# 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?

• I'll let someone else explain what's happening at the silicon level for a complete answer. But, no, you're not misunderstanding. Those websites you linked to are a little misleading. There are definitely BJTs that have greater than 1V Vce(sat). In fact, any good distributor website (digikey, for example) will let you filter by Vce(sat) when searching for a BJT. Commented Apr 22, 2015 at 5:03
• There are many properties of transistors. Optimizing one may degrade another. You don't say how much current you are switching, but probably the most common transistor in the US is the 3904. Vce sat of 0.3V when Ic = 50ma, and using a forced beta of 10. If that isn't good enough for your application, consider a MOSFET. Commented Apr 22, 2015 at 5:15

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).
• @jacobsa In the common emitter configuration, you must include a resistor between the GPIO pin and base. The general form for base current is $I_b=\dfrac{V_{OH}}{R_b}$. Commented Apr 22, 2015 at 21:37