Electronic – Why might the collector-emitter voltage for this transistor be so high


I'm attempting to use an NPN transistor as a switch, to drive an LED load. So I am only interested in the off and saturated states.

My understanding is that when the base-emitter junction is saturated (in my case due to the emitter being tied to ground and the base voltage being high enough), the collector-emitter voltage is close to zero. Multiple places on the web (e.g. here and here) say the same thing, that the collector-emitter voltage should be at most a few tenths of a volt.

However I'm looking at the ULN2003A, which appears to be a very common IC for arrays of NPN transistors, and the electrical characteristics table in its data sheet says that the collector-emitter saturation voltage is typically 1.1 V, and can be up to 1.3 V. Why is this so high? Am I misunderstanding something?

Best Answer

The ULN2003A is a Darlington array, made up of Darlington pairs. The Darlington pair has higher current gain than an individual bipolar transistor. The side effects include a higher \$V_{be}\$ (in common emitter configuration, it will appear as about two diode drops instead of one) and \$V_{ce}\$ (\$V_{be}\$ of one transistor, plus \$V_{ce}\$ of the other).