How is a Darlington pair biased?

[warriorpoet]

So I get the theory behind Darlington pairs, and successfully bread boarded a few schematics, but I really want to understand the proper math behind the resistors for a Darlington pair. Right now all I can find are suggested resistor values, and I'd really like to be able to calculate these values on my own.

Thanks!

[Technician]

A darlington pair is biased very much like a single transistor, except that a single transistor needs approx. 0.6-0.7 volts d.c. (from base to emitter) to bias it into the "on" state. A darlington pair, because it's 2 transistors connected emitter of transistor #1 to the base of transistor #2, will need twice as much voltage to bias it into the "on" state; around 1.2-1.4 volts d.c. (from the base of transistor #1 to the emitter of transistor #2). Use 2 resistors in series as a "voltage divider" to provide the desired bias. Here is one way to bias a darlington pair: If you use NPN transistors; for a darlington pair, connect one end of resistor #1 to the collector of transistor #1 and to the + (plus) end of your battery or d.c. power supply, and the other end of that resistor to the base of the first transistor. Connect a second resistor from the base of that same transistor to the emitter of the second transistor, and the battery - (minus) end. The simplest thing to do is use 2 equal value resistors, so that half the battery voltage will be across each resistor. Then, the darlington pair will be biased "on" at 4.5 volts. Use fairly high resistor values (at least 18K ohms each), to make the battery last longer and limit the current from the transistor #1 base through the transistor #2 emitter (too much current will destroy both transistors).

[Technician]

Calculate bias resistor values using Ohm's Law: R=E/I; I=E/R; E=I*R. Two fixed value resistors are connected in series from