The catch: it will drop nearly a volt and run warm, so it needs a heat sink.
Beta times beta gives huge gain — at the cost of a doubled turn-on voltage and more heat.
Do not expect a Darlington to saturate down to 0.2 V like a single transistor — its minimum on-voltage is about 0.7 V or more. For a low-loss, fully-on switch handling big current, a power MOSFET is usually the better choice.