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?

  • \$\begingroup\$ 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. \$\endgroup\$
    – Dan Laks
    Commented Apr 22, 2015 at 5:03
  • \$\begingroup\$ 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. \$\endgroup\$
    – user57037
    Commented Apr 22, 2015 at 5:15

1 Answer 1


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).

  • \$\begingroup\$ Thank you, that makes a lot of sense. So when choosing the current-limiting resistor for my LED, I'll want to calculate the voltage across it as 5V (my supply voltage) minus LED voltage minus V_CE (1.1 V), is that right? \$\endgroup\$
    – jacobsa
    Commented Apr 22, 2015 at 5:12
  • 1
    \$\begingroup\$ @jacobsa You didn't specify a base current or load current, but you don't usually use the typical figure for design. You use the worst case. If you're looking at the 200mA collector current spec, use 1.3V, and make sure your base current is at least 350uA. \$\endgroup\$
    – Matt Young
    Commented Apr 22, 2015 at 12:28
  • \$\begingroup\$ The load current should be only about 20 mA, for an LED. How do I control the base current, given that I'm hooking up the base pin to a logic IC? \$\endgroup\$
    – jacobsa
    Commented Apr 22, 2015 at 21:32
  • \$\begingroup\$ @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}\$. \$\endgroup\$
    – Matt Young
    Commented Apr 22, 2015 at 21:37
  • \$\begingroup\$ Right, but I believe the ULN2003A has a built in resistor protecting the base. \$\endgroup\$
    – jacobsa
    Commented Apr 23, 2015 at 2:10

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