simulate this circuit – Schematic created using CircuitLab

I am using an NPN transistor (2N4401).

In the above schematic, charge is only coming in through the base, and not the collector.

My LED is dim, but still lit. Is this the expected behavior of a transistor, or am I doing something wrong?

My expectation was that there should be no charge flowing through the emitter since the collector is not connected. Obviously, when the collector has charge applied, but the base does not, then the LED does not light up. This is expected, but it's also happening the other way around.

  • 1
    \$\begingroup\$ Your schematic says the transistor is a 2N3904 but your text says its a 2N4401. It doesn't matter in this case, but this sort of confusion and sloppiness can get your question downvoted and/or closed under different circumstances. \$\endgroup\$ Commented Jul 16, 2014 at 12:32
  • 3
    \$\begingroup\$ If current is going in the base, where is it coming out? \$\endgroup\$
    – Phil Frost
    Commented Jul 16, 2014 at 12:40
  • \$\begingroup\$ Also probably insightful: Why would one drive LEDs with a common emitter? \$\endgroup\$
    – Phil Frost
    Commented Jul 16, 2014 at 12:41

2 Answers 2


In such a circuit the emitter current equals the collector current PLUS the base current:

Ie = Ic + Ib

In normal situations the collector current is much larger than the base current, hence by good approximation

Ie = Ic

But in your case the collector current is zero, so the first equation degrades to

Ie = Ib

Which is exactly what you see: a very small current through the LED.

Another way to look at your circuit is to replace the b-e of the transistor with a diode. This gives the same results.

  • \$\begingroup\$ +1 for the equation explanation. So in other words, the behavior I see is completely expected based on how the transistor is supposed to function? \$\endgroup\$
    – Josh Beam
    Commented Jul 16, 2014 at 11:40
  • \$\begingroup\$ totally expected \$\endgroup\$ Commented Jul 16, 2014 at 12:20
  • \$\begingroup\$ Using an NFET instead of NPN would reduce the base current to essentially zero. \$\endgroup\$
    – Ben Voigt
    Commented Jul 16, 2014 at 16:35
  • \$\begingroup\$ True, but totally irrelevant: the question is about a BJT. \$\endgroup\$ Commented Jul 16, 2014 at 16:50

The base-emitter junction of a transistor is just a diode. When the collector is disconnected, current can flow through R1, the base-emitter junction, D1 and R2, with a current of roughly 6.5V / 25270Ω = 260µA, which is enough to see.

When the collector is connected, R1 no longer limits the current to the LED, so you'll get more like 6.5V / 270Ω = 24mA through the LED.


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